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p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N 100010
#define gc getchar()
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define lint long long
#define mod 1000000007
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std;
inline int inn()
{
int x,ch;while((ch=gc)<'0'||ch>'9');
x=ch^'0';while((ch=gc)>='0'&&ch<='9')
x=(x<<1)+(x<<3)+(ch^'0');return x;
}
int fac[N],facinv[N],f[N],onc[N],sz[N],p[N],cnt[N];
int cyc_cnt,vis[N],l[N],pre[N],jhs[N],in[N],mi[N],it[N];
inline int fast_pow(int x,int k,int ans=1)
{ for(;k;k>>=1,x=(lint)x*x%mod) (k&1)?ans=(lint)ans*x%mod:0;return ans; }
inline int prelude(int n)
{
rep(i,fac[0]=1,n) fac[i]=(lint)fac[i-1]*i%mod;
facinv[n]=fast_pow(fac[n],mod-2);
for(int i=n-1;i>=0;i--) facinv[i]=facinv[i+1]*(i+1ll)%mod;
rep(i,mi[0]=1,n) mi[i]=mi[i-1]*2,(mi[i]>=mod?mi[i]-=mod:0);
rep(i,f[0]=1,n/2) f[i]=f[i-1]*(2*i-1ll)%mod;return 0;
}
inline int C(int n,int m)
{
if(n<0||m<0||n<m) return 0;
return (lint)fac[n]*facinv[m]%mod*facinv[n-m]%mod;
}
int main()
{
int n=inn();rep(i,1,n) p[i]=inn(),in[p[i]]++;
//unsigned int QwQ=1,t=1;rep(i,1,n) QwQ+=p[i]*t,t*=998244353;cout<<QwQ<<endl;
for(int i=1,x;i<=n;i++)
{
for(x=i;!vis[x];x=p[x]) vis[x]=i;
if(vis[x]^i) continue;
for(cyc_cnt++;!onc[x];x=p[x])
onc[x]=cyc_cnt,sz[cyc_cnt]++;
}
// debug(cyc_cnt)ln;
// rep(i,1,cyc_cnt) debug(i)sp,debug(sz[i])ln;
// rep(i,1,n) debug(i)sp,debug(onc[i])sp,debug(in[i])ln;cerr ln;
memset(vis,0,sizeof(int)*(n+1));
for(int i=1,x,c;i<=n;i++) if(!in[i])
{
for(x=i,c=0;!onc[x];vis[x]=1,x=p[x],c++)
if(vis[x]) return !printf("0\n");
if(vis[x]) return !printf("0\n");vis[x]=1,l[x]=c;
}
rep(i,1,n) if(onc[i]&&l[i]) jhs[onc[i]]=1;
rep(i,1,cyc_cnt) if(!jhs[i]) cnt[sz[i]]++;
// rep(i,1,cyc_cnt) debug(i)sp,debug(jhs[i])sp,debug(sz[i])ln;
// rep(i,1,n) debug(i)sp,debug(l[i])sp,debug(cnt[i])ln;cerr ln;
prelude(n);int ans=1;
for(int i=1,k,t,s,v;i<=n;i++) if(cnt[i])
{
rep(j,it[0]=1,cnt[i]/2) it[j]=(lint)it[j-1]*i%mod;
for(k=cnt[i],t=0,s=0;t<=cnt[i]/2;t++)
v=(lint)C(k,2*t)*f[t]%mod*it[t]%mod,
(((i&1)&&i>1)?v=(lint)v*mi[k-2*t]%mod:0),
s+=v,(s>=mod?s-=mod:0);
ans=(lint)ans*s%mod;
}
for(int i=1,x,c;i<=n;i++) if(onc[i]&&l[i])
{ for(x=i,c=1;!l[p[x]];x=p[x],c++);pre[p[x]]=c; }
// rep(i,1,n) debug(i)sp,debug(pre[i])ln;
rep(i,1,n) if(onc[i]&&l[i])
{
if(l[i]>pre[i]) return !printf("0\n");
else if(l[i]<pre[i]) ans=ans*2,(ans>=mod?ans-=mod:0);
}
return !printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define mo 1000000007
using namespace std;
int inn[1000100],a[1000100],num[1000100],wei[1000100],mp[1000100],f[1000100];
bool vis[1000100];
int work(int l,int n)
{
f[0]=1;f[1]=1+(l&1)-(l==1);
for (int i=2;i<=n;i++)
f[i]=((1+(l&1)-(l==1))*f[i-1]+(long long)l*(i-1)%mo*f[i-2])%mo;
return f[n];
}
int main()
{
//freopen("in.txt","r",stdin);
int n;scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);inn[a[i]]++;
if (inn[a[i]]>2) {printf("0\n");return 0;}
}
for (int i=1;i<=n;i++) if (!inn[i])
{
int js=0,now=i;
while (inn[now]!=2) {vis[now]=true;now=a[now];js++;}
num[now]=js;//cerr<<now<<' '<<js<<endl;
}
int ans=1;
for (int i=1;i<=n;i++) if (!vis[i])
{
int len=0,now=i;
while (!vis[now]) {wei[++len]=now;vis[now]=true;now=a[now];}//cerr<<len<<endl;
if (i!=now) {printf("0\n");return 0;}
int lst=1;for (int j=1;j<=len;j++) if (num[wei[j]]) lst=j-len;cerr<<lst<<endl;
if (lst==1) {mp[len]++;continue;}
for (int j=1;j<=len;j++) if (num[wei[j]])
{
if (j-lst>num[wei[j]]) ans=ans*2%mo;
if (j-lst<num[wei[j]]) {printf("0\n");return 0;}
lst=j;
}
}
for (int i=1;i<=n;i++) if (mp[i]) ans=(long long)ans*work(i,mp[i])%mo;
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef double db;
int get(){
char ch;
while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
if (ch=='-'){
int s=0;
while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
return -s;
}
int s=ch-'0';
while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
return s;
}
const int N = 2e+5+5;
const int mo = 1e+9+7;
int cnt[N];
LL inv[N],js[N],miv[N];
int n;
int a[N];
int fa[N];
struct edge{
int x,nxt;
}e[N*2];
int h[N],tot,key[N],k;
bool bz[N];
int pt[N],u;
int siz[N];
int pre[N],suf[N];
bool dfs(int x){
bz[x]=1;
siz[x]=1;
int s=0;
for(int p=h[x];p;p=e[p].nxt)
if (!bz[e[p].x]){
s++;
if (dfs(e[p].x))return 1;
siz[x]+=siz[e[p].x];
}
return s>1;
}
LL quickmi(LL x,LL tim){LL ans=1;for(;tim;tim/=2,x=x*x%mo)if (tim&1)ans=ans*x%mo;return ans;}
int getfather(int x){return fa[x]==x?x:fa[x]=getfather(fa[x]);}
void inse(int x,int y){
e[++tot].x=y;
e[tot].nxt=h[x];
h[x]=tot;
}
int main(){
n=get();
inv[0]=js[0]=inv[1]=js[1]=miv[0]=1;
miv[1]=(mo+1)/2;
fo(i,2,n){
inv[i]=1ll*(mo-mo/i)*inv[mo%i]%mo;
js[i]=js[i-1]*i%mo;
miv[i]=miv[i-1]*miv[1]%mo;
}
fo(i,2,n)inv[i]=inv[i]*inv[i-1]%mo;
fo(i,1,n)fa[i]=i;
fo(i,1,n){
int x=a[i]=get();
inse(x,i);
int tx=getfather(x),ti=getfather(i);
if (tx!=ti)fa[ti]=tx;
else key[++k]=i;
}
int ans=1;
fo(i,1,k){
int x=key[i];
bz[pt[u=1]=x]=1;
for(x=a[x];x!=key[i];x=a[x])bz[pt[++u]=x]=1;
fo(j,1,u)if (dfs(pt[j])){ans=0;break;}if (!ans)break;
int ps=0;
fd(j,u,1){ps=max(0,ps-1);if (siz[pt[j]]>1){if (ps){ans=0;break;}ps=siz[pt[j]]-1;}}if (!ans)break;
fd(j,u,1){ps=max(0,ps-1);if (siz[pt[j]]>1){if (ps){ans=0;break;}ps=siz[pt[j]]-1;}}if (!ans)break;
int mx=0;
fo(j,1,u)mx=max(mx,siz[pt[j]]);
if (mx==1)cnt[u]++;
fo(j,1,u)pre[j]=pre[j-1]+siz[pt[j]]-1;
fo(j,1,u)pre[j+u]=pre[j+u-1]+siz[pt[j]]-1;
fo(j,1,u)
if (siz[pt[j]]>1){
if (pre[j+u-1]==pre[j+u-siz[pt[j]]])ans=ans*2%mo;
}
}
if (!ans)return printf("0\n"),0;
fo(i,1,n)
if (cnt[i]){
LL now=1,tmp=0;
fo(j,0,cnt[i]/2){
if (i%2==1&&i>1)tmp=(tmp+js[cnt[i]]*inv[j]%mo*inv[cnt[i]-j*2]%mo*miv[j]%mo*now%mo*quickmi(2,cnt[i]-j*2)%mo)%mo;
else tmp=(tmp+js[cnt[i]]*inv[j]%mo*inv[cnt[i]-j*2]%mo*miv[j]%mo*now%mo)%mo;
now=now*i%mo;
}
ans=tmp*ans%mo;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define ll long long
char buf[1<<20],*p1,*p2;
#define GC (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?0:*p1++)
template<class T> inline void read(T &n){
char ch=GC;T w=1,x=0;
while(!isdigit(ch)){if(ch=='-') w=-1;ch=GC;}
while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=GC;}
n=x*w;
}
const int maxn=200005;
int n,a[maxn],deg[maxn];
int vis[maxn],isc[maxn],link[maxn];
int cnt[maxn],ans=1;
const int mod=1e9+7;
inline int add(int x,int y){
x+=y;return x>=mod?x-mod:x;
}
inline int mul(int x,int y){
return 1ll*x*y%mod;
}
void solve(int x){
int now=0,fir=0,firl=0,lst=0;
for(;isc[x];x=a[x]){
now++,isc[x]=0;
if(link[x]){
if(!fir) fir=lst=now,firl=link[x];
else{
int tmp=now-lst;
if(tmp<link[x]) puts("0"),exit(0);
if(tmp>link[x]) ans=add(ans,ans);
lst=now;
}
}
}
if(!fir) cnt[now]++;
else{
int tmp=fir+now-lst;
if(tmp<firl) puts("0"),exit(0);
if(tmp>firl) ans=add(ans,ans);
}
}
int f[maxn]={1};
int main(){
read(n);
for(int i=1;i<=n;i++) read(a[i]),deg[a[i]]++;
for(int i=1,j;i<=n;i++){
if(vis[i]) continue;
for(j=i;!vis[j];j=a[j]) vis[j]=i;
if(vis[j]!=i) continue;
for(;!isc[j];j=a[j]) isc[j]=1;
}
for(int i=1;i<=n;i++)
if((isc[i]&°[i]>2)||(!isc[i]&°[i]>1)) puts("0"),exit(0);
for(int i=1,j;i<=n;i++){
if(deg[i]) continue;
int len=0;
for(j=i;!isc[j];j=a[j]) len++;
link[j]=len;
}
for(int i=1;i<=n;i++) if(isc[i]) solve(i);
for(int i=1;i<=n;i++){
if(!cnt[i]) continue;
int flag=((i>1)&&(i&1));
for(int j=1;j<=cnt[i];j++){
flag?f[j]=add(f[j-1],f[j-1]):f[j]=f[j-1];
if(j>1) f[j]=add(f[j],mul(f[j-2],mul(j-1,i)));
}
ans=mul(ans,f[cnt[i]]);
}
cout<<ans<<endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<algorithm>
#define N 200010
#define ll long long
#define mod 1000000007
using namespace std;
struct edge{int x, y, next;}a[N];
int n, p[N], x, l, flag[N], sum, s[N], cl, d[N], fa[N], ans, def[N], f[N], ff, last, g[N];
inline void add(int x, int y){a[++l].x=x; a[l].y=y; a[l].next=p[x]; p[x]=l;}
inline void dfs(int x, int FA){
flag[x]=1; sum++;
for(int i=p[x]; i; i=a[i].next)if(i!=(FA^1)){
if(flag[a[i].y]){cl=1; d[0]=x; d[1]=a[i].y;}
else{fa[a[i].y]=x; dfs(a[i].y, i);}
}
}
inline void dfs1(int x, int fa){
def[x]=0;
int sum=0;
for(int i=p[x]; i; i=a[i].next)if(fa!=a[i].y&&!f[a[i].y]){
sum++; dfs1(a[i].y, x); def[x]=def[a[i].y]+1;
}
if(sum>=2)ff=0;
}
int main(){
scanf("%d", &n);
l=1; memset(p, 0, sizeof(p));
for(int i=1; i<=n; i++){scanf("%d", &x); add(x, i); add(i, x);}
ans=1; memset(flag, 0, sizeof(flag)); memset(s, 0, sizeof(s)); memset(f, 0, sizeof(f));
for(int i=1; i<=n; i++)if(!flag[i]){
sum=0; dfs(i, 0);
x=d[1]; while(x!=d[0]){x=fa[x]; d[++cl]=x;}
if(sum==cl){s[sum]++; continue;}
for(int j=p[d[1]]; j; j=a[j].next)if(a[j].y==d[0])
if(j&1){for(int k=1; k<=cl/2; k++)swap(d[k], d[cl-k+1]); d[0]=d[cl];}
for(int j=1; j<=cl; j++)f[d[j]]=1;
ff=1;
for(int j=1; j<=cl&&ff; j++)dfs1(d[j], 0);
if(!ff){printf("0"); return 0;}
last=cl; while(!def[d[last]])last--;
for(int j=1; j<=cl; j++)if(def[d[j]]){
x=(j-last-1+cl)%cl;
if(def[d[j]]<=x)ans=ans*2%mod;
if(def[d[j]]>=x+2){printf("0"); return 0;}
last=j;
}
}
for(int i=1; i<=n; i++)if(s[i]){
x=1; if((i&1)&&i>=3)x++;
g[0]=1; g[1]=x;
for(int j=2; j<=s[i]; j++)g[j]=((ll)x*g[j-1]%mod+(ll)(j-1)*i%mod*g[j-2]%mod)%mod;
ans=(ll)ans*g[s[i]]%mod;
}
printf("%d", ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstring>
#define gc getchar()
using namespace std;
const int N=3e5+10,mod=1e9+7;
inline void qr(int &x)
{
x=0;char c=gc;int f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=gc;}
while(c>='0'&&c<='9'){x=x*10+(c^48);c=gc;}
x*=f;
}
void qw(int x)
{
if(x<0)x=-x,putchar('-');
if(x/10)qw(x/10);
putchar(x%10+48);
}
struct edge{int y,next;}e[N<<1];int len,last[N];
void ins(int x,int y){e[++len]=(edge){y,last[x]};last[x]=len;}
int n,fa[N];
int get(int x){return x==fa[x]?x:fa[x]=get(fa[x]);}
int ring[N],tot,num[N],a[N];bool v[N];
bool fring(int x,int ed)
{
if(x==ed){v[x]=1;ring[++tot]=x;return 1;}
for(int k=last[x],y;k;k=e[k].next)
{
y=e[k].y;
if(fring(y,ed)){v[x]=1;ring[++tot]=x;return 1;}
}
return 0;
}
int cnt,c[N],P[N],Q[N],sum[N];
void dfs(int x)
{
int sz=0;
for(int k=last[x],y;k;k=e[k].next)
{
y=e[k].y;
if(!v[y])++sz,++cnt,dfs(y);
}
if(sz>1){puts("0");exit(0);}
}
int main()
{
//freopen("dayinf.in","r",stdin);
//freopen("dayinf.out","w",stdout);
qr(n);
for(int i=1;i<=n;i++)qr(a[i]),fa[i]=i;
for(int i=1;i<=n;i++)
{
int p=get(i),q=get(a[i]);
if(p!=q)fa[p]=q,P[q]|=P[p],Q[q]|=Q[p],ins(a[i],i);
else P[q]=i,Q[q]=a[i];
}
int ans=1;
for(int i=1;i<=n;i++)if(fa[i]==i)
{
tot=0;fring(P[i],Q[i]);
for(int j=1;j<=tot;j++)cnt=0,dfs(ring[j]),num[j]=cnt;
reverse(num+1,num+tot+1);
int fi=-1,s1=1,s2=0,cal=0;
for(int j=1;j<=tot;j++)if(num[j]){fi=j;break;}
if(fi==-1){++sum[tot];continue;}
if(tot==1)
{
if(num[fi]>1)return puts("0"),0;
continue;
}
for(int j=tot+1;j<=3*tot;j++)num[j]=num[j-tot];
for(int j=fi+1;j<fi+num[fi];j++)if(num[j])return puts("0"),0;
for(int j=fi+num[fi],nxt=j+1;j<fi+tot;j=nxt,nxt=j+1)if(num[j])
{
if(j+num[j]-tot>fi){s1=0;return puts("0"),0;}
for(;nxt<j+num[j];nxt++)if(num[nxt]){s1=0;return puts("0"),0;}
if(num[j+num[j]])continue;
s1=1ll*2*s1%mod;
}
cal=(cal+s1)%mod;
if(!num[fi+num[fi]])
{
s2=1;
for(int j=fi+num[fi]+1,nxt=j+1;j<fi+tot;j=nxt,nxt=j+1)if(num[j])
{
if(j+num[j]-tot>fi+1){s2=0;return puts("0"),0;}
for(;nxt<j+num[j];nxt++)if(num[nxt]){s2=0;return puts("0"),0;}
if(num[j+num[j]])continue;
s2=1ll*2*s2%mod;
}
}
cal=(cal+s2)%mod;
ans=1ll*ans*cal%mod;
}
for(int i=1;i<=n;i++)if(sum[i])
{
if((i&1)&&i!=1)
{
c[0]=1;c[1]=2;
for(int j=2;j<=sum[i];j++)
c[j]=(c[j-1]*2+1ll*c[j-2]*(j-1)%mod*i%mod)%mod;
ans=1ll*ans*c[sum[i]]%mod;continue;
}
c[0]=1;c[1]=1;
for(int j=2;j<=sum[i];j++)
c[j]=(c[j-1]+1ll*c[j-2]*(j-1)%mod*i%mod)%mod;
ans=1ll*ans*c[sum[i]]%mod;
}
qw(ans);puts("");
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#define ll long long
#define P 1000000007
int a[100005],in[100005],col[100005],cnt[100005],n,foot[100005];bool cir[100005];long long dp[100005],ans=1;
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;++i) scanf("%d",&a[i]),++in[a[i]];
for (int i=1;i<=n;++i)
{
if (col[i]) continue;int x=i;
for(;!col[x];x=a[x]) col[x]=i;
if (col[x]!=i) continue;
for (;!cir[x];x=a[x]) cir[x]=1;
}
for (int i=1;i<=n;++i) if ((cir[i]&&in[i]>2)||(!cir[i]&&in[i]>1)) {printf("0");return 0;}
for (int i=1;i<=n;++i)
{
if (in[i]) continue;int x=i,tmp=0;
while(!cir[x]) x=a[x],tmp++;foot[x]=tmp;
}
for (int i=1;i<=n;++i)
{
if (!cir[i]) continue;int x=i,st=0,first=0,id=0,len=0;
while (cir[x])
{
++id;cir[x]=0;
if (foot[x])
{
if (!first){st=first=id;len=foot[x];x=a[x];continue;}
else
{
int numm=(foot[x]<(id-st))+(foot[x]<=(id-st));
(ans*=numm)%=P;if (!ans) {printf("0");return 0;}st=id;x=a[x];continue;
}
}
x=a[x];
}
if (first)
{
int numm=(len<(id+first-st))+(len<=(id+first-st));
(ans*=numm)%=P;if (!ans) {printf("0");return 0;}
}
else cnt[id]++;
}
for (int i=1;i<=n;++i)
{
dp[0]=1;if (!cnt[i]) continue;
if (i>1&&(i%2)) for (int j=1;j<=cnt[i];++j) {dp[j]=dp[j-1]*2%P;if (j>1) (dp[j]+=dp[j-2]*(j-1)*i%P)%=P;}
else for (int j=1;j<=cnt[i];++j){dp[j]=dp[j-1];if (j>1) (dp[j]+=dp[j-2]*(j-1)*i%P)%=P;}
(ans*=dp[cnt[i]])%=P;if (!ans){printf("0");return 0;}
}
printf("%lld",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#define SF scanf
#define PF printf
#define MAXN 100010
#define MOD 1000000007
typedef long long ll;
using namespace std;
int a[MAXN],pre[MAXN];
int d[MAXN],n,cnt[MAXN];
ll ans=1;
ll f[MAXN];
bool vis[MAXN];
int main(){
SF("%d",&n);
for(int i=1;i<=n;i++){
SF("%d",&a[i]);
d[a[i]]++;
}
for(int i=1;i<=n;i++){
if(d[i]>2){
PF("0");
return 0;
}
if(d[i]<2||vis[i]==1)
continue;
int p=a[i];
vis[i]=1;
pre[p]=i;
while(p!=i){
if(vis[p]==1){
PF("0");
return 0;
}
vis[p]=1;
pre[a[p]]=p;
p=a[p];
}
}
for(int i=1;i<=n;i++)
if(d[i]==0){
int p=i,l1=0,l2=1;
while(vis[p]==0){
vis[p]=1;
p=a[p];
l1++;
}
p=pre[p];
while(d[p]!=2){
p=pre[p];
l2++;
}
if(l1<l2)
ans=ans*2ll%MOD;
else if(l1>l2){
PF("0");
return 0;
}
}
for(int i=1;i<=n;i++)
if(vis[i]==0){
int len=0;
int p=i;
while(vis[p]==0){
len++;
vis[p]=1;
p=a[p];
}
cnt[len]++;
}
for(int i=1;i<=n;i++){
f[0]=1;
ll mul=i%2+1;
if(i==1)
mul=1;
f[1]=mul;
for(int j=2;j<=cnt[i];j++)
f[j]=(1ll*f[j-2]*(j-1)%MOD*i%MOD+f[j-1]*mul%MOD)%MOD;
ans=ans*f[cnt[i]]%MOD;
}
PF("%lld",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | import java.io.*;
import java.util.*;
public class Main {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
int solve(int[] p) {
int n = p.length;
int[] in = new int[n];
for (int x : p) {
in[x]++;
}
int[] vis = new int[n];
Arrays.fill(vis, -1);
boolean[] onCycle = new boolean[n];
for (int i = 0; i < n; i++) {
int v = i;
while (vis[v] == -1) {
vis[v] = i;
v = p[v];
}
if (vis[v] != i) {
continue;
}
onCycle[v] = true;
for (int u = p[v]; u != v; u = p[u]) {
onCycle[u] = true;
}
}
for (int i = 0; i < n; i++) {
if ((!onCycle[i] && in[i] > 1) || (onCycle[i] && (in[i] < 1 || in[i] > 2))) {
return 0;
}
}
int[] lenIn = new int[n];
for (int i = 0; i < n; i++) {
if (!onCycle[i] && in[i] == 0) {
int cnt = 0, v = i;
while (!onCycle[v]) {
cnt++;
v = p[v];
}
lenIn[v] = cnt;
}
}
int ans = 1;
for (int i = 0; i < n; i++) {
if (lenIn[i] != 0) {
int v = p[i], len = 1;
while (lenIn[v] == 0) {
v = p[v];
len++;
}
if (len < lenIn[v]) {
return 0;
}
if (len > lenIn[v]) {
ans = 2 * ans % P;
}
}
}
int[] cnt = new int[n + 1];
for (int i = 0; i < n; i++) {
if (onCycle[i]) {
boolean justCycle = lenIn[i] == 0;
int v = p[i], len = 1;
onCycle[v] = false;
while (v != i) {
justCycle &= lenIn[v] == 0;
v = p[v];
len++;
onCycle[v] = false;
}
if (justCycle) {
cnt[len]++;
}
}
}
for (int len = 1; len <= n; len++) {
ans = (int)((long)ans * go(len, cnt[len]) % P);
}
return ans;
}
int go(int len, int cnt) {
int[] dp = new int[cnt + 1];
dp[0] = 1;
int mult = len % 2 == 1 && len > 1 ? 2 : 1;
for (int i = 1; i <= cnt; i++) {
dp[i] = mult * dp[i - 1] % P;
if (i > 1) {
dp[i] += (int)((long)dp[i - 2] * (i - 1) % P * len % P);
if (dp[i] >= P) {
dp[i] -= P;
}
}
}
return dp[cnt];
}
static final int P = 1_000_000_007;
static final Random rng = new Random();
void test() {
int n = 9;
for (int i = 0; i < Integer.MAX_VALUE; i++) {
int[] a = new int[n];
for (int j = 0; j < n; j++) {
a[j] = rng.nextInt(n);
}
if (solve(a) != stupid(a)) {
throw new AssertionError(Arrays.toString(a));
}
System.err.println(i);
}
// int[] a = {2, 0, 1, 2, 3};
// System.err.println(solve(a));
// System.err.println(stupid(a));
}
int stupid(int[] need) {
int n = need.length;
int[] p = new int[n];
for (int i = 0; i < n; i++) {
p[i] = i;
}
int cnt = 0;
do {
boolean good = true;
for (int i = 0; i < n; i++) {
good &= (p[i] == need[i] || p[p[i]] == need[i]);
}
if (good) {
System.err.println(Arrays.toString(p));
}
cnt += good ? 1 : 0;
} while (nextPermutation(p));
return cnt;
}
static boolean nextPermutation(int[] a) {
int n = a.length;
int ptr = n - 2;
while (ptr >= 0 && a[ptr] > a[ptr + 1]) {
ptr--;
}
if (ptr < 0) {
return false;
}
for (int i = ptr + 1, j = n - 1; i < j; i++, j--) {
int tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
for (int i = ptr + 1;; i++) {
if (a[ptr] < a[i]) {
int tmp = a[ptr];
a[ptr] = a[i];
a[i] = tmp;
return true;
}
}
}
void submit() throws IOException {
int n = nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt() - 1;
}
out.println(solve(a));
}
Main() throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
submit();
// test();
out.close();
}
public static void main(String[] args) throws IOException {
new Main();
}
String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
eof = true;
return null;
}
}
return st.nextToken();
}
String nextString() {
try {
return br.readLine();
} catch (IOException e) {
eof = true;
return null;
}
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
} | JAVA |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#include<ctime>
#include<set>
#include<map>
#include<queue>
#include<stack>
#define sqr(x) ((x)*(x))
#define fz1(i,n) for ((i)=1;(i)<=(n);(i)++)
#define fd1(i,n) for ((i)=(n);(i)>=1;(i)--)
#define fz0g(i,n) for ((i)=0;(i)<=(n);(i)++)
#define fd0g(i,n) for ((i)=(n);(i)>=0;(i)--)
#define fz0k(i,n) for ((i)=0;(i)<(n);(i)++)
#define fd0k(i,n) for ((i)=(long long)((n)-1);(i)>=0;(i)--)
#define fz(i,x,y) for ((i)=(x);(i)<=(y);(i)++)
#define fd(i,y,x) for ((i)=(y);(i)>=(x);(i)--)
#define fzin fz1(i,n)
#define fzim fz1(i,m)
#define fzjn fz1(j,n)
#define fzjm fz1(j,m)
#define ff(c,itr) for (__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define rdst(st,len){static char ss[len];scanf(" %s",ss);(st)=ss;}
#define inc(x,y) {x+=(y);if(x>=mod)x-=mod;}
#define dec(x,y) {x-=(y);if(x<0)x+=mod;}
#define spln(i,n) (i==n?'\n':' ')
using namespace std;
int n,m,i,j,mod=1e9+7,vis[100005],icr[100005],a[100005],cnt[100005],ans,f[100005],g[100005];
vector<int> bi[100005];
vector<int> seq,nxt;
int qp(int x,int y)
{
int z=1;
while(y){
if(y&1){
z=1ll*z*x%mod;
}
y/=2;
x=1ll*x*x%mod;
}
return z;
}
void solve(int x)
{
int st=x;
for(;;){
int y=x,z,lst=0,l1=0,l2=0;
for(;;){
l1++;lst=y;y=a[y];vis[y]=1;
if(bi[y].size()==2) break;
}
z=bi[y][0]^bi[y][1]^lst;vis[z]=1;
l2=1;
while(!bi[z].empty()){
z=bi[z][0];l2++;vis[z]=1;
}
if(l2>l1){
puts("0");
exit(0);
}
if(l2<l1) ans=2ll*ans%mod;
if(y==st) return;x=y;
}
}
int main()
{
scanf("%d",&n);
fz1(i,n){
scanf("%d",&a[i]);
bi[a[i]].push_back(i);
}
fz1(i,n)if(!vis[i]){
int x=i;while(!vis[x]){
vis[x]=i;x=a[x];
}
if(vis[x]!=i) continue;
while(!icr[x]){
icr[x]=1;x=a[x];
}
}
fz1(i,n){
if(icr[i]){
if(bi[i].size()>2){
puts("0");
return 0;
}
}
else{
if(bi[i].size()>1){
puts("0");
return 0;
}
}
}
ans=1;
memset(vis,0,sizeof(vis));
fz1(i,n)if(bi[i].size()==2&&!vis[i]){
solve(i);
}
fz1(i,n)if(!vis[i]){
int x=i,c=0;while(!vis[x]){
vis[x]=1;x=a[x];c++;
}
cnt[c]++;
}
fz1(i,n){
f[0]=1;if((i&1)&&i>1) f[1]=2; else f[1]=1;
fz(j,2,cnt[i]){
if((i&1)&&i>1) f[j]=(f[j-1]+f[j-1]+1ll*f[j-2]*(j-1)%mod*i)%mod; else f[j]=(f[j-1]+1ll*f[j-2]*(j-1)%mod*i)%mod;
}
ans=1ll*ans*f[cnt[i]]%mod;
}
cout<<ans<<endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=1e5+5,mod=1e9+7;
int n,i,j,sz,cir;
int a[N],vis[N],dgr[N],foot[N],cnt[N],dis[N],nc[N];
long long ans,g[N];
void dfs(int x) {
vis[x]=1; sz++; if (dgr[a[x]]!=1 || nc[a[x]]) cir=0;
if (vis[a[x]]) return ;
dfs(a[x]);
}
void dfs2(int x,int step) {
if (dgr[a[x]]>1) {foot[a[x]]=step; return ;}
dfs2(a[x],step+1);
}
void dfs3(int x,int step) {
if (dgr[a[x]]==2) {dis[a[x]]=step; return;}
dfs3(a[x],step+1);
}
void dp(int n,int k)
{
int i;
g[0]=1;
rep(i,1,n)
{
if ((k%2) && (k>1)) g[i]=g[i-1]*2%mod;
else g[i]=g[i-1];
if (i>1) g[i]=(g[i]+g[i-2]*(i-1)%mod *k%mod)%mod;
}
ans=ans*g[n]%mod;
}
void calc(int x)
{
if (foot[x]<dis[x]) ans=ans*2%mod;
else if (foot[x]>dis[x]) ans=0;
}
int main()
{
//freopen("cycle.in","r",stdin);
//freopen("cycle.out","w",stdout);
scanf("%d",&n); ans=1;
rep(i,1,n) scanf("%d",&a[i]),dgr[a[i]]++;
rep(i,1,n) if (dgr[i]>2) { printf("0\n"); return 0; }
rep(i,1,n)
if (!vis[i])
{
sz=0; cir=1;
dfs(i);
if (cir) cnt[sz]++;
else nc[i]=1;
}
rep(i,1,n) if (!dgr[i]) dfs2(i,1);
rep(i,1,n) if (dgr[i]==2) dfs3(i,1);
rep(i,1,n) if (cnt[i]) dp(cnt[i],i);
rep(i,1,n) if (dgr[i]==2) calc(i);
printf("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
const
int Mod=(int)1e9+7;
using namespace std;
int deg[100001],a[100001],num[100001],l[100001],d[100001];
bool vis[100001];
int Fact[100001],Inv[100001];
inline int Pow(int a,int x)
{
int res=1;
for(;x;x>>=1,a=a*1ll*a%Mod)
if(x&1)res=a*1ll*res%Mod;
return res;
}
queue<int>Q;
inline int C(int x,int y){return Fact[x]*1ll*Inv[y]%Mod*Inv[x-y]%Mod;}
#define x first
#define y second
#define mp make_pair
#define pb push_back
vector<pair<int,int> >Cir;
inline int pr(int n)
{
return Fact[n<<1]*1ll*Inv[n]%Mod*Pow(Mod+1>>1,n)%Mod;
}
int main()
{
Fact[0]=1;
for(int i=1;i<=100000;i++)Fact[i]=Fact[i-1]*1ll*i%Mod;
Inv[100000]=Pow(Fact[100000],Mod-2);
for(int i=99999;~i;i--)Inv[i]=Inv[i+1]*(i+1ll)%Mod;
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",a+i),deg[a[i]]++;
for(int i=1;i<=n;i++)if(deg[i]>2)return puts("0"),0;
for(int i=1;i<=n;i++)if(!(d[i]=deg[i]))Q.push(i);
while(!Q.empty())
{
int u=Q.front();Q.pop();
vis[u]=1;
if(!--d[a[u]])Q.push(a[u]);
}
for(int i=1;i<=n;i++)if(vis[i]&°[i]>1)return puts("0"),0;
for(int i=1;i<=n;i++)if(vis[i]&&!deg[i])
{
int x=i,cnt=0;
for(;vis[x];x=a[x])cnt++;
l[x]=cnt;
}
int ans=1;
for(int i=1;i<=n;i++)if(!vis[i])
{
int x=i,fl=0,tp=1,tot=0;
for(;!vis[x];x=a[x])++tot,fl|=l[x],vis[x]=1;
if(!fl){num[tot]++;continue;}
Cir.clear();
int p=1;x=i;
do
{
if(l[x])Cir.pb(mp(l[x],p));
x=a[x];p++;
}while(x^i);
Cir.pb(mp(Cir[0].x,Cir[0].y+tot));
for(int j=1;j<Cir.size();j++)
if(Cir[j].x>Cir[j].y-Cir[j-1].y)tp=0;
else if(Cir[j].x^(Cir[j].y-Cir[j-1].y))tp=tp*2ll%Mod;
ans=ans*1ll*tp%Mod;
}
// printf("%d\n",ans);
for(int i=1;i<=n;i++)if(num[i])
{
int m=num[i],tp=0;
for(int j=0;j<=m/2;j++)
{
int tp2=1ll*C(m,2*j)*pr(j)%Mod*Pow(i,j)%Mod;
if(i>1&&(i&1))tp2=tp2*1ll*Pow(2,m-2*j)%Mod;
tp=(tp+tp2)%Mod;
}
ans=ans*1ll*tp%Mod;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100100
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define drp(x, a, b) for(int x=a; x>=b; x--)
int deg[N], d[N], n, vis[N], a[N], tot, num[N], l[N], ans=1, fac[N], inv[N];
vector<pair<int, int> > fu;
queue<int> q;
int power(int a, int k){
int ret=1;
while(k)
{
if(k&1) ret=1ll*ret*a%mod;
a=1ll*a*a%mod; k>>=1;
}
return ret;
}
int C(int n, int m){
return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int pr(int n){
return 1ll*fac[2*n]*inv[n]%mod*power((mod+1)>>1, n)%mod;
}
void init(){
fac[0]=1;
rep(i, 1, n) fac[i]=1ll*fac[i-1]*i%mod;
inv[n]=power(fac[n], mod-2);
drp(i, n, 1) inv[i-1]=1ll*inv[i]*i%mod;
}
int st[N<<1];
void renew(int &x, const int y){
x += y;
if(x < 0) x += mod;
if(x >= mod) x -= mod;
}
int getdp(){
int fr = 0;
rep(i, 1, tot) st[i + tot] = st[i];
rep(i, 1, tot) if(l[st[i]]){
fr = i;
break;
}
reverse(st + fr + 1, st + fr + tot);
int a = fr + l[st[fr]] - 1, b = a + 1, c = 1;
rep(i, fr + 1, fr + tot - 1)if(l[st[i]]){
int na = i + l[st[i]] - 1, nb = na + 1, cc = 0;
if(a < i) renew(cc, c);
if(b < i) renew(cc, c);
a = na;
b = nb;
c = cc;
}
int ans = 0;
if(a < fr + tot) renew(ans, c);
if(b < fr + tot) renew(ans, c);
return ans;
}
int que[N], h, t;
int main(){
scanf("%d", &n);
init();
rep(i, 1, n)
{
scanf("%d", a+i);
deg[a[i]]++;
}
rep(i, 1, n) if(deg[i]>2) return puts("0"), 0;
rep(i, 1, n) d[i]=deg[i];
rep(i, 1, n) if(!d[i]) q.push(i);
while(!q.empty())
{
int u=q.front(); q.pop();
vis[u]=1;
if(--d[a[u]]==0) q.push(a[u]);
}
rep(i, 1, n) if(vis[i] && deg[i]>1) return puts("0"), 0;
rep(i, 1, n) if(vis[i] && !deg[i])
{
int x=i, cnt=0;
for(; vis[x]; x=a[x]) cnt++;
l[x]=cnt;
}
rep(i, 1, n) if(!vis[i])
{
int x=i, fl=0, ass=1; tot=0;
for(; !vis[x]; x=a[x]) st[++tot]=x, fl|=bool(l[x]), vis[x]=1;
if(!fl) num[tot]++;
else
{/*
int p=0; fu.clear();
if(l[i]) fu.pb(mp(l[i], 0));
for(p++, x=a[i]; x!=i; x=a[x]) if(l[x]) fu.pb(mp(l[x], p));
fu.pb(mp(fu[0].fi, fu[0].se+tot));
int sz=fu.size();
rep(i, 1, sz-1)
{
if(fu[i].fi>fu[i].se-fu[i-1].se) ass=0;
if(fu[i].fi<fu[i].se-fu[i-1].se) ass=2ll*ass%mod;
}*/
ass=getdp();
}
ans=1ll*ans*ass%mod;
}
rep(i, 1, n) if(num[i])
{
int m=num[i], ass=0;
rep(j, 0, m/2)
{
int tmp=1;
tmp=1ll*tmp*power(i, j)%mod;
tmp=1ll*tmp*C(m, 2*j)%mod;
tmp=1ll*tmp*fac[2*j]%mod;
tmp=1ll*tmp*inv[j]%mod;
tmp=1ll*tmp*power((mod+1)>>1, j)%mod;
if(i>1 && (i&1))
tmp=1ll*tmp*power(2, m-2*j)%mod;
ass=(ass+tmp)%mod;
}
ans=1ll*ans*ass%mod;
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10,mod=1e9+7;
int n,a[maxn],cnt[maxn];
int dfsclock,dfn[maxn],low[maxn],stk[maxn],top,scccnt,scc[maxn],size[maxn];
ll ans,fac[maxn],inv[maxn],f[maxn],pow2[maxn];
int type[maxn],weight[maxn];
vector<int> g[maxn];
inline ll qpow(ll a,ll n){
ll res=1;
while(n){
if(n&1ll)
res=res*a%mod;
a=a*a%mod;
n>>=1;
}
return res;
}
inline void init(){
fac[0]=1;for(int i=1;i<maxn;++i)fac[i]=fac[i-1]*i%mod;
inv[maxn-1]=qpow(fac[maxn-1],mod-2);for(int i=maxn-1;i;--i)inv[i-1]=inv[i]*i%mod;
f[0]=1;for(int i=1;i<maxn;++i)f[i]=f[i-1]*(2*i-1)%mod;
pow2[0]=1;for(int i=1;i<maxn;++i)pow2[i]=pow2[i-1]*2%mod;
}
void dfs(int pos){
dfn[pos]=low[pos]=++dfsclock;
stk[++top]=pos;
if(!dfn[a[pos]]){
dfs(a[pos]);
low[pos]=min(low[pos],low[a[pos]]);
}
else if(!scc[a[pos]])
low[pos]=min(low[pos],dfn[a[pos]]);
if(dfn[pos]<=low[pos]){
++scccnt;
do{
scc[stk[top]]=scccnt;
++size[scccnt];
}
while(stk[top--]!=pos);
}
}
int main(){
init();
cin>>n;
for(int i=1;i<=n;++i)
cin>>a[i];
for(int i=1;i<=n;++i)
if(!dfn[i])
dfs(i);
for(int i=1;i<=n;++i)
if(scc[a[i]]==scc[i])
type[scc[i]]=1;
else
g[a[i]].push_back(i),weight[scc[a[i]]]=true;
for(int i=1;i<=n;++i)
if(g[i].size()>1){
puts("0");
return 0;
}
for(int i=1;i<=scccnt;++i)
if(type[i]==1){
if(weight[i])
type[i]=2;
else
++cnt[size[i]];
}
ans=1;
for(int i=1;i<=n;++i)
if(type[scc[i]]==2){
type[scc[i]]=0;
static int ring[maxn];
int len;ll tmp=1;
ring[len=1]=i;
for(int p=a[i];p!=i;p=a[p])
ring[++len]=p;
for(int j=1;j<=len;++j)
if(g[ring[j]].size()){
int p=ring[j],val=0;
while(g[p].size())
++val,p=g[p][0];
int nxt=1;
for(int k=j==1?len:j-1;g[ring[k]].empty();k=k==1?len:k-1)
++nxt;
if(nxt>val)
tmp=tmp*2%mod;
if(nxt<val){
puts("0");
return 0;
}
}
ans=ans*tmp%mod;
}
for(int i=1;i<=n;++i)
if((i&1)&&i>1){
ll tmp=0;
for(int j=0;2*j<=cnt[i];++j)
tmp=(tmp+fac[cnt[i]]*inv[2*j]%mod*inv[cnt[i]-2*j]%mod*f[j]%mod*qpow(i,j)%mod*pow2[cnt[i]-2*j])%mod;
ans=ans*tmp%mod;
}
else{
ll tmp=0;
for(int j=0;2*j<=cnt[i];++j)
tmp=(tmp+fac[cnt[i]]*inv[2*j]%mod*inv[cnt[i]-2*j]%mod*f[j]%mod*qpow(i,j))%mod;
ans=ans*tmp%mod;
}
cout<<ans<<endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100005, mod = 1e9 + 7;
inline int gi()
{
char c = getchar();
while (c < '0' || c > '9') c = getchar();
int sum = 0;
while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
return sum;
}
inline int add(int x, int y) {return x + y >= mod ? x + y - mod : x + y;}
int n, ans, a[maxn], cir[maxn], flen[maxn], sum[maxn], f[maxn], deg[maxn], vis[maxn];
void solvecir(int x)
{
int len = 0, s = 0, lst = 0, slen = 0;
while (cir[x]) {
++len; cir[x] = 0;
if (flen[x]) {
if (!s) s = lst = len, slen = flen[x];
else {
ans = (ll)ans * ((flen[x] < len - lst) + (flen[x] <= len - lst)) % mod;
lst = len;
}
}
x = a[x];
}
if (!s) ++sum[len];
else ans = (ll)ans * ((slen < len - lst + s) + (slen <= len - lst + s)) % mod;
}
void solve()
{
for (int i = 1; i <= n; ++i) {
if (deg[i]) continue;
int x = i, len = 0;
while (!cir[x]) x = a[x], ++len;
flen[x] = len;
}
ans = 1;
for (int i = 1; i <= n; ++i) if (cir[i]) solvecir(i);
for (int i = 1; i <= n; ++i) {
if (!sum[i]) continue;
f[0] = 1;
for (int j = 1; j <= sum[i]; ++j) {
if (i > 1 && (i & 1)) f[j] = add(f[j - 1], f[j - 1]);
else f[j] = f[j - 1];
if (j > 1) f[j] = (f[j] + (ll)f[j - 2] * (j - 1) % mod * i) % mod;
}
ans = (ll)ans * f[sum[i]] % mod;
}
}
int main()
{
n = gi();
for (int i = 1; i <= n; ++i) a[i] = gi(), ++deg[a[i]];
for (int i = 1; i <= n; ++i) {
if (vis[i]) continue;
int x = i; while (!vis[x]) vis[x] = i, x = a[x];
if (vis[x] != i) continue;
while (!cir[x]) cir[x] = 1, x = a[x];
}
for (int i = 1; i <= n; ++i)
if ((cir[i] && deg[i] > 2) || (!cir[i] && deg[i] > 1)) return puts("0"), 0;
solve();
printf("%d\n", ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e5 + 50;
const LL mod = 1e9 + 7;
vector <int> G[N], cir;
int n, a[N], sz, pt;
int vis[N], cur;
int len[N], cnt[N];
LL fac[N], inv[N];
inline void Dfs(int x) {
if (vis[x]) {
if (vis[x] != cur) sz = 0;
else pt = x;
return;
}
vis[x] = cur; sz ++;
Dfs(a[x]);
}
inline void Dfs1(int x) {
if (vis[x] == cur) return;
cir.push_back(x);
vis[x] = cur;
Dfs1(a[x]);
}
inline int iDfs(int x) {
if (G[x].size() > 1u) return (int) -1e9;
if (!G[x].size()) return 1;
return 1 + iDfs(G[x][0]);
}
inline LL Pow(LL x, LL exp) {
LL r = 1;
for (; exp; exp >>= 1, x = x * x % mod)
if (exp & 1) r = r * x % mod;
return r;
}
inline void Init() {
for (int i = fac[0] = 1; i < N; i ++)
fac[i] = fac[i - 1] * i % mod;
inv[N - 1] = Pow(fac[N - 1], mod - 2);
for (int i = N - 1; i; i --)
inv[i - 1] = inv[i] * i % mod;
}
inline LL C(int x, int y) {
return fac[x] * inv[y] % mod * inv[x - y] % mod;
}
int main() {
Init();
scanf("%d", &n);
for (int i = 1; i <= n; i ++) {
scanf("%d", &a[i]);
G[a[i]].push_back(i);
}
LL res = 1;
cir.reserve(n);
for (int i = 1; i <= n; i ++)
if (!vis[i]) {
sz = 1; cur ++;
Dfs(i); if (!sz) continue;
cur ++; cir.clear();
Dfs1(pt);
reverse(cir.begin(), cir.end());
int tot = 0, le = cir.size();
for (int j = 0; j < le; j ++) {
vector <int> &Gi = G[cir[j]];
if (Gi.size() > 2u) return 0 * puts("0");
if (Gi.size() == 1u) len[j + 1] = 0;
else tot += (len[j + 1] = iDfs(Gi[vis[Gi[0]] == cur]));
if (len[j + 1] < 0) return 0 * puts("0");
}
if (!tot) {
cnt[le] ++; continue;
}
int dis = 0;
for (int j = 1; j <= le; j ++)
if (len[j]) {
dis = j - 1; break;
}
for (int j = le; j; j --) {
dis ++;
if (!len[j]) continue;
if (dis > len[j]) {
res <<= 1;
if (res >= mod) res -= mod;
}
else if (dis < len[j]) return 0 * puts("0");
dis = 0;
}
}
for (int i = 1; i <= n; i ++) {
if (!cnt[i]) continue;
LL t = ((i & 1) && i != 1) + 1;
LL tmp = Pow(t, cnt[i]);
for (int j = (cnt[i] >> 1); j; j --)
tmp += C(cnt[i], j << 1) * C(j << 1, j) % mod * Pow((mod + 1) >> 1, j) % mod * fac[j] % mod * Pow(i, j) % mod * Pow(t, (LL) cnt[i] - (j << 1)) % mod;
res = res * (tmp % mod) % mod;
}
printf("%lld\n", res);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005,MOD=1000000007;
int n,A[MAXN],deg[MAXN],pre[MAXN];
int cnt[MAXN],dp[MAXN];
int ans=1;
bool vis[MAXN];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&A[i]);
deg[A[i]]++;
}
for(int i=1;i<=n;i++)
{
if(deg[i]>2)
{puts("0");return 0;}
if(deg[i]<2||vis[i])
continue;
int u=i;
do
{
if(vis[u])
{puts("0");return 0;}
vis[u]=true;
pre[A[u]]=u;
u=A[u];
}while(u!=i);
}
for(int i=1;i<=n;i++)
if(deg[i]==0)
{
int u=i;
int len1=0,len2=0;
while(!vis[u])
{
len1++;
vis[u]=true;
u=A[u];
}
do
{
len2++;
u=pre[u];
}while(deg[u]!=2);
if(len2<len1)
{puts("0");return 0;}
else if(len2>len1)
ans=ans*2%MOD;
}
for(int i=1;i<=n;i++)
if(!vis[i])
{
int len=0;
int u=i;
do
{
len++;
vis[u]=true;
u=A[u];
}while(u!=i);
cnt[len]++;
}
for(int i=1;i<=n;i++)
{
int mul=1;
if(i!=1&&i%2==1)
mul=2;
dp[0]=1;
dp[1]=mul;
for(int j=2;j<=cnt[i];j++)
dp[j]=(1LL*dp[j-2]*(j-1)%MOD*i%MOD+1LL*dp[j-1]*mul%MOD)%MOD;
ans=1LL*ans*dp[cnt[i]]%MOD;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | //In the Name of God
//Ya Ali
#include<bits/stdc++.h>
#define int long long
#define pb push_back
#define err(A) cout<<#A<<" = "<<(A)<<endl
using namespace std;
const int M=1e9+7;
const int maxn=100100;
int n;
int a[maxn];
vector<int> g[maxn];
bool vis[maxn];
int cnt[maxn];
int dp[maxn];
int ans=1;
void gn()
{
cout<<ans<<endl;
exit(0);
}
int len(int v)
{
int ret=1;
int c=0;
vis[v]=true;
for(int u:g[v])
if(!vis[u])
{
c++;
ret+=len(u);
}
if(c>1)
ans=0;
return ret;
}
void solve(int v)
{
vector<int> q,p;
while(!vis[v])
{
vis[v]=true;
q.pb(v);
v=a[v];
}
for(int u:q)
vis[u]=false;
q.resize(0);
while(!vis[v])
{
vis[v]=true;
q.pb(v);
v=a[v];
}
p.resize(q.size());
int prv=-1;
for(int i=0;i<q.size();i++)
if(g[q[i]].size()==2)
prv=i;
if(prv==-1)
{
cnt[q.size()]++;
return;
}
for(int i=0;i<q.size();i++)
{
p[i]=prv;
if(g[q[i]].size()==2)
{
int l1=len(q[i])-1;
int l2=(i-prv+q.size())%q.size();
if(l2==0) l2=q.size();
if(l1>l2)
ans=0;
if(l1<l2)
ans=ans*2%M;
prv=i;
}
}
}
int32_t main()
{
ios::sync_with_stdio(0);cin.tie(0);
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i],a[i]--;
for(int i=0;i<n;i++)
g[a[i]].pb(i);
for(int i=0;i<n;i++)
if(g[i].size()>2)
ans=0,gn();
for(int i=0;i<n;i++)
if(!vis[i])
solve(i);
for(int i=1;i<=n;i++)
{
dp[0]=1;
dp[1]=1+(i>2 and i&1);
for(int j=2;j<=cnt[i];j++)
dp[j]=(dp[j-1]*dp[1]+dp[j-2]*(j-1)*i)%M;
ans=ans*dp[cnt[i]]%M;
}
gn();
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
typedef long long LL;
const int N=1e5+100;
int gi() {
int w=0;bool q=1;char c=getchar();
while ((c<'0'||c>'9') && c!='-') c=getchar();
if (c=='-') q=0,c=getchar();
while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
return q? w:-w;
}
int p[N],vis[N],st[N],len[N],tot[N];
int head[N],nxt[N];
bool rt[N];
int f[N],g[N];
inline int dfs(int k) {
int s=0;
for (int i=head[k];i;i=nxt[i])
if (!rt[i]) {
if (s) puts("0"),exit(0);
s=dfs(i)+1;
}
return s;
}
int main()
{
const int mod=1e9+7;
int n=gi(),i,k,t,s,cnt=0,top,ans=1,p2=0;
bool is;
for (i=1;i<=n;i++) p[i]=gi(),nxt[i]=head[p[i]],head[p[i]]=i;
for (i=1;i<=n;i++)
if (!vis[i]) {
++cnt;
for (k=i;!vis[k];k=p[k])
vis[k]=cnt;
if (vis[k]!=cnt) continue;
for (t=k,top=0;st[1]!=t;t=p[t]) rt[st[++top]=t]=true;
reverse(st+1,st+1+top);
is=false;
for (k=1;k<=top;k++)
is|=len[k]=dfs(st[k]);
if (is) {
for (k=1;k<=top;k++)
if (len[k]) {
for (t=k%top+1,s=1;!len[t];t=t%top+1,s++);
if (s<len[k]) return puts("0"),0;
if (s!=len[k]) p2++;
}
} else tot[top]++,p2+=top>1&&(top&1);
}
const int inv4=1LL*(mod+1)*(mod+1)/4%mod;
for (i=1;i<=n;i++) {
for (k=f[0]=f[1]=g[0]=g[1]=1;k<=tot[i];k++) {
f[k+1]=(1LL*f[k-1]*k%mod*inv4%mod*i+f[k])%mod;
g[k+1]=(1LL*g[k-1]*k%mod*i+g[k])%mod;
}
ans=1LL*ans*(i==1||i%2==0?g[tot[i]]:f[tot[i]])%mod;
}
while (p2--) (ans<<=1)%=mod;
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define MAXN 100005
#define P 1000000007
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
int n, a[MAXN], d[MAXN], cnt[MAXN], p[MAXN], len[MAXN];
bool visited[MAXN], used[MAXN], special[MAXN];
deque <int> Q;
long long dp[MAXN];
void add(long long &x, long long y) {
x = (x + y) % P;
}
void times(long long &x, long long y) {
x = x * y % P;
}
int main() {
read(n);
for (int i = 1; i <= n; i++)
read(a[i]), d[a[i]]++;
for (int i = 1; i <= n; i++)
if (d[i] == 0) {
visited[i] = true;
Q.push_back(i);
}
while (!Q.empty()) {
int tmp = Q.front();
Q.pop_front();
d[a[tmp]]--;
if (d[a[tmp]] == 0) {
visited[a[tmp]] = true;
Q.push_back(a[tmp]);
}
}
memset(d, 0, sizeof(d));
for (int i = 1; i <= n; i++)
d[a[i]]++;
for (int i = 1; i <= n; i++)
if (visited[i] && d[i] >= 2 || !visited[i] && d[i] >= 3) {
printf("0\n");
return 0;
}
for (int i = 1; i <= n; i++)
if (!visited[i]) {
p[a[i]] = i;
if (!used[i]) {
int pos = a[i], mx = d[i], len = 1;
used[i] = true;
while (pos != i) {
mx = max(mx, d[pos]);
len++; used[pos] = true;
pos = a[pos];
}
if (mx <= 1) cnt[len]++;
else {
pos = a[i]; special[i] = true;
while (pos != i) {
special[pos] = true;
pos = a[pos];
}
}
}
}
long long ans = 1;
for (int i = 1; i <= n; i++) {
if (cnt[i] == 0) continue;
dp[0] = 1;
for (int j = 1; j <= cnt[i]; j++) {
dp[j] = dp[j - 1];
if (i & 1 && i > 1) add(dp[j], dp[j - 1]);
if (j > 1) add(dp[j], dp[j - 2] * (j - 1) % P * i);
}
times(ans, dp[cnt[i]]);
}
for (int i = 1; i <= n; i++) {
if (d[i]) continue;
int pos = i, l = 0;
while (!special[pos]) {
l++, pos = a[pos];
}
len[pos] = l;
}
for (int i = 1; i <= n; i++) {
if (len[i] == 0) continue;
int pos = p[i], l = 1;
while (!len[pos]) {
l++, pos = p[pos];
}
if (l > len[i]) times(ans, 2);
else if (l < len[i]) ans = 0;
}
cout << ans << endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define N 120000
const LL mod=1000000007;
LL n,ans,a[N],b[N],d[N],cnt[N],f[N];
bool vis[N];
int main(){
scanf("%lld",&n);
for (LL i=1;i<=n;++i){
scanf("%lld",a+i);
++d[a[i]];
}
for (LL i=1;i<=n;++i){
if (d[i]>2){puts("0"); return 0;}
if (d[i]<2||vis[i]) continue;
vis[i]=1; b[a[i]]=i;
for (LL j=a[i];j!=i;j=a[j]){
if (vis[j]){puts("0"); return 0;}
vis[j]=1; b[a[j]]=j;
}
}
ans=1;
for (LL i=1;i<=n;++i)
if (!d[i]){
LL j=i,len1=0,len2=1;
for (;!vis[j];j=a[j]){vis[j]=1; ++len1;}
for (j=b[j];d[j]<2;j=b[j]) ++len2;
if (len1>len2){puts("0"); return 0;}
if (len1<len2) ans=ans*2%mod;
}
for (LL i=1;i<=n;++i)
if (!vis[i]){
vis[i]=1;
LL len=1;
for (LL j=a[i];!vis[j];j=a[j]){vis[j]=1; ++len;}
++cnt[len];
}
for (LL i=1;i<=n;++i){
LL t=i!=1&&(i&1)?2:1;
f[0]=1; f[1]=t;
for (int j=2;j<=cnt[i];++j) f[j]=(f[j-1]*t+f[j-2]*(j-1)%mod*i)%mod;
ans=ans*f[cnt[i]]%mod;
}
printf("%lld\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int N=100100,mod=1000000007;
typedef long long ll;
int a[N],d[N],n,vis[N],ans=1,cnt[N],b[N];
void gofail(){cout<<0<<'\n';exit(0);}
int main(){
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin>>n;
for(int i=1;i<=n;++i)cin>>a[i],d[a[i]]++;//,b[a[i]]=i;
for(int i=1;i<=n;++i){
if(d[i]>2)gofail();
if(d[i]<2||vis[i])continue;
vis[i]=1;b[a[i]]=i;
for(int j=a[i];j!=i;j=a[j]){
if(vis[j])gofail();
vis[j]=1;b[a[j]]=j;
}
}
for(int i=1;i<=n;++i)
if(!d[i]){
int cha=0,cir=1,j=i;
for(;!vis[j];j=a[j])vis[j]=1,++cha;
for(j=b[j];d[j]<2;j=b[j])++cir;
if(cha<cir)ans=ans*2%mod;
else if(cha>cir)gofail();
}
for(int i=1;i<=n;++i)
if(!vis[i]){
int L=1;vis[i]=1;
for(int j=a[i];j!=i;j=a[j])vis[j]=1,++L;
cnt[L]++;
}
for(int i=1;i<=n;++i)if(cnt[i]){
int t=2;if(i==1||(i%2==0))t=1;
int u=1,v=t;
for(int j=2;j<=cnt[i];++j){
int w=((ll)u*(j-1)%mod*i+(ll)v*t)%mod;
u=v;v=w;
}
ans=(ll)ans*v%mod;
}
cout<<ans<<'\n';
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | //Heaplax
//别让自己后悔
#include<bits/stdc++.h>
#define N 100005
#define LL long long
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
while(ch=getchar(),ch<33);
if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
while(ch=getchar(),ch>33)x=x*10+ch-48;
if(fs)x=-x;
}
using namespace std;
const int mod=1000000007;
vector<int> cir[N];
int n,ans=1,cnt,a[N],d[N],sy[N],fl[N],sum[N],f[N];
int vis[N];
int main()
{
re(n);
for(int i=1;i<=n;++i)
re(a[i]),++d[a[i]];
for(int i=1;i<=n;++i)
if(!d[i])
{
static int st[N];
int top=0,now=i;
while(!vis[now])
{
st[++top]=now;
vis[now]=i;
now=a[now];
}
if(sy[now])
{
if(fl[now])return puts("0"),0;
fl[now]=top;
}
else if(vis[now]!=i)return puts("0"),0;
else
{
++cnt;
while(st[top]!=now)
{
cir[cnt].push_back(st[top]);
sy[st[top]]=cnt;
--top;
}
--top;
cir[cnt].push_back(now);
sy[now]=cnt;
fl[now]=top;
}
}
for(int i=1;i<=n;++i)
if(!vis[i])
{
int now=i,len=0;
do
{
++len;
vis[now]=1;
now=a[now];
}
while(now!=i);
++sum[len];
}
// LOG(ans);
for(int i=1;i<=n;++i)
{
f[0]=1;
for(int j=1;j<=sum[i];++j)
{
f[j]=f[j-1];
if((i&1) && i>1)f[j]=(f[j]+f[j-1])%mod;
if(j>=2)f[j]=(f[j]+(j-1ll)*i%mod*f[j-2])%mod;
}
ans=1ll*ans*f[sum[i]]%mod;
// LOG(ans);
}
// LOG(ans);
// LOG(cnt);
for(int i=1;i<=cnt;++i)
{
// for(int j:cir[i])cerr<<fl[j]<<" ";
// cerr<<endl;
reverse(cir[i].begin(),cir[i].end());
int last=0;
for(int j=1;j<cir[i].size();++j)if(fl[cir[i][j]])
{
if(j-last>fl[cir[i][j]])ans=ans*2%mod;
else if(j-last<fl[cir[i][j]])return puts("0"),0;
last=j;
}
// LOG(ans);
if(cir[i].size()-last>fl[cir[i][0]])ans=ans*2%mod;
else if(cir[i].size()-last<fl[cir[i][0]])return puts("0"),0;
}
printf("%d\n",ans);
}
/*
1 2
2 1
3 4
4 3
5 6
6 7
7 5
8 9
9 10
10 8
11 8
12 9
13 11
*/
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstring>
using namespace std;
const int N=1e5+10,mod=1e9+7;
int n,a[N],d[N],cir[N],f[N],vis[N],sum[N],foot[N],Ans=1;
void GetTree(int x)
{
int t=0,bs=1,fl=0;
for(vis[x]=1,x=a[x];!fl;x=a[x])
{
if(vis[x]) fl=1;vis[x]=1;t++;
if(foot[x]) bs=bs*(t<foot[x]?0:(t==foot[x]?1:2))%mod,t=0;
}
Ans=1ll*Ans*bs%mod;
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++) cin>>a[i],d[a[i]]++;
for(int i=1;i<=n;i++)
{
if(vis[i]) continue;
int x=i;while(!vis[x]) vis[x]=i,x=a[x];
if(vis[x]!=i) continue;
while(!cir[x]) cir[x]=1,x=a[x];
}
for(int i=1;i<=n;i++)
if((cir[i]&&d[i]>2)||(!cir[i]&&d[i]>1)) return puts("0"),0;
for(int i=1;i<=n;i++)
if(d[i]==0)
{
int x=i,t=0;
while(!cir[x]) t++,x=a[x];
foot[x]=t;
}
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
if(cir[i]&&foot[i]&&!vis[i]) GetTree(i);
for(int i=1;i<=n;i++)
if(!vis[i]&&cir[i])
{
int x=i,t=0;
while(!vis[x]) vis[x]=1,t++,x=a[x];
sum[t]++;
}
for(int w=1,u;w<=n;w++)
{
f[0]=1;u=sum[w];
if(!u) continue;
for(int i=1;i<=u;i++)
{
f[i]=f[i-1]*(((w&1)&&w>1)?2:1)%mod;
if(i>1) f[i]=(f[i]+1ll*(i-1)*f[i-2]%mod*w%mod)%mod;
}
Ans=1ll*Ans*f[u]%mod;
}
cout<<Ans<<endl;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
int w[101000], Deg[101000], n, chk[101000], Q[101000], tail, head, C[101000];
bool v[101000];
long long Res = 1, D[101000], Mod = 1000000007;
vector<int>E[101000];
int main(){
int i, x, y, j;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d",&w[i]);
E[w[i]].push_back(i);
Deg[w[i]]++;
if(Deg[w[i]] == 3) goto fail;
}
for(i=1;i<=n;i++)if(!Deg[i])Q[++tail] = i;
while(head < tail){
x = Q[++head];
Deg[w[x]]--;
if(!Deg[w[x]]) Q[++tail] = w[x];
}
for(i=1;i<=tail;i++)chk[Q[i]] = 1;
for(i=1;i<=n;i++){
if(chk[i] && E[i].size() == 2)goto fail;
if(chk[i]){
x=i;
while(!v[x]){v[x]=true;x=w[x];}
}
if(E[i].size() == 2){
x = E[i][0], y = E[i][1];
while(E[x].size()*E[y].size()==1) x = E[x][0], y = E[y][0];
if(E[x].size() + E[y].size() == 2)continue;
if(E[x].size() + E[y].size() > 2)goto fail;
Res = Res * 2 % Mod;
}
}
for(i=1;i<=n;i++){
if(!v[i]){
x=i;y=0;
while(!v[x]){y++;v[x]=true;x=w[x];}
C[y]++;
}
}
for(i=1;i<=n;i++){
if(!C[i])continue;
D[0] = 1;
for(j=1;j<=C[i];j++){
D[j] = D[j-1];
if(i!=1&&i%2==1)D[j]=D[j]*2%Mod;
if(j!=1)D[j] = (D[j] + D[j-2] * (j-1) % Mod * i)%Mod;
}
Res = Res * D[C[i]] % Mod;
}
printf("%lld\n",Res);
return 0;
fail:
puts("0");
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head
const int N=101000;
VI vp,e[N],s[N];
int vis[N],cyc[N],inc[N],a[N],pre[N],n,cycnt[N];
ll ret,f[N];
void dfs(int u) {
vp.pb(u); vis[u]=1;
for (auto v:e[u]) if (!vis[v]) dfs(v);
}
bool check(VI &v) {
int c=v[0];
for (auto u:v) vis[u]=0;
int m=0;
while (1) {
vis[c]=1;
c=a[c];
if (vis[c]==1) break;
}
int d=c;
while (1) {
cyc[m++]=c;
inc[c]=1;
c=a[c];
if (c==d) break;
}
for (auto u:v) vis[u]=1;
if (m==SZ(v)) {
cycnt[m]++;
return 1;
}
for (auto u:v) {
if (SZ(s[u])>2) return 0;
if (SZ(s[u])==2&&!inc[u]) return 0;
}
rep(i,0,m) pre[cyc[i]]=cyc[(i+m-1)%m];
rep(i,0,m) {
int u=cyc[i];
if (SZ(s[u])==1) continue;
int nbr=(s[u][0]==pre[u])?s[u][1]:s[u][0];
VI chain; chain.pb(nbr);
while (1) {
if (SZ(s[nbr])==0) break;
assert(SZ(s[nbr])==1);
nbr=s[nbr][0];
chain.pb(nbr);
}
bool v1=1,v2=1; nbr=pre[u];
rep(j,0,SZ(chain)-1) {
if (SZ(s[nbr])==2) v1=0,v2=0;
nbr=pre[nbr];
}
if (SZ(s[nbr])==2) v2=0;
if (v1==0) return 0;
if (v2==1) ret=ret*2%mod;
}
return 1;
}
ll fac[N],fnv[N];
ll comb(int x,int y) {
return fac[x]*fnv[y]%mod*fnv[x-y]%mod;
}
int main() {
ret=1;
scanf("%d",&n);
rep(i,1,n+1) scanf("%d",a+i);
fac[0]=fnv[0]=1;
rep(i,1,n+1) fac[i]=fac[i-1]*i%mod,fnv[i]=powmod(fac[i],mod-2);
rep(i,1,n+1) {
e[i].pb(a[i]); e[a[i]].pb(i);
s[a[i]].pb(i);
}
rep(i,1,n+1) if (!vis[i]) {
vp.clear();
dfs(i);
if (!check(vp)) {
puts("0");
return 0;
}
}
// deal with cycle
f[0]=1;
rep(i,1,n+1) f[i]=f[i-1]*(2*i-1)%mod;
rep(i,1,n+1) if (cycnt[i]!=0) {
ll r=0;
for (int j=0;2*j<=cycnt[i];j++) {
r=(r+comb(cycnt[i],2*j)*f[j]%mod*powmod(i,j)%mod*powmod((i%2==1&&i!=1)?2:1,cycnt[i]-2*j))%mod;
}
ret=ret*r%mod;
}
printf("%lld\n",ret);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int maxn=100010,mod=1000000007,inv2=500000004;
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline int read(){
int x=0,f=0;char ch=getchar();
while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
return f?-x:x;
}
int n,a[maxn],deg[maxn],ans=1,stk[maxn],tp,q[maxn],h,r,len[maxn],seq[maxn],tot,cnt[maxn];
int fac[maxn],inv[maxn],invfac[maxn];
bool vis[maxn],ins[maxn],cyc[maxn];
void dfs(int u){
if(vis[u]){
if(ins[u]){
ROF(i,tp,1){
cyc[stk[i]]=true;
if(stk[i]==u) break;
}
}
return;
}
vis[u]=ins[u]=true;
stk[++tp]=u;
dfs(a[u]);
ins[u]=false;
}
bool dfs2(int u){
if(vis[u]) return true;
vis[u]=true;
seq[++tot]=len[u];
return dfs2(a[u]) && !len[u];
}
inline int C(int n,int m){
return 1ll*fac[n]*invfac[m]%mod*invfac[n-m]%mod;
}
inline int qpow(int a,int b){
int ans=1;
for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) ans=1ll*ans*a%mod;
return ans;
}
int main(){
n=read();
FOR(i,1,n) a[i]=read(),deg[a[i]]++;
FOR(i,1,n) dfs(i);
FOR(i,1,n) if(cyc[i] && deg[i]>=3|| !cyc[i] && deg[i]>=2) return puts("0"),0;
h=1;r=0;
FOR(i,1,n) if(!deg[i]) q[++r]=i;
while(h<=r){
int u=q[h++];
len[a[u]]=len[u]+1;
if(!cyc[a[u]]) q[++r]=a[u];
}
MEM(vis,0);
FOR(i,1,n) if(cyc[i] && !vis[i]){
tot=0;
if(dfs2(i)) cnt[tot]++;
else{
int pre=0;
FOR(j,1,tot) if(seq[j]){
if(pre){
int at=j-seq[j];
if(at<pre) return puts("0"),0;
if(at>pre && tot>=2) ans=2*ans%mod;
}
pre=j;
}
FOR(j,1,tot) if(seq[j]){
int at=j-seq[j]+tot;
if(at<pre) return puts("0"),0;
if(at>pre && tot>=2) ans=2*ans%mod;
break;
}
}
}
fac[0]=fac[1]=inv[1]=invfac[0]=invfac[1]=1;
FOR(i,2,n){
fac[i]=1ll*fac[i-1]*i%mod;
inv[i]=mod-1ll*(mod/i)*inv[mod%i]%mod;
invfac[i]=1ll*invfac[i-1]*inv[i]%mod;
}
FOR(i,1,n) if(cnt[i]){
int s=0;
FOR(j,0,cnt[i]/2){
int x=1ll*C(cnt[i],2*j)*C(2*j,j)%mod*fac[j]%mod;
if(j) x=1ll*x*qpow(inv2,j)%mod*qpow(i,j)%mod;
if(i%2==1 && i!=1) x=1ll*x*qpow(2,cnt[i]-2*j)%mod;
s=(s+x)%mod;
}
ans=1ll*ans*s%mod;
}
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define iter(i, n) for (int i = 1; i <= n; ++i)
#define NR 101000
const int mod = 1e9 + 7;
int n, a[NR], in[NR], in_x[NR], cnt[NR], f[NR], cc[NR], sz;
bool vis[NR], ring[NR];
vector<int> R[NR];
int main() {
scanf("%d", &n);
iter(i, n) {
scanf("%d", &a[i]);
++in[a[i]];
in_x[a[i]] = i;
}
iter(i, n) if (in[i] > 2) { puts("0"); return 0; }
int ans = 1;
iter(i, n) if (!vis[i]) {
int x = i; bool is_ring = true;
for (; !vis[x]; x = a[x]) {
vis[x] = true;
}
if (!ring[x]) {
++sz;
for (; !ring[x]; x = a[x]) {
if (in[x] != 1) is_ring = false;
ring[x] = true;
R[sz].push_back(x);
}
}
if (x != i) {
bool is_chain = true;
x = i;
while (in[x] == 1) x = in_x[x];
if (in[x] != 0) { puts("0"); return 0; }
int len = 1;
for (vis[x] = true, x = a[x]; !ring[x]; x = a[x]) {
++len;
vis[x] = true;
if (in[x] != 1) { puts("0"); return 0; }
}
cc[x] = len;
} else {
if (is_ring) ++cnt[R[sz].size()];
}
}
iter(i, sz) {
reverse(R[i].begin(), R[i].end());
int rem = 0;
for (int x : R[i]) {
if (cc[x] != 0) {
if (rem > 1) { puts("0"); return 0; }
else rem = cc[x];
} else if (rem != 0) {
--rem;
if (rem == 0) ans = ans * 2 % mod;
}
}
if (rem != 0) {
for (int x : R[i]) {
if (cc[x] != 0) {
if (rem > 1) { puts("0"); return 0; }
else break;
} else {
--rem;
if (rem == 0) {
ans = ans * 2 % mod;
break;
}
}
}
}
}
iter(k, n) {
f[0] = 1;
iter(i, cnt[k]) {
f[i] = f[i - 1] * ((k % 2 && k != 1) ? 2 : 1) % mod;
if (i > 1) f[i] = (f[i] + 1ll * (i - 1) * f[i - 2] % mod * k) % mod;
}
ans = 1ll * ans * f[cnt[k]] % mod;
}
printf("%d\n", ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#define ll long long
#define P 1000000007
int a[100005],in[100005],col[100005],cnt[100005],n,L[100005];bool cir[100005];long long f[100005],ans=1;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;++i) scanf("%d",&a[i]),++in[a[i]];
for(int i=1;i<=n;++i)
{
if(col[i]) continue;int x=i;for(;!col[x];x=a[x]) col[x]=i;
if(col[x]!=i) continue;for(;!cir[x];x=a[x]) cir[x]=1;
}
for(int i=1;i<=n;++i) if((cir[i]&&in[i]>2)||(!cir[i]&&in[i]>1)) return puts("0"),0;
for(int i=1;i<=n;++i){if(in[i]) continue;int x=i,tmp=0;while(!cir[x]) x=a[x],tmp++;L[x]=tmp;}
for(int i=1;i<=n;++i)
{
if(!cir[i]) continue;int x=i,st=0,fir=0,id=0,len=0;
while(cir[x])
{
++id;cir[x]=0;
if(L[x])
{
if(!fir){st=fir=id;len=L[x];x=a[x];continue;}
else{(ans*=(L[x]<(id-st))+(L[x]<=(id-st)))%=P;if(!ans)return puts("0"),0;st=id;x=a[x];continue;}
}
x=a[x];
}
if(fir){(ans*=(len<(id+fir-st))+(len<=(id+fir-st)))%=P;if(!ans)return puts("0"),0;}
else cnt[id]++;
}
for(int i=1;i<=n;++i)
{
f[0]=1;if(!cnt[i]) continue;
if(i>1&&(i%2)) for(int j=1;j<=cnt[i];++j) {f[j]=f[j-1]*2%P;if(j>1) (f[j]+=f[j-2]*(j-1)*i%P)%=P;}
else for(int j=1;j<=cnt[i];++j){f[j]=f[j-1];if(j>1) (f[j]+=f[j-2]*(j-1)*i%P)%=P;}
(ans*=f[cnt[i]])%=P;if(!ans)return puts("0"),0;
}
printf("%lld",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=100100,P=1000000007;
int i,j,k,n,m,ch,ff,ans;
int a[N],b[N],z[N],fg[N],num[N],f[N],Jc[N],Jc_[N],tm2[N],d[N],dp[N];
void R(int &x) {
ff=x=0;ch=getchar();
while (ch<'0' || '9'<ch) { if (ch=='-') ff=1;ch=getchar();}
while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
if (ff) x=-x;
}
int ksm(int x,int y) {
int z=1;
for (;y;y>>=1,x=(ll) x*x%P) if (y&1) z=(ll) z*x%P;
return z;
}
void pre(int n) {
int i;
Jc[0]=Jc_[0]=f[0]=tm2[0]=1;
for (i=1;i<=n;i++) tm2[i]=(tm2[i-1]+tm2[i-1])%P;
for (i=1;i<=n;i++) f[i]=(ll) f[i-1]*(2*i-1)%P;
for (i=1;i<=n;i++) Jc[i]=(ll) Jc[i-1]*i%P;
Jc_[n]=ksm(Jc[n],P-2);
for (i=n-1;i;i--) Jc_[i]=(ll) Jc_[i+1]*(i+1)%P;
}
int C(int n,int m) {
return (ll) Jc[n]*Jc_[m]%P*Jc_[n-m]%P;
}
int Js(int n,int m) {
int i,k,ans=0;
if ((m&1) && m>1) {
for (i=0,k=1;i+i<=n;i++,k=(ll) k*m%P) ans=((ll) f[i]*k%P*C(n,i+i)%P*tm2[n-i-i]%P+ans)%P;
}
else {
for (i=0,k=1;i+i<=n;i++,k=(ll) k*m%P) ans=((ll) f[i]*k%P*C(n,i+i)%P+ans)%P;
}
return ans;
}
int main() {
R(n);
pre(n);
for (i=1;i<=n;i++) R(a[i]);
for (i=1;i<=n;i++) d[a[i]]++;
for (i=1;i<=n;i++) if (!z[i]) {
for (j=i;z[j]!=i && !z[j];j=a[j]) z[j]=i;
if (z[j]==i) {
fg[j]=1;
for (k=a[j];k!=j;k=a[k]) fg[k]=1;
}
}
for (i=1;i<=n;i++) if (d[i]>fg[i]+1) return puts("0"),0;
for (i=1;i<=n;i++) if (!fg[i]) b[a[i]]=i;
for (i=1;i<=n;i++) if (fg[i]) {
for (j=b[i];j;j=b[j]) dp[i]++;
}
memset(z,0,sizeof z);
ans=1;
for (i=1;i<=n;i++) if (!z[i] && fg[i]) {
int nm=0,Fg=0;
for (j=i;!z[j];j=a[j]) {
z[j]=1;
nm++;
if (dp[j]) Fg=j;
}
if (!Fg) num[nm]++;
else {
k=0;
for (j=a[Fg];j!=Fg;j=a[j]) {
k++;
if (dp[j]) {
if (dp[j]>k) ans=0;
if (dp[j]<k) ans=(ans+ans)%P;
k=0;
}
}
k++;
if (dp[Fg]>k) ans=0;
if (dp[Fg]<k) ans=(ans+ans)%P;
}
}
for (i=1;i<=n;i++) if (num[i]) ans=(ll) ans*Js(num[i],i)%P;
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
using namespace std;
#define N 200127
#define mod 1000000007
inline int M(int x){return (x>=mod)?(x-mod):x;}
inline int M1(int x){return (x<0)?(x+mod):x;}
int n,a[N],d[N],c[N],vis[N],ans,f[N],len[N],s[N];
//c:color,vis:oncircle
int main(){
scanf("%d",&n);int i,j,x,y,p,st,la,lal;ans=1;
for(i=1;i<=n;i++)scanf("%d",&a[i]),++d[a[i]];
for(i=1;i<=n;i++)
if(!c[i])
{
x=i;while(!c[x]){c[x]=i;x=a[x];}
if(c[x]!=i)continue;while(!vis[x]){vis[x]=1;x=a[x];}
}
for(i=1;i<=n;i++)
if((vis[i]&&(d[i]>2))||((!vis[i])&&(d[i]>1)))
{puts("0");return 0;}
for(i=1;i<=n;i++)
if(!d[i]){p=0;x=i;while(!vis[x]){++p;x=a[x];}len[x]=p;}
for(i=1;i<=n;i++)//基环树
if(vis[i])
{
x=i;p=st=la=lal=0;
while(vis[x])
{
++p;vis[x]=0;
if(len[x])
{
if(!st){st=la=p;lal=len[x];}
else{ans=1ll*ans*((len[x]<(p-la))+(len[x]<=(p-la)))%mod;la=p;}
}x=a[x];
}
if(st)ans=1ll*ans*((lal<(p+st-la))+(lal<=(p+st-la)))%mod;//是基环树
else ++s[p];
}
for(i=1;i<=n;i++)//正常环
if(s[i])
{
f[0]=1;const int up=s[i];
if((i>1)&&(i&1))
{
for(j=1;j<=up;j++)
f[j]=M(f[j-1]+f[j-1]),f[j]=M(f[j]+((j>1)?(1ll*f[j-2]*(j-1)%mod*i%mod):0));//就像PKUSC中的sg一样
}
else
{
for(j=1;j<=up;j++)
f[j]=f[j-1],f[j]=M(f[j]+((j>1)?(1ll*f[j-2]*(j-1)%mod*i%mod):0));//有i种方式
}ans=1ll*ans*f[up]%mod;
}printf("%d",ans);return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9, P = 1e9 + 7;
int a[N], n, Gr[N], s[N], t, la, h[N], sh, f[N], ans = 1;
bool v[N], inh[N], is[N];
vector<int>V[N];
unordered_map<int, int>T;
void Ext () { puts("0"); exit(0); }
void Dfs (int x, int fr) {
is[x] = 1; s[++t] = x; v[x] = 1;
for (auto v : V[x]) if (v != fr) {
if (is[v]) {
la = v;
break ;
} else {
Dfs(v, x);
if (la != -2) break ;
}
} else fr = -1;
if (la > 0) {
h[sh++] = x;
inh[x] = sh;
if (Gr[x] > 3) Ext();
if (x == la) la = -1;
}
is[x] = 0; --t;
}
namespace T1 {
void Dfs (int u, int fa) {
if (!inh[u]) {
v[u] = 1;
if (Gr[u] > 2) Ext();
}
for (auto v : V[u]) if (!inh[v] && v != fa) {
Dfs(v, u);
f[u] = f[v] + 1;
}
}
int main () {
for (int i = 0; i < sh; ++i) Dfs(h[i], 0);
int ans = 1, i = 0, j, k;
while (i < sh && !f[h[i]]) ++i; if (i == sh) return ++T[sh], 1;
do if (f[h[i]]) {
for (j = (i + 1) % sh, k = 1; !f[h[j]]; j = (j + 1) % sh, ++k);
int ret = 0;
if (f[h[i]] < k) ++ret;
if (f[h[i]] <= k) ++ret;
ans = 1ll * ans * ret % P;
if (j <= i) break ;
} while (i = j, 1);
return ans;
}
}
int Sol (int m, int n) {
f[0] = f[1] = 1; int b = 1 + (m > 1 && (m & 1));
for (int i = 1; i <= n; ++i) f[i] = (1ll * f[i - 2] * m % P * (i - 1) + f[i - 1] * b) % P;
return f[n];
}
int main () {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
V[i].emplace_back(a[i]), V[a[i]].emplace_back(i);
++Gr[i]; ++Gr[a[i]];
}
for (int i = 1; i <= n; ++i) if (!v[i]) {
la = -2, sh = 0, Dfs(i, 0);
if (sh > 1 && a[h[0]] == h[1]) reverse(h, h + sh);
ans = 1ll * ans * T1::main() % P;
}
for (auto v : T) ans = 1ll * ans * Sol(v.first, v.second) % P;
printf("%d\n", ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <list>
using namespace std;
typedef pair<int, int> Pi;
typedef long long ll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define sz(x) (int)x.size()
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) x.begin(), x.end()
int N, A[100010];
int deg[100010];
int cyc[100010];
map <int, int> OC;
int p[100010], z[100010];
int Find(int x){return p[x] == x ? x : p[x] = Find(p[x]); }
int chk[100010];
const int MOD = 1e9 + 7;
ll D[100010];
int d[100010];
int get(int X){
int res = 1;
vector <int> v; v.pb(d[X]);
int t = A[X];
while(t != X){
v.pb(d[t]);
t = A[t];
}
int L = sz(v);
for(int S=0;S<L;S++)if(v[S] != 1){
for(int j=0;j<L;){
int t = v[(S-j+L)%L];
if(t == 0){j++; continue; }
for(int k=1;k<t;k++){
if(v[(S-j-k+L+L)%L] > 0)return 0;
}
if(v[(S-j-t+L+L)%L] > 0){
j += t;
}
else{
j += t + 1;
res = res * 2 % MOD;
}
}
break;
}
return res;
}
void solve(){
scanf("%d", &N);
for(int i=1;i<=N;i++)scanf("%d", A+i);
for(int i=1;i<=N;i++)deg[A[i]]++;
queue <int> que;
for(int i=1;i<=N;i++)if(deg[i] == 0)que.push(i);
while(!que.empty()){
int t = que.front(); que.pop();
deg[A[t]]--;
d[A[t]] = d[t] + 1;
if(deg[A[t]] == 0){
que.push(A[t]);
}
}
for(int i=1;i<=N;i++)cyc[i] = deg[i], deg[i] = 0;
for(int i=1;i<=N;i++)deg[A[i]]++;
for(int i=1;i<=N;i++){
if(deg[i] > 2 || (deg[i] == 2 && cyc[i] == 0)){
printf("0");
return;
}
}
for(int i=1;i<=N;i++)p[i] = i, z[i] = 1;
for(int i=1;i<=N;i++){
int x = Find(i), y = Find(A[i]);
if(x != y)p[x] = y, z[y] += z[x];
}
for(int i=1;i<=N;i++)if(deg[i] == 2)chk[Find(i)] = 1;
for(int i=1;i<=N;i++)if(Find(i) == i && chk[i] == 0)OC[z[i]]++;
ll ans = 1;
for(auto e : OC){
D[0] = 1;
int t = ((e.Fi % 2 == 1 && e.Fi > 1) ? 2 : 1);
for(int i=1;i<=e.Se;i++){
D[i] = t * D[i-1] + (i > 1 ? ((ll)(i-1) * e.Fi % MOD) * D[i-2] : 0);
D[i] %= MOD;
}
ans = ans * D[e.Se] % MOD;
}
for(int i=1;i<=N;i++)if(Find(i) == i && chk[i] == 1){
ans = ans * get(i) % MOD;
}
printf("%lld\n", ans);
}
int main(){
int Tc = 1; //scanf("%d", &Tc);
for(int tc=1;tc<=Tc;tc++){
solve();
}
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long i64;
const int MAXN = 100000 + 5;
const int MOD = 1e9 + 7;
int N;
i64 answer, dp[MAXN];
int degree[MAXN], lenCnt[MAXN];
int pre[MAXN], nxt[MAXN];
bool vis[MAXN];
bool noSolution = false;
void findCycle()
{
for (int i = 1; i <= N; i++)
{
if (degree[i] > 2)
return noSolution = true, (void) 0;
if (degree[i] != 2 || vis[i])
continue;
int cur = i;
do
{
if (vis[cur])
return noSolution = true, (void) 0;
vis[cur] = true;
pre[nxt[cur]] = cur;
cur = nxt[cur];
} while (cur != i);
}
}
void processCircleBasedTree()
{
for (int i = 1; i <= N; i++)
{
if (degree[i])
continue;
int cur = i, footLen = 0, cycleLen = 0;
while (!vis[cur])
vis[cur] = true,
cur = nxt[cur],
footLen++;
do
cur = pre[cur],
cycleLen++;
while (degree[cur] == 1);
if (footLen < cycleLen)
answer = answer * 2 % MOD;
if (footLen > cycleLen)
return noSolution = true, (void) 0;
}
}
void countCycle()
{
for (int i = 1; i <= N; i++)
{
if (vis[i])
continue;
int cur = i, len = 0;
do
{
vis[cur] = true;
cur = nxt[cur];
len++;
} while (cur != i);
lenCnt[len]++;
}
}
int main()
{
answer = 1;
ios::sync_with_stdio(false), cin.tie(nullptr);
cin >> N;
for (int i = 1; i <= N; i++)
cin >> nxt[i], degree[nxt[i]]++;
findCycle();
if (noSolution)
return cout << 0 << endl, 0;
processCircleBasedTree();
if (noSolution)
return cout << 0 << endl, 0;
countCycle();
if (noSolution)
return cout << 0 << endl, 0;
for (int i = 1; i <= N; i++)
{
dp[0] = 1LL;
for (int j = 1; j <= lenCnt[i]; j++)
{
dp[j] = dp[j - 1] * (((i % 2 == 0) || i == 1) ? 1 : 2);
if (j >= 2)
dp[j] = (dp[j] + ((((dp[j - 2] * (j - 1)) % MOD) * i) % MOD))
% MOD;
}
answer = answer * dp[lenCnt[i]] % MOD;
}
cout << answer << endl;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
const int N=100005,mo=1000000007;
int deg[N],a[N],vis[N],b[N],cnt[N],f[N];
bool u[N];
int calc(int x,int y){return (x<y)+(x<=y);}
int main(){
int n,ans=1;
scanf("%d\n",&n);
for (int i=1;i<=n;i++){
scanf("%d",&a[i]);
deg[a[i]]++;
}
for (int i=1;i<=n;i++)
if (!vis[i]){
int j=i;
for (;!vis[j];j=a[j]) vis[j]=i;
if (vis[j]==i)
for (;!u[j];j=a[j]) u[j]=1;
}
for (int i=1;i<=n;i++)
if (!deg[i]){
int j=i,k=0;
for (;!u[j];j=a[j]) k++;
b[j]=k;
}
else if (deg[i]>u[i]+1) ans=0;
for (int i=1;i<=n && ans;i++)
if (u[i]){
int x=0,y=0,k=0;
for (int j=i;u[j];j=a[j]){
k++;u[j]=0;
if (b[j]){
if (!x) x=b[j],y=k;
else ans=ans*calc(b[j],k)%mo;
k=0;
}
}
if (!x) cnt[k]++;
else ans=ans*calc(x,y+k)%mo;
}
for (int i=1;i<=n && ans;i++){
int x=(i&1)+(i>1);f[0]=1;
for (int j=1;j<=cnt[i];j++){
f[j]=f[j-1]*x%mo;
if (j>1) f[j]=(f[j]+1ll*(j-1)*i%mo*f[j-2])%mo;
}
ans=1ll*ans*f[cnt[i]]%mo;
}
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define mo 1000000007
#define pi 3.1415926535898
#define eps 1e-9
using namespace std;
long long read(){
long long xx=0,flagg=1;
char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')
ch=getchar();
if(ch=='-'){
flagg=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
xx=xx*10+ch-'0';
ch=getchar();
}
return xx*flagg;
}
void pus(long long xx,long long flagg){
if(xx<0){
putchar('-');
xx=-xx;
}
if(xx>=10)
pus(xx/10,0);
putchar(xx%10+'0');
if(flagg==1)
putchar(' ');
if(flagg==2)
putchar('\n');
return;
}
long long n,i,j,zz[100005],x,sum[100005],book[100005],book2[100005];
long long top2,sum2,zhan[100005],zhan2[100005],book3[100005],ans;
long long jc[100005],ny[100005],flag;
long long top,nex[100005],to[100005],fir[100005];
void lj(int u,int v){
top++;
nex[top]=fir[u];
fir[u]=top;
to[top]=v;
}
int ss(int v){
int sum3=0;
sum[v]=1-book2[v];
for(int top1=fir[v];top1;top1=nex[top1])
if(book2[to[top1]]==0){
sum[v]+=ss(to[top1]);
sum3++;
}
if(sum3>=2)
flag=1;
return sum[v];
}
long long ksm(long long u,long long v){
long long o=1;
while(v){
if(v&1)
o=o*u%mo;
u=u*u%mo;
v>>=1;
}
return o;
}
long long C(long long u,long long v){
return jc[u]*ny[v]%mo*ny[u-v]%mo;
}
int main(){
n=read();
for(i=1;i<=n;i++){
zz[i]=read();
lj(zz[i],i);
}
jc[0]=1;
for(i=1;i<=n;i++)
jc[i]=jc[i-1]*i%mo;
ny[n]=ksm(jc[n],mo-2);
for(i=n-1;i>=0;i--)
ny[i]=ny[i+1]*(i+1)%mo;
for(i=1;i<=n;i++)
if(book[i]==0){
x=i;
while(book[x]==0){
book[x]=i;
x=zz[x];
}
if(book[x]==i){
while(book2[x]==0){
book2[x]=1;
x=zz[x];
}
}
}
ans=1;
for(i=1;i<=n;i++)
if(book2[i]==1)
ss(i);
if(flag==1){
pus(0,2);
return 0;
}
for(i=1;i<=n;i++)
if(book2[i]==1){
x=i;sum2=0;top2=0;
while(book2[x]!=2){
book2[x]=2;
sum2++;
if(sum[x]!=0){
top2++;
zhan[top2]=sum[x];
zhan2[top2]=sum2;
}
x=zz[x];
}
if(top2!=0){
zhan[top2+1]=zhan[1];
zhan2[top2+1]=zhan2[1]+sum2;
for(j=1;j<=top2;j++)
if(zhan[j+1]<zhan2[j+1]-zhan2[j])
ans=ans*2%mo;
else if(zhan[j+1]>zhan2[j+1]-zhan2[j])
ans=0;
}
else
book3[sum2]++;
}
for(j=1;j<=n;j++){
sum2=0;
for(i=0;(i<<1)<=book3[j];i++)
if((j&1)&&j!=1)
sum2=(sum2+C(book3[j],i<<1)*jc[i<<1]%mo*ny[i]%mo*ksm(ksm(2,i),mo-2)%mo*ksm(j,i)%mo*ksm(2,book3[j]-(i<<1)))%mo;
else
sum2=(sum2+C(book3[j],i<<1)*jc[i<<1]%mo*ny[i]%mo*ksm(ksm(2,i),mo-2)%mo*ksm(j,i))%mo;
ans=ans*sum2%mo;
}
pus(ans,2);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 100000
#define MO 1000000007
using namespace std;
int N,a[MAXN+5],d[MAXN+5],pre[MAXN+5];
int cnt[MAXN+5],f[MAXN+5];
bool vis[MAXN+5];
int main()
{
scanf("%d",&N);
for(int i=1;i<=N;i++)
scanf("%d",&a[i]),d[a[i]]++;
for(int i=1;i<=N;i++)
if(d[i]>2)
{
printf("0\n");
return 0;
}
for(int i=1;i<=N;i++)
if(d[i]==2&&vis[i]==false)
{
int p=i;
do
{
if(vis[p]==true)
{
printf("0\n");
return 0;
}
vis[p]=true;
pre[a[p]]=p;
p=a[p];
}while(p!=i);
}
int ans=1;
for(int i=1;i<=N;i++)//处理基环内向树
if(d[i]==0)
{
int p=i,len1=0,len2=0;
do
{
vis[p]=true;
p=a[p];
len1++;
}while(vis[p]==false);
do
{
p=pre[p];
len2++;
}while(d[p]==1);
if(len1>len2)
{
printf("0\n");
return 0;
}
if(len2>len1)
ans=2LL*ans%MO;
}
for(int i=1;i<=N;i++)//处理单个的循环
if(vis[i]==false)
{
int len=0,p=i;
do
{
p=a[p];
vis[p]=true;
len++;
}while(p!=i);
cnt[len]++;
}
for(int i=1;i<=N;i++)
if(cnt[i])
{
int mul=1;
if(i%2==1&&i!=1)
mul=2;
f[0]=1,f[1]=mul;
for(int j=2;j<=cnt[i];j++)
f[j]=(1LL*f[j-2]*(j-1)%MO*i%MO+1LL*f[j-1]*mul%MO)%MO;
ans=1LL*ans*f[cnt[i]]%MO;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#define MOD 1000000007
using namespace std;
typedef long long ll;
ll dp(int n,int m) {
ll s1=1,s2=0;
for(int i=1;i<=m;i++) {
ll f=(s1*(((n&1)&&n>1)?2LL:1LL)+s2*(i-1)%MOD*n)%MOD;
s2=s1;
s1=f;
}
return s1;
}
int a[100005],sum[100005];
bool vis[100005],cir[100005];
vector <int> son[100005];
int getlen(int x) {
vis[x]=1;
if (!cir[x]&&son[x].size()>=2) {
puts("0");
exit(0);
}
for(int i=0;i<son[x].size();i++)
if (!cir[son[x][i]]) return getlen(son[x][i])+1;
return 1;
}
int now[100005],l[100005],pos[100005];
ll calc(int x) {
do {
vis[x]=1;
x=a[x];
} while (!vis[x]);
int d=0,tot=0;
do {
cir[x]=1;
x=a[x];
} while (!cir[x]);
int t=x;
do {
pos[x]=++d;
l[x]=getlen(x)-1;
if (l[x]) now[++tot]=x;
x=a[x];
} while (x!=t);
if (!tot) {
sum[d]++;
return 1;
}
ll ans=1;
for(int i=1;i<=tot;i++) {
int p=now[i],q=now[i%tot+1];
int dis=(pos[q]-pos[p]+d)%d;
if (!dis) dis=d;
if (dis<l[q]) {
puts("0");
exit(0);
}
if (dis>l[q]) ans=ans*2LL%MOD;
}
return ans;
}
int main() {
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++) {
scanf("%d",&a[i]);
son[a[i]].push_back(i);
if (son[a[i]].size()>2) {
puts("0");
exit(0);
}
}
ll ans=1;
for(int i=1;i<=n;i++)
if (!vis[i]) ans=(ans*calc(i))%MOD;
for(int i=1;i<=n;i++)
if (sum[i]) ans=ans*dp(i,sum[i])%MOD;
printf("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007
#define maxn 200010
inline ll read()
{
ll x=0; char c=getchar(),f=1;
for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
return x*f;
}
inline void write(ll x)
{
static char buf[20];
int len=0;
if(x<0)putchar('-'),x=-x;
for(;x;x/=10)buf[len++]=x%10+'0';
if(!len)putchar('0');
else while(len)putchar(buf[--len]);
}
inline void writesp(ll x){write(x); putchar(' ');}
inline void writeln(ll x){write(x); putchar('\n');}
int p[maxn],vis[maxn],col[maxn],mark[maxn],dep[maxn],rt[maxn],deg[maxn];
int len[maxn],dis[maxn];
int cnt[maxn],f[maxn];
std::vector<int>a;
int n,tot;
int main()
{
// freopen("agc008E.in","r",stdin);
// freopen("agc008E.out","w",stdout);
n=read();
for(int i=1;i<=n;i++)
p[i]=read();
for(int i=1;i<=n;i++)
if(!vis[i]){
int now=i;
while(!vis[now]){
a.push_back(now); vis[now]=1;
now=p[now];
}
if(col[now]){
int sz=a.size();
for(int j=sz-1;j>=0;j--)
col[a[j]]=col[now],rt[a[j]]=rt[now],dep[a[j]]=dep[now]+(sz-j);
}
else{
int sz=a.size(),flag=0;
++tot;
for(int j=sz-1;j>=0;j--){
col[a[j]]=tot;
if(!flag)mark[a[j]]=1,rt[a[j]]=a[j],dep[a[j]]=0;
else rt[a[j]]=now,dep[a[j]]=dep[a[j+1]]+1;
flag|=(a[j]==now);
}
}
a.clear();
}
for(int i=1;i<=n;i++)
if(!mark[i])++deg[p[i]];
for(int i=1;i<=n;i++)
if(deg[i]>=2){
puts("0"); return 0;
}
for(int i=1;i<=n;i++)
len[rt[i]]=std::max(len[rt[i]],dep[i]);
memset(vis,0,sizeof(vis));
ll ans=1;
for(int i=1;i<=n;i++)
if(mark[i]&&!vis[i]){
a.push_back(i); vis[i]=1;
int now=p[i];
while(now!=i){
a.push_back(now); vis[now]=1;
now=p[now];
}
int flag=0,sz=a.size();
for(int j=0;j<sz;j++)
if(len[a[j]]){
flag=1; break;
}
if(flag){
int last=-1,first=-1;
for(int j=0;j<sz;j++)
if(len[a[j]]){
if(~first)dis[a[j]]=j-last;
else first=j;
last=j;
}
dis[a[first]]=first+sz-last;
for(int j=0;j<sz;j++)
if(len[a[j]]){
if(len[a[j]]<dis[a[j]])ans=ans*2%mod;
else if(len[a[j]]>dis[a[j]]){
puts("0"); return 0;
}
}
}
else ++cnt[sz];
a.clear();
}
for(int i=1;i<=n;i++){
int w=((i&1)&&(i>1))?2:1;
f[0]=1; f[1]=w;
for(int j=2;j<=cnt[i];j++)
f[j]=(f[j-1]*w+(ll)(j-1)*f[j-2]%mod*i)%mod;
ans=ans*f[cnt[i]]%mod;
}
writeln(ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
#define For(i, j, k) for(int i = j; i <= k; i++)
using namespace std;
const int N = 100010;
const int Mod = 1e9 + 7;
int n;
int to[N], deg[N], pre[N];
int cnt[N], f[N];
bool vis[N];
int main(){
scanf("%d", &n);
For(i, 1, n) scanf("%d", &to[i]), deg[to[i]]++;
For(i, 1, n) if(deg[i] > 2){
puts("0");
return 0;
}
For(i, 1, n) if(deg[i] == 2 && !vis[i]){
int o = i;
do{
pre[to[o]] = o;
if(vis[o]){
puts("0");
return 0;
}
vis[o] = true;
o = to[o];
}while(o != i);
}
int ans = 1;
For(i, 1, n) if(!deg[i]){
int o = i, len = 0;
while(!vis[o]) vis[o] = true, o = to[o], ++len;
int len2 = 0;
do{
o = pre[o];
++len2;
}while(deg[o] == 1);
if(len < len2) ans = ans * 2 % Mod;
else if(len > len2){
puts("0");
return 0;
}
}
For(i, 1, n) if(!vis[i]){
int o = i, len = 0;
do{
vis[o] = true;
o = to[o];
++len;
}while(o != i);
cnt[len]++;
}
For(i, 1, n) if(cnt[i]){
int x = i > 1 && i % 2 == 1 ? 2 : 1;
f[0] = 1, f[1] = x;
For(j, 2, cnt[i]) f[j] = (1ll * (j - 1) * i * f[j - 2] + f[j - 1] * x) % Mod;
ans = 1ll * ans * f[cnt[i]] % Mod;
}
printf("%d\n", ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long
using namespace std;
const int N=1e5+10,mod=1e9+7;
int n,cnt,head[N],vis[N],a[N];
int tot,cir_tot,q[N],fa[N],dep[N],mk[N],len[N];
int sum[N],ans,f[N];
struct edge{int to,nxt;}e[N<<1];
void adde(int x,int y){
e[++cnt].to=y; e[cnt].nxt=head[x]; head[x]=cnt;
}
void dfs(int u,int par){
vis[u]=1; tot++;
for (int i=head[u],v;i;i=e[i].nxt)
if ((i^1)!=par){
v=e[i].to;
if (!vis[v]) fa[v]=u,dep[v]=dep[u]+1,dfs(v,i);
else if (dep[v]<=dep[u]){
for (int i=u;i!=v;i=fa[i]) q[++cir_tot]=i,mk[i]=1;
q[++cir_tot]=v,mk[v]=1;
}
}
}
int tmp;
void dfs1(int u,int par){
if (par) tmp++;
int son=0;
for (int i=head[u],v;i;i=e[i].nxt)
if (v=e[i].to,v!=par&&!mk[v]) dfs1(v,u),son++;
if (son>1){puts("0"); exit(0);}
}
int main(){
scanf("%d",&n);
cnt=1;
rep (i,1,n) scanf("%d",&a[i]),adde(a[i],i),adde(i,a[i]);
ans=1;
rep (i,1,n) if (!vis[i]){
tot=cir_tot=0,dfs(i,-1);
if (cir_tot==2&&q[1]==q[2]) cir_tot--;
if (tot==cir_tot) sum[tot]++;
else{
if (a[q[1]]!=q[2]) reverse(q+1,q+1+cir_tot);
int st=1;
rep (j,1,cir_tot){
tmp=0,dfs1(q[j],0),len[j]=tmp;
if (tmp) st=j;
}
int len_2=0;
for (int j=st%cir_tot+1;;j=j%cir_tot+1){
len_2++;
if (len[j]){
ans=(ll)ans*((len[j]<=len_2)+(len[j]<len_2))%mod;
len_2=0;
}
if (j==st) break;
}
}
}
rep (i,1,n){
if (!sum[i]) continue;
f[0]=1;
rep (j,1,sum[i]){
f[j]=f[j-1];
if (i>1&&(i&1)) f[j]=(f[j]+f[j-1])%mod;
if (j>1) f[j]=(f[j]+(ll)f[j-2]*(j-1)%mod*i%mod)%mod;
}
ans=(ll)ans*f[sum[i]]%mod;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <bitset>
#include <vector>
#include <string>
#include <cassert>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define rep(i, x, y) for(int i = (int)x; i <= (int)y; i ++)
#define fi first
#define se second
#define pb push_back
#define mk make_pair
using namespace std;
typedef long long LL;
const int N = 100005;
const int P = (int)1e9 + 7;
int st[N<<1], deg[N], d[N], n, vis[N], a[N], tot, ct[N], sz[N], ans=1, fac[N], inv[N];
int power(int x, int y){
int an = 1;
for(; y; y >>= 1, x = (LL)x * x % P) if(y & 1) an = (LL)an * x % P;
return an;
}
int C(int x, int y){
if(x < 0 || y < 0 || x < y) return 0;
return (LL)fac[x] * inv[y] % P * inv[x - y] % P;
}
void renew(int &x, const int y){
x += y;
if(x < 0) x += P;
if(x >= P) x -= P;
}
void init(){
fac[0] = inv[0] = fac[1] = inv[1] = 1;
rep(i, 2, n){
fac[i] = (LL)fac[i - 1] * i % P;
inv[i] = power(fac[i], P - 2);
}
}
int getdp(){
int fr = 0;
rep(i, 1, tot) st[i + tot] = st[i];
rep(i, 1, tot) if(sz[st[i]]){
fr = i;
break;
}
reverse(st + fr + 1, st + fr + tot);
int a = fr + sz[st[fr]] - 1, b = a + 1, c = 1;
rep(i, fr + 1, fr + tot - 1)if(sz[st[i]]){
int na = i + sz[st[i]] - 1, nb = na + 1, cc = 0;
if(a < i) renew(cc, c);
if(b < i) renew(cc, c);
a = na;
b = nb;
c = cc;
}
int ans = 0;
if(a < fr + tot) renew(ans, c);
if(b < fr + tot) renew(ans, c);
return ans;
}
int que[N], t;
int main(){
scanf("%d", &n);
rep(i, 1, n) scanf("%d", a + i);
rep(i, 1, n){
deg[a[i]] ++;
if(deg[a[i]] > 2){ puts("0"); exit(0); }
}
rep(i, 1, n) d[i] = deg[i];
rep(i, 1, n) if(!d[i]) que[++t] = i;
rep(h, 1, t){
int u = que[h];
--d[a[u]];
vis[u] = 1;
if(!d[a[u]]) que[++t] = a[u];
}
rep(i, 1, n) if(vis[i] && deg[i] >= 2){
puts("0");
exit(0);
}
rep(i, 1, n) if(vis[i] && !deg[i]){
int x = i, cnt = 0;
for(x = i; vis[x]; x = a[x])cnt++;
sz[x] = cnt;
}
init();
ans=1;
rep(i, 1, n)if(!vis[i])
{
int u = i, cu = 1;
tot = 0;
int res = 0;
do{
st[++tot] = u;
cu &= !sz[u];
vis[u] = 1;
u = a[u];
}while(u != i);
if(cu){
ct[tot] ++;
res = 1;
}else{
res = getdp();
}
ans = (LL)ans * res % P;
}
rep(i, 1, n) if(ct[i])
{
int tot = ct[i];
int res = 0;
for(int x = 0; x <= tot; x += 2){
int cb = 1;
cb = (LL)cb * power(i, (x >> 1)) % P;
cb = (LL)cb * C(tot, x) % P;
cb = (LL)cb * fac[x] % P;
cb = (LL)cb * inv[x >> 1] % P;
cb = (LL)cb * power((P + 1) >> 1, x >> 1) % P;
if((i > 1) && (i & 1)){
cb = (LL)cb * power(2, tot - x) % P;
}
renew(res, cb);
}
ans = (LL)ans * res % P;
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define SZ 666666
int n,a[SZ],ff[SZ],sz[SZ]; Edg
int gf(int x) {return ff[x]?ff[x]=gf(ff[x]):x;}
void uni(int a,int b)
{
a=gf(a),b=gf(b);
(a^b)?(ff[a]=b):0;
}
map<int,int> cp;
bool ic[SZ],vv[SZ];
int dfs(int x)
{
int ans=0,chh=0; vv[x]=1;
for esb(x,e,b) if(ic[b]||vv[b]);else
{
int t=dfs(b);++chh;
if(t<0) return -1;
ans=max(ans,t+1);
}
if(chh>=2) return -1;
return ans;
}
int py(int a,int b)
{
if(b>a) return 0;
if(a==b) return 1;
return 2;
}
const ll MOD=1e9+7;
ll f[SZ];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d",a+i),adde(i,a[i]),uni(i,a[i]);
for(int i=1;i<=n;++i) ++sz[gf(i)];
ll ans=1;
for(int i=1;i<=n;++i) if(!ff[i])
{
int g=i;
for(int j=1;j<=sz[i];++j) g=a[g];
vector<int> cyc; cyc.pb(g);
for(int h=a[g];h!=g;h=a[h]) cyc.pb(h);
if(cyc.size()==sz[i])
{++cp[sz[i]]; continue;}
for(auto r:cyc) ic[r]=1;
vector<int> tmp;
for(auto r:cyc)
{
int t=dfs(r);
if(t<0) {puts("0"); return 0;}
tmp.pb(t);
}
int cl=0,pv=-1,bl=0;
for(int i=0;i<tmp.size();++i)
{
++cl;
if(!tmp[i]) continue;
if(pv==-1) bl=cl,pv=tmp[i];
else ans*=py(cl,tmp[i]),ans%=MOD;
cl=0;
}
ans*=py(bl+cl,pv); ans%=MOD;
}
for(auto g:cp)
{
int x=g.fi,y=g.se;
f[0]=1;
for(int i=1;i<=y;++i)
f[i]=(((i>=2)?(f[i-2]*(i-1)%MOD*x):0)+(1+(x%2==1&&x>1))*f[i-1])%MOD;
ans*=f[y]; ans%=MOD;
}
cout<<ans<<"\n";
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 100010
#define MO 1000000007
#define LL long long
int a[MAXN],n,vis[MAXN],du[MAXN],pre[MAXN],cnt[MAXN];
LL fac[MAXN],inv[MAXN],Pow2[MAXN],P[MAXN],ans=1;
LL PowMod(LL a,int b)
{
LL ret=1;
while(b)
{
if(b&1) ret=ret*a%MO;
a=a*a%MO;
b>>=1;
}
return ret;
}
LL C(int n,int m)
{
return fac[n]*inv[m]%MO*inv[n-m]%MO;
}
void Pre()
{
fac[0]=1;
for(int i=1;i<MAXN;i++)
fac[i]=fac[i-1]*i%MO;
inv[MAXN-1]=PowMod(fac[MAXN-1],MO-2);
for(int i=MAXN-2;i>=0;i--)
inv[i]=inv[i+1]*(i+1)%MO;
Pow2[0]=1;
for(int i=1;i<MAXN;i++)
Pow2[i]=2*Pow2[i-1]%MO;
P[1]=1;
for(int i=2;i<MAXN;i++)
P[i]=P[i-1]*(2*i-1)%MO;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
du[a[i]]++;
}
for(int i=1;i<=n;i++)
{
if(du[i]>2)
{
printf("0\n");
return 0;
}
if(du[i]<2||vis[i]) continue;
int p=i;
do
{
if(vis[p])
{
printf("0\n");
return 0;
}
vis[p]=1,pre[a[p]]=p;p=a[p];
}while(p!=i);
}
for(int i=1;i<=n;i++)
if(!du[i])
{
int l1=0,l2=0,p=i;
while(!vis[p])
{
vis[p]=1,p=a[p],l1++;
}
do
{
p=pre[p];l2++;
}while(du[p]!=2);
if(l1<l2) ans=ans*2%MO;
else if(l1>l2)
{
printf("0\n");
return 0;
}
}
for(int i=1;i<=n;i++)
if(!vis[i])
{
int p=i,c=0;
while(!vis[p])
{
c++;vis[p]=1;
p=a[p];
}
cnt[c]++;
}
Pre();
for(int i=1;i<=n;i++)
if(cnt[i])
{
if((i&1)&&i!=1)
{
LL tmp=Pow2[cnt[i]];
LL x=i;
for(int j=1;2*j<=cnt[i];(x*=i)%=MO,j++)
(tmp+=C(cnt[i],2*j)*P[j]%MO*x%MO*Pow2[cnt[i]-2*j]%MO)%=MO;
ans*=tmp;
ans%=MO;
}
else
{
LL tmp=1;
LL x=i;
for(int j=1;2*j<=cnt[i];(x*=i)%=MO,j++)
(tmp+=C(cnt[i],2*j)*P[j]%MO*x%MO)%=MO;
ans*=tmp;
ans%=MO;
}
}
printf("%lld\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ll long long
#define MOD 1000000007
using namespace std;
inline int read(){
int re=0,flag=1;char ch=getchar();
while(!isdigit(ch)){
if(ch=='-') flag=-1;
ch=getchar();
}
while(isdigit(ch)) re=(re<<1)+(re<<3)+ch-'0',ch=getchar();
return re*flag;
}
ll f[200010],finv[200010],meth[200010];
ll qpow(ll a,ll b){
ll re=1;
while(b){
if(b&1) re=(re*a)%MOD;
a=a*a%MOD;b>>=1;
}
return re;
}
void init(){
ll i,len=200000;
f[0]=f[1]=finv[0]=finv[1]=1;
for(i=2;i<=len;i++) f[i]=f[i-1]*i%MOD;
finv[len]=qpow(f[len],MOD-2);
for(i=len;i>2;i--) finv[i-1]=finv[i]*i%MOD;
}
int n,a[200010],vis[200010],cir[200010],cntcir=0,in[200010],bst[200010],siz[200010];
ll ans=1;
vector<int>s;
vector<int>nd[200010];
bool cmp(int l,int r){
return siz[l]<siz[r];
}
ll C(ll x,ll y){
return f[x]*finv[y]%MOD*finv[x-y]%MOD;
}
int main(){
n=read();int i,j;ll tmp,c,cc;
init();
meth[0]=1;
for(i=1;i<=n;i++) meth[i]=C(i*2,i)*f[i]%MOD*qpow(qpow(2,MOD-2),i)%MOD;
for(i=1;i<=n;i++) a[i]=read(),in[a[i]]++;
for(i=1;i<=n;i++){//取出环
j=i;
while(!vis[j]) vis[j]=i,j=a[j];
if(vis[j]^i) continue;
cntcir++;
while(!cir[j]){
cir[j]=cntcir,nd[cntcir].push_back(j),siz[cntcir]++,j=a[j];
}
}
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++){//判断环套树，以及外挂树是不是都是链
if(in[i]) continue;
j=i;
while(!cir[j]&&!vis[j]) j=a[j];
if(vis[j]){
puts("0");return 0;
}
bst[cir[j]]=1;tmp=cir[j];
j=i;c=0;
while(!cir[j]) cir[j]=tmp,c++,vis[j]=1,j=a[j];
vis[j]=c;
}
for(i=1;i<=cntcir;i++){//环套树处理
if(!bst[i]){s.push_back(i);continue;}
for(j=0;j<nd[i].size();j++) if(vis[nd[i][j]]) break;
tmp=j;
do{
c=j;cc=1;
j--;
(j+=(int)nd[i].size());
j%=(int)nd[i].size();
for(;!vis[nd[i][j]];j--,j=((j<0)?j+nd[i].size():j)) cc++;
if(vis[nd[i][c]]>cc){
puts("0");return 0;
}
if(vis[nd[i][c]]<cc) (ans*=2)%=MOD;
}while(tmp!=j);
}
sort(s.begin(),s.end(),cmp);
for(i=0;i<s.size();i+=c){
j=i;tmp=0;
while(siz[s[j]]==siz[s[i]]&&j<s.size()) j++;
c=j-i;
for(j=0;j<=c/2;j++){
(tmp+=C(c,2*j)*meth[j]%MOD*qpow(siz[s[i]],j)%MOD*qpow(2,(c-2*j)*(siz[s[i]]!=1)*(siz[s[i]]&1))%MOD)%=MOD;
}
ans=ans*tmp%MOD;
}
printf("%lld\n",ans);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 100000
#define MO 1000000007
using namespace std;
int N,a[MAXN+5],d[MAXN+5],pre[MAXN+5];
int cnt[MAXN+5],f[MAXN+5];
bool vis[MAXN+5];
int main()
{
// freopen("test.in","r",stdin);
// freopen("test.out","w",stdout);
scanf("%d",&N);
for(int i=1;i<=N;i++)
scanf("%d",&a[i]),d[a[i]]++;
for(int i=1;i<=N;i++)
if(d[i]>2)
{
printf("0\n");
return 0;
}
for(int i=1;i<=N;i++)
if(d[i]==2&&vis[i]==false)
{
int p=i;
do
{
if(vis[p]==true)
{
printf("0\n");
return 0;
}
vis[p]=true;
pre[a[p]]=p;
p=a[p];
}while(p!=i);
}
int ans=1;
for(int i=1;i<=N;i++)//处理基环内向树
if(d[i]==0)
{
int p=i,len1=0,len2=0;
do
{
vis[p]=true;
p=a[p];
len1++;
}while(vis[p]==false);
do
{
p=pre[p];
len2++;
}while(d[p]==1);
if(len1>len2)
{
printf("0\n");
return 0;
}
if(len2>len1)
ans=2LL*ans%MO;
}
for(int i=1;i<=N;i++)//处理单个的循环
if(vis[i]==false)
{
int len=0,p=i;
do
{
p=a[p];
vis[p]=true;
len++;
}while(p!=i);
cnt[len]++;
}
for(int i=1;i<=N;i++)
if(cnt[i])
{
int mul=1;
if(i%2==1&&i!=1)
mul=2;
f[0]=1,f[1]=mul;
for(int j=2;j<=cnt[i];j++)
f[j]=(1LL*f[j-2]*(j-1)%MO*i%MO+1LL*f[j-1]*mul%MO)%MO;
ans=1LL*ans*f[cnt[i]]%MO;
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <cstdlib>
#define ref(i,x,y)for(int i=x;i<=y;++i)
const int N=100005,mod=1e9+7;
void print(int x){printf("%d\n",x);exit(0);}
int n,m,l,res,a[N],cnt,head[N],h[N],w[N*2],p[N],f[N];bool b[N],vis[N];
struct edge{int to,nxt;}e[N];
void add(int x,int y){e[++cnt]=(edge){y,head[x]};head[x]=cnt;}
int dfs(int f,int x){
vis[x]=1;
int Y=0,t=0;
for(int i=head[x];i;i=e[i].nxt){
int y=e[i].to;if(y!=f)Y=y,t++;
}
if(t>1)return 0;
if(t==0)return 1;
int s=dfs(x,Y);if(s)++s;
return s;
}
int main(){
scanf("%d",&n);
ref(i,1,n)scanf("%d",&a[i]),add(a[i],i);
res=1;
ref(i,1,n)if(!vis[i]){
m=l=0;
for(int x=i;;b[x]=1,x=a[x]){
h[++m]=x;if(b[x])break;
}
bool fg=0;
for(int x=m;x==m||h[x]!=h[m];x--){
w[++l]=dfs(h[x-1],h[x]);
if(!w[l])print(0);w[l]--;
}
ref(i,1,l)w[i+l]=w[i];
ref(i,1,l)if(w[i]){
fg=1;int tt=0;
ref(j,i+1,2*l)if(!w[j])tt++;else break;
if(tt<w[i]-1)print(0);
if(tt>w[i]-1)(res*=2)%=mod;
}
if(!fg)p[l]++;
}
ref(i,1,n)if(p[i]){
f[0]=1;if((i&1)&&i>1)f[1]=2;else f[1]=1;
if((i&1)&&i>1) ref(j,2,p[i])f[j]=(1LL*f[j-2]*(j-1)%mod*i+2*f[j-1])%mod;
else ref(j,2,p[i])f[j]=(1LL*f[j-2]*(j-1)%mod*i+f[j-1])%mod;
res=1LL*res*f[p[i]]%mod;
}
print(res);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define re(i,a,b) for(int i=(a);i<(b);i++)
#define repd(i,a,b) for(int i=(a);i>=(b);i--)
#define clr(a) memset(a,0,sizeof(a));
#define il inline
#define sz(a) ((int)o.size())
#define run(x) for(int k=head[x];k;k=e[k].ne)
#define v e[k].t
#define all(a) o.begin(),o.end()
#define mp make_pair
#define pb push_back
#define w1 first
#define w2 second
#define adm(a,b,c) {a=a+b;if(a>=c)a-=c;else if(a<0)a+=c;}
typedef double db;
typedef long long ll;typedef long double ld;typedef unsigned long long ull;
typedef pair<int,int> pa;
const int N=1e6+5,M=1e7+5,INF=1e9,mod=1e9+7;
const ll linf=1e18;const double eps=1e-8,pi=acos(-1);
il int gmin(int &a,int b){if(a>b)a=b;}il ll gmin(ll &a,ll b){if(a>b)a=b;}il int gmax(int &a,int b){if(a<b)a=b;}il ll gmax(ll &a,ll b){if(a<b)a=b;}
il void read(ll&x){ll f=1,t=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){t=t*10+ch-'0';ch=getchar();}x=t*f;}il ll read(ll&x,ll&y){read(x);read(y);}
il void read(int&x){int f=1,t=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){t=t*10+ch-'0';ch=getchar();}x=t*f;}il int read(int&x,int&y){read(x);read(y);}
il void read(int&a,int&b,int&c){read(a);read(b);read(c);}il void read(ll&a,ll&b,ll&c){read(a);read(b);read(c);}
il int read(){int x;read(x);return x;}
il ll qpow(ll a,ll b,ll p){ll ret=1;for(;b;b>>=1,a=a*a%p)if(b&1)ret=ret*a%p;return ret;}il ll qpow(ll a,ll b){ll ret=1;for(;b;b>>=1,a=a*a%mod)if(b&1)ret=ret*a%mod;return ret;}
il ll qmul(ll a,ll b,ll p){ll ret=0;for(;b;b>>=1,a=(a<<1)%p)if(b&1)adm(ret,a,p);return ret;}il ll qmul(ll a,ll b){ll ret=0;for(;b;b>>=1,a=(a<<1)%mod)if(b&1)adm(ret,a,mod);return ret;}
il void judge(){
freopen("dato.in","r",stdin);
freopen("dato.out","w",stdout);
}
int n,res=1,a[N],vis[N],d[N],len[N],cnt[N];
ll f[N];
void gao(int n,int m){
ll f1=1;if(n>1&&n%2==1)f1++;
f[0]=1;f[1]=f1;
rep(i,2,m)(f[i]=f1*f[i-1]%mod+f[i-2]*(i-1)%mod*n%mod)%=mod;
res=res*f[m]%mod;
}
int main(){
read(n);rep(i,1,n)read(a[i]),d[a[i]]++;
rep(i,1,n)if(d[i]>2)return puts("0"),0;
rep(i,1,n)if(!d[i]){
int x=i,sz=0;
for(;d[x]<2;x=a[x])vis[x]=1,sz++;
len[x]=sz;
}
rep(i,1,n)if(!vis[i]){
int x;vector<int>o;
for(x=i;!vis[x];x=a[x])vis[x]=1,o.pb(x);
if(x!=i)return puts("0"),0;
int isc=1;re(j,0,o.size())if(len[o[j]])isc=0;
if(isc)cnt[o.size()]++;else{
int tmp=1;
repd(j,o.size()-1,0)if(!len[o[j]])tmp++;else break;
re(j,0,o.size())if(len[o[j]]){
if(len[o[j]]<tmp)res=res*2%mod;
if(len[o[j]]>tmp)return puts("0"),0;
tmp=1;
}else tmp++;
}
}rep(i,1,n)if(cnt[i])gao(i,cnt[i]);
cout<<res<<endl;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int N;
vector<int> G[100010],ele;
int A[100010];
bool vis[100010];
bool loop[100010];
int len[100010];
int par[100010];
int num[100010];
int nxt[100010];
int cnt;
ll ans = 1;
ll dp[100010];
ll mod = 1000000007;
void DFS(int v)
{
if(vis[v])return;
vis[v] = true;
ele.push_back(v);
for(int i = 0; i < G[v].size(); i++)DFS(G[v][i]);
DFS(par[v]);
return;
}
int main()
{
scanf("%d",&N);
for(int i = 0; i < N; i++)
{
int a;
scanf("%d",&a);
G[a].push_back(i + 1);
par[i + 1] = a;
}
for(int i = 1; i <= N; i++)if(!vis[i])
{
cnt = 0;
ele.clear();
DFS(i);
int now = i;
for(int j = 0; j < ele.size(); j++)now = par[now];
for(int j = 0; j < ele.size(); j++)
{
if(!loop[now])cnt++;
loop[now] = true;
now = par[now];
}
if(cnt == ele.size())
{
num[cnt]++;
continue;
}
int sum = 1;
for(int j = 0; j < ele.size() * 2; j++)
{
nxt[now] = sum;
sum++;
if(G[now].size() == 2)sum = 1;
now = par[now];
}
for(int j = 0; j < ele.size(); j++)
{
if(G[ele[j]].size() >= 3)ans = 0;
if(!loop[ele[j]] && G[ele[j]].size() >= 2)ans = 0;
if(G[ele[j]].size() == 0)
{
int l = 0;
int tmp = ele[j];
while(!loop[tmp])
{
l++;
tmp = par[tmp];
}
len[tmp] = l;
}
}
for(int j = 0; j < ele.size(); j++)
{
int tmp = ele[j];
if(!loop[tmp] || G[tmp].size() == 1)continue;
if(nxt[tmp] < len[tmp])ans = 0;
if(nxt[tmp] > len[tmp])ans *= 2;
ans %= mod;
}
}
for(int i = 1; i <= N; i++)
{
for(int j = 0; j <= num[i]; j++)dp[j] = 0;
dp[num[i]] = 1;
for(int j = num[i]; j > 0; j--)
{
dp[j - 1] += dp[j];
dp[j - 1] %= mod;
if(j - 2 >= 0)dp[j - 2] += dp[j] * (ll)(j - 1) % mod * (ll)i % mod;
dp[j - 2] %= mod;
if(i % 2 && i > 1)dp[j - 1] += dp[j];
dp[j - 1] %= mod;
}
ans *= dp[0];
ans %= mod;
}
printf("%lld\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
template<typename T>inline void read(T &x)
{
char c=x=0;
for(c=getchar();!isdigit(c);c=getchar());
for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48);
}
namespace gg
{
typedef long long ll;
const int N=101000,MOD=1000000007;
ll qpow(ll a,ll b){ll c=1;for(;b;b>>=1,a=a*a%MOD)if(b&1)c=c*a%MOD;return c;}
struct disjoint_union_set
{
int q[N];
inline void init(int n){for(int i=1;i<=n;i++)q[i]=i;}
inline int ask(int p){return p==q[p]?p:q[p]=ask(q[p]);}
inline void link(int u,int v){q[ask(u)]=ask(v);}
inline bool uni(int u,int v){return ask(u)==ask(v);}
}d;
ll fact[N],ifact[N];
bool is_cir[N],not_leave[N],vis[N];
int dep[N],cnt[N];
int s[N];
int n;
ll initialize()
{
read(n),d.init(n);
fact[0]=1;
for(int i=1;i<=n;i++)
{
read(s[i]);
not_leave[s[i]]=1;
fact[i]=fact[i-1]*i%MOD;
}
ifact[n]=qpow(fact[n],MOD-2);
for(int i=n;i;i--)ifact[i-1]=ifact[i]*i%MOD;
for(int i=1;i<=n;i++)
{
if(d.uni(i,s[i]))
{
for(int p=i;!is_cir[p];p=s[p])
vis[p]=is_cir[p]=1;
}
d.link(i,s[i]);
}
for(int i=1;i<=n;i++)
if(!not_leave[i])
{
int p=i,w=0;
for(;!vis[p];p=s[p])
vis[p]=1,w++;
if(!is_cir[p] || dep[p])return 0;
dep[p]=w;
}
ll ret=1;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
if(dep[i] && !vis[i])
{
int p=s[i],w=1;
for(;!vis[p];vis[p]=1,p=s[p])
{
if(dep[p]>w)return 0;
else if(dep[p])
{
if(dep[p]<w)ret=ret*2%MOD;
w=0;
}
w++;
}
}
for(int i=1;i<=n;i++)
if(!vis[i] && is_cir[i])
{
int p=i,w=0;
for(;!vis[p];vis[p]=1,p=s[p])w++;
cnt[w]++;
}
return ret;
}
void solve()
{
ll ans,ret,res,tmp;
if(!(ans=initialize())){printf("0\n");return;}
for(int x=1;x<=n;x++)
{
int K=cnt[x];if(!K)continue;
// printf("%d : %d\n",x,K);
res=0;
for(int i=0;i*2<=K;i++)
{
ret=fact[K]*ifact[K-i*2]%MOD*ifact[i]%MOD*qpow(x,i)%MOD;
tmp=qpow(2,i),tmp=qpow(tmp,MOD-2);
ret=ret*tmp%MOD;
if(x>1 && x%2)ret=ret*qpow(2,K-2*i)%MOD;
res=(res+ret)%MOD;
}
ans=ans*res%MOD;
}
printf("%lld\n",ans);
}
}
int main()
{
// freopen("E.in","r",stdin);
// freopen("E.out","w",stdout);
gg::solve();
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int N = 200000;
const int MOD = 1e9 + 7;
int n;
int ai[N], vis[N], cir[N], cnt[N], V[N];
vector <int> bi[N];
int ans = 1;
int f[N], fac[N], inv[N];
int powi(int a, int b)
{
int c = 1;
for (; b; b >>= 1, a = 1ll * a * a % MOD)
if (b & 1) c = 1ll * c * a % MOD; return c;
}
int C(int a, int b)
{
return 1ll * fac[a] * inv[b] % MOD * inv[a - b] % MOD;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; ++ i)
scanf("%d", &ai[i]), bi[ai[i]].push_back(i);
for (int i = 1; i <= n; ++ i)
if (!vis[i])
{
int p = i;
for (; !vis[p]; p = ai[p]) vis[p] = i;
if (vis[p] == i)
for (; !cir[p]; p = ai[p]) cir[p] = 1;
}
for (int i = 1; i <= n; ++ i)
if (bi[i].size() - cir[i] > 1)
return puts("0"), 0;
for (int i = 1; i <= n; ++ i)
if (bi[i].size() > 1 && !cir[bi[i][1]])
swap(bi[i][0], bi[i][1]);
for (int i = 1; i <= n; ++ i) vis[i] = 0;
for (int i = 1; i <= n; ++ i)
if (cir[i] && !vis[i])
{
int l = 0, G = 0;
for (int j = i; !vis[j]; j = ai[j])
{
l ++;
vis[j] = 1;
cnt[l] = 0;
if (!cir[bi[j][0]])
for (int k = j; bi[k].size(); k = bi[k][0])
cnt[l] ++, G = 1;
}
if (!G)
{
V[l] ++;
continue;
}
for (int j = 1; j <= l; ++ j) cnt[j + l] = cnt[j];
int x = 1;
while (!cnt[x]) x ++;
for (int j = x + 1, k = x; j <= x + l; ++ j)
if (cnt[j])
{
if (cnt[j] > j - k) return puts("0"), 0;
else if (cnt[j] < j - k) ans = 2ll * ans % MOD;
k = j;
}
}
f[0] = 1;
for (int i = 1; i <= n; ++ i) f[i] = 1ll * f[i - 1] * (2 * i - 1) % MOD;
fac[0] = 1;
for (int i = 1; i <= n; ++ i) fac[i] = 1ll * fac[i - 1] * i % MOD;
inv[n] = powi(fac[n], MOD - 2);
for (int i = n - 1; ~i; -- i) inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
for (int i = 1; i <= n; ++ i)
{
int res = 0;
for (int j = 0, k = 1; j <= V[i] / 2; ++ j, k = 1ll * k * i % MOD)
res = (res + 1ll * C(V[i], j * 2) * f[j] % MOD * k % MOD * (i > 1 && (i & 1)? powi(2, V[i] - j * 2): 1)) % MOD;
ans = 1ll * ans * res % MOD;
}
printf("%d\n", ans);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100, MOD = 1000 * 1000 * 1000 + 7;
int sum(int a, int b) {
a += b;
if (a < 0)
a += MOD;
else if (a >= MOD)
a -= MOD;
return a;
}
int mul(int a, int b) {
return 1LL * a * b % MOD;
}
void _sum(int &a, int b) {
a = sum(a, b);
}
void _mul(int &a, int b) {
a = mul(a, b);
}
int n, ans = 1, out[N], cnt[N], fac[N], inv[N], val[N], have[N], power[3][N];
bool mark[N];
vector<int> in[N];
void ZERO() {
cout << "0\n";
exit(0);
}
void pre_pro() {
fac[0] = 1;
for (int i = 1; i < N; i++)
fac[i] = mul(fac[i - 1], i);
inv[0] = inv[1] = 1;
for (int i = 2; i < N; i++)
inv[i] = mul(MOD - MOD / i, inv[MOD % i]);
for (int i = 2; i < N; i++)
_mul(inv[i], inv[i - 1]);
val[0] = 1;
for (int i = 1; i < N; i++)
val[i] = mul(2 * i - 1, val[i - 1]);
for (int i = 1; i <= 2; i++) {
power[i][0] = 1;
for (int j = 1; j < N; j++)
power[i][j] = mul(power[i][j - 1], i);
}
}
int c(int m, int n) {
return (m > n? 0: mul(fac[n], mul(inv[m], inv[n - m])));
}
void input() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> out[i];
in[--out[i]].push_back(i);
}
}
void check() {
for (int i = 0; i < n; i++)
if (in[i].size() > 2)
ZERO();
}
void go(int v, int now = 0) {
if (cnt[v])
ZERO();
cnt[v] = now;
if (in[v].size() > 1)
return;
mark[v] = true;
go(out[v], now + 1);
}
void set_len() {
for (int i = 0; i < n; i++)
if (in[i].empty())
go(i);
}
void dfs(int v, vector<int> *vec) {
mark[v] = true;
vec->push_back(v);
if (mark[out[v]] == false)
dfs(out[v], vec);
}
bool cycle(vector<int> vec) {
for (int v: vec)
if (cnt[v])
return false;
return true;
}
void f(vector<int> vec) {
vector<int> X;
for (int i = 0; i < vec.size(); i++)
if (cnt[vec[i]])
X.push_back(i);
X.push_back(X[0]);
for (int i = 1; i < X.size(); i++) {
int a = (X[i] - X[i - 1] + vec.size()), b = cnt[vec[X[i]]];
if (a > vec.size())
a -= vec.size();
if (b > a)
ZERO();
else if (b < a)
_mul(ans, 2);
}
}
void find_cycles() {
for (int v = 0; v < n; v++)
if (mark[v] == false) {
vector<int> vec;
dfs(v, &vec);
if (cycle(vec))
have[vec.size()]++;
else
f(vec);
}
}
int handle(int l) {
int res = 0;
int p = 1 + ((l & 1) && (l > 1));
int now = 1;
for (int i = 0; 2 * i <= have[l]; i++) {
_sum(res, mul(mul(c(2 * i, have[l]), val[i]), mul(power[p][have[l] - 2 * i], now)));
_mul(now, l);
}
return res;
}
void check_cycles() {
for (int i = 1; i < N; i++)
_mul(ans, handle(i));
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
pre_pro();
input();
check();
set_len();
find_cycles();
check_cycles();
cout << ans;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
#pragma GCC optimize ("O2,unroll-loops")
//#pragma GCC optimize("no-stack-protector,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef pair<ll, ll> pll;
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define debugp(x) cerr<<#x<<"= {"<<(x.first)<<", "<<(x.second)<<"}"<<endl;
#define debug2(x, y) cerr<<"{"<<#x<<", "<<#y<<"} = {"<<(x)<<", "<<(y)<<"}"<<endl;
#define debugv(v) {cerr<<#v<<" : ";for (auto x:v) cerr<<x<<' ';cerr<<endl;}
#define all(x) x.begin(), x.end()
#define pb push_back
#define kill(x) return cout<<x<<'\n', 0;
const int inf=1000000010;
const ll INF=10000000000000010LL;
const int mod=1000000007;
const int MAXN=100010, LOG=20;
int n, m, k, u, v, x, y, t, a, b, ans=1;
int A[MAXN], B[MAXN], ted[MAXN], deg[MAXN];
int mark[MAXN], last[MAXN], dor[MAXN];
ll dp[MAXN];
int main(){
ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
cin>>n;
for (int i=1; i<=n; i++) cin>>A[i], deg[A[i]]++;
for (int i=1; i<=n; i++) if (!mark[i]){
int v=i;
while (mark[v]<=2){
if (last[v]!=i) last[v]=i, mark[v]=1;
else mark[v]++;
if (mark[v]==2) dor[v]=1;
v=A[v];
}
}
for (int i=1; i<=n; i++) if (deg[i]>2 || (deg[i]==2 && !dor[i])) kill(0);
for (int i=1; i<=n; i++) B[A[i]]^=i;
memset(mark, 0, sizeof(mark));
for (int i=1; i<=n; i++) if (dor[i] && !mark[i]){
int v=i, bad=0;
while (!mark[v]){
mark[v]++;
B[A[v]]^=v;
v=A[v];
}
vector<int> vec;
while (mark[v]<2){
mark[v]++;
int len=0, u=v;
while (!dor[B[u]] && B[u]){
u=B[u];
len++;
}
if (len) bad=1;
vec.pb(len);
v=A[v];
}
// debugv(vec)
if (!bad){
ted[vec.size()]++;
continue ;
}
int c0=0;
while (vec.back()==0) c0++, vec.pop_back();
for (int x:vec){
if (x==0) c0++;
else{
if (x<=c0) ans=ans*2%mod;
if (x-1>c0) kill(0)
c0=0;
}
}
}
// debug(ans)
for (int i=1; i<=n; i++) if (ted[i]){
ll z=1+(i>1 && i%2);
dp[0]=1;
dp[1]=z;
for (int j=2; j<=ted[i]; j++) dp[j]=(dp[j-1]*z + (j-1ll)*i%mod*dp[j-2])%mod;
ans=ans*dp[ted[i]]%mod;
}
cout<<ans<<"\n";
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100100
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define drp(x, a, b) for(int x=a; x>=b; x--)
int deg[N], d[N], n, vis[N], a[N], tot, num[N], l[N], ans=1, fac[N], inv[N];
vector<pair<int, int> > fu;
queue<int> q;
int power(int a, int k){
int ret=1;
while(k)
{
if(k&1) ret=1ll*ret*a%mod;
a=1ll*a*a%mod; k>>=1;
}
return ret;
}
int C(int n, int m){
return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int pr(int n){
return 1ll*fac[2*n]*inv[n]%mod*power((mod+1)>>1, n)%mod;
}
void init(){
fac[0]=1;
rep(i, 1, n) fac[i]=1ll*fac[i-1]*i%mod;
inv[n]=power(fac[n], mod-2);
drp(i, n, 1) inv[i-1]=1ll*inv[i]*i%mod;
}
int st[N<<1];
void renew(int &x, const int y){
x += y;
if(x < 0) x += mod;
if(x >= mod) x -= mod;
}
int getdp(){
int fr = 0;
rep(i, 1, tot) st[i + tot] = st[i];
rep(i, 1, tot) if(l[st[i]]){
fr = i;
break;
}
reverse(st + fr + 1, st + fr + tot);
int a = fr + l[st[fr]] - 1, b = a + 1, c = 1;
rep(i, fr + 1, fr + tot - 1)if(l[st[i]]){
int na = i + l[st[i]] - 1, nb = na + 1, cc = 0;
if(a < i) renew(cc, c);
if(b < i) renew(cc, c);
a = na;
b = nb;
c = cc;
}
int ans = 0;
if(a < fr + tot) renew(ans, c);
if(b < fr + tot) renew(ans, c);
return ans;
}
int main(){
scanf("%d", &n);
init();
rep(i, 1, n)
{
scanf("%d", a+i);
deg[a[i]]++;
}
rep(i, 1, n) if(deg[i]>2) return puts("0"), 0;
rep(i, 1, n) d[i]=deg[i];
rep(i, 1, n) if(!d[i]) q.push(i);
while(!q.empty())
{
int u=q.front(); q.pop();
vis[u]=1;
if(--d[a[u]]==0) q.push(a[u]);
}
rep(i, 1, n) if(vis[i] && deg[i]>1) return puts("0"), 0;
rep(i, 1, n) if(vis[i] && !deg[i])
{
int x=i, cnt=0;
for(; vis[x]; x=a[x]) cnt++;
l[x]=cnt;
}
rep(i, 1, n) if(!vis[i])
{
int x=i, fl=0, ass=1; tot=0;
for(; !vis[x]; x=a[x]) st[++tot]=x, fl|=bool(l[x]), vis[x]=1;
if(!fl) num[tot]++;
else
{/*
int p=0; fu.clear();
if(l[i]) fu.pb(mp(l[i], 0));
for(p++, x=a[i]; x!=i; x=a[x]) if(l[x]) fu.pb(mp(l[x], p));
fu.pb(mp(fu[0].fi, fu[0].se+tot));
int sz=fu.size();
rep(i, 1, sz-1)
{
if(fu[i].fi>fu[i].se-fu[i-1].se) ass=0;
if(fu[i].fi<fu[i].se-fu[i-1].se) ass=2ll*ass%mod;
}*/
ass=getdp();
}
ans=1ll*ans*ass%mod;
}
rep(i, 1, n) if(num[i])
{
int m=num[i], ass=0;
rep(j, 0, m/2)
{
int tmp=1ll*C(m, 2*j)*pr(j)%mod*power(i, j)%mod;
if(i>1 && (i&1)) tmp=1ll*tmp*power(2, m-2*j)%mod;
ass=(ass+tmp)%mod;
}
ans=1ll*ans*ass%mod;
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <algorithm>
using namespace std;
const int md=1000000007,MX=100100;
int n,i,j,k,cc,d,r=1,cur,all[MX],fi[MX],la[MX],a[MX],b[MX],c[MX],cnt[MX],cyc[MX],comp[MX],v[MX],gv[MX];
long long f[MX],g[MX];
bool u[MX],w[MX],ic[MX],was;
int dfs(int i, int d) {
b[d]=i;
u[i]=true;
w[i]=true;
if (w[a[i]]) {
c[i]=++cc;
cyc[cc]=1;
fi[cc]=cur;
ic[a[i]]=true;
all[cur++]=a[i];
for (int j=d; b[j]!=a[i]; j--, cyc[cc]++) {
ic[b[j]]=true;
all[cur++]=b[j];
}
la[cc]=cur;
} else if (u[a[i]]) c[i]=c[a[i]]; else c[i]=dfs(a[i],d+1);
w[i]=false;
comp[c[i]]++;
return c[i];
}
int main() {
scanf("%d",&n);
for (f[0]=g[0]=i=1; i<=n; i++) {
scanf("%d",&a[i]);
v[a[i]]++;
f[i]=(((i>1)?(i-1)*f[i-2]:0LL)+f[i-1]*2)%md;
g[i]=(((i>1)?(i-1)*g[i-2]:0LL)+g[i-1])%md;
}
for (i=1; i<=n; i++) if (v[i]==0) dfs(i,1);
for (i=1; i<=n; i++) if (u[i]) {
if (v[i]>2) { puts("0"); return 0; }
if (!ic[i]) {
if (v[i]>1) { puts("0"); return 0; }
gv[a[i]]=i;
}
}
for (i=1; i<=cc; i++) {
if (comp[i]>2*cyc[i]) { puts("0"); return 0; }
was=false;
for (cur=0, j=fi[i]; j<la[i]; j++, cur--) {
k=all[j];
int cz=0;
for (int e=gv[k]; e!=0; e=gv[e]) cz++;
if (cz>0) {
if (was) {
if (cur>0) { puts("0"); return 0; }
if (cur<0) r=(r*2)%md;
} else was=true;
cur=cz;
}
}
for (j=fi[i]; j<la[i]; j++, cur--) {
k=all[j];
int cz=0;
for (int e=gv[k]; e!=0; e=gv[e]) cz++;
if (cz>0) {
if (cur>0) { puts("0"); return 0; }
if (cur<0) r=(r*2)%md;
cur=cz;
break;
}
}
}
for (i=1; i<=n; i++) if (!u[i]) {
for (j=i, k=0; !u[j]; j=a[j], k++) u[j]=true;
cnt[k]++;
}
r=(r*g[cnt[1]])%md;
for (j=2; j<=n; j+=2) {
for (i=1; i<=cnt[j]; i++) g[i]=(((i>1)?((i-1)*g[i-2])%md*j:0LL)+g[i-1])%md;
r=(r*g[cnt[j]])%md;
}
for (j=3; j<=n; j+=2) {
for (i=1; i<=cnt[j]; i++) f[i]=(((i>1)?((i-1)*f[i-2])%md*j:0LL)+f[i-1]*2)%md;
r=(r*f[cnt[j]])%md;
}
printf("%d\n",r);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cmath>
#include<stdio.h>
#include<algorithm>
#define ll long long
using namespace std;
const int N=1000100,P=1e9+7;
inline int read(){
int x=0,f=0,c=getchar();
for(;c>'9'||c<'0';f=c=='-',c=getchar());
for(;c>='0'&&c<='9';c=getchar())
x=(x<<1)+(x<<3)+c-'0';return f?-x:x;
}
inline int ksm(int x,int y){
int z=1;for(;y;y>>=1,x=1ll*x*x%P)
if(y&1)z=1ll*z*x%P;return z;
}
int q[N],r,n,a[N],du[N],f[N],ans,s[N],fac[N],inv[N],pw[N],ipw[N];
void dfs(int x){du[x]--;q[++r]=x;if(du[a[x]])dfs(a[x]);}
inline int C(int n,int m){return 1ll*fac[n]*inv[m]%P*inv[n-m]%P;}
int main(){
int i,l,x,la,fr,re;
n=read();ans=1;
for(i=1;i<=n;i++)a[i]=read(),du[a[i]]++;
for(i=1;i<=n;i++){
if(!du[i])q[++r]=i;
if(du[i]>2)goto fuck;
}
for(l=1;l<=r;l++){
x=q[l];if(f[a[x]])goto fuck;
f[a[x]]=f[x]+1;if(!(--du[a[x]]))q[++r]=a[x];
}
for(l=1;l<=n;l++)if(du[l]){
r=la=0;dfs(l);re=1;
for(i=1;i<=r;i++)
if(f[x=q[i]]){
if(!la){fr=la=i;continue;}
if(i-la<f[x])goto fuck;
if(i-la>f[x])re=(re+re)%P;la=i;
}
if(!la){s[r]++;continue;}
x=q[fr];fr=fr+r;
if(fr-la<f[x])goto fuck;
if(fr-la>f[x])re=(re+re)%P;
ans=1ll*ans*re%P;
}
fac[0]=inv[0]=pw[0]=ipw[0]=1;
pw[1]=2;ipw[1]=(P+1)/2;
for(i=1;i<=n;i++){
fac[i]=1ll*fac[i-1]*i%P;
inv[i]=ksm(fac[i],P-2);
pw[i]=1ll*pw[i-1]*pw[1]%P;
ipw[i]=1ll*ipw[i-1]*ipw[1]%P;
}
for(l=1;l<=n;l++)if(s[l]){
re=0;for(i=0;i+i<=s[l];i++){
x=1ll*C(s[l],i+i)*fac[i+i]%P*inv[i]%P*ipw[i]%P*ksm(l,i)%P;
if(l>1&&(l&1)){x=1ll*x*pw[s[l]-i-i]%P;}re=(re+x)%P;
}ans=1ll*ans*re%P;
}printf("%d\n",ans);
return 0;fuck:puts("0");;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<stdio.h>
#include<queue>
#include<assert.h>
#include<tuple>
#include<string>
#include<algorithm>
#include<iostream>
#include<map>
#include<string.h>
#include<vector>
#include<math.h>
#include<stdlib.h>
#include<set>
#include<ctype.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef tuple<int,int,int> t3;
const int MX = 1 << 20;
const int MM = 1000000007;
int D[MX], vst[MX], cnt[MX], cycle[MX];
ll T[MX], U[MX], V[MX];
int N;
void dfs(int x, int c = 1){
if( vst[x] ){
if( vst[x] == -1 ) cnt[x] = c - 1;
else{
int t = x;
do{
cycle[D[t]] = t;
t = D[t];
}while(t != x);
cnt[x] = vst[x] - 1;
}
return;
}
vst[x] = c;
dfs(D[x], c+1);
vst[x] = -1;
}
int indeg[MX];
int main()
{
scanf("%d", &N);
for(int i = 1; i <= N; i++) scanf("%d", D+i), indeg[D[i]]++;
for(int i = 1; i <= N; i++){
if( indeg[i] >= 3 ) return !printf("0\n");
}
for(int i = 1; i <= N; i++){
if( vst[i] || indeg[i] ) continue;
dfs(i);
}
for(int i = 1; i <= N; i++){
if(vst[i]) continue;
dfs(i);
}
for(int i = 1; i <= N; i++) if( !cycle[i] && cnt[i] ) return !printf("0\n");
ll ans = 1;
for(int i = 1; i <= N; i++){
if( !cnt[i] ) continue;
int x = i;
for(int j = 1; j < cnt[i]; j++){
x = cycle[x];
if( cnt[x] ) return !printf("0\n");
}
if( !cnt[cycle[x]] ) ans = ans*2 % MM;
}
for(int i = 1; i <= N; i++) vst[i] = 0;
map<int, int> L;
int c1 = 0;
for(int i = 1; i <= N; i++){
if( vst[i] || !cycle[i] ) continue;
int tt = 0, ch = 1;
int x = i;
do{
x = D[x]; vst[x] = 1;
if( cnt[x] ) ch = 0;
tt++;
}while(x != i);
if( ch ) L[tt] += 1;
}
for(auto c : L){
T[0] = 1; T[1] = c.first >= 2 && c.first%2 == 1 ? 2 : 1;
for(int i = 2; i <= c.second; i++){
T[i] = (T[i-1] * T[1] + T[i-2] * c.first % MM * (i-1)) % MM;
}
ans = ans * T[c.second] % MM;
}
printf("%lld\n", ans);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef long long in;
typedef vector<in> VI;
typedef vector<VI> VVI;
const in mdl=1000000007LL;
in p2(in a){
return (1LL<<a);
}
in pw(in a, in b, in lm=62){
a%=mdl;
if(a<0)
a+=mdl;
in r=1;
for(in i=lm;i>=0;--i){
r=r*r%mdl;
if(b&p2(i))
r=r*a%mdl;
}
return r;
}
in inv(in a){
a%=mdl;
if(a<0)
a+=mdl;
assert(a!=0);
return pw(a,mdl-2,30);
}
VI fc,invfc;
in ncr(in a, in b){
if(b==0 || b==a)
return 1;//even if a<0
if(b<0 || b>a)
return 0;
return fc[a]*invfc[b]%mdl*invfc[a-b]%mdl;
}
void inifc(){
const in mxfc=1001000;
fc.resize(mxfc);
invfc.resize(mxfc);
fc[0]=fc[1]=invfc[0]=invfc[1]=1;
for(in i=2;i<mxfc;++i){
fc[i]=fc[i-1]*i%mdl;
invfc[i]=invfc[mdl%i]*(mdl-mdl/i)%mdl;
}
for(in i=2;i<mxfc;++i){
invfc[i]*=invfc[i-1];
invfc[i]%=mdl;
}
}
void fl(){
cout<<0<<endl;
exit(0);
}
in tw(in cnt, in wmix, in wsing){
in sm=0;
in cmix=0;
in wsl=1;
while(cmix<=cnt){
if(wsing)
sm+=pw(wmix,cmix/2)*pw(2,cnt-cmix)%mdl*wsl%mdl*invfc[cmix/2]%mdl;
else
sm+=pw(wmix,cmix/2)*wsl%mdl*invfc[cmix/2]%mdl;
wsl*=cnt-cmix;
wsl%=mdl;
++cmix;
wsl*=cnt-cmix;
wsl%=mdl;
++cmix;
wsl*=inv(2);
wsl%=mdl;
}
sm%=mdl;
return sm;
}
VI pcyc;
VI a;
VI dg;
VI itsz;
VI vis;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
inifc();
in n;
cin>>n;
a.resize(n);
pcyc.resize(n+1);
dg.resize(n);
forn(i,n){
cin>>a[i];
--a[i];
++dg[a[i]];
if(dg[a[i]]>2)
fl();
}
itsz.resize(n);
vis.resize(n);
queue<in> q;
forn(i,n){
if(dg[i]==0){
vis[i]=1;
q.push(i);
}
}
while(!q.empty()){
in u=q.front();
q.pop();
if(itsz[a[u]])
fl();
itsz[a[u]]=itsz[u]+1;
if(dg[a[u]]==1){
vis[a[u]]=1;
q.push(a[u]);
}
}
VI tit;
in wys=1;
forn(i,n){
tit.clear();
if(vis[i])
continue;
in u=i;
in ft=-1;
do{
vis[u]=1;
tit.PB(itsz[u]);
if(itsz[u])
ft=sz(tit)-1;
u=a[u];
}while(u!=i);
if(ft==-1){
++pcyc[sz(tit)];
continue;
}
in cf=ft;
do{
in cnt=0;
do{
++cnt;
cf=(cf+1)%sz(tit);
}while(tit[cf]==0);
if(cnt<tit[cf])
fl();
if(cnt>tit[cf])
wys=wys*2%mdl;
}while(cf!=ft);
}
forn(i,n+1){
if(pcyc[i]){
wys=wys*tw(pcyc[i],i,((i%2==1 && i!=1)?1:0))%mdl;
}
}
cout<<wys<<endl;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define ha 1000000007
#define fuck {puts("0");return 0;}
using namespace std;
const int N=100010;
int nxt[N],n;
int pre[N][2];
int ring[N],f[N];
bool vis[N];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",nxt+i);
if(!pre[nxt[i]][0]) pre[nxt[i]][0]=i;
else if(!pre[nxt[i]][1]) pre[nxt[i]][1]=i;
else fuck
}
int ans=1;
for(int i=1;i<=n;i++)
{
if(vis[i]) continue;
int j=i,sz=0;
while(!vis[j])
vis[j]=1,j=nxt[j];
int p=j,q;
do{
sz++;q=nxt[p];
if(pre[q][0]!=p)
swap(pre[q][0],pre[q][1]);
}while((p=q)!=j);
bool flag=1;int cur=1;
do{
if(!pre[p][1]) continue;
int a=pre[p][0],b=pre[p][1];
flag=0;vis[b]=1;
while(pre[b][0])
{
if(pre[a][1]||pre[b][1]) fuck
a=pre[a][0];b=pre[b][0];vis[b]=1;
}
if(!pre[a][1]) (cur<<=1)%=ha;
}while((p=nxt[p])!=j);
if(flag) ring[sz]++;
else ans=1ll*ans*cur%ha;
}
for(int i=1;i<=n;i++)
{
bool flag=(i&1)&&(i!=1);
f[0]=1;f[1]=flag+1;
for(int j=2;j<=ring[i];j++)
f[j]=(1ll*f[j-2]*(j-1)%ha*i+(f[j-1]<<flag))%ha;
ans=1ll*ans*f[ring[i]]%ha;
}
cout<<ans<<endl;
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cctype>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long int64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
return f*x;
}
const int N = 1e5+5, P = 1e9+7;
int n, a[N], d[N], rec[N], cir[N], size[N], c[N];
int64 ans = 1, f[N];
int main()
{
n = read(); for(int i = 1; i <= n; ++i) a[i] = read(), ++d[a[i]];
for(int i = 1, p; i <= n; ++i)
{
if(rec[i]) continue;
p = i; while(!rec[p]) rec[p] = i, p = a[p];
if(rec[p] == i) while(!cir[p]) cir[p] = 1, p = a[p];
}
for(int i = 1, k, p; i <= n; ++i)
{
if((cir[i]&&d[i] > 2)||(!cir[i]&&d[i] > 1)) return puts("0"), 0;
if(d[i]) continue; p = i, k = 0;
while(!cir[p]) ++k, p = a[p]; size[p] = k;
}
for(int i = 1, p, fir, j, k, firs; i <= n; ++i)
{
if(!cir[i]) continue;
firs = fir = j = k = 0, p = i;
while(cir[p])
{
++k, cir[p] = 0;
if(size[p]&&!fir) j = fir = k, firs = size[p];
else if(size[p]) ans = ans*((size[p] < k-j)+(size[p] <= k-j))%P, j = k;
p = a[p];
}
if(!fir) ++c[k];
else ans = ans*((firs < k-j+fir)+(firs <= k-j+fir))%P;
}
for(int i = 1; i <= n; ++i)
{
if(!c[i]) continue;
f[0] = 1;
for(int j = 1; j <= c[i]; ++j)
{
if(i > 1&&(i&1)) f[j] = (f[j-1]<<1)%P;
else f[j] = f[j-1];
if(j > 1) f[j] = (f[j]+f[j-2]*(j-1)%P*i%P)%P;
}
ans = ans*f[c[i]]%P;
}
printf("%lld\n", ans);
return 0;
}
/*
1
1 1
2
1 2 2
3
1 2 3 4
4
1 2 3 4 10
5
1 2 3 4 5 26
6
1 2 3 4 5 6 76
7
1 2 3 4 5 6 7 232
9
2 1 4 5 6 3 8 7 9
*/ | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=110000,mod=1000000007;
int n,ans;
int jc[N],njc[N],bir[N],nbir[N];
int a[N],fa[N],du[N],num[N],vis[N],tail[N],rem[N];
int f[N][2][2];
vector<int>vec[N],v1;
int qpow(int x,int y)
{
int ret=1;
while(y)
{
if(y&1)ret=(ll)ret*x%mod;
x=(ll)x*x%mod;y>>=1;
}
return ret;
}
void init()
{
for(int i=1;i<=n;i++)fa[i]=i;
jc[0]=njc[0]=bir[0]=nbir[0]=1;
for(int i=1;i<=n;i++)
{
jc[i]=(ll)jc[i-1]*i%mod;
bir[i]=bir[i-1]*2%mod;
}
njc[n]=qpow(jc[n],mod-2);
nbir[n]=qpow(bir[n],mod-2);
for(int i=n-1;i>=1;i--)
{
njc[i]=(ll)njc[i+1]*(i+1)%mod;
nbir[i]=nbir[i+1]*2%mod;
}
}
int find(int x){return fa[x]==x ? x : fa[x]=find(fa[x]);}
void quit(){puts("0");exit(0);}
void ins(int x)
{
if(!tail[x])return;
if(rem[x]<tail[x])quit();
else if(rem[x]>tail[x])ans=ans*2%mod;
}
int main()
{
//freopen("tt.in","r",stdin);
scanf("%d",&n);
init();
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(find(i)!=find(a[i]))
fa[find(i)]=find(a[i]);
du[a[i]]++;
}
for(int i=1;i<=n;i++)
{
if(du[i]>2)quit();
vec[find(i)].push_back(i);
}
ans=1;
for(int i=1,sz;i<=n;i++)if(sz=vec[i].size())
{
int p=-1;
for(int j=0;j<sz;j++)
if(du[vec[i][j]]==2)p=vec[i][j];
if(p==-1)num[sz]++;
else
{
int t=0,p1=a[p];
vis[p]=++t;
while(p1!=p)
{
if(vis[p1])quit();
vis[p1]=++t;
p1=a[p1];
}
for(int j=0,cnt;j<sz;j++)
if(du[p1=vec[i][j]]==0)
{
cnt=0;
while(!vis[p1])
{
cnt++;
if(du[p1]==2)quit();
p1=a[p1];
}
tail[p1]=cnt;
}
int now=0;p1=p;
for(int j=1;j<=t*2;j++,p1=a[p1])
{
if(tail[p1])
rem[p1]=max(rem[p1],now+1),now=0;
else now++;
}
v1.clear();
ins(p);p1=a[p];
while(p1!=p)
{
ins(p1);
p1=a[p1];
}
}
}
for(int i=1;i<=n;i++)if(num[i])
{
int t=0,now=1;
for(int j=0;j<=num[i];j+=2)
{
int t1=(ll)jc[num[i]]*njc[num[i]-j]%mod*njc[j>>1]%mod*nbir[j>>1]%mod*now%mod;
if((i&1)&&i!=1)t1=(ll)t1*bir[num[i]-j]%mod;
t=(t+t1)%mod;
now=(ll)now*i%mod;
}
ans=(ll)ans*t%mod;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
const long long mod = 1000000007;
long long inv[100100]={0,1},fact[100100]={1,1},ifact[100100]={1,1};
int N,P[100100],D[100100],L[100100],G[100100],X[100100],S[200200];
queue<int> Q;
int main()
{
scanf ("%d",&N);
for (int i=1;i<=N;i++) scanf ("%d",&P[i]), D[P[i]]++;
for (int i=1;i<=N;i++){
if (D[i] >= 3){
puts("0");
return 0;
}
if (D[i] == 0) Q.push(i);
}
while (!Q.empty()){
int x = Q.front(); Q.pop();
if (L[P[x]]){
puts("0");
return 0;
}
L[P[x]] = L[x] + 1;
if (--D[P[x]] == 0) Q.push(P[x]);
}
long long ans = 1;
for (int i=1;i<=N;i++) if (D[i] == 1){
int x = i, c = 0;
while (D[x] == 1){
X[c] = x;
S[c] = (L[x] > 0);
c++;
D[x] = 0;
x = P[x];
}
reverse(X,X+c);
reverse(S,S+c);
for (int i=0;i<c;i++) S[i+c] = S[i];
for (int i=1;i<c*2;i++) S[i] += S[i-1];
if (S[c-1] == 0) G[c]++;
else{
for (int i=0;i<c;i++){
int x = X[i], v = 0;
if (L[x]){
if (L[x] <= c && S[i+L[x]-1] - S[i] == 0) v++;
if (L[x] <= c-1 && S[i+L[x]] - S[i] == 0) v++;
ans = ans * v % mod;
}
}
}
}
for (int i=2;i<=N;i++){
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
fact[i] = fact[i-1] * i % mod;
ifact[i] = ifact[i-1] * inv[i] % mod;
}
for (int i=1;i<=N;i++) if (G[i]){
long long coeff = (i % 2 == 1 && i > 1 ? 2 : 1);
long long sum = 0, now = 1;
for (int j=0;j<G[i];j++) now = now * coeff % mod;
for (int j=0;j<=G[i]/2;j++){
sum = (sum + now * fact[G[i]] % mod * ifact[j*2] % mod * ifact[G[i]-j*2]) % mod;
now = now * (j * 2 + 1) % mod;
now = now * i % mod;
now = now * inv[coeff] % mod;
now = now * inv[coeff] % mod;
}
ans = ans * sum % mod;
}
printf ("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define mp make_pair
#define PI pair<int,int>
#define lowbit(i) i&-i
#define For(i,l,r) for(int i=(int)(l);i<=(int)(r);i++)
#define Rep(i,r,l) for(int i=(int)(r);i>=(int)(l);i--)
#define pb push_back
#define fi first
#define se second
inline char gc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline ll read(){
ll x = 0; char ch = gc(); bool positive = 1;
for (; !isdigit(ch); ch = gc()) if (ch == '-') positive = 0;
for (; isdigit(ch); ch = gc()) x = x * 10 + ch - '0';
return positive ? x : -x;
}
inline void write(ll a){
if(a<0){
a=-a; putchar('-');
}
if(a>=10)write(a/10);
putchar('0'+a%10);
}
inline void writeln(ll a){write(a); puts("");}
inline void wri(ll a){write(a); putchar(' ');}
inline ull rnd(){
ull ans=0; For(i,0,5)ans=ans<<15^rand(); return ans;
}
const int N=100005,mod=1000000007;
int a[N],f[N],fa[N],TO[N];
vector<int> pre[N],v[N];
ll fac[N],ni[N],ans=1;
bool vis[N];
inline int gf(int x){
return fa[x]==x?x:fa[x]=gf(fa[x]);
}
void GG(int frog){
if(!frog){
puts("0"); exit(0);
}
}
int getpre(int x,int id){
For(i,0,1)if(vis[pre[x][i]]==id)return pre[x][i];
assert(0);
}
ll ksm(ll a,int b){
int ans=1;
for(;b;b>>=1){
if(b&1)ans=ans*a%mod;
a=a*a%mod;
}
return ans;
}
void mer(int x,int y){
fa[gf(x)]=gf(y);
}
int main(){
int n=read();
For(i,fac[0]=1,n)fac[i]=fac[i-1]*i%mod;
ni[n]=ksm(fac[n],mod-2); Rep(i,n,1)ni[i-1]=ni[i]*i%mod;
For(i,1,n)pre[a[fa[i]=i]=read()].pb(i);
For(i,1,n)mer(i,a[i]);
For(i,1,n)v[gf(i)].pb(i);
For(i,1,n)if(fa[i]==i){
int f=1;
for(auto j:v[i])if(pre[j].size()!=1)f=0;
if(f)TO[v[i].size()]++;
else{
for(int j=i;;j=a[j]){
if(!vis[j])vis[j]=1;
else{
for(auto k:v[i])vis[k]=0;
vis[j]=1; for(int k=a[j];k!=j;k=a[k])vis[k]=1;
bool t=0; int sz=0;
for(int k=j;t==0||k!=j;k=a[k]){
t=1;
int f=0,g=0;
if(pre[k].size()>=2){//cout<<pre[k].size()<<endl;
GG(pre[k].size()==2);
for(int l=getpre(k,1);pre[l].size()==1;l=pre[l][0])f++;
for(int l=getpre(k,0);pre[l].size();l=pre[l][0]){GG(pre[l].size()==1);g++;}
g++;
//cout<<g<<" "<<f<<endl;
GG(g<=f+1);
if(g<=f)ans=ans*2%mod;
//cout<<g<<" "<<f<<" "<<v[k].size()<<endl;
}
}
break;
}
}
}
}
For(i,1,n)if(TO[i]){
int d=((i&1)&&i>1)?2:1;
int sum=ksm(d,TO[i]);
For(j,1,TO[i]/2)sum=(sum+fac[TO[i]]*ni[j]%mod*ni[TO[i]-j-j]%mod*ksm(d,TO[i]-j-j)%mod*ksm((ll)i*(mod+1)/2%mod,j))%mod;
//cout<<sum<<" "<<TO[i]<<endl;
ans=ans*sum%mod;
}
cout<<ans<<endl;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#define ll long long
#define mod 1000000007
#define N 110000
int a[N],num[N],col[N],mrk[N],cnt[N],n,lng[N];
bool cir[N];
ll dp[N];
int main()
{
scanf("%d", &n);
ll ans=1;
for (int i=1;i<=n;++i)
{
scanf("%d", &a[i]);
++num[a[i]];
}
for (int i = 1; i <= n; i++)
{
if (col[i])
continue;
int x = i;
for (; !col[x]; x = a[x])
col[x] = i;
if (col[x] != i)
continue;
for (;!cir[x];x=a[x])
cir[x]=1;
}
for (int i = 1; i <= n; i++)
if ((cir[i]&&num[i]>2)||(!cir[i]&&num[i]>1)) {printf("0");return 0;}
for (int i=1;i<=n;++i){
if (num[i]) continue;int x=i,tmp=0;
while(!cir[x]) x=a[x],tmp++;lng[x]=tmp;
}
for (int i=1;i<=n;++i){
if (!cir[i]) continue;int x=i,st=0,first=0,id=0,len=0;
while (cir[x]){
++id;cir[x]=0;
if (lng[x]){
if (!first){
st=first=id;len=lng[x];x=a[x];continue;
}else{
int numm=(lng[x]<(id-st))+(lng[x]<=(id-st));
(ans*=numm)%=mod;if (!ans) {printf("0");return 0;}st=id;x=a[x];continue;
}
}x=a[x];
}if (first){
int numm=(len<(id+first-st))+(len<=(id+first-st));
(ans*=numm)%=mod;if (!ans) {printf("0");return 0;}
}else cnt[id]++;
}
for (int i=1;i<=n;++i){
dp[0]=1;if (!cnt[i]) continue;
if (i>1&&(i%2)){
for (int j=1;j<=cnt[i];++j) {
dp[j]=dp[j-1]*2%mod;if (j>1) (dp[j]+=dp[j-2]*(j-1)*i%mod)%=mod;
}
}else{
for (int j=1;j<=cnt[i];++j){
dp[j]=dp[j-1];if (j>1) (dp[j]+=dp[j-2]*(j-1)*i%mod)%=mod;
}
}
(ans*=dp[cnt[i]])%=mod;if (!ans){printf("0");return 0;}
}printf("%lld",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define maxn 100010
#define mod 1000000007
using namespace std;
int read()
{
int x=0,f=1;
char ch=getchar();
while(ch-'0'<0||ch-'0'>9){if(ch=='-') f=-1;ch=getchar();}
while(ch-'0'>=0&&ch-'0'<=9){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n;
int a[maxn],deg[maxn],vis[maxn],cir[maxn],len[maxn],sum[maxn];
int ans=1;
void work(int x)
{
int now=0,fr=0,ed=0,l=0;
while(cir[x])
{
++now;cir[x]=0;
if(len[x])
{
if(!fr) fr=ed=now,l=len[x];
else{
ans=1ll*ans*((len[x]<now-ed)+(len[x]<=now-ed))%mod;
ed=now;
}
}
x=a[x];
}
if(!fr) sum[now]++;
else{
ans=1ll*ans*((l<now-ed+fr)+(l<=now-ed+fr))%mod;
}
}
int f[maxn];
void solve()
{
for(int i=1;i<=n;i++)
{
if(deg[i]) continue;
int x=i,l=0;
while(!cir[x])
{
x=a[x];
l++;
}
len[x]=l;
}
for(int i=1;i<=n;i++) if(cir[i]) work(i);
for(int i=1;i<=n;i++)
{
if(!sum[i]) continue;
f[0]=1;
for(int j=1;j<=sum[i];j++)
{
if(i>1&&(i&1)) f[j]=(f[j-1]+f[j-1])%mod;
else f[j]=f[j-1];
if(j>1) f[j]=(f[j]+1ll*f[j-2]*(j-1)%mod*i%mod)%mod;
}
ans=1ll*ans*f[sum[i]]%mod;
}
}
int main()
{
n=read();
for(int i=1;i<=n;i++) a[i]=read(),deg[a[i]]++;
for(int i=1;i<=n;i++)
{
if(vis[i]) continue;
int x=i;
while(!vis[x]) vis[x]=i,x=a[x];
if(vis[x]!=i) continue;
while(!cir[x]) cir[x]=1,x=a[x];
}
for(int i=1;i<=n;i++)
{
if((cir[i]&°[i]>2)||(!cir[i]&°[i]>1))
{
puts("0");
return 0;
}
}
solve();
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MX=100105,md=1000000007;
int i,j,k;
int f[MX],fac[MX],facinv[MX],M[MX];
int pw(int x,int y){
int z=1;
for(;y;y>>=1,x=(LL)x*x%md)if(y&1)z=(LL)z*x%md;
return z;
}
int C(int n,int m){return (LL)fac[n]*facinv[m]%md*facinv[n-m]%md;}
void ini(int n){
int i;
fac[0]=facinv[0]=f[0]=M[0]=1;
for(i=1;i<=n;i++)M[i]=(M[i-1]+M[i-1])%md;
for(i=1;i<=n;i++)f[i]=(LL)f[i-1]*(2*i-1)%md;
for(i=1;i<=n;i++)fac[i]=(LL)fac[i-1]*i%md;
facinv[n]=pw(fac[n],md-2);
for(i=n-1;i;i--)facinv[i]=(LL)facinv[i+1]*(i+1)%md;
}
int n,m,ch,ff,ans;
int a[MX],b[MX],z[MX],fg[MX],num[MX],d[MX],dp[MX];
int cal(int n,int m){
int i,k,ans=0;
if((m&1)&& m>1){
for(i=0,k=1;i+i<=n;i++,k=(LL)k*m%md)ans=((LL)f[i]*k%md*C(n,i+i)%md*M[n-i-i]%md+ans)%md;
}else{
for(i=0,k=1;i+i<=n;i++,k=(LL)k*m%md)ans=((LL)f[i]*k%md*C(n,i+i)%md+ans)%md;
}
return ans;
}
int main(){
cin>>n;
ini(n);
for(i=1;i<=n;i++)cin>>a[i];
for(i=1;i<=n;i++)d[a[i]]++;
for(i=1;i<=n;i++)if(!z[i]){
for(j=i;z[j]!=i && !z[j];j=a[j])z[j]=i;
if(z[j]==i){
fg[j]=1;
for(k=a[j];k!=j;k=a[k])fg[k]=1;
}
}
for(i=1;i<=n;i++)if(d[i]>fg[i]+1)return cout<<0<<endl,0;
for(i=1;i<=n;i++)if(!fg[i])b[a[i]]=i;
for(i=1;i<=n;i++)if(fg[i]){
for(j=b[i];j;j=b[j])dp[i]++;
}
memset(z,0,sizeof z);
ans=1;
for(i=1;i<=n;i++)if(!z[i] && fg[i]){
int nm=0,Fg=0;
for(j=i;!z[j];j=a[j]){
z[j]=1;
nm++;
if(dp[j])Fg=j;
}
if(!Fg)num[nm]++;
else {
k=0;
for(j=a[Fg];j!=Fg;j=a[j]){
k++;
if(dp[j]){
if(dp[j]>k)ans=0;
if(dp[j]<k)ans=(ans+ans)%md;
k=0;
}
}
k++;
if(dp[Fg]>k)ans=0;
if(dp[Fg]<k)ans=(ans+ans)%md;
}
}
for(i=1;i<=n;i++)if(num[i])ans=(LL)ans*cal(num[i],i)%md;
cout<<ans<<endl;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int N=200005,md=1000000007;
vector<int> vct[N];
int n,i,j,k,a[N],p[N],d[N],fa[N],ans=1,cnt,size[N],huan[N],f[N],c[N],num[N],head[N],adj[N],nxt[N],pre[N],u[N];
bool oh[N],v[N];
int find(int x)
{
return !fa[x]?x:fa[x]=find(fa[x]);
}
void dfs(int x)
{
v[x]=true;
p[++cnt]=x;
if(!v[a[x]])
dfs(a[x]);
else
{
int i;
for(i=cnt;i>=1;--i)
if(p[i]==a[x])
break;
if(i>=1)
{
int fst=0,lst=0,st=i;
while(i<=cnt)
{
if(d[p[i]]>1)
{
if(!fst)
fst=i;
pre[p[i]]=i-lst;
lst=i;
}
oh[p[i]]=true;
huan[find(x)]++;
++i;
}
pre[p[fst]]=fst-st+1+cnt-lst;
}
}
}
int work(int x)
{
int rtn=1;
for(int y=head[x];y;y=nxt[y])
if(!oh[adj[y]])
rtn+=work(adj[y]);
return rtn;
}
int main()
{
scanf("%d",&n);
for(i=1;i<=n;++i)
{
scanf("%d",a+i);
adj[i]=i;
nxt[i]=head[a[i]];
head[a[i]]=i;
++d[a[i]];
if(find(i)!=find(a[i]))
fa[find(i)]=find(a[i]);
}
for(i=1;i<=n;++i)
vct[find(i)].push_back(i);
for(i=1;i<=n;++i)
if(d[i]>2)
{
printf("0");
return 0;
}
for(i=1;i<=n;++i)
{
size[find(i)]++;
if(!v[i])
{
cnt=0;
dfs(i);
}
}
for(i=1;i<=n;++i)
if(!oh[i]&&d[i]>1)
{
printf("0");
return 0;
}
for(i=1;i<=n;++i)
if(d[i]>1)
c[find(i)]++;
for(i=1;i<=n;++i)
if(d[i]>1)
u[i]=work(i)-1;
for(i=1;i<=n;++i)
if(d[i]>1)
{
if(u[i]<pre[i])
ans=ans*2%md;
else if(u[i]>pre[i])
{
printf("0");
return 0;
}
}
for(i=1;i<=n;++i)
if(find(i)==i)
{
if(c[i]==0)
++num[size[i]];
}
for(i=1;i<=n;++i)
if(num[i])
{
f[0]=1,f[1]=(i==1?1:i&1?2:1);
for(j=2;j<=num[i];++j)
f[j]=(1ll*f[j-1]*(i==1?1:i&1?2:1)%md+1ll*(j-1)*f[j-2]%md*i%md)%md;
ans=1ll*ans*f[num[i]]%md;
}
printf("%d",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
#define N 100005
#define ll long long
#define For(i,x,y) for(int i=(x);i<=(y);++i)
#define Rof(i,x,y) for(int i=(x);i>=(y);--i)
#define p_b push_back
#define pii pair<int,int>
#define mp make_pair
#define fr first
#define sd second
#define mod 1000000007
using namespace std;
ll _2[N],_2_[N],fac[N],invf[N];
int nxt[N],id[N],len[N],deg[N],a[N],tot[N],cir[N],d[N],pos;
bool vis[N],inq[N];
vector<pii > v[N];
void dfs(int x,int dep){
if(inq[x] || cir[x]){
if(!cir[x]){
len[x]=dep-d[x],pos=x,cir[x]=x,id[x]=1;
if(d[x]-1) v[x].push_back(mp(1,d[x]-1));
} else if(dep-1) v[cir[x]].push_back(mp(id[x],dep-1));
return;
}
d[x]=dep,inq[x]=1,vis[x]=1;
dfs(nxt[x],dep+1),inq[x]=0;
if(pos && dep>d[pos]) cir[x]=pos,id[x]=id[nxt[x]]+1;
}
int ___(int x,int y){ if(x>y) return 2; if(x<y) return 0;return 1; }
ll qpow(ll x,int y){
ll tmp=1;
for(;y;y>>=1,(x*=x)%=mod) if(y&1) (tmp*=x)%=mod;
return tmp;
}
int main(){
int n;ll ans=1,ans2=1;
scanf("%d",&n);
_2[0]=_2_[0]=1;ll fuck=qpow(2ll,mod-2);
fac[0]=fac[1]=invf[0]=invf[1]=1,_2[1]=2,_2_[1]=fuck;
For(i,1,n) scanf("%d",&a[i]),nxt[i]=a[i],deg[nxt[i]]++;
For(i,2,n){
_2[i]=(_2[i-1]+_2[i-1])%mod;
_2_[i]=_2_[i-1]*fuck%mod;
fac[i]=1ll*fac[i-1]*i%mod;
invf[i]=1ll*(mod-mod/i)*invf[mod%i]%mod;
}
For(i,1,n) (invf[i]*=invf[i-1])%=mod;
For(i,1,n)
if(!deg[i]) pos=0,dfs(i,1);
For(i,1,n) if(!vis[i]) pos=0,dfs(i,1);
For(i,1,n){
if(deg[i]>2) return puts("0"),0;
else if(!cir[i] && deg[i]>=2) return puts("0"),0;
}
For(i,1,n) sort(v[i].begin(),v[i].end());
For(i,1,n){
if(!v[i].size()){ if(len[i]) tot[len[i]]++;continue; }
if(v[i].size()==1){
if(len[i]<v[i][0].sd) return puts("0"),0;
else if(len[i]>v[i][0].sd)(ans*=2ll)%=mod;
} else{
For(j,0,(int)v[i].size()-2){
ll p=1ll*___(v[i][j+1].fr-v[i][j].fr,v[i][j].sd);
if(!p) return puts("0"),0;
else (ans*=p)%=mod;
}
int p=v[i].size()-1;
if(len[i]-v[i][p].fr+v[i][0].fr<v[i][p].sd) return puts("0"),0;
else if(len[i]-v[i][p].fr+v[i][0].fr>v[i][p].sd) (ans*=2ll)%=mod;
}
}
For(i,1,n){
if(!tot[i]) continue;
ll qwq=0;
For(j,0,tot[i]/2){
int x=j*2;
ll tmp=qpow(1ll*i,j)*fac[tot[i]]%mod*invf[tot[i]-x]%mod*_2_[j]%mod*invf[j]%mod;
(tmp*=_2[((i%2) && i!=1)*(tot[i]-x)])%=mod;
(qwq+=tmp)%=mod;
}
(ans2*=qwq)%=mod;
}
printf("%lld\n",ans*ans2%mod);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define maxn 200005
#define mod 1000000007
using namespace std;
int deg[maxn],a[maxn],incy[maxn],vis[maxn],svis[maxn],n;
int chlen[maxn],cnt[maxn];
int ans=1;
int add(int a,int b){a+=b; return a>=mod?a-mod:a;}
void SolveCyc(int x)
{
int t=x; x=a[x],svis[t]=svis[x]=1;
int fi=0,fpos=0,la=0,len=1;
while(1)
{
while(!chlen[x]&&x!=t) x=a[x],svis[x]=1,len++;
if(chlen[x])
{
if(!fi) fi=x,la=len,fpos=len;
else
{
int coef=(len-la>=chlen[x])+(len-la>chlen[x]);
ans=1ll*ans*coef%mod,la=len;
}
}
if(x==t) break;
x=a[x],svis[x]=1,len++;
}
if(chlen[t])
{
if(!fi) fi=t;
else la=len;
}
if(!fi) cnt[len]++;
else
{
int coef=(len-la+fpos>=chlen[fi])+(len-la+fpos>chlen[fi]);
ans=1ll*ans*coef%mod;
}
}
int dp[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),deg[a[i]]++;
for(int i=1;i<=n;i++)
{
if(vis[i]) continue;
int nw=i; while(!vis[nw]) vis[nw]=i,nw=a[nw];
if(vis[nw]!=i) continue;
int t=nw; incy[nw]=1,nw=a[nw];
while(nw!=t) incy[nw]=1,nw=a[nw];
}
for(int i=1;i<=n;i++)
if(((!incy[i])&°[i]>=2)||deg[i]>=3) return puts("0"),0;
for(int i=1;i<=n;i++)
{
if(deg[i]) continue;
int nw=i; while(!incy[nw]) chlen[a[nw]]=chlen[nw]+1,nw=a[nw];
}
for(int i=1;i<=n;i++)
if(incy[i]&&!svis[i]) SolveCyc(i);
for(int i=1;i<=n;i++)
{
if(!cnt[i]) continue;
dp[0]=1;
for(int j=1;j<=cnt[i];j++)
{
if(i>1&&(i&1)) dp[j]=add(dp[j-1],dp[j-1]);
else dp[j]=dp[j-1];
if(j>1) dp[j]=add(dp[j],1ll*dp[j-2]*(j-1)%mod*i%mod);
}
ans=1ll*ans*dp[cnt[i]]%mod;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <iostream>
#include <cstdio>
#include <cstring>
#include <cassert>
#include <cctype>
using namespace std;
typedef long long lint;
#define cout cerr
#define ni (next_num<int>())
template<class T>inline T next_num(){
T i=0;char c;
while(!isdigit(c=getchar())&&c!='-');
bool flag=c=='-';
flag?c=getchar():0;
while(i=i*10-'0'+c,isdigit(c=getchar()));
return flag?-i:i;
}
template<class T1,class T2>inline void apmax(T1 &a,const T2 &b){if(a<b)a=b;}
template<class T1,class T2>inline void apmin(T1 &a,const T2 &b){if(b<a)a=b;}
const int N=100010,O=1000000007;
int pre[N][2],nxt[N];
bool vis[N];
int ring[N];
int f[N];
inline int Main(){
int n=ni;
memset(pre,0,sizeof(pre));
for(int i=1;i<=n;i++){
nxt[i]=ni;
if(pre[nxt[i]][0]==0){
pre[nxt[i]][0]=i;
}else if(pre[nxt[i]][1]==0){
pre[nxt[i]][1]=i;
}else return 0;
}
memset(ring,0,sizeof(ring));
lint ans=1;
for(int i=1;i<=n;i++){
if(!vis[i]){
int j=i,sz=0;
for(;!vis[j];vis[j]=true,j=nxt[j]);
for(int p=j,q;sz++,q=nxt[p],pre[q][0]!=p?swap(pre[q][0],pre[q][1]):void(),p=q,p!=j;);
int p=j;
bool flag=true;
int cur=1;
do if(pre[p][1]){
flag=false;
int a=pre[p][0],b=pre[p][1];
for(;vis[b]=true,pre[b][0];a=pre[a][0],b=pre[b][0]){
if(pre[a][1]||pre[b][1])return 0;
}
if(pre[a][1]==0){
cur=(cur<<1)%O;
}
}while(p=nxt[p],p!=j);
if(flag){
assert(cur==1);
ring[sz]++;
}else{
(ans*=cur)%=O;
}
}
}
for(int i=1;i<=n;i++){
bool flag=(i&1)&&i!=1;
f[0]=1,f[1]=flag+1;
for(int j=2;j<=ring[i];j++){
f[j]=((lint)f[j-2]*(j-1)%O*i%O+(f[j-1]<<flag))%O;
}
(ans*=f[ring[i]])%=O;
}
return ans;
}
int main(){
printf("%d\n",Main());
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define LL long long
#define db double
using namespace std;
const int N=2e5+10,mod=1e9+7,inf=1<<30;
int rd()
{
int x=0,w=1;char ch=0;
while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
return x*w;
}
void ad(int &x,int y){x+=y,x-=x>=mod?mod:0;}
int fpow(int a,int b){int an=1;while(b){if(b&1) an=1ll*an*a%mod;a=1ll*a*a%mod,b>>=1;}return an;}
int ginv(int a){return fpow(a,mod-2);}
vector<int> e[N];
int n,a[N],v1[N],ti,cr[N],st[N],tp,f[N],g[N],b[N],h[N],fac[N],iac[N];
int C(int a,int b){return b<0||a<b?0:1ll*fac[a]*iac[b]%mod*iac[a-b]%mod;}
bool mk[N];
void dfs(int x)
{
int nn=e[x].size();
for(int i=0;i<nn;++i)
{
int y=e[x][i];
dfs(y),f[x]=f[x]?-inf:f[y]+1;
}
}
int main()
{
//iee
fac[0]=1;
for(int i=1;i<=N-5;++i) fac[i]=1ll*fac[i-1]*i%mod;
iac[N-5]=ginv(fac[N-5]);
for(int i=N-5;i;--i) iac[i-1]=1ll*iac[i]*i%mod;
n=rd();
for(int i=1;i<=n;++i) a[i]=rd();
for(int i=1;i<=n;++i)
if(!v1[i])
{
++ti;
int x=i;
tp=0;
while(!v1[x]) v1[x]=ti,st[++tp]=x,x=a[x];
if(cr[x]) {mk[x]=1;continue;}
if(v1[x]<ti) continue;
int cn=0;
for(int j=tp;;--j)
{
++cn;
if(st[j]==x) break;
}
for(int j=tp;;--j)
{
cr[st[j]]=cn;
if(st[j]==x) break;
}
mk[x]=tp>cn;
}
++ti;
for(int i=1;i<=n;++i)
if(cr[i]&&v1[i]<ti)
{
int x=i;
for(int j=1;j<=cr[i]*2;++j) v1[x]=ti,mk[a[x]]|=mk[x],x=a[x];
}
for(int i=1;i<=n;++i)
if(!cr[i]) e[a[i]].push_back(i);
int ans=1;
for(int i=1;i<=n;++i)
if(cr[i])
{
if(mk[i])
{
int x=i;
tp=0;
for(int j=1;j<=cr[i];++j) st[++tp]=x,x=a[x];
for(int j=1;j<=cr[i];++j) dfs(st[j]);
for(int j=1;j<=cr[i];++j)
if(f[st[j]]<0||f[st[j]]>cr[i]){puts("0");return 0;}
for(int j=1;j<=cr[i];++j) g[j]=g[j-1]+(bool)f[st[j]];
for(int j=1;j<=cr[i];++j) g[j+cr[i]]=g[j-1+cr[i]]+(bool)f[st[j]];
for(int j=cr[i]+1;j<=cr[i]+cr[i];++j)
{
int y=st[j-cr[i]];
if(!f[y]) continue;
if(cr[i]>1&&g[j-1]-g[j-f[y]]){puts("0");return 0;}
ans=1ll*ans*(2-(bool)(g[j-1]-g[j-f[y]-1]))%mod;
}
}
else ++b[cr[i]];
int x=i,lm=cr[i];
for(int j=1;j<=lm;++j) cr[x]=0,x=a[x];
}
h[0]=h[1]=1;
for(int i=2;i<=n;++i) h[i]=1ll*h[i-1]*(2*i-1)%mod;
for(int i=1;i<=n;++i)
{
if(!b[i]) continue;
int sm=0,dx=1+(i>1&&(i&1));
for(int j=0;j+j<=b[i];++j)
ad(sm,1ll*C(b[i],j+j)*h[j]%mod*fpow(i,j)%mod*fpow(dx,b[i]-j-j)%mod);
ans=1ll*ans*sm%mod;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define FL "a"
using namespace std;
typedef vector<int> VI;
typedef long long ll;
typedef double dd;
const int N=1e5+10;
const int mod=1e9+7;
inline ll read(){
ll data=0,w=1;char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
inline void file(){
freopen(FL".in","r",stdin);
freopen(FL".out","w",stdout);
}
inline void upd(int &a,int b){a+=b;if(a>=mod)a-=mod;}
inline void dec(int &a,int b){a-=b;if(a<0)a+=mod;}
inline int poww(int a,int b){
int res=1;
for(;b;b>>=1,a=1ll*a*a%mod)
if(b&1)res=1ll*res*a%mod;
return res;
}
inline void er(){puts("0");exit(0);}
int n,a[N],ans;
int d[N],head[N],nxt[N<<1],to[N<<1],cnt;
inline void add(int u,int v){to[++cnt]=v;nxt[cnt]=head[u];head[u]=cnt;}
int t[N],f[N];
int vis[N],line[N],dis[N],cal[N],top,cir[N],len;
void dfs(int u){
vis[u]=1;cal[++top]=u;
for(int i=head[u],v;i;i=nxt[i])
if(~i&1){
if(!vis[v=to[i]])dfs(v);
else
for(int j=top;cal[j]!=v;j--)
cir[++len]=cal[j];
}
top--;
}
inline void solve(int x){
top=len=0;
for(int p=0;!vis[x];vis[x]=1,cal[++top]=x,p=x,x=a[x]);
for(int i=top,k=1;i;i--){
k?cir[++len]=cal[i],dis[cal[i]]=len:vis[cal[i]]=0;
if(cal[i]==x)k=0;
}
reverse(cir+1,cir+len+1);
for(int i=1;i<=len;i++)dis[cir[i]]=len-dis[cir[i]]+1;
//for(int i=1;i<=len;i++)printf("%d ",cir[i]);puts("");
for(int k=1,u;k<=len;k++){
u=cir[k];if(d[u]>2)er();line[u]=0;
for(int i=head[u],v;i;i=nxt[i])
if(!vis[v=to[i]]){
line[u]++;vis[v]=1;
for(v;d[v];v=to[head[v]],vis[v]=1,line[u]++)
if(d[v]>1)er();
}
}
for(int k=1;!line[cir[k]];k++)
if(k==len){t[len]++;return;}
if(len==1&&line[cir[1]]<=1)return;
top=0;
for(int k=1;k<=len;k++)
if(line[cir[k]])cal[++top]=cir[k];
/*for(int i=1;i<=top;i++)
printf("(%d,%d,%d)",cal[i],dis[cal[i]],line[cal[i]]);puts("");*/
if(line[cal[1]]>dis[cal[1]]+(len-dis[cal[top]]))er();
else ans=1ll*min(2,dis[cal[1]]+(len-dis[cal[top]])-line[cal[1]]+1)*ans%mod;
for(int i=2;i<=top;i++)
if(line[cal[i]]>dis[cal[i]]-dis[cal[i-1]])er();
else ans=1ll*min(2,dis[cal[i]]-dis[cal[i-1]]-line[cal[i]]+1)*ans%mod;
}
int main()
{
n=read();cnt=1;ans=1;
for(int i=1;i<=n;i++){
a[i]=read();d[a[i]]++;add(a[i],i);
//printf("%d %d\n",i,a[i]);
if(a[i]>n)er();
}
for(int i=1;i<=n;i++)
if(!vis[i])solve(i);
for(int i=1;i<=n;i++)
if(t[i]){
f[0]=1;
for(int j=1;j<=t[i];j++){
f[j]=1ll*((i&1)&&i!=1?2:1)*f[j-1]%mod;
upd(f[j],1ll*(j-1)*i%mod*f[j-2]%mod);
}
ans=1ll*ans*f[t[i]]%mod;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) int(a.size())
const int N=1e5+10;
const int mod=1e9+7;
int gi() {
int x=0,o=1;char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
if(ch=='-') o=-1,ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*o;
}
int n,a[N],in[N],ans=1,cnt[N],fa[N],f[N],pre[N];
bool vis[N];
vi vec[N];
int getf(int x) {
return x==fa[x]?x:fa[x]=getf(fa[x]);
}
void No() {
cout<<0;exit(0);
}
int main() {
cin>>n;
for(int i=1;i<=n;i++) fa[i]=i;
for(int i=1;i<=n;i++) a[i]=gi(),++in[a[i]],fa[getf(i)]=getf(a[i]);
for(int i=1;i<=n;i++) if(in[i]>2) No();
for(int i=1;i<=n;i++) vec[getf(i)].pb(i);
for(int u=1;u<=n;u++)
if(getf(u)==u) {
bool fl=1;int sz=0;
for(auto x:vec[u]) {
++sz;
if(in[x]!=1) fl=0;
}
//cerr<<u<<' '<<fl<<' '<<sz<<'\n';
if(fl) ++cnt[sz];
else {
int now=u;
while(!vis[now]) vis[now]=1,now=a[now];
for(auto x:vec[u]) vis[x]=0;
while(!vis[now]) vis[now]=1,pre[a[now]]=now,now=a[now];
for(auto x:vec[u])
if(in[x]>1&&!vis[x]) No();
for(auto x:vec[u])
if(!in[x]) {
int now=x,l1=0,l2=0;
while(!vis[now]) ++l1,now=a[now];
++l2;now=pre[now];
while(in[now]==1) ++l2,now=pre[now];
//cerr<<l1<<' '<<l2<<'\n';
if(l1>l2) No();
if(l1<l2) ans=2ll*ans%mod;
}
}
}
for(int k=1;k<=n;k++) {
for(int i=0;i<=cnt[k];i++) f[i]=0;
f[0]=1;
for(int i=1;i<=cnt[k];i++) {
f[i]=f[i-1]*((k&1)&&k>1?2:1)%mod;
if(i>1) f[i]=(f[i]+1ll*f[i-2]*(i-1)%mod*k)%mod;
}
ans=1ll*ans*f[cnt[k]]%mod;
//cerr<<"k,cnt[k]="<<k<<' '<<cnt[k]<<'\n';
}
cout<<ans;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef long double louble;
template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}
const char lf = '\n';
namespace ae86
{
const int bufl = 1 << 15;
char buf[bufl],*s=buf,*t=buf;
inline int fetch()
{
if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
return *s++;
}
inline int ty()
{
int a=0;int b=1,c=fetch();
while(!isdigit(c))b^=c=='-',c=fetch();
while(isdigit(c))a=a*10+c-48,c=fetch();
return b?a:-a;
}
}
using ae86::ty;
const int _ = 100007 , mo = 1000000007;
template<typename T1,typename T2> inline T1 ad(T1 &a,T2 b){return a=a+b>=mo?a+b-mo:a+b;}
template<typename T1,typename T2> inline T1 dl(T1 &a,T2 b){return a=a>=b?a-b:a-b+mo;}
template<typename T1,typename T2> inline T1 add(T1 a,T2 b){return a+b>=mo?a+b-mo:a+b;}
template<typename T1,typename T2> inline T1 del(T1 a,T2 b){return a>=b?a-b:a-b+mo;}
lint powa(lint a,lint t)
{
lint b=1;a=(a+mo)%mo;
while(t){if(t&1)b=b*a%mo;a=a*a%mo,t>>=1;}
return b;
}
inline lint inva(lint a)
{
return powa(a,mo-2);
}
int n,nex[_]={0},din[_]={0},ed[_]={0};
int isc[_]={0},dlen[_]={0},got[_]={0};
int ccnt[_]={0};
lint ans=1;
void dfs(int x)
{
int len=0,lasloc=0,firloc=0,firlen=0;
while(!got[x])
{
len++,got[x]=1;
if(dlen[x])
{
if(!firloc)firloc=len,firlen=dlen[x];
else ans=ans*add<int,int>(dlen[x]<len-lasloc,dlen[x]<=len-lasloc)%mo;
lasloc=len;
}
x=nex[x];
}
if(!firloc)ccnt[len]++;
else ans=ans*add<int,int>(firlen<len-lasloc+firloc,firlen<=len-lasloc+firloc)%mo;
}
lint f[_]={0};
int main()
{
ios::sync_with_stdio(0),cout.tie(nullptr);
n=ty();
for(int i=1;i<=n;i++)nex[i]=ty(),din[nex[i]]++;
for(int i=1;i<=n;i++)
{
if(ed[i])continue;
int now=i;
while(!ed[now])ed[now]=i,now=nex[now];
if(ed[now]!=i)continue;
while(!isc[now])isc[now]=1,now=nex[now];
}
for(int i=1;i<=n;i++)if(din[i]>1+isc[i]){cout<<0<<lf;return 0;}
for(int i=1;i<=n;i++)
{
if(din[i])continue;
int x=i,len=0;
while(!isc[x])len++,x=nex[x];
dlen[x]=len;
}
for(int i=1;i<=n;i++)if(isc[i] && !got[i])dfs(i);
for(int i=1;i<=n;i++)
{
if(!ccnt[i])continue;
f[0]=1;
for(int j=1;j<=ccnt[i];j++)
{
if(i>1 && (i&1))f[j]=add(f[j-1],f[j-1]);
else f[j]=f[j-1];
if(j>1)ad(f[j],f[j-2]*(j-1)%mo*i%mo);
}
ans=ans*f[ccnt[i]]%mo;
}
cout<<ans<<lf;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | # include <bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
int n;
const int maxn = 100010;
int p[maxn];
int ind[maxn];
bool vis[maxn];
int pos[maxn];
int feet[maxn];
map<int, int> cir;
int dp[maxn];
int DP(int m,int n) {
long long s = 1 + ((m > 1) & (m & 1));
long long p = m;
dp[0] = 1;
dp[1] = s;
for(int i = 2; i <= n; ++i) {
dp[i] = (s * dp[i-1] + p * (i-1) % mod * dp[i-2]) % mod;
}
return dp[n];
}
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) { scanf("%d", &p[i]); ind[p[i]] += 1; }
for(int i = 1; i <= n; ++i) if(ind[i] > 2) { puts("0"); return 0; }
for(int i = 1; i <= n; ++i) if(ind[i] == 0) {
int j, l = 0;
for(j = i; ind[j] < 2; j = p[j]) vis[j] = true, l += 1;
feet[j] = l;
}
int ans = 1;
for(int i = 1; i <= n; ++i) if(!vis[i]) {
int l = 0, j;
for(j = i; !vis[j]; j = p[j]) vis[j] = true, pos[l++] = j;
if(i!=j) ans = 0;
int last = 0;
for(int j = 0; j < l; ++j) if(feet[pos[j]] > 0) last = j - l;
if(last == 0) { cir[l]++; continue; }
for(int j = 0; j < l; ++j) if(feet[pos[j]] > 0) {
if(j - last > feet[pos[j]]) ans = (ans << 1) % mod;
else if(j - last < feet[pos[j]]) ans = 0;
last = j;
}
}
for(auto e: cir) ans = 1ll * ans * DP(e.first, e.second) % mod;
cout << ans << endl;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int mod=1e9+7;
int gi() {
int x=0,o=1;char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
if(ch=='-') o=-1,ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*o;
}
int n,ans=1,a[N],du[N],pre[N],cnt[N],f[N];
bool vis[N];
int main() {
cin>>n;
for(int i=1;i<=n;i++) a[i]=gi(),du[a[i]]++;
for(int i=1;i<=n;i++) {
if(du[i]>2) return puts("0"),0;
if(du[i]<2||vis[i]) continue;
int p=i;
do {
if(vis[p]) return puts("0"),0;
vis[p]=1;pre[a[p]]=p;p=a[p];
} while(p!=i);
}
for(int i=1;i<=n;i++)
if(!du[i]) {
int p=i,l1=0,l2=0;
while(!vis[p]) vis[p]=1,p=a[p],l1++;
do l2++,p=pre[p]; while(du[p]!=2);
if(l1<l2) ans=2ll*ans%mod;
else if(l1>l2) return puts("0"),0;
}
for(int i=1;i<=n;i++)
if(!vis[i]) {
int p=i,l=0;
do p=a[p],l++,vis[p]=1; while(p!=i);
cnt[l]++;
}
for(int i=1;i<=n;i++) {
int mul=1;if(i!=1&&(i&1)) mul++;
f[0]=1;f[1]=mul;
for(int j=2;j<=cnt[i];j++) f[j]=(1ll*f[j-2]*(j-1)%mod*i%mod+1ll*f[j-1]*mul%mod)%mod;
ans=1ll*ans*f[cnt[i]]%mod;
}
printf("%d\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define L long long
using namespace std;
const int q=1000000007;
int n,a[100010],b[100010],x[100010],f[100010],p=1,w[100010],u[100010];
vector<int> y[100010];
inline int C(int n,int m)
{
return (L)a[n]*b[m/2]%q*b[n-m]%q;
}
inline void calc(int i)
{
int j,k=0,l,v,e;
for(j=i;w[j]<2;j=x[j])
{
w[j]=2;
u[++k]=j;
}
if(j!=i || k==0)
return;
for(i=2;i<k+2-i;i++)
swap(u[i],u[k+2-i]);
u[k+1]=u[1];
for(j=1,l=0,v=1;j<=k;j++,l--)
{
if(y[u[j]].size()>2)
v=0;
if(y[u[j]].size()==2)
{
if(l>0)
v=0;
if(l<0)
v=(v<<1)%q;
for(l=1,e=y[u[j]][0]^y[u[j]][1]^u[j+1];y[e].size()==1;e=y[e][0],l++);
if(y[e].size()>1)
v=0;
}
}
if(l>0)
v=0;
if(l<0)
v=(v<<1)%q;
p=(L)p*v%q;
}
int main()
{
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
int i,j,k,l;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&x[i]);
y[x[i]].push_back(i);
}
b[1]=1;
for(i=2;i<=n;i++)
b[i]=q-(L)q/i*b[q%i]%q;
a[0]=b[0]=1;
for(i=1;i<=n;i++)
{
a[i]=(L)a[i-1]*i%q;
b[i]=(L)b[i-1]*b[i]%q;
}
for(i=1;i<=n;i++)
if(!w[i])
{
for(j=i,k=0,l=0;!w[j];j=x[j],k++)
{
w[j]=1;
if(y[j].size()>1)
l=1;
}
if(j==i)
{
if(!l)
f[k]++;
}
else
calc(j);
}
for(i=1;i<=n;i++)
{
if(i&1 && i>1)
for(j=1,l=1;j<=f[i];j++)
l=(l<<1)%q;
else
l=1;
for(j=0,k=0;j*2<=f[i];j++,l=(L)l*i%q)
{
k=(k+(L)C(f[i],2*j)*l)%q;
l=(L)l*(q+1>>1)%q;
if(i&1 && i>1)
l=(L)l*(q+1>>1)%q*(q+1>>1)%q;
}
p=(L)p*k%q;
}
printf("%d\n",p);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int maxn=100010,mod=1000000007,inv2=500000004;
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline int read(){
int x=0,f=0;char ch=getchar();
while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
return f?-x:x;
}
int n,a[maxn],deg[maxn],ans=1,stk[maxn],tp,q[maxn],h,r,len[maxn],seq[maxn],tot,cnt[maxn];
int fac[maxn],inv[maxn],invfac[maxn];
bool vis[maxn],ins[maxn],cyc[maxn];
void dfs(int u){
if(vis[u]){
if(ins[u]){
ROF(i,tp,1){
cyc[stk[i]]=true;
if(stk[i]==u) break;
}
}
return;
}
vis[u]=ins[u]=true;
stk[++tp]=u;
dfs(a[u]);
ins[u]=false;
}
bool dfs2(int u){
if(vis[u]) return true;
vis[u]=true;
seq[++tot]=len[u];
return dfs2(a[u]) && !len[u];
}
inline int C(int n,int m){
return 1ll*fac[n]*invfac[m]%mod*invfac[n-m]%mod;
}
inline int qpow(int a,int b){
int ans=1;
for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) ans=1ll*ans*a%mod;
return ans;
}
int main(){
n=read();
FOR(i,1,n) a[i]=read(),deg[a[i]]++;
FOR(i,1,n) dfs(i);
FOR(i,1,n) if(cyc[i] && deg[i]>=3|| !cyc[i] && deg[i]>=2) return puts("0"),0;
h=1;r=0;
FOR(i,1,n) if(!deg[i]) q[++r]=i;
while(h<=r){
int u=q[h++];
len[a[u]]=len[u]+1;
if(!cyc[a[u]]) q[++r]=a[u];
}
MEM(vis,0);
FOR(i,1,n) if(cyc[i] && !vis[i]){
tot=0;
if(dfs2(i)) cnt[tot]++;
else{
int pre=0;
FOR(j,1,tot) if(seq[j]){
if(pre){
int at=j-seq[j];
if(at<pre) return puts("0"),0;
if(at>pre && tot>=2) ans=2*ans%mod;
}
pre=j;
}
FOR(j,1,tot) if(seq[j]){
int at=j-seq[j]+tot;
if(at<pre) return puts("0"),0;
if(at>pre && tot>=2) ans=2*ans%mod;
break;
}
}
}
fac[0]=fac[1]=inv[1]=invfac[0]=invfac[1]=1;
FOR(i,2,n){
fac[i]=1ll*fac[i-1]*i%mod;
inv[i]=mod-1ll*(mod/i)*inv[mod%i]%mod;
invfac[i]=1ll*invfac[i-1]*inv[i]%mod;
}
FOR(i,1,n) if(cnt[i]){
int s=0;
FOR(j,0,cnt[i]/2){
int x=1ll*C(cnt[i],2*j)*C(2*j,j)%mod*fac[j]%mod;
if(j) x=1ll*x*qpow(inv2,j)%mod*qpow(i,j)%mod;
if(i%2==1 && i!=1) x=1ll*x*qpow(2,cnt[i]-2*j)%mod;
s=(s+x)%mod;
}
ans=1ll*ans*s%mod;
}
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100100
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define drp(x, a, b) for(int x=a; x>=b; x--)
int deg[N], d[N], n, vis[N], a[N], num[N], l[N], ans=1, fac[N], inv[N];
vector<pair<int, int> > fu;
queue<int> q;
int power(int a, int k){
int ret=1;
while(k)
{
if(k&1) ret=1ll*ret*a%mod;
a=1ll*a*a%mod; k>>=1;
}
return ret;
}
int C(int n, int m){
return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int pr(int n){
return 1ll*fac[2*n]*inv[n]%mod*power((mod+1)>>1, n)%mod;
}
void init(){
fac[0]=1;
rep(i, 1, n) fac[i]=1ll*fac[i-1]*i%mod;
inv[n]=power(fac[n], mod-2);
drp(i, n, 1) inv[i-1]=1ll*inv[i]*i%mod;
}
int st[N<<1];
int main(){
scanf("%d", &n);
init();
rep(i, 1, n)
{
scanf("%d", a+i);
deg[a[i]]++;
}
rep(i, 1, n) if(deg[i]>2) return puts("0"), 0;
rep(i, 1, n) d[i]=deg[i];
rep(i, 1, n) if(!d[i]) q.push(i);
while(!q.empty())
{
int u=q.front(); q.pop();
vis[u]=1;
if(--d[a[u]]==0) q.push(a[u]);
}
rep(i, 1, n) if(vis[i] && deg[i]>1) return puts("0"), 0;
rep(i, 1, n) if(vis[i] && !deg[i])
{
int x=i, cnt=0;
for(; vis[x]; x=a[x]) cnt++;
l[x]=cnt;
}
rep(i, 1, n) if(!vis[i])
{
int x=i, fl=0, ass=1, tot=0;
for(; !vis[x]; x=a[x]) st[++tot]=x, fl|=bool(l[x]), vis[x]=1;
if(!fl){ num[tot]++; continue; }
fu.clear();
// if(l[i]) fu.pb(mp(l[i], 1));
// for(int p=2, x=a[i]; x!=i; x=a[x]) if(l[x]) fu.pb(mp(l[x], p));
int p=1; x=i;
do{
if(l[x]) fu.pb(mp(l[x], p));
x=a[x]; p++;
}while(x!=i);
//rep(p, 1, tot) if(l[st[p]]) fu.pb(mp(l[st[p]], p));
fu.pb(mp(fu[0].fi, fu[0].se+tot));
int sz=fu.size();
rep(j, 1, sz-1)
{
if(fu[j].fi>fu[j].se-fu[j-1].se) ass=0;
if(fu[j].fi<fu[j].se-fu[j-1].se) ass=2ll*ass%mod;
}
ans=1ll*ans*ass%mod;
}
rep(i, 1, n) if(num[i])
{
int m=num[i], ass=0;
rep(j, 0, m/2)
{
int tmp=1ll*C(m, 2*j)*pr(j)%mod*power(i, j)%mod;
if(i>1 && (i&1)) tmp=1ll*tmp*power(2, m-2*j)%mod;
ass=(ass+tmp)%mod;
}
ans=1ll*ans*ass%mod;
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=(a),_ed=(b);i<=_ed;++i)
#define DREP(i,a,b) for(int i=(a),_ed=(b);i>=_ed;--i)
#define mp(x,y) make_pair((x),(y))
#define sz(x) (int)(x).size()
#define pb push_back
typedef long long ll;
typedef pair<int,int> pii;
inline int read(){
register int x=0,f=1;register char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}
while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}
return f?x:-x;
}
const int N=1e5+5,mod=1e9+7;
int ans=1,n,a[N],d[N],vis[N],is[N],cnt[N],dis[N],f[N];
void work(int u){
int len=0,s=-1,sd,las;
while(is[u]){
is[u]=0,++len;
if(dis[u]){
if(!~s)s=len,sd=dis[u];
else ans=1ll*ans*((dis[u]<=len-las)+(dis[u]<len-las))%mod;
las=len;
}
u=a[u];
}
if(!~s)++cnt[len];
else ans=1ll*ans*((sd<=s+len-las)+(sd<s+len-las))%mod;
}
int main(){
//freopen("in.in","r",stdin);
REP(i,1,n=read())a[i]=read(),++d[a[i]];
REP(u,1,n){
if(vis[u])continue;
int p=u;while(!vis[p])vis[p]=u,p=a[p];
if(vis[p]!=u)continue;
while(!is[p])is[p]=1,p=a[p];
}
REP(i,1,n)if((is[i]&&d[i]>2)||(!is[i]&&d[i]>1))return puts("0"),0;
REP(i,1,n)if(!d[i]){
int p=i,l=0;
while(!is[p])++l,p=a[p];
dis[p]=l;
}
REP(i,1,n)if(is[i])work(i);
REP(l,1,n){
if(!cnt[l])continue;
f[0]=1;
REP(i,1,cnt[l]){
if(l>1&&l&1)f[i]=2ll*f[i-1]%mod;//i->p_i + i->p_{p_i}(同构)
else f[i]=f[i-1];//i->p_i
if(i>1)f[i]=(f[i]+1ll*f[i-2]*(i-1)%mod*l%mod)%mod;//i->p_{p_i}(变成2个环)
}
ans=1ll*ans*f[cnt[l]]%mod;
}
printf("%d\n",ans);
return 0;
}
/*
对于一个排列p_i,i向p_i连边,形成若干简单环(每个点入度出度都为1)
a_i是p_i或p_{p_i},反映到现图上就是点i的下一个点或下下个点
把p_i的边擦除,连i->a_i的边
会有4中种情况,分别讨论一下
现在就是给定a_i的图还原成p_i的图
*/
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int mod=1e9+7;
int gi() {
int x=0,o=1;char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
if(ch=='-') o=-1,ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*o;
}
int n,ans=1,a[N],du[N],pre[N],cnt[N],f[N];
bool vis[N];
int main() {
cin>>n;
for(int i=1;i<=n;i++) a[i]=gi(),du[a[i]]++;
for(int i=1;i<=n;i++) {
if(du[i]>2) return puts("0"),0;
if(du[i]<2||vis[i]) continue;
int p=i;
do {
if(vis[p]) return puts("0"),0;
vis[p]=1;pre[a[p]]=p;p=a[p];
} while(p!=i);
}
for(int i=1;i<=n;i++)
if(!du[i]) {
int p=i,l1=0,l2=0;
while(!vis[p]) vis[p]=1,p=a[p],l1++;
do l2++,p=pre[p]; while(du[p]!=2);
if(l1<l2) ans=2ll*ans%mod;
else if(l1>l2) return puts("0"),0;
}
for(int i=1;i<=n;i++)
if(!vis[i]) {
int p=i,l=0;
do p=a[p],l++,vis[p]=1; while(p!=i);
cnt[l]++;
}
for(int i=1;i<=n;i++) {
int mul=1;if(i!=1&&(i&1)) mul++;
f[0]=1;f[1]=mul;
for(int j=2;j<=cnt[i];j++) f[j]=(1ll*f[j-2]*(j-1)%mod*i%mod+1ll*f[j-1]*mul%mod)%mod;
ans=1ll*ans*f[cnt[i]]%mod;
}
printf("%d\n",ans);
return 0;
}
//orzgzy
//鸡贼明年进队超稳
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
int in[100010];
int a[100010];
int vis[100010],line[100010],cycc[100010];
vector<vector<int> > cyc;
bool oncyc[100010];
long long int dp[100010];
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i],in[a[i]]++;
for(int i=1;i<=n;i++)
if(!in[i]){
int u=i;vis[i]=i;while(!vis[a[u]]) u=a[u],vis[u]=i;
if(vis[a[u]]==i){
cyc.push_back({});
cyc.back().push_back(u);
for(int j=u;a[j]!=u;j=a[j])
cyc.back().push_back(a[j]);
for(auto it:cyc.back())
oncyc[it]=true;
}
else if(!oncyc[a[u]]||line[a[u]]){
cout<<0<<endl;
return 0;
}
int cnt=0;
for(int j=i;j!=a[u];j=a[j]) cnt++;
line[a[u]]=cnt;
}
for(int i=1;i<=n;i++)
if(!vis[i]){
int cnt=0;
for(int j=i;!vis[j];j=a[j]) cnt++,vis[j]=i;
cycc[cnt]++;
}
long long int ans=1;
for(int i=1;i<=n;i++){
int cnt=cycc[i];
dp[0]=1;
dp[1]=1;
if(i>1&&(i&1))
dp[1]++;
for(int j=2;j<=cnt;j++)
dp[j]=(dp[j-1]*dp[1]+dp[j-2]*(j-1)*i)%mod;
ans=ans*dp[cnt]%mod;
}
for(auto it:cyc){
reverse(it.begin(),it.end());
int sz=it.size();
int last=sz;
for(int i=0;i<sz;i++,last++)
if(line[it[i]]) break;
for(int i=sz-1;i>=0;i--)
if(line[it[i]]){
int l1=line[it[i]],l2=last-i;
if(l1>l2)
ans=0;
else if(l1<l2)
ans=ans*2%mod;
last=i;
}
}
cout<<ans<<endl;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
const int maxn=2e5,mod=1e9+7;
int n,ctt,tmp,a[maxn],d[maxn],cnt[maxn],cir_cnt[maxn];
long long ans=1,f[maxn];
bool vis[maxn];
vector <int> to[maxn];
queue <int> q;
void dfs(int now){
if(cnt[now]!=1) tmp=now;
vis[now]=true;ctt++;
int siz=to[now].size();
for(int i=0;i<siz;i++){
int u=to[now][i];
if(vis[u]==true) continue;
dfs(u);
}
}
int work(int now){
int siz=to[now].size(),len;
for(int i=0;i<siz;i++){
int u=to[now][i];
if(u==tmp){len=cnt[tmp]-2;continue;}
len=work(u);
}
if(cnt[now]==1){return len-1;}
else{
if(len>=1){ans=0;}
if(len<=-1){ans=ans*2%mod;}
return cnt[now]-2;
}
}
void Go(int now){
ctt=0;tmp=0;
dfs(now);
if(!tmp){cir_cnt[ctt]++;}
else{work(tmp);}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){cnt[i]=1;scanf("%d",&a[i]);to[i].push_back(a[i]);d[a[i]]++;}
for(int i=1;i<=n;i++){if(d[i]==0) q.push(i);}
while(!q.empty()){
int now=q.front();q.pop();
int siz=to[now].size();
vis[now]=true;
for(int i=0;i<siz;i++){
int u=to[now][i];
d[u]--;
if(cnt[u]!=1) ans=0;
else cnt[u]+=cnt[now];
if(d[u]==0) q.push(u);
}
}
for(int i=1;i<=n;i++){if(!vis[i]){Go(i);}}
for(int i=1;i<=n;i++){
f[0]=1;
int opt=((i&1)&&(i!=1))?2:1,k=cir_cnt[i];
for(int j=1;j<=k;j++){f[j]=0;}
for(int j=0;j<k;j++){
f[j+1]=(f[j+1]+f[j]*opt)%mod;
f[j+2]=(f[j+2]+f[j]*(k-j-1)%mod*i%mod)%mod;
}
ans=ans*f[k]%mod;
}
printf("%lld\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mxN=1e5, M=1e9+7;
int n, c[mxN+1];
vector<int> a[mxN];
bool b[mxN];
ll ans=1, d[mxN+1];
void fk() {
cout << 0;
exit(0);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for(int i=0, b; i<n; ++i) {
cin >> b, --b;
a[b].push_back(i);
}
for(int i=0; i<n; ++i)
if(a[i].size()>2)
fk();
for(int i=0; i<n; ++i) {
if(b[i]||a[i].size()<2)
continue;
int u1=a[i][0], u2=a[i][1];
while(1) {
if(u2==i)
swap(u1, u2);
if(u1==i)
break;
b[u1]=1;
vector<int> v=a[u1];
if(~u2) {
v.insert(v.end(), a[u2].begin(), a[u2].end());
b[u2]=1;
}
if(v.size()>2||!v.size())
fk();
u1=v[0];
if(v.size()>1)
u2=v[1];
else {
ans=ans*(1+(u2>=0))%M;
u2=-1;
}
}
b[u1]=1;
if(~u2) {
if(a[u2].size())
fk();
b[u2]=1;
}
}
for(int i=0; i<n; ++i) {
if(b[i])
continue;
int s=0;
for(int u=i; !b[u]; u=a[u][0]) {
b[u]=1;
++s;
}
++c[s];
}
d[0]=1;
for(int i=1; i<=n; ++i) {
for(int j=1; j<=c[i]; ++j)
d[j]=(d[j-1]*(i&1&&i>1?2:1)+(j>1?d[j-2]*(j-1)%M*i:0))%M;
ans=ans*d[c[i]]%M;
}
cout << ans;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, MOD = 1e9 + 7;
int a[N], vis[N], chain[N], in[N], len[N];
bool in_circle[N];
int main() {
int n;
scanf("%d", &n);
bool flag = 0;
for (int i = 1; i <= n; ++ i) {
scanf("%d", &a[i]);
in[a[i]] ++;
if (in[a[i]] >= 3) flag = 1;
}
if (flag) return 0 * puts("0");
vector<vector<int>> circles;
int xxx = 0;
for (int i = 1; i <= n; ++ i) {
if (vis[i]) continue;
++ xxx;
vis[i] = xxx;
vector<int> st;
st.push_back(i);
int j = a[i];
while (!vis[j]) {
st.push_back(j);
vis[j] = xxx;
j = a[j];
}
if (vis[j] < xxx) continue;
int end = j;
vector<int> t;
while (t.empty() || t.back() != end) {
t.push_back(st.back());
st.pop_back();
}
for (int x : t) in_circle[x] = 1;
circles.push_back(t);
}
for (int i = 1; i <= n; ++ i) {
if (in_circle[i]) continue;
if (in[i] == 2) flag = 1;
if (in[i]) continue;
int cnt = 0, j = i;
while (!in_circle[j]) {
j = a[j];
cnt ++;
}
chain[j] = cnt;
}
if (flag) return 0 * puts("0");
unordered_map<int, int> nofeet;
vector<vector<int>> feet;
for (auto &t : circles) {
bool has = 0;
for (int x : t) {
if (chain[x]) has = 1;
}
if (has) feet.push_back(t); else nofeet[(int) t.size()] ++;
}
int ans = 1;
for (auto xy : nofeet) {
int n = xy.first, cnt = xy.second;
//printf("no feet : n=%d cnt=%d\n", n, cnt);
static int dp[N];
dp[0] = 1;
for (int i = 1; i <= cnt; ++ i) {
dp[i] = 1LL * dp[i - 1] * ((n & 1) && n > 1 ? 2 : 1) % MOD;
if (i > 1) (dp[i] += 1LL * (i - 1) * n % MOD * dp[i - 2] % MOD) %= MOD;
}
ans = 1LL * ans * cnt[dp] % MOD;
}
for (auto &t : feet) {
int n = (int) t.size(), st;
for (int i = 0; i < n; ++ i) if (chain[t[i]]) {
st = i;
break;
}
int res = 0;
for (int i = 0; i < n; ++ i) {
if (chain[t[(st + n - i) % n]]) {
len[t[(st + n - i) % n]] = res;
res = 0;
}
else ++ res;
}
len[t[st]] = res;
for (int i : t) {
if (!chain[i]) continue;
if (chain[i] > len[i] + 1) ans = 0;
if (chain[i] < len[i] + 1) (ans *= 2) %= MOD;
}
}
printf("%d\n", ans);
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+10,p=1e9+7;
int inc(int x,int y){x+=y;return x>=p?x-p:x;}
int mul(int x,int y){return (ll)x*y%p;}
int n,a[N],cir[N],link[N],len[N],top[N];
bool ok[N];
int length(int x){
if (top[x]) return link[x];
link[x]=length(a[x])+1;
top[x]=top[a[x]];
return link[x];
}
int cnt[N];
void init(){
static int vis[N],deg[N];
scanf("%d",&n);
for (int i=1;i<=n;i++) scanf("%d",&a[i]),deg[a[i]]++;
for (int i=1;i<=n;i++)
if (!vis[i]){
int j=i;
while (!vis[j]) vis[j]=i,j=a[j];
if (vis[j]!=i) continue;
int p=j;
for (cir[p]=p,j=a[j];j!=p;cir[j]=p,j=a[j]);
}
for (int i=1;i<=n;i++)
if (cir[i]) len[cir[i]]++,top[i]=i;
for (int i=1;i<=n;i++) length(i);
for (int i=1;i<=n;i++){
if (deg[i]>2) puts("0"),exit(0);
if (deg[i]>1&&!cir[i]) puts("0"),exit(0);
}
for (int i=1;i<=n;i++) link[top[i]]=max(link[top[i]],link[i]);
for (int i=1;i<=n;i++)
if (link[i]) ok[cir[top[i]]]=1;
for (int i=1;i<=n;i++)
if (!ok[i]&&len[i]) cnt[len[i]]++;
}
typedef pair<int,int> pii;
int calc(int L,int n){
static map<pii,int> M;
if (!n) return 1;
pii now(L,n);
if (M.count(now)) return M[now];
int ans=calc(L,n-1);
if ((L&1)&&L>1) ans=inc(ans,ans);
if (n>1) ans=inc(ans,mul(mul(L,(n-1)),calc(L,n-2)));
return M[now]=ans;
}
int q[N],size;
int Calc(){
static int cnt[N];
for (int i=0;i<=size*2;i++) cnt[i]=0;
for (int i=1;i<=size;i++)
if (q[i]>size) return 0;
for (int i=1;i<=size;i++) cnt[size+i-q[i]]++,cnt[size+i]--;
for (int i=1;i<=size*2;i++) cnt[i]+=cnt[i-1];
for (int i=1;i<=size;i++) cnt[i]=cnt[i+size]=cnt[i]+cnt[i+size];
cnt[0]=cnt[size];
for (int i=1;i<=size;i++)
if (cnt[i]>1) return 0;
int ans=1;
for (int i=1;i<=size;i++)
if (q[i]&&!cnt[size+i-q[i]-1]) ans=inc(ans,ans);
return ans;
}
int main()
{
init();
int ans=1;
for (int i=1;i<=n;i++)
if (cnt[i]) ans=mul(ans,calc(i,cnt[i]));
for (int i=1;i<=n;i++)
if (ok[i]&&len[i]){
q[size=1]=link[i];
for (int j=a[i];j!=i;j=a[j]) q[++size]=link[j];
ans=mul(ans,Calc());
}
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | // Next or Nextnext
// * frank_c1
// * 2017 / 09 / 29
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL pow_mod(LL b, LL p, LL k) {
LL ret = 1;
for (; p; p >>= 1) {
if (p & 1) ret = ret * b % k;
b = b * b % k;
} return ret;
}
const int maxn = (int)(1e5) + 5;
const int mo = (int)(1e9) + 7;
int a[maxn], in[maxn], w[maxn], cnt[maxn], vi[maxn], s[maxn];
int q[maxn], l = 1, r, len;
LL fac[maxn], ivf[maxn], pw[maxn];
inline LL C(int n, int m) {
if (m < 0 || n - m < 0) return 0;
return (fac[n] * ivf[m] % mo) * ivf[n - m] % mo;
}
inline LL F(int n, int m) {
return (C(n, m << 1) * C(m << 1, m) % mo) * (pow_mod(pw[m], mo - 2, mo) * fac[m] % mo) % mo;
}
int calc() {
LL now = 1;
for (int i = 1; i <= len; ++i) if (cnt[w[i]]) {
int ps = i, v = 0;
do {
++v; --ps; if (ps < 1) ps = len;
} while (!cnt[w[ps]]);
if (v < cnt[w[i]]) return 0;
if (v > cnt[w[i]]) now = now * 2 % mo;
} return now;
}
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), ++in[a[i]];
fac[0] = ivf[0] = pw[0] = 1;
for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % mo, pw[i] = pw[i - 1] * 2 % mo;
ivf[n] = pow_mod(fac[n], mo - 2, mo);
for (int i = n - 1; i >= 1; --i) ivf[i] = ivf[i + 1] * (i + 1) % mo;
memcpy(w, in, sizeof(in));
for (int i = 1; i <= n; ++i) if (!in[i]) q[++r] = i;
for (; l <= r; ) {
int x = q[l++];
if (!(--in[a[x]])) q[++r] = a[x]; cnt[a[x]] = cnt[x] + 1;
}
for (int i = 1; i <= n; ++i) if (w[i] - in[i] > 1) return printf("0\n"), 0;
LL ret = 1;
for (int i = 1; i <= n; ++i) if (in[i] && !vi[i]) {
int x = i, c = 0; len = 0;
while (!vi[x]) {
w[++len] = x; vi[x] = 1; if (cnt[x]) c = 1; x = a[x];
}
if (!c) ++s[len]; else ret = ret * calc() % mo;
}
for (int i = 1; i <= n; ++i) if (s[i]) {
LL tmp = 0, ww = 1;
for (int j = 0; j <= (s[i] >> 1); ++j) {
LL buf = 1;
(buf *= (F(s[i], j) * ww % mo)) %= mo;
if ((i & 1) && i > 1) (buf *= pw[s[i] - (j << 1)]) %= mo;
ww = ww * i % mo; (tmp += buf) %= mo;
} (ret *= tmp) %= mo;
}
return printf("%lld\n", ret), 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#define maxn 100005
#define mod 1000000007
int n,a[maxn],pre[maxn],deg[maxn],cnt[maxn],f[maxn],ans=1;
bool vis[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) { scanf("%d",&a[i]); deg[a[i]]++; }
for(int i=1;i<=n;i++)
{
if(deg[i]>2) { printf("0\n"); return 0; }
if(deg[i]<2||vis[i]) continue;
int p=i;
do
{
if(vis[p]) { printf("0\n"); return 0; }
vis[p]=true,pre[a[p]]=p,p=a[p];
}while(p!=i);
}
for(int i=1;i<=n;i++)
if(!deg[i])
{
int p=i,l1=0,l2=0;
while(!vis[p]) vis[p]=true,p=a[p],l1++;
do l2++,p=pre[p]; while(deg[p]!=2);
if(l1<l2) ans=ans*2%mod;
else if(l1>l2) { printf("0\n"); return 0; }
}
for(int i=1;i<=n;i++)
if(!vis[i])
{
int p=i,l=0;
do l++,p=a[p],vis[p]=true; while(p!=i);
cnt[l]++;
}
for(int i=1;i<=n;i++)
{
int mul=1;
if(i!=1&&(i&1)) mul++;
f[0]=1,f[1]=mul;
for(int j=2;j<=cnt[i];j++) f[j]=(1ll*f[j-2]*(j-1)%mod*i%mod+1ll*f[j-1]*mul%mod)%mod;
ans=1ll*ans*f[cnt[i]]%mod;
}
printf("%d\n",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const long long mod=1000000007ll;
int main()
{
int n;
scanf("%d",&n);
static int a[100000];
for(int i=0;i<n;i++){
scanf("%d",a+i);
a[i]--;
}
vector<vector<int> > e(n);
for(int i=0;i<n;i++){
e[a[i]].push_back(i);
}
static int b[100000];
static int T[100000];
for(int i=0;i<n;i++){
b[i]=-1;
}
int C[100001]={0};
long long ans=1ll;
for(int t=0;t<n;t++){
if(b[t]!=-1){
continue;
}
int i=t;
while(b[i]==-1){
b[i]=t;
i=a[i];
}
if(b[i]!=t){
continue;
}
int p=i;
i=a[i];
int ini=i;
int N=0;
bool F=0;
do{
if(e[i].size()>2){
ans=0ll;
goto ZERO;
}
if(e[i].size()==1){
T[N]=0;
}
else{
F=1;
int q=e[i][0]+e[i][1]-p;
T[N]=1;
while(e[q].size()>0){
if(e[q].size()>1){
ans=0ll;
goto ZERO;
}
q=e[q][0];
T[N]++;
}
}
N++;
p=i;
i=a[i];
}while(i!=ini);
if(!F){
C[N]++;
}
else{
for(int i=0;i<N;i++){
if(T[i]==0){
continue;
}
int j=1;
while(T[(i+N-j)%N]==0){
j++;
}
if(j>=T[i]+1){
ans*=2;
ans%=mod;
}
else if(j<T[i]){
ans=0;
goto ZERO;
}
}
}
}
for(int i=1;i<=n;i++){
if(C[i]==0){
continue;
}
vector<long long> dp(C[i]+1);
dp[0]=1ll;
dp[1]=(i>1&&i%2==1?2:1);
for(int j=2;j<=C[i];j++){
dp[j]=(dp[j-1]*dp[1]%mod+dp[j-2]*(j-1)%mod*i%mod)%mod;
}
ans*=dp[C[i]];
ans%=mod;
}
ZERO:
printf("%lld\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100100
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define drp(x, a, b) for(int x=a; x>=b; x--)
int deg[N], d[N], n, vis[N], a[N], num[N], l[N], ans=1, fac[N], inv[N];
vector<pair<int, int> > fu;
queue<int> q;
int power(int a, int k){
int ret=1;
while(k)
{
if(k&1) ret=1ll*ret*a%mod;
a=1ll*a*a%mod; k>>=1;
}
return ret;
}
int C(int n, int m){
return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int pr(int n){
return 1ll*fac[2*n]*inv[n]%mod*power((mod+1)>>1, n)%mod;
}
void init(){
fac[0]=1;
rep(i, 1, n) fac[i]=1ll*fac[i-1]*i%mod;
inv[n]=power(fac[n], mod-2);
drp(i, n, 1) inv[i-1]=1ll*inv[i]*i%mod;
}
int main(){
scanf("%d", &n);
init();
rep(i, 1, n)
{
scanf("%d", a+i);
deg[a[i]]++;
}
rep(i, 1, n) if(deg[i]>2) return puts("0"), 0;
rep(i, 1, n) d[i]=deg[i];
rep(i, 1, n) if(!d[i]) q.push(i);
while(!q.empty())
{
int u=q.front(); q.pop();
vis[u]=1;
if(--d[a[u]]==0) q.push(a[u]);
}
rep(i, 1, n) if(vis[i] && deg[i]>1) return puts("0"), 0;
rep(i, 1, n) if(vis[i] && !deg[i])
{
int x=i, cnt=0;
for(; vis[x]; x=a[x]) cnt++;
l[x]=cnt;
}
rep(i, 1, n) if(!vis[i])
{
int x=i, fl=0, ass=1, tot=0;
for(; !vis[x]; x=a[x]) ++tot, fl|=bool(l[x]), vis[x]=1;
if(!fl){ num[tot]++; continue; }
fu.clear();
int p=1; x=i;
do{
if(l[x]) fu.pb(mp(l[x], p));
x=a[x]; p++;
}while(x!=i);
fu.pb(mp(fu[0].fi, fu[0].se+tot));
int sz=fu.size();
rep(j, 1, sz-1)
{
if(fu[j].fi>fu[j].se-fu[j-1].se) ass=0;
if(fu[j].fi<fu[j].se-fu[j-1].se) ass=2ll*ass%mod;
}
ans=1ll*ans*ass%mod;
}
rep(i, 1, n) if(num[i])
{
int m=num[i], ass=0;
rep(j, 0, m/2)
{
int tmp=1ll*C(m, 2*j)*pr(j)%mod*power(i, j)%mod;
if(i>1 && (i&1)) tmp=1ll*tmp*power(2, m-2*j)%mod;
ass=(ass+tmp)%mod;
}
ans=1ll*ans*ass%mod;
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <cassert>
#include <cstdlib>
#include <vector>
#include <map>
const int64_t MOD=1e9+7;
const int64_t INV2=5e8+4;
int vis[100005];
int next[100005];
std::vector<int> prev[100005];
int buddy[100005];
int root[100005];
bool cycle[100005];
int64_t inv[100005];
void fail(){
printf("0\n");
exit(0);
}
void dfs_root(int node){
root[node]=node;
if(!root[next[node]]) dfs_root(next[node]);
root[node]=root[next[node]];
}
int main(){
int N;
scanf("%d",&N);
inv[1]=1;
for(int i=2;i<=N;i++){
inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD;
}
for(int i=1;i<=N;i++){
scanf("%d",&next[i]);
prev[next[i]].push_back(i);
}
for(int i=1;i<=N;i++){
if(prev[i].size()>2) fail();
}
for(int i=1;i<=N;i++){
int layer[4];
int layer_cnt=0;
for(int x:prev[i]) layer[layer_cnt++]=x;
while(layer_cnt==2){
int a=layer[0],b=layer[1];
//printf("LAYER %d:%d\n",a,b);
if(buddy[a]||buddy[b]){
assert(buddy[a]==b);
assert(buddy[b]==a);
break;
}
buddy[a]=b,buddy[b]=a;
layer_cnt=0;
for(int x:prev[a]) layer[layer_cnt++]=x;
for(int x:prev[b]) layer[layer_cnt++]=x;
}
if(layer_cnt>2) fail();
}
std::map<int,int> pures;
for(int i=1;i<=N;i++){
if(!root[i]) dfs_root(i);
}
for(int i=1;i<=N;i++){
if(root[i]==i){
bool ispure=true;
int length=0;
int j=i;
do{
if(prev[j].size()>1) ispure=false;
cycle[j]=true;
length++;
j=next[j];
}while(j!=i);
if(ispure){
pures[length]++;
}
}
}
int64_t ans=1;
for(int i=1;i<=N;i++){
if(cycle[i]&&buddy[next[i]]&&!buddy[i]){
ans=ans*2%MOD;
}
}
for(auto pair:pures){
int64_t len=pair.first,cnt=pair.second;
//printf("PURE %ld x%ld\n",len,cnt);
int64_t ac=1;
if(len%2==1&&len>1){
for(int i=0;i<cnt;i++){
ac=ac*2%MOD;
}
}
int64_t sum=ac;
//printf(" %ld",ac);
int pairs=0;
while(cnt>1){
ac=(ac*len)%MOD*((cnt*(cnt-1)/2)%MOD)%MOD*inv[++pairs]%MOD;
cnt-=2;
if(len%2==1&&len>1){
ac=ac*INV2%MOD*INV2%MOD;
}
sum=(sum+ac)%MOD;
//printf("+%ld",ac);
}
//printf(" => %ld\n",sum);
ans=ans*sum%MOD;
}
printf("%ld\n",ans);
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define RI register int
#define g getchar()
using namespace std;
typedef long long ll;
const int mod=1e9+7,N=1e5+10;
void qr(int &x) {
char c=g;x=0;
while(!isdigit(c))c=g;
while(isdigit(c))x=x*10+c-'0',c=g;
}
int n; ll ans;
int a[N],deg[N],vis[N],footL[N],sum[N],f[N];
bool cir[N];
int qm(int x) {return x>=mod?x-mod:x;}
void workcir(int x) {
int now=0,fr=0,frL=0,ed=0;
//now是当前环的长度，fr是第一个有脚点的位置,frL是位置对应的脚长,ed为上一个有脚点的位置。
while(cir[x]) {
++now; cir[x]=0;
if(footL[x]) {
if(!fr) ed=fr=now,frL=footL[x];
else {
ans=ans*((footL[x]<now-ed)+(footL[x]<=now-ed))%mod;//脚的放法
ed=now;
}
}
x=a[x];
}
if(!ed) ++sum[now];
else ans=ans*((frL<now-ed+fr)+(frL<=now-ed+fr))%mod;
}
void work() {
for(RI i=1;i<=n;i++) {//脚对环上交接点贡献
if(deg[i]) continue;
int x=i,len=0; while(!cir[x])x=a[x],++len;
footL[x]=len;//脚长
}
ans=f[0]=1;
for(RI i=1;i<=n;i++) if(cir[i]) workcir(i);//以环为基准计算基环内向树的情况
for(RI i=1;i<=n;i++) {//计算简单环
for(RI j=1;j<=sum[i];j++) {
if(i>1&&(i&1)) f[j]=qm(f[j-1]<<1);
else f[j]=f[j-1];
if(j>1) f[j]=qm(f[j]+(ll)f[j-2]*(j-1)%mod*i%mod);
}
ans=ans*f[sum[i]]%mod;
}
}
int main() {
qr(n);
for(RI i=1;i<=n;i++) qr(a[i]),++deg[a[i]];
for(RI i=1;i<=n;i++) if(!vis[i]) {
int x=i;while(!vis[x]) vis[x]=i,x=a[x];
if(vis[x]!=i) continue;//脚
while(!cir[x]) cir[x]=1,x=a[x];//标记环上点
}
for(RI i=1;i<=n;i++)
if((cir[i]&°[i]>2)||(!cir[i]&°[i]>1)) return puts("0"),0;//开叉就不行
work(); printf("%lld\n",ans); return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<cstdio>
#include<vector>
using namespace std;
const int MAXN = 100000;
const int MOD = int(1E9) + 7;
vector<int>rev[MAXN + 5];
int a[MAXN + 5], f[MAXN + 5], dis[MAXN + 5], dep[MAXN + 5];
bool vis[MAXN + 5], vis2[MAXN + 5], tag[MAXN + 5];
int siz, del, ans = 1, tmp;
void dfs1(int x) {
siz++;
if( rev[x].size() >= 2 ) tmp = x;
vis[x] = true;
for(int i=0;i<rev[x].size();i++)
if( !vis[rev[x][i]] ) dfs1(rev[x][i]);
if( !vis[a[x]] ) dfs1(a[x]);
}
int dfs2(int x) {
if( rev[x].size() == 0 ) return 1;
else if( rev[x].size() == 1 ) {
int d = dfs2(rev[x][0]);
if( d == -1 ) return -1;
else return d + 1;
}
else return -1;
}
int fact[MAXN + 5], inv[MAXN + 5], fun[MAXN + 5];
int pow_mod(int b, int p) {
int ret = 1;
while( p ) {
if( p & 1 ) ret = 1LL*ret*b%MOD;
b = 1LL*b*b%MOD;
p >>= 1;
}
return ret;
}
void init() {
fact[0] = 1;
for(int i=1;i<=MAXN;i++)
fact[i] = 1LL*fact[i-1]*i%MOD;
inv[MAXN] = pow_mod(fact[MAXN], MOD-2);
for(int i=MAXN-1;i>=0;i--)
inv[i] = 1LL*inv[i+1]*(i+1)%MOD;
fun[0] = 1;
for(int i=1;i<=MAXN;i++)
fun[i] = 1LL*fun[i-1]*(2*i-1)%MOD;
}
int comb(int n, int m) {
return 1LL*fact[n]*inv[m]%MOD*inv[n-m]%MOD;
}
int main() {
init(); int N;
scanf("%d", &N);
for(int i=1;i<=N;i++) {
scanf("%d", &a[i]);
rev[a[i]].push_back(i);
}
for(int i=1;i<=N;i++)
if( !vis[i] ) {
tmp = -1; siz = 0; dfs1(i);
if( tmp != -1 ) {
int nw = tmp;
do {
vis2[nw] = true;
nw = a[nw];
}while( !vis2[nw] );
if( nw != tmp ) {
ans = 0;
continue;
}
int p = nw; del = 1;
do {
tag[p] = true;
if( rev[p].size() > 2 )
del = 0;
p = a[p];
}while( p != nw );
if( del == 0 ) {
ans = 0;
continue;
}
int tmp;
do {
if( rev[p].size() == 2 ) {
dep[p] = tag[rev[p][0]] ? dfs2(rev[p][1]) : dfs2(rev[p][0]);
if( dep[p] == -1 ) del = 0;
tmp = p;
}
p = a[p];
}while( p != nw );
p = nw = tmp;
int lst = 0;
do {
p = a[p]; lst++;
dis[p] = lst;
if( rev[p].size() == 2 ) {
lst = 0;
if( dis[p] > dep[p] ) del = 2LL*del%MOD;
else if( dis[p] < dep[p] ) del = 0;
}
}while( p != nw );
ans = 1LL*ans*del%MOD;
}
else f[siz]++;
}
for(int i=1;i<=N;i++) {
if( (i & 1) && i != 1 ) {
del = 0;
for(int j=0;j<=f[i];j+=2)
del = (del + 1LL*comb(f[i], j)*fun[j/2]%MOD*pow_mod(i, j/2)%MOD*pow_mod(2, f[i]-j)%MOD)%MOD;
ans = 1LL*del*ans%MOD;
}
else {
del = 0;
for(int j=0;j<=f[i];j+=2)
del = (del + 1LL*comb(f[i], j)*fun[j/2]%MOD*pow_mod(i, j/2)%MOD)%MOD;
ans = 1LL*del*ans%MOD;
}
}
printf("%d\n", ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<stdio.h>
typedef long long ll;
const int mod=1000000007;
int mul(int a,int b){return(ll)a*b%mod;}
int a[100010],d[100010],p[100010],f[100010],c[100010];
bool v[100010];
int main(){
int n,i,j,c1,c2,ans;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d",a+i);
d[a[i]]++;
}
#define wrong {putchar('0');return 0;}
ans=1;
for(i=1;i<=n;i++){
if(d[i]>2)wrong
if(v[i]||d[i]<2)continue;
j=i;
do{
if(v[j])wrong
v[j]=1;
p[a[j]]=j;
j=a[j];
}while(j!=i);
}
for(i=1;i<=n;i++){
if(d[i]==0){
c1=0;
for(j=i;d[j]!=2;j=a[j]){
c1++;
v[j]=1;
}
c2=1;
for(j=p[j];d[j]!=2;j=p[j])c2++;
if(c1>c2)wrong
if(c1<c2)ans=mul(ans,2);
}
}
for(i=1;i<=n;i++){
if(!v[i]){
c1=0;
j=i;
do{
v[j]=1;
j=a[j];
c1++;
}while(j!=i);
c[c1]++;
}
}
for(i=1;i<=n;i++){
c1=1+(i!=1&&(i&1));
f[0]=1;
f[1]=c1;
for(j=2;j<=c[i];j++)f[j]=(mul(f[j-2],mul(j-1,i))+mul(f[j-1],c1))%mod;
ans=mul(ans,f[c[i]]);
}
printf("%d",ans);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include"bits/stdc++.h"
#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
#define iout(x) printf("%d\n",x)
#define lout(x) printf("%lld\n",x)
#define REP(x,l,u) for(int x = (l);x<=(u);x++)
#define RREP(x,l,u) for(int x = (l);x>=(u);x--)
#define mst(x,a) memset(x,a,sizeof(x))
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define MP make_pair
#define se second
#define fi first
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
#define sz(x) ((int)x.size())
#define cl(x) x.clear()
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
using namespace std;
const int maxn = 100010;
const int mod = 1e9+7;
const int MAX = 1000000010;
const double eps = 1e-6;
const double PI = acos(-1);
template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}
template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}
template<typename A,typename B> inline A fexp(A x,B p){A ans=1;for(;p;p>>=1,x=1LL*x*x%mod)if(p&1)ans=1LL*ans*x%mod;return ans;}
template<typename A,typename B> inline A fexp(A x,B p,A mo){A ans=1;for(;p;p>>=1,x=1LL*x*x%mo)if(p&1)ans=1LL*ans*x%mo;return ans;}
int n,ans=1;
int A[maxn],vi[maxn],Is[maxn],ch[maxn],has[maxn],f[maxn],deg[maxn];
int calc(int x,int y){
if(x<y)return 2;
if(x>y)return 0;
return 1;
}
void Work(){
REP(i,1,n)if(deg[i]>2||(deg[i]==2&&!Is[i]))return void(puts("0"));
REP(i,1,n)if(!deg[i]){
int x=i,t=0;while(!Is[x])x=A[x],t++;
ch[x]=t;
}
REP(i,1,n)if(Is[i]){
int x=i;
do x=A[x];while(!ch[x]&&x!=i);
if(x==i&&!ch[x]){
int t=0;
do x=A[x],t++,Is[x]=0;while(x!=i);
has[t]++;
continue;
}
int j=x,now=1;x=A[x];
for(;;){
if(ch[x]){
ans=1ll*ans*calc(ch[x],now)%mod;
now=0;
}
if(x==j)break;
now++;x=A[x];
}
while(Is[x])Is[x]=0,x=A[x];
}
REP(i,1,n)if(has[i]){
int x=(i>1&&(i&1))+1;
f[0]=1;
REP(j,1,has[i]){
f[j]=1ll*f[j-1]*x%mod;
if(j>=2)f[j]=(1ll*f[j-2]*(j-1)%mod*i+f[j])%mod;
}
ans=1ll*ans*f[has[i]]%mod;
}
iout(ans);
}
void Init(){
read(n);
REP(i,1,n)read(A[i]);
REP(i,1,n)deg[A[i]]++;
REP(i,1,n)if(!vi[i]){
int x=i;
while(!vi[x])vi[x]=i,x=A[x];
if(vi[x]==i){
while(!Is[x])Is[x]=1,x=A[x];
}
}
}
int main(){
Init();
Work();
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pa;
typedef unsigned int uint;
typedef unsigned long long ull;
#define w1 first
#define ls (x<<1)
#define w2 second
#define rs (x<<1|1)
#define mp make_pair
#define pb push_back
#define mid ((l+r)>>1)
#define SZ(x) (int((x).size()))
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define rep2(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define per(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define Rep(p,x) for(int (p)=head[(x)];(p);(p)=nxt[(p)])
#define Rep2(p,x) for(int (p)=cur[(x)];(p);(p)=nxt[(p)])
template<class T>void read(T&num){
num=0;T f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
num*=f;
}
const int maxn=1e5+5,mod=1e9+7;
int n,ans=1;
int a[maxn],d[maxn],vis[maxn],len[maxn],q[maxn],cnt[maxn],f[maxn];
void solve(int n,int m){
int now=1+(n>1&&(n&1));
f[0]=1;
f[1]=now;
rep(i,2,m)f[i]=(1ll*now*f[i-1]%mod+1ll*(i-1)*n%mod*f[i-2]%mod)%mod;
ans=1ll*ans*f[m]%mod;
}
int main(){
read(n);
rep(i,1,n)read(a[i]),d[a[i]]++;
rep(i,1,n)if(d[i]>2)ans=0;
rep(i,1,n)if(!d[i]){
int j=i,l=0;
for(;d[j]<2;j=a[j],l++)vis[j]=1;
len[j]=l;
}
rep(i,1,n)if(!vis[i]){
int tot=0,j;
for(j=i;!vis[j];j=a[j])vis[j]=1,q[++tot]=j;
if(j!=i)ans=0;
int cir=1;
rep(j,1,tot)if(len[q[j]])cir=0;
if(cir)cnt[tot]++;
else{
int maxl=1;
per(j,tot,1)
if(!len[q[j]])maxl++;
else break;
rep(j,1,tot)
if(len[q[j]]){
if(len[q[j]]<maxl)ans=2ll*ans%mod;
if(len[q[j]]>maxl)ans=0;
maxl=1;
}else maxl++;
}
}
rep(i,1,n)if(cnt[i])solve(i,cnt[i]);
printf("%d\n",ans);
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<vector>
#include<set>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
long long fact[310000];
long long inv[310000];
long long finv[310000];
int p[110000];
int cnt[110000];
vector<int>g[110000];
vector<int>rev[110000];
int v[110000];
int cur;
int num[110000];
long long pw(long long a,long long b){
long long ret=1;
while(b){
if(b%2)ret=ret*a%mod;
b/=2;a=a*a%mod;
}
return ret;
}
void dfs(int a){
for(int i=0;i<g[a].size();i++){
if(v[g[a][i]])continue;
v[g[a][i]]=1;
dfs(g[a][i]);
}
num[cur++]=a;
}
int scc[110000];
void dfs2(int a){
scc[a]=cur;
for(int i=0;i<rev[a].size();i++){
if(v[rev[a][i]])continue;
v[rev[a][i]]=1;
dfs2(rev[a][i]);
}
}
int dfs3(int a,int b){
int ret=1;
for(int i=0;i<rev[a].size();i++){
if(b==scc[rev[a][i]])continue;
ret+=dfs3(rev[a][i],b);
}
return ret;
}
int cy[110000];
int in[110000];
int out[110000];
int sz[110000];
int fi[110000];
int ren[110000];
long long dp[110000];
int main(){
fact[0]=finv[0]=1;
inv[1]=1;
for(int i=1;i<310000;i++){
fact[i]=fact[i-1]*i%mod;
}
for(int i=2;i<310000;i++){
inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
}
for(int i=1;i<310000;i++){
finv[i]=finv[i-1]*inv[i]%mod;
}
int a;scanf("%d",&a);
for(int i=0;i<a;i++){
scanf("%d",p+i);p[i]--;
cnt[p[i]]++;
}
for(int i=0;i<a;i++)if(cnt[i]>2){
printf("0\n");return 0;
}
for(int i=0;i<a;i++){
g[i].push_back(p[i]);
rev[p[i]].push_back(i);
}
for(int i=0;i<a;i++){
if(v[i])continue;
v[i]=1;
dfs(i);
}
cur=0;
for(int i=0;i<a;i++)v[i]=0;
for(int i=a-1;i>=0;i--){
if(v[num[i]])continue;
v[num[i]]=1;
fi[cur]=num[i];
dfs2(num[i]);
cur++;
}
for(int i=0;i<a;i++){
in[scc[p[i]]]++;
sz[scc[i]]++;
if(scc[p[i]]!=scc[i])out[scc[i]]++;
}
for(int i=0;i<a;i++){
if(cnt[i]>1&&scc[i]!=scc[p[i]]){
printf("0\n");return 0;
}
}
long long ret=1;
for(int i=0;i<cur;i++){
//printf("%d %d\n",in[i],sz[i]);
if(out[i])continue;
if(in[i]==sz[i]){
cy[sz[i]]++;
continue;
}
if(out[i])continue;
vector<int>L;
int at=fi[i];
while(1){
L.push_back(dfs3(at,scc[at])-1);
at=p[at];
if(at==fi[i])break;
}
int now=0;
for(int j=0;j<L.size()*2;j++){
ren[j%L.size()]=now;
if(L[j%L.size()]==0)now++;
else now=1;
}
for(int j=0;j<L.size();j++){
if(L[j]==0)continue;
if(ren[j]<L[j])ret*=0;
else if(ren[j]>L[j])ret=ret*2%mod;
}
}
for(int i=1;i<=a;i++){
long long tmp=0;
for(int j=0;j*2<=cy[i];j++){
long long ks=pw(i,j)*fact[cy[i]]%mod*finv[cy[i]-j*2]%mod*finv[j]%mod*pw(500000004,j)%mod;
if(i%2&&i!=1){
ks=ks*pw(2,cy[i]-j*2)%mod;
}
tmp=(tmp+ks)%mod;
}
ret=ret*tmp%mod;
}
printf("%lld\n",ret);
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int Q=1<<17,MOD=1e9+7;
inline int add(int a,int b)
{a+=b;return a>=MOD?a-MOD:a;}
inline int mul(int a,int b)
{return 1LL*a*b%MOD;}
int a[Q];
int vis[Q];
int pi=1,maxn=0;
int val[Q];
int cir[Q];
void dfs(int x,int Time)
{
if(vis[x]){
if(vis[x]!=Time||cir[x])return;
cir[x]=1;
for(int t=a[x];t!=x;t=a[t])
cir[t]=1;
return;
}
vis[x]=Time;
dfs(a[x],Time);
}
int in[Q];
#define GG return puts("0")&0
int f[Q];
int Ga(int x)
{return cir[x]?x:(a[x]=Ga(a[x]));}
int len[Q];
int l[Q];
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),++in[a[i]];
for(int i=1;i<=n;i++)
if(in[i]>2)GG;
for(int i=1;i<=n;i++)
if(!vis[i])dfs(i,i);
for(int i=1;i<=n;i++){
if(!cir[i])
if(in[i]>1)GG;
++len[Ga(i)];
}
for(int i=1;i<=n;i++)
if(cir[i]&&vis[i]!=-1){
int mx=0;
for(int t=i;vis[t]!=-1;t=a[t]){
vis[t]=-1;
l[++mx]=len[t]-1;
}
for(int i=1;i<=mx;i++)
l[mx+i]=l[i];
int lst=998244353,mus=0;
for(int i=(mx<<1);i;--i)
if(l[i]){
if(lst-i<mus)GG;
if(lst-i>mus&&lst>mx&&lst<=(mx<<1))pi=mul(pi,2);
lst=i;
mus=l[i];
}
if(lst>n)val[++maxn]=mx;
}
sort(val+1,val+maxn+1);
f[0]=1;
for(int i=1,had=0;i<=maxn;i++){
if(val[i]==val[i-1])++had;
else had=0;
f[i]=add(mul(f[i-1],((val[i]&1)&(val[i]>1))+1),i==1?0:mul(mul(had,val[i]),f[i-2]));
}
printf("%d",mul(f[maxn],pi));
return 0;
} | CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | #include<bits/stdc++.h>
#define LL long long
using namespace std;
const int mod=1e9+7;
int fastpow(int x,int a){
int ret=1;
while(a){
if(a&1) ret=(LL)ret*x%mod;
x=(LL)x*x%mod;a>>=1;
}
return ret;
}
struct Combin{
#define N 200005
int fac[N],rv[N],facrv[N];
Combin(){
fac[0]=rv[1]=facrv[0]=1;
for(int i=2;i<N;i++) rv[i]=((-(LL)(mod/i)*rv[mod%i]%mod)+mod)%mod;
for(int i=1;i<N;i++) fac[i]=(LL)fac[i-1]*i%mod;
for(int i=1;i<N;i++) facrv[i]=(LL)facrv[i-1]*rv[i]%mod;
}
int C(int r1,int n1){
if(r1>n1) return 0;
return fac[n1]*(LL)facrv[r1]%mod*facrv[n1-r1]%mod;
}
#undef N
}C;
#define N 100005
int p[N],n,cnt,in[N],len[N],vis[N];
int ans=1;
void tp(int x){
if(len[p[x]]!=0) ans=0;
len[p[x]]=len[x]+1;
len[x]=-1;
in[p[x]]--;
if(in[p[x]]==0) tp(p[x]);
}
int sz[N];
void solve(int id){
vis[id]=1;
if(p[id]==id&&len[id]==0){sz[1]++;return;}
deque<int> v;
bool iscir=true;
v.push_back(len[id]);
for(int i=p[id];i!=id;i=p[i])
vis[i]=1,v.push_back(len[i]);
for(int i:v) if(i!=0) iscir=false;
if(iscir){
sz[v.size()]++;
}else{
while(v.back()==0) v.push_front(0),v.pop_back();
while(!v.empty()){
int val=v.back();v.pop_back();
while(!v.empty() && v.back()==0) v.pop_back(),val--;
if(val==1) continue;
else if(val<1) ans=ans*2%mod;
else ans=0;
}
}
}
int f(int j){//j = 2k
//how many ways to matchs
//if there k matchs in 2k points
//ans = C(2,2k)*C(2,2k-2)...*C(2,4)*C(2,2)/(k!)
// =C(2k-2,2k)*C(2k-4,2k-2)...C(0,2)/(k!)
// =(2k)!/((2!)^k)/(k!)
return C.fac[j]*1LL*C.facrv[j/2]%mod*fastpow(C.facrv[2],j/2)%mod;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&p[i]),in[p[i]]++;
for(int i=1;i<=n;i++)
if(in[i]==0&&len[i]==0)
tp(i);
if(ans==0) return puts("0"),0;
for(int i=1;i<=n;i++)
if(vis[i]==0&&len[i]>=0)
solve(i);
for(int i=1;i<=n;i++){
int ansnow=0;
if(i==1||i%2==0){
for(int j=0;j<=sz[i];j+=2)
ansnow = (ansnow + C.C(j,sz[i]) *1LL* f(j)%mod *fastpow(i,j/2)%mod * fastpow(1,sz[i]-j) )%mod;
}else{
for(int j=0;j<=sz[i];j+=2)
ansnow = (ansnow + C.C(j,sz[i]) *1LL* f(j)%mod *fastpow(i,j/2)%mod * fastpow(2,sz[i]-j) )%mod;
}
ans = ans*1LL*ansnow%mod;
}
cout<<ans<<endl;
return 0;
}
| CPP |
p03842 AtCoder Grand Contest 008 - Next or Nextnext | You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?
* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.
Find the count modulo 10^9 + 7.
Constraints
* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.
Examples
Input
3
1 2 3
Output
4
Input
2
1 1
Output
1
Input
3
2 1 1
Output
2
Input
3
1 1 1
Output
0
Input
13
2 1 4 3 6 7 5 9 10 8 8 9 11
Output
6 | 6 | 0 | /*
https://blog.csdn.net/litble/article/details/83118814
图论神啊，好题好题好题！！！
i 向 pi 连条边，构成的图简化以后就都是 “i 向 ai 连条边” 这个图了！！！
注意：是由一个环和若干条指向环的链组成的基环内向树，至于为什么是链，详见样例4
现在手上只有 a，那么就**倒推**有多少种可能图吧
*/
#include <bits/stdc++.h>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;
const int mod = 1e9 + 7, N = 1e5 + 10;
typedef long long ll;
int n, a[N], deg[N], vis[N], cir[N], L[N], sum[N];
ll ans, f[N];
void calc(int x) {
int now = 0, fr = 0, ed = 0, l = 0;
// fr: 第一个脚的位置 ed: 上一个脚的位置 l: 第一个脚的长度 now: 当前节点是从 x 开始走环走到的第几个点
while (cir[x]) {
++now, cir[x] = 0;
if (L[x]) {
if (!fr) ed = fr = now, l = L[x];
else {
int tmp = (L[x] < now - ed) + (L[x] <= now - ed);
ans = 1ll * ans * tmp % mod, ed = now;
}
}
x = a[x];
}
if (!fr) ++sum[now];
else {
int tmp = (l < now - ed + fr) + (l <= now - ed + fr);
ans = 1ll * ans * tmp % mod;
}
}
void solve() {
rep(i, 1, n) {
if (deg[i]) continue;
int x = i, len = 0; while (!cir[x]) ++len, x = a[x];
L[x] = len;
}
ans = 1;
rep(i, 1, n) if (cir[i]) calc(i);
rep(i, 1, n) {
f[0] = 1;
rep(j, 1, sum[i]) {
if (i > 1 && (i & 1)) f[j] = f[j - 1] * 2 % mod;
else f[j] = f[j - 1];
if (j > 1) f[j] = (f[j] + f[j - 2] * (j - 1) % mod * i % mod) % mod;
}
ans = 1ll * ans * f[sum[i]] % mod;
}
}
int main() {
cin >> n;
rep(i, 1, n) scanf("%d", &a[i]), ++deg[a[i]];
rep(i, 1, n) {
if (vis[i]) continue;
int x = i;
while (!vis[x]) vis[x] = i, x = a[x];
if (vis[x] != i) continue;
while (!cir[x]) cir[x] = 1, x = a[x];
}
rep(i, 1, n)
if ((!cir[i] && deg[i] > 1) || (cir[i] && deg[i] > 2)) return puts("0"), 0;
solve();
printf("%lld\n", ans);
return 0;
}
| CPP |
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