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p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); ++i)
const int MOD = int(1e9)+7;
struct mint {
ll x;
mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
mint& operator+=(const mint a) { if ((x += a.x) >= MOD) x -= MOD; return *this; }
mint operator+(const mint a) const { return mint(*this) += a; }
};
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
const int M = 60;
mint dp[M+1][2][2][2];
int main() {
ll L, R; cin >> L >> R;
dp[0][0][0][0] = 1;
rep(i, M) {
int Li = L>>(M-1-i)&1; // L の上から i 桁目
int Ri = R>>(M-1-i)&1; // R の上から i 桁目
rep(j, 2) rep(l, 2) rep(r, 2) rep(xi, 2) rep(yi, 2) {
int j2 = j, l2 = l, r2 = r; // 遷移前の状態で初期化
if (xi > yi) continue; // (xi, yi) = (1, 0) なら無視
if (!j && (xi < yi)) continue; // (1, 1) が現れる前に (0, 1) が現れたら無視
if (xi&yi) j2 = 1; // (1, 1) が現れたら j2 を更新
// 条件 x ≧ L の処理
if (!l && (xi < Li)) continue;
if (xi > Li) l2 = 1;
// 条件 y ≦ R の処理
if (!r && (yi > Ri)) continue;
if (yi < Ri) r2 = 1;
dp[i+1][j2][l2][r2] += dp[i][j][l][r]; // 遷移式
}
}
mint ans = 0;
rep(l, 2) rep(r, 2) ans += dp[M][1][l][r];
cout << ans << '\n';
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define int long long
#define pi pair<int,int>
#define ff first
#define ss second
#define endl '\n'
#define boost ios::sync_with_stdio(false);cin.tie(nullptr)
#include "string"
const int M = 1e9 + 7;
int mod( int m ,int mod = M) {
m %= mod;
return (m + mod) % mod;
}
// find y - x = x ^ y && L <= x <= y <= R
int dp[60][2][2][2];
int L,R;
int solve(int id, int l, int r, int cur) {
if(id < 0)
return 1;
int &tmp = dp[id][l][r][cur];
if(tmp != -1)
return tmp;
tmp = 0;
if(l or !((L >> id) & 1ll))
tmp = mod(tmp + solve(id - 1, l, r | ((R >> id) & 1ll), cur) );
if((l or !((L >> id) & 1LL)) && (r or ((R >> id) & 1ll)) && cur)
tmp = mod(tmp + solve(id - 1, l, r, cur));
if(r or ((R >> id) & 1ll))
tmp = mod(tmp + solve(id - 1, l | !((L >> id) & 1ll) , r, 1) );
return tmp;
}
int32_t main(){
boost;
int t = 1;
//cin >> t;
while(t--) {
cin >> L >> R;
memset(dp, -1, sizeof dp);
int ans = solve(59, 0, 0, 0);
cout << ans << endl;
}
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007ll;
const int N_DIGITS = 60;
ll L[N_DIGITS], R[N_DIGITS];
ll Dp[1 + N_DIGITS][2][2][2]; // これまでで L<x, y<R, x,y>0 が 不成立/成立
const bool Bs[2] = {false, true};
void go(ll &tgt, ll src) { (tgt += src) %= MOD; }
int main() {
ll l, r;
cin >> l >> r;
for (int i = N_DIGITS - 1; i >= 0; --i) {
L[i] = l % 2;
R[i] = r % 2;
l /= 2;
r /= 2;
}
Dp[0][0][0][0] = 1;
for (int i = 0; i < N_DIGITS; ++i) {
for (bool overL : Bs) {
for (bool underR : Bs) {
for (bool some : Bs) {
ll src = Dp[i][overL][underR][some];
// 0,0
if (overL || L[i] == 0) {
go(Dp[i + 1][overL][underR || R[i] == 1][some], src);
}
// 0,1
if ((overL || L[i] == 0) && (underR || R[i] == 1) && some) {
go(Dp[i + 1][overL][underR][true], src);
}
// 1,1
if (underR || R[i] == 1) {
go(Dp[i + 1][overL || L[i] == 0][underR][true], src);
}
}
}
}
/*
cout << "l " << L[i] << " r " << R[i] << " :";
for (bool overL : Bs) {
for (bool underR : Bs) {
for (bool some : Bs) {
cout << " " << Dp[i][overL][underR][some];
}
}
}
cout << endl;
*/
}
ll res = 0;
for (bool overL : Bs) {
for (bool underR : Bs) {
for (bool some : Bs) {
(res += Dp[N_DIGITS][overL][underR][some]) %= MOD;
}
}
}
cout << res << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define N 200005
#define LL long long
const int mo = 1e9+7;
LL dp[70][2][2][2],num1[70],num2[70] ;
LL dfs(int pos,bool L,bool R,bool flg) {
if(!pos) return 1;
if(~dp[pos][L][R][flg]) return dp[pos][L][R][flg];
LL res = 0;
int up1 = L? num1[pos]:0,up2 = R? num2[pos]:1;
if(up2) res += dfs(pos-1,num1[pos] == 1 && L,num2[pos] == 1 && R,1),res %= mo;
if(up1==0 && up2 && flg) res += dfs(pos-1,num1[pos] == 0 && L,num2[pos] == 1 && R,1),res %= mo;
if(!up1) res += dfs(pos-1,num1[pos] == 0 && L,num2[pos] == 0 && R,flg),res %= mo;
return dp[pos][L][R][flg] = res;
}
LL Solve(LL L,LL R) {
int cot1 = 0,cot2 = 0;
while(L) num1[++cot1] = L&1,L >>= 1;
while(R) num2[++cot2] = R&1,R >>= 1;
return dfs(cot2,1,1,0);
}
int main() {
memset(dp,-1,sizeof(dp));
LL L,R;
cin >>L>>R;
cout<<Solve(L,R);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | /*input
*/
#include <bits/stdc++.h>
#define up(i,a,b) for(int (i) = (a);(i)<=(b);++(i))
#define down(i,b,a) for(int (i) = (b);i>=(a);--i)
#define debug(x) cerr << (x) << '\n';
#define bits(x,i) ((x >> i) & 1)
#define mid ((l+r)/2)
#define pr pair<int,int>
using namespace std;
const int logA = 60;
const int mod = 1e9 + 7;
long long dp[logA][2][2][2];
long long l,r;
long long solve(int pos,int isLow, int isHigh,int firstBit){
if (pos < 0) return 1;
long long& res = dp[pos][isLow][isHigh][firstBit];
if (res != -1) return res;
res = 0;
int low = (isLow ? 0 : bits(l, pos)), high = (isHigh ? 1 : bits(r, pos));
for(int choice = low; choice <= high;++choice){
res +=
solve(pos - 1,isLow | (choice != low),isHigh | (choice != high), firstBit | choice == 1);
if (choice == 1 && low == 0 && firstBit == 1)
res += solve(pos-1, isLow, isHigh,1);
res %= mod;
}
return res;
}
int main(){
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> l >> r;
memset(dp, -1, sizeof(dp));
cout << solve(59, 0,0,0) << '\n';
// debug
int ans = 0;
// for(int i=l;i<=r;++i){
// for(int j=i;j<=r;++j) if (j % i == (j ^ i)) {
// ans++;
// // cout << j << ' ' << i << '\n';
// }
// }
// cout << ans << '\n';
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fo(i, n) for(int i = 1; i <= n; ++i)
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1e9 + 7;
const int N = 1005;
ll L, R;
ll a[N], b[N];
ll dp[70][2][2][2];
ll calc(ll pos, int is_l, int is_r, int ch = 0) {
if(pos < 0) return 1;
if(dp[pos][is_l][is_r][ch] != -1) return dp[pos][is_l][is_r][ch];
ll &res = dp[pos][is_l][is_r][ch];
res = 0;
// 1 1
if(!is_r or b[pos])
(res += calc(pos - 1, is_l && a[pos], is_r, 1)) %= mod;
// 0 1
if((!is_r or b[pos]) and (!is_l or !a[pos]) and ch)
(res += calc(pos - 1, is_l && !a[pos], is_r, 1)) %= mod;
// 0 0
if(!is_l or !a[pos])
(res += calc(pos - 1, is_l && !a[pos], is_r and !b[pos], ch)) %= mod;
return res;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> L >> R;
for(int i = 61; i >= 0; --i) {
a[i] = (L >> i) & 1;
b[i] = (R >> i) & 1;
}
memset(dp, -1, sizeof dp);
cout << calc(61, 1, 1);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (n); ++i)
const int MAX = 1e5 + 10;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int main(){
long long L; cin >> L;
long long R; cin >> R;
long long dp[61][2][2][2] = {}; // Zero initialization.
dp[60][0][0][0] = 1;
for(int i = 59; i >= 0; --i){
int r = R >> i & 1;
int l = L >> i & 1;
rep(j, 2)rep(k, 2)rep(s, 2){
rep(x, 2)rep(y, 2){
if(y == 0 && x == 1) continue;
int nj, nk, ns;
nj = j, nk = k, ns = s;
if(s == 0 && y == 1 && x == 0) continue;
if(x == 1 && y == 1) ns = 1;
if(k == 0 && r == 0 && y == 1) continue;
if(r == 1 && y == 0) nk = 1;
if(j == 0 && l == 1 && x == 0) continue;
if(l == 0 && x == 1) nj = 1;
dp[i][nj][nk][ns] += dp[i + 1][j][k][s];
dp[i][nj][nk][ns] %= MOD;
}
}
}
long long ans = 0;
rep(j, 2)rep(k, 2)rep(s, 2){
ans += dp[0][j][k][s];
ans %= MOD;
}
cout << ans << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define MAXN (100000)
#define MOD (1000000007)
uint64_t dp[61][2][2][2];
int main(void)
{
uint64_t l,r,ans;
scanf("%ld %ld",&l, &r);
dp[60][0][0][0] = 1;
for(int i=59;i>=0;i--)
{
int lb = (l>>i)&1;
int rb = (r>>i)&1;
for(int j=0;j<2;j++)for(int k=0;k<2;k++)for(int s=0;s<2;s++)
{
for(int x=0;x<2;x++)for(int y=0;y<2;y++)
{
int nj = j, nk = k, ns = s;
if(x && !y) continue;
if(!s && (x!=y)) continue;
if(!s && x && y) ns = 1;
if(!j && !x && lb) continue;
if(!j && x && !lb) nj = 1;
if(!k && y && !rb) continue;
if(!k && !y && rb) nk = 1;
dp[i][nj][nk][ns] += dp[i+1][j][k][s];
dp[i][nj][nk][ns] %= MOD;
}
}
}
ans = 0;
for(int j=0;j<2;j++)for(int k=0;k<2;k++)for(int s=0;s<2;s++)
{
ans = (ans + dp[0][j][k][s])%MOD;
}
printf("%ld\n",ans);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
//#define int long long
typedef long long ll;
const int M=1e9+7;
int f[65][2][2],dl[65],dr[65];
ll l,r;
int dfs(int pos,bool fl,bool fr){
if (!pos) return 1;
int &t=f[pos][fl][fr];
if (~t) return t;
t=0;int up=fr?dr[pos]:1,dn=fl?dl[pos]:0;
for (int y=dn;y<=up;y++)
for (int x=dn;x<=y;x++) (t+=dfs(pos-1,fl&(x==dn),fr&(y==up)))%=M;
return t;
}
int solve(ll l,ll r){
int lenl=0,lenr=0;
for (;l;l>>=1) dl[++lenl]=l&1;
for (;r;r>>=1) dr[++lenr]=r&1;
int ans=0;
for (int i=lenl;i<=lenr;i++) (ans+=dfs(i-1,i==lenl,i==lenr))%=M;
return ans;
}
signed main(){
memset(f,-1,sizeof(f));
scanf("%lld%lld",&l,&r);
printf("%d",solve(l,r));
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
input=lambda :sys.stdin.buffer.readline().rstrip()
class ModInt(object):
__MOD=10**9+7
def __init__(self,x):
self.__x=x%self.__MOD
def __repr__(self):
return str(self.__x)
def __add__(self,other):
return ModInt(self.__x+other.__x) if isinstance(other,ModInt) else ModInt(self.__x+other)
def __sub__(self,other):
return ModInt(self.__x-other.__x) if isinstance(other,ModInt) else ModInt(self.__x-other)
def __mul__(self,other):
return ModInt(self.__x*other.__x) if isinstance(other,ModInt) else ModInt(self.__x*other)
def __truediv__(self,other):
return ModInt(self.__x*pow(other.__x,self.__MOD-2,self.__MOD)) if isinstance(other, ModInt) else ModInt(self.__x*pow(other, self.__MOD-2,self.__MOD))
def __pow__(self,other):
return ModInt(pow(self.__x,other.__x,self.__MOD)) if isinstance(other,ModInt) else ModInt(pow(self.__x,other,self.__MOD))
__radd__=__add__
def __rsub__(self,other):
return ModInt(other.__x-self.__x) if isinstance(other,ModInt) else ModInt(other-self.__x)
__rmul__=__mul__
def __rtruediv__(self,other):
return ModInt(other.__x*pow(self.__x,self.__MOD-2,self.__MOD)) if isinstance(other,ModInt) else ModInt(other*pow(self.__x,self.__MOD-2,self.__MOD))
def __rpow__(self,other):
return ModInt(pow(other.__x,self.__x,self.__MOD)) if isinstance(other,ModInt) else ModInt(pow(other,self.__x,self.__MOD))
def resolve():
from itertools import product
L,R=map(int,input().split())
D=61
dp=[[[[ModInt(0)]*2 for _ in range(2)] for _ in range(2)] for _ in range(D+1)]
dp[D][0][0][0]=ModInt(1)
for d in range(D-1,-1,-1):
lb=L>>d&1; rb=R>>d&1
for i,j,m,x,y in product([0,1],repeat=5):
ni,nj,nm=i,j,m
if(x>y): continue
# i:L<=X
if(i==0 and lb>x): continue
if(lb<x): ni=1
# j:Y<=R
if(j==0 and y>rb): continue
if(y<rb): nj=1
# m:MSB
if(m==0 and x!=y): continue
if(x==1 and y==1): nm=1
dp[d][ni][nj][nm]+=dp[d+1][i][j][m];
print(sum(dp[0][i][j][m] for i,j,m in product([0,1],repeat=3)))
resolve() | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include "bits/stdc++.h"
#define MAXN 100009
#define INF 1000000007
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(__typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
#define debug(x) cerr<< #x <<" = "<< x<<endl;
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
int dp[60][2][2][2][2][2];
string A,B;
int f(int x,int y,int e){
if(!e)
return (!x and y);
return !(x and !y);
}
int rec(int pos,int a,int b,int c,int d,int e){//a-y,b-x,y-x
if(pos==60)
return (!a and !b and !c and !f(e,0,d));
int &ret=dp[pos][a][b][c][d][e];
if(~ret)
return ret;ret=0;
for(int i=0;i<2;i++)//y
for(int j=0;j<2;j++)//x
if(!c){
ret+=rec(pos+1,f(A[pos]-'0',i,a),f(B[pos]-'0',j,b),f(i,j,c),
f(e,i,d),j);
if(ret>=INF)
ret-=INF;
}
return ret;
}
string men(ll x){
string res;
while(x>0){
res+=(x%2)+'0';
x/=2;
}
while(res.size()<60)
res+='0';
return res;
}
int f(ll a,ll b){
memset(dp,-1,sizeof dp);
A=men(a);B=men(b);
return rec(0,0,0,0,0,0);
}
int main(){
//freopen("file.in", "r", stdin);
ll l,r;
scanf("%lld%lld",&l,&r);
int ans=f(r,r)-f(r,l-1);
if(ans<0)
ans+=INF;
printf("%d\n",ans);
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
const ll MOD=1e9+7;
int main()
{
ll l, r; cin>>l>>r;
ll dp[2][2][60]={};
for(int i=0; i<60; i++){
if(l>=(1ll<<(i+1)) || r<(1ll<<i)) continue;
int p=0, q=0;
if((1ll<<i)>l) p=1;
if(r>=(1ll<<(i+1))) q=1;
dp[p][q][i]=1;
}
for(int i=59; i>=1; i--){
(dp[1][1][i-1]+=3*dp[1][1][i])%=MOD;
if((r&(1ll<<(i-1)))==0){
(dp[1][0][i-1]+=dp[1][0][i])%=MOD;
}else{
(dp[1][1][i-1]+=dp[1][0][i])%=MOD;
(dp[1][0][i-1]+=2*dp[1][0][i])%=MOD;
}
if((l&(1ll<<(i-1)))!=0){
(dp[0][1][i-1]+=dp[0][1][i])%=MOD;
}else{
(dp[1][1][i-1]+=dp[0][1][i])%=MOD;
(dp[0][1][i-1]+=2*dp[0][1][i])%=MOD;
}
if((r&(1ll<<(i-1)))==0 && (l&(1ll<<(i-1)))!=0){
}else if((r&(1ll<<(i-1)))==0){
(dp[0][0][i-1]+=dp[0][0][i])%=MOD;
}else if((l&(1ll<<(i-1)))!=0){
(dp[0][0][i-1]+=dp[0][0][i])%=MOD;
}else{
(dp[0][1][i-1]+=dp[0][0][i])%=MOD;
(dp[1][0][i-1]+=dp[0][0][i])%=MOD;
(dp[0][0][i-1]+=dp[0][0][i])%=MOD;
}
}
ll ans=0;
for(int i=0; i<2; i++) for(int j=0; j<2; j++) (ans+=dp[i][j][0])%=MOD;
cout<<ans<<endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<iostream>
#include<bitset>
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
bitset<64> L, R;
ll memo[60][2][2][2];
ll dfs(int pos, int flagX, int flagY, int flagZ) {
if (pos == -1) return 1;
if (memo[pos][flagX][flagY][flagZ] != -1) {
return memo[pos][flagX][flagY][flagZ];
}
ll ret = 0;
if (flagX || L[pos] == 0) {
ret = (ret + dfs(pos - 1, flagX, (R[pos] == 1 ? 1 : flagY), flagZ)) % MOD;
}
if ((flagX || L[pos] == 0) && (flagY || R[pos] == 1) && flagZ) {
ret = (ret + dfs(pos - 1, flagX, flagY, flagZ)) % MOD;
}
if (flagY || R[pos] == 1) {
ret = (ret + dfs(pos - 1, (L[pos] == 0 ? 1 : flagX), flagY, 1)) % MOD;
}
return memo[pos][flagX][flagY][flagZ] = ret;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
ll l, r;
cin >> l >> r;
L = bitset<64>(l);
R = bitset<64>(r);
for (int i = 0; i < 60; ++i) {
for (int j = 0; j < 2; ++j) {
for (int k = 0; k < 2; ++k) {
for (int l = 0; l < 2; ++l) {
memo[i][j][k][l] = -1;
}
}
}
}
cout << dfs(59, 0, 0, 0) << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <cstdio>
#include <iostream>
using namespace std;
#define Mod 1000000007
#define ll long long
ll l,r;
ll p[130];
int main(){
cin>>l>>r;
p[0]=1;
for(int i=1;i<=120;i++){
p[i]=p[i-1]*3%Mod;
}
for(int i=62;i>=0;i--){
if((((1ll<<i)&l)^((1ll<<i)&r))!=0){
break;
}
if(l&(1ll<<i)){
l^=(1ll<<i);
r^=(1ll<<i);
}
}
ll ans=0;
ll high,low;
for(int i=62;i>=0;i--){
if((1ll<<i)&r){
high=i;
break;
}
}
for(int i=62;i>=0;i--){
if((1ll<<i)&l){
low=i;
break;
}
}
if(l==r&&l==0){
puts("1");
return 0;
}
if(l==0){
ll now=1;
for(int i=high;i>=0;i--){
if((1ll<<i)&r){
ans+=p[i]*now%Mod;
ans%=Mod;
now<<=1;
now%=Mod;
}
}
ans+=now;
ans%=Mod;
cout<<ans<<endl;
return 0;
}
ll now=1;
for(int i=high-1;i>=0;i--){
if((1ll<<i)&r){
ans+=p[i]*now%Mod;
ans%=Mod;
now<<=1;
now%=Mod;
}
}
ans+=now;
ans%=Mod;
now=1;
for(int i=low-1;i>=0;i--){
if(!((1ll<<i)&l)){
ans+=p[i]*now%Mod;
ans%=Mod;
now<<=1;
now%=Mod;
}
}
ans+=now;
ans%=Mod;
for(int i=low+1;i<high;i++){
ans+=p[i];
ans%=Mod;
}
cout<<ans<<endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <algorithm>
long long int l, r;
long long int mod = 1e9 + 7;
long long int memo[65][3][3][3] = { 0 };
int getBit(long long int nr, long long int pos)
{
return ((1LL << pos) & nr) > 0;
}
long long int f(int pos, int flagx, int flagy, int msb)
{
if (pos == -1)
return 1;
if (memo[pos][flagx][flagy][msb] != -1)
return memo[pos][flagx][flagy][msb];
long long int cur = 0;
//x0 y0
int lbit = getBit(l, pos);
int rbit = getBit(r, pos);
if (flagx || lbit == 0)
cur += f(pos - 1, flagx, rbit ? rbit : flagy, msb);
//x0 y1
if ((flagx || lbit == 0) && (flagy || rbit == 1) && msb)
cur += f(pos - 1, flagx, flagy, msb);
//x1 y1
if (flagy || rbit == 1)
cur += f(pos - 1, !lbit ? 1 : flagx, flagy, 1);
cur %= mod;
memo[pos][flagx][flagy][msb] = cur;
return cur;
}
int main()
{
std::cin >> l >> r;
for (int i = 0; i < 64; ++i)
{
for (int j = 0; j < 2; ++j)
{
for (int k = 0; k < 2; ++k)
{
for (int q = 0; q < 2; ++q)
memo[i][j][k][q] = -1;
}
}
}
std::cout << f(62, 0, 0, 0);
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include"bits/stdc++.h"
using namespace std;
using ll = int64_t;
ll L, R;
constexpr ll MAX_BIT = 64;
constexpr ll MOD = 1e9 + 7;
bitset<MAX_BIT> L_B, R_B;
map<tuple<ll, bool, bool, bool>, pair<bool, ll>> memo;
ll solve(ll i, bool X_ge_L, bool Y_le_R, bool appear_LSB) {
if (i == MAX_BIT) {
return 1;
}
auto curr_tuple = make_tuple(i, X_ge_L, Y_le_R, appear_LSB);
if (memo[curr_tuple].first) {
return memo[curr_tuple].second;
}
memo[curr_tuple].first = true;
ll result = 0;
//(1)x 0, y 0とする
// こうすることができるのはすでにx >= Lが確定している場合か、Lの対応位置が0である場合
if (X_ge_L || !L_B[MAX_BIT - 1 - i]) {
//Rの対応位置が1だったらy <= Rが確定する
(result += solve(i + 1, X_ge_L, R_B[MAX_BIT - 1 - i] || Y_le_R, appear_LSB)) %= MOD;
}
//(2)x 0, y 1とする
// こうすることができるのはすでにx >= Lが確定している場合か、Lの対応位置が0である場合 かつ
// すでにy <= Rが確定している場合か、Rの対応位置が1である場合 かつ
// すでにx 1, y 1となる部分が最左ビットとして現れている
if ((X_ge_L || !L_B[MAX_BIT - 1 - i]) && (Y_le_R || R_B[MAX_BIT - 1 - i]) && appear_LSB) {
(result += solve(i + 1, X_ge_L, Y_le_R, appear_LSB)) %= MOD;
}
//(3)x 1, y 1とする
// こうすることができるのはすでにy <= Rが確定している場合か、Rの対応位置が1である場合
if (Y_le_R || R_B[MAX_BIT - 1 - i]) {
//x 1, y 1という部分が現れたとしてappear_LSBを立てる
(result += solve(i + 1, !L_B[MAX_BIT - 1 - i] || X_ge_L, Y_le_R, true)) %= MOD;
}
//制約の関係からx 1, y 0となることはないので以上で全部
return memo[curr_tuple].second = result;
}
int main() {
cin >> L >> R;
L_B = bitset<MAX_BIT>(L);
R_B = bitset<MAX_BIT>(R);
cout << solve(0, false, false, false) << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int i = 0; i < (int)(n); i++)
uint64_t p3[100];
uint64_t mod = 1000000007;
map<pair<uint64_t, uint64_t>, uint64_t> memo;
uint64_t solve(uint64_t l, uint64_t r, int b){
if(l > r) return 0;
if(l == r) return 1;
if(l == 0 && r == (1ull << b) - 1){
return p3[b];
}
auto it = memo.find(make_pair(l, r));
if(it != memo.end()){
return it->second;
}
uint64_t h = 1ull << (b - 1);
uint64_t ret = 0;
// 0, 0
if(l < h){
ret += solve(l, min(r, h - 1), b - 1);
}
// 0, 1
if(l < h && r >= h){
ret += solve(l, r - h, b - 1);
}
// 1, 1
if(r >= h){
ret += solve(max(l, h) - h, r - h, b - 1);
}
ret %= mod;
memo[make_pair(l, r)] = ret;
return ret;
}
int main(){
uint64_t l, r;
cin >> l >> r;
p3[0] = 1;
REP(i, 100) if(i > 0){
p3[i] = p3[i - 1] * 3 % mod;
}
uint64_t ret = 0;
REP(b, 62){
uint64_t m = 1ull << b;
uint64_t n = m * 2 - 1;
if(l <= n && r >= m){
ret += solve(max(l, m) - m, min(r, n) - m, b);
ret %= mod;
}
}
cout << ret << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <vector>
#include <iostream>
#include <bitset>
typedef long long int ll;
using namespace std;
ll l, r;
bitset<60> L, R;
ll memo[60][2][2][2];
void init(ll l, ll r){
L = bitset<60> (l);
R = bitset<60> (r);
//std::cout << L << '\n';
//std::cout << R << '\n';
for(int i=0;i<60;i++) for(int j=0;j<2;j++) for(int k=0;k<2;k++)for(int t=0;t<2;t++) memo[i][j][k][t] = -1;
}
ll f(ll pos, int flagX, int flagY, int flagZ){
if(pos==-1) return 1;
if(memo[pos][flagX][flagY][flagZ]!=-1) return memo[pos][flagX][flagY][flagZ];
ll ret = 0;
if(flagX||!L[pos]) ret += f(pos-1, flagX, (R[pos]?1:flagY), flagZ);
if((flagX||!L[pos])&&(flagY||R[pos])&&flagZ) ret += f(pos-1, flagX, flagY, flagZ);
if(flagY||R[pos]) ret += f(pos-1, (!L[pos]?1:flagX), flagY, 1);
ret %= 1000000007;
return memo[pos][flagX][flagY][flagZ] = ret;
}
int main(int argc, char const *argv[]) {
std::cin >> l >> r;
init(l, r);
std::cout << (f(59, 0, 0, 0)) << '\n';
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using Int = int_fast64_t;
constexpr Int mod = 1e9+7;
const vector<vector<vector<Int>>> a = {
{
{1, 0, 0, 0},
{0, 1, 0, 0},
{0, 0, 2, 0},
{0, 0, 1, 3}
},
{
{1, 0, 0, 0},
{1, 2, 0, 0},
{1, 0, 2, 0},
{0, 1, 1, 3}
},
{
{0, 0, 0, 0},
{0, 1, 0, 0},
{0, 0, 1, 0},
{0, 0, 0, 3}
},
{
{1, 0, 0, 0},
{0, 2, 0, 0},
{0, 0, 1, 0},
{0, 1, 0, 3}
}
};
Int l, r;
vector<vector<Int>> dp;
vector<Int> p(vector<vector<Int>> b, vector<Int> v){
vector<Int> res(v.size(), 0);
for(size_t i=0; i<v.size(); ++i)
for(size_t j=0; j<b[i].size(); ++j)
res[i] = (res[i] + (b[i][j] * v[j]) % mod) % mod;
return res;
}
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
cin >> l >> r;
Int cl = 0;
while((l >> cl) > 0) ++cl;
Int cr = 0;
while((r >> cr) > 0) ++cr;
// cerr << cr << " " << cl << "\n";
dp.resize(cr, vector<Int>(4, 0));
if(cl == cr){
dp[0][0] = 1;
}else{
dp[0][1] = 1;
for(size_t i=1; (Int)i<cr-cl; ++i) dp[i][3] = 1;
dp[cr-cl][2] = 1;
}
// for(size_t i=0; i<dp.size(); ++i)
// for(size_t j=0; j<4; ++j)
// cerr << dp[i][j] << " \n"[j==3];
for(size_t i=1; i<dp.size(); ++i){
size_t j = dp.size() - i - 1;
size_t k = ((l >> j) & 1) * 2 + ((r >> j) & 1);
// cerr << j << " " << k << "\n";
vector<Int> t = p(a[k], dp[i-1]);
for(size_t l=0; l<4; ++l) dp[i][l] = (dp[i][l] + t[l]) % mod;
}
Int ans = 0;
for(auto i:dp.back()) ans = (ans + i) % mod;
cout << ans << "\n";
// for(size_t i=0; i<dp.size(); ++i)
// for(size_t j=0; j<4; ++j)
// cerr << dp[i][j] << " \n"[j==3];
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
using namespace std;
long long L, R;
long long dp[65][2][2][2][2];
long long mod = 1000000007;
long long solve(int pos, int bit1, int gir1, int gir2, int gir3) {
// pos : 現在の桁数
// bit1 : 大きい方の前のビット
// gir1 : x がギリギリかどうか
// gir2 : y がギリギリかどうか
// gir3 : y <= 2x がギリギリかどうか
if (pos == -1) {
if (bit1 == 1 && gir3 == 1) return 0;
return 1;
}
if (dp[pos][bit1][gir1][gir2][gir3] >= 1) return dp[pos][bit1][gir1][gir2][gir3] - 1LL;
long long ret = 0;
// ケース 1 : 両方 0 にするパターン
if ((gir1 == 0 || (L & (1LL << pos)) == 0) && (gir3 == 0 || bit1 == 0)) {
long long c1 = pos - 1;
long long c2 = 0;
long long c3 = gir1;
long long c4 = gir2; if ((R & (1LL << pos)) != 0) c4 = 0;
long long c5 = gir3;
ret += solve(c1, c2, c3, c4, c5);
ret %= mod;
}
// ケース 1 : 両方 1 にするパターン
if ((gir2 == 0 || (R & (1LL << pos)) != 0)) {
long long c1 = pos - 1;
long long c2 = 1;
long long c3 = gir1; if ((L & (1LL << pos)) == 0) c3 = 0;
long long c4 = gir2;
long long c5 = gir3; if (bit1 == 0) c5 = 0;
ret += solve(c1, c2, c3, c4, c5);
ret %= mod;
}
// ケース 3 : 小さい方を 0、大きい方を 1 にするパターン
if ((gir1 == 0 || (L & (1LL << pos)) == 0) && (gir2 == 0 || (R & (1LL << pos)) != 0) && (gir3 == 0 || bit1 == 0)) {
long long c1 = pos - 1;
long long c2 = 1;
long long c3 = gir1;
long long c4 = gir2;
long long c5 = gir3;
ret += solve(c1, c2, c3, c4, c5);
ret %= mod;
}
dp[pos][bit1][gir1][gir2][gir3] = ret + 1;
return ret;
}
int main() {
cin >> L >> R;
long long ans = solve(60, 0, 1, 1, 1);
cout << ans << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define ts cout<<"ok"<<endl
#define int long long
#define hh puts("")
#define pc putchar
#define mo 1000000007
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
//char buf[1<<21],*p1=buf,*p2=buf;
using namespace std;
const int N=65;
int l,r,lenl,lenr,pl[N],pr[N],dp[N][2][2],ans;
inline int read(){
int ret=0,ff=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-') ff=-1;ch=getchar();}
while(isdigit(ch)){ret=ret*10+(ch^48);ch=getchar();}
return ret*ff;
}
void write(int x){if(x<0){x=-x,pc('-');}if(x>9) write(x/10);pc(x%10+48);}
void writeln(int x){write(x),hh;}
void writesp(int x){write(x),pc(' ');}
int dfs(int pos,int x1,int x2){
if(!pos) return 1;
if(dp[pos][x1][x2]!=-1) return dp[pos][x1][x2];
int &res=dp[pos][x1][x2];res=0;
int t1=x1?pl[pos]:0,t2=x2?pr[pos]:1;
for(int y=t1;y<=t2;y++)
for(int x=t1;x<=y;x++)
res=(res+dfs(pos-1,x1&(x==t1),x2&(y==t2)))%mo;
return res;
}
signed main(){
memset(dp,-1,sizeof(dp));
l=read(),r=read();
while(l){
pl[++lenl]=l&1;
l>>=1;
}
while(r){
pr[++lenr]=r&1;
r>>=1;
}
for(int i=lenl;i<=lenr;i++) ans=(ans+dfs(i-1,i==lenl,i==lenr))%mo;//枚举哪个最高位为1
write(ans);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
const LL mod=1000000007;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
LL l,r;
LL dp[100][2][2][2];
LL rec(int k,bool yr,bool xl,bool f){
if(k==-1) return f;
if(~dp[k][yr][xl][f]) return dp[k][yr][xl][f];
LL ret=0;
int y=yr?(r>>k&1):1;
int x=xl?(l>>k&1):0;
if(!f){
ret = (ret + rec(k-1,yr,x&xl,true))%mod;
if(x==0) ret = (ret + rec(k-1,false,xl,false))%mod;
}
else{
for (int i = 0; i <= y; i++) {
if(i==0){
if(x==0){
ret = (ret + rec(k-1,(y==i)&yr,xl,f))%mod;
}
}
else{
if(x==1){
ret = (ret + rec(k-1,y&yr,xl,f))%mod;
}
else{
ret = (ret + rec(k-1,y&yr,xl,f))%mod;
ret = (ret + rec(k-1,y&yr,false,f))%mod;
}
}
}
}
return dp[k][yr][xl][f] = ret;
}
int main(){
cin >> l >> r;
memset(dp,-1,sizeof(dp));
LL t = r;
int s=0;
while(t){
s++;
t/=2;
}
cout << rec(s-1,true,true,false) << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
mod = 10 ** 9 +7
l = '{:060b}'.format(L)
r = '{:060b}'.format(R)
memo = [[[[-1]*2 for i in range(2)] for j in range(2)] for k in range(60)]
def f(pos, flagL, flagR, flagM):
if pos == 60:
return 1
if memo[pos][flagL][flagR][flagM] != -1:
return memo[pos][flagL][flagR][flagM]
ret = 0
#(0, 0)
if flagL or l[pos] == '0':
ret+=f(pos+1, flagL, 1 if r[pos]=='1' else flagR, flagM)
#(0, 1)
if flagM and (flagL or l[pos]=='0') and (flagR or r[pos]=='1'):
ret += f(pos+1, flagL, flagR, flagM)
#(1, 1)
if flagR or r[pos]=='1':
ret += f(pos+1, 1 if l[pos]=='0' else flagL, flagR, 1)
memo[pos][flagL][flagR][flagM] = ret%mod
return ret%mod
print(f(0, 0, 0, 0)%mod) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define ts cout<<"ok"<<endl
#define int long long
#define hh puts("")
#define pc putchar
#define mo 1000000007
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
//char buf[1<<21],*p1=buf,*p2=buf;
using namespace std;
const int N=65;
int l,r,lenl,lenr,pl[N],pr[N],dp[N][2][2],ans;
inline int read(){
int ret=0,ff=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-') ff=-1;ch=getchar();}
while(isdigit(ch)){ret=ret*10+(ch^48);ch=getchar();}
return ret*ff;
}
void write(int x){if(x<0){x=-x,pc('-');}if(x>9) write(x/10);pc(x%10+48);}
void writeln(int x){write(x),hh;}
void writesp(int x){write(x),pc(' ');}
//问题可以转化成y-x=y^x的对数,那么y二进制下为1,x为0或1,y二进制下为0,x必为0
int dfs(int pos,int x1,int x2){
if(!pos) return 1;
if(dp[pos][x1][x2]!=-1) return dp[pos][x1][x2];
int &res=dp[pos][x1][x2];res=0;
int t1=x1?pl[pos]:0,t2=x2?pr[pos]:1;
for(int y=t1;y<=t2;y++)
for(int x=t1;x<=y;x++)
res=(res+dfs(pos-1,x1&(x==t1),x2&(y==t2)))%mo;
return res;
}
signed main(){
memset(dp,-1,sizeof(dp));
l=read(),r=read();
while(l){
pl[++lenl]=l&1;
l>>=1;
}
while(r){
pr[++lenr]=r&1;
r>>=1;
}
for(int i=lenl;i<=lenr;i++) ans=(ans+dfs(i-1,i==lenl,i==lenr))%mo;//枚举哪个最高位为1
write(ans);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;
ll dp[62][2][2][2];
ll solve(ll L,ll R){
dp[61][0][0][0]=1;
for(int i=60;i>=0;i--)rep(j,2)rep(k,2)rep(t,2){
int bit=R>>i&1;
int lim=(k?1:bit);
for(int l=0;l<=lim;l++){
int bit2=L>>i&1;
int lim2=(t?0:bit2);
if(l==1&&j==0)lim2=1;
for(int m=lim2;m<=l;m++){
(dp[i][j||l][k||l<bit][t||bit2<m]+=dp[i+1][j][k][t])%=MOD;
}
}
}
ll ans=0;
rep(j,2)rep(k,2)rep(t,2){
(ans+=dp[0][j][k][t])%=MOD;
}
return ans;
}
int main(){
ll l,r;cin>>l>>r;
cout<<solve(l,r)<<endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
using namespace std;
template <int32_t MOD>
struct mint {
int32_t value;
mint() : value() {}
mint(int64_t value_) : value(value_ < 0 ? value_ % MOD + MOD : value_ >= MOD ? value_ % MOD : value_) {}
inline mint<MOD> operator + (mint<MOD> other) const { int32_t c = this->value + other.value; return mint<MOD>(c >= MOD ? c - MOD : c); }
inline mint<MOD> & operator += (mint<MOD> other) { this->value += other.value; if (this->value >= MOD) this->value -= MOD; return *this; }
};
int64_t msb(int64_t n) {
if (not n) return 0;
return 1ull << (63 - __builtin_clzll(n));
}
constexpr int MOD = 1e9 + 7;
mint<MOD> solve(int64_t l, int64_t r) {
array<array<mint<MOD>, 2>, 2> cur = {};
for (int64_t i = 1ull << 62; i; i >>= 1) {
array<array<mint<MOD>, 2>, 2> prv = cur;
cur = {};
if (msb(l) <= i and i <= msb(r)) {
cur[i == msb(l)][i == msb(r)] += 1;
}
REP (p, 2) REP (q, 2) REP (xy, 3) {
bool x = (xy >= 2);
bool y = (xy >= 1);
if (p and (l & i) and not x) continue;
if (q and not (r & i) and y) continue;
bool np = p and ((bool)(l & i) == x);
bool nq = q and ((bool)(r & i) == y);
cur[np][nq] += prv[p][q];
}
}
mint<MOD> cnt = 0;
REP (p, 2) REP (q, 2) {
cnt += cur[p][q];
}
return cnt;
}
int main() {
int64_t l, r; cin >> l >> r;
cout << solve(l, r).value << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
MOD = 10**9 + 7
*BL, = map(int, bin(L)[2:])
*BR, = map(int, bin(R)[2:])
a = len(BL)
b = len(BR)
BL = [0]*(len(BR) - len(BL)) + BL
BL.reverse()
BR.reverse()
memo = [[[-1]*2 for i in range(2)] for j in range(70)]
def dfs(i, p, q):
if memo[i][p][q] != -1:
return memo[i][p][q]
if i == -1:
memo[i][p][q] = 1
return 1
r = 0
if p == q == 0:
if BL[i] == 1 and BR[i] == 0:
memo[i][0][0] = 0
return 0
if BL[i] == 1 or BR[i] == 0:
r = dfs(i-1, 0, 0)
else:
r = (dfs(i-1, 0, 1) + dfs(i-1, 1, 0) + dfs(i-1, 0, 0)) % MOD
elif p == 0:
if BL[i] == 1:
r = dfs(i-1, 0, 1)
else:
r = (2*dfs(i-1, 0, 1) + dfs(i-1, 1, 1)) % MOD
elif q == 0:
if BR[i] == 0:
r = dfs(i-1, 1, 0)
else:
r = (2*dfs(i-1, 1, 0) + dfs(i-1, 1, 1)) % MOD
else:
r = 3 * dfs(i-1, 1, 1)
memo[i][p][q] = r
return r
ans = 0
for l in range(a-1, b):
ans += dfs(l-1, +(not a-1 == l), +(not b-1 == l))
print(ans % MOD) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
#define Mod 1000000007
LL l, r;
int wr[100], wl[100], Dp[100][8];
inline int Inc(int u, int v)
{
return u + v - (u + v >= Mod ? Mod : 0);
}
bool CheckR(int state, int w, int br)
{
return (state & 1) || (br <= wr[w]);
}
bool CheckL(int state, int w, int bl)
{
return (state & 2) || (bl >= wl[w]);
}
bool CheckS(int state, int bl, int br)
{
return (state & 4) || (bl == br);
}
int Transfer(int state, int w, int bl, int br)
{
int res = 0;
if ((state & 1) || br < wr[w])
res |= 1;
if ((state & 2) || bl > wl[w])
res |= 2;
if ((state & 4) || (bl == 1))
res |= 4;
return res;
}
int Solve()
{
int sz = 0;
for (LL y = r; y; y >>= 1)
wr[++ sz] = (y & 1);
for (int i = 1; i <= sz; i ++)
wl[i] = (l >> (i - 1) & 1);
Dp[sz + 1][0] = 1;
for (int w = sz + 1; w; w --)
for (int state = 0; state < 8; state ++)
{
if (!Dp[w][state]) continue ;
for (int br = 0; br < 2; br ++)
for (int bl = 0; bl <= br; bl ++)
{
if (!CheckR(state, w - 1, br) || !CheckL(state, w - 1, bl) || !CheckS(state, bl, br))
continue ;
int _state = Transfer(state, w - 1, bl, br);
Dp[w - 1][_state] = Inc(Dp[w - 1][_state], Dp[w][state]);
}
}
int res = 0;
for (int state = 0; state < 8; state ++)
res = Inc(res, Dp[1][state]);
return res;
}
int main()
{
scanf("%lld%lld", &l, &r);
printf("%d\n", Solve());
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | def solve(L,R):
n=R.bit_length()
dp=[[0 for i in range(8)] for i in range(n)]
if R&1==1 and L&1==1:
dp[0][0]=1
dp[0][1]=1
dp[0][2]=2
dp[0][3]=3
dp[0][4]=1
dp[0][5]=1
dp[0][6]=2
dp[0][7]=3
elif R&1==1 and L&1==0:
dp[0][0]=2
dp[0][1]=3
dp[0][2]=2
dp[0][3]=3
dp[0][4]=2
dp[0][5]=3
dp[0][6]=2
dp[0][7]=3
elif R&1==0 and L&1==1:
dp[0][0]=0
dp[0][1]=0
dp[0][2]=1
dp[0][3]=1
dp[0][4]=1
dp[0][5]=1
dp[0][6]=2
dp[0][7]=3
else:
dp[0][0]=1
dp[0][1]=1
dp[0][2]=1
dp[0][3]=1
dp[0][4]=2
dp[0][5]=3
dp[0][6]=2
dp[0][7]=3
for i in range(1,n):
for j in range(8):
if j==0:
if R>>i &1==1 and L>>i &1==1:
dp[i][j]=dp[i-1][1]
elif R>>i &1==1 and L>>i &1==0:
dp[i][j]=dp[i-1][4]+dp[i-1][3]
elif R>>i&1 ==0 and L>>i&1 ==0:
dp[i][j]=dp[i-1][0]
elif j==1:
if R>>i &1==1 and L>>i &1==1:
dp[i][j]=dp[i-1][1]
elif R>>i &1==1 and L>>i &1==0:
dp[i][j]=dp[i-1][5]+dp[i-1][3]+dp[i-1][1]
elif R>>i&1==0 and L>>i&1==0:
dp[i][j]=dp[i-1][1]
elif j==2:
if R>>i&1==1:
dp[i][j]=dp[i-1][6]+dp[i-1][3]
else:
dp[i][j]=dp[i-1][2]
elif j==3:
if R>>i&1==1:
dp[i][j]=dp[i-1][7]+2*dp[i-1][3]
else:
dp[i][j]=dp[i-1][3]
elif j==4:
if L>>i&1==1:
dp[i][j]=dp[i-1][5]
else:
dp[i][j]=dp[i-1][7]+dp[i-1][4]
elif j==5:
if L>>i&1==1:
dp[i][j]=dp[i-1][5]
else:
dp[i][j]=2*dp[i-1][5]+dp[i-1][7]
elif j==6:
dp[i][j]=dp[i-1][6]+dp[i-1][7]
else:
dp[i][j]=3*dp[i-1][7]
return dp[-1][0]
mod=10**9+7
L,R=map(int,input().split())
print(solve(L,R)%mod) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import sys
import math
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
def main():
md = 10 ** 9 + 7
l, r = map(int, input().split())
n = int(math.log2(r)) + 1
# dp[i][t][f][g] i桁目まで見て最上位が決定済みか(t)、L<xが確定(f)、y<Rが確定
dp = [[[[0] * 2 for _ in range(2)] for __ in range(2)] for ___ in range(n + 1)]
dp[0][0][0][0] = 1
ans = 0
for i in range(n):
lk = l >> (n - 1 - i) & 1
rk = r >> (n - 1 - i) & 1
for t in range(2):
for f in range(2):
for g in range(2):
pre = dp[i][t][f][g]
for x, y in [(0, 0), (0, 1), (1, 1)]:
nt, nf, ng = t, f, g
if t == 0 and (x, y) == (0, 1): continue
if f == 0 and lk > x: continue
if g == 0 and y > rk: continue
if (x, y) == (1, 1): nt = 1
if lk == 0 and x == 1: nf = 1
if y == 0 and rk == 1: ng = 1
if i == n - 1:
ans = (ans + pre) % md
else:
dp[i + 1][nt][nf][ng] = (dp[i + 1][nt][nf][ng] + pre) % md
print(ans)
main()
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <memory.h>
#include <iostream>
#define ll long long
#define N 63
#define mod 1000000007
int dp[N][2][2][2];
ll L, R;
int add(int x, int y) {
int ret = (x + y);
if (ret >= mod) {
ret -= mod;
}
return ret;
}
int solveDp(int ind, int msb, int xeq, int yeq) {
if (ind < 0) {
if (msb) {
return 1;
}
return 0;
}
int &ret = dp[ind][msb][xeq][yeq];
if (ret != -1) {
return ret;
}
ret = 0;
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
if (i > j) {
continue;
}
if (xeq && i == 0 && L&(1LL << ind)) {
continue;
}
if (yeq && j == 1 && (R&(1LL << ind)) == 0)
{
continue;
}
int nxeq, nyeq;
nxeq = xeq;
nyeq = yeq;
if (xeq && i == 1 && (L&(1LL << ind)) ==0 ) {
nxeq = 0;
}
if (yeq && j == 0 && (R&(1LL << ind))) {
nyeq = 0;
}
if (msb == 0) {
if (i != j) {
continue;
}
if (i == 1) {
ret = add(ret, solveDp(ind - 1, 1, nxeq, nyeq));
}
else {
ret = add(ret, solveDp(ind - 1, 0, nxeq, nyeq));
}
}
else {
ret = add(ret, solveDp(ind - 1, 1, nxeq, nyeq));
}
}
}
return ret;
}
void solve() {
scanf("%lld %lld", &L, &R);
memset(dp, -1, sizeof(dp));
printf("%d\n", solveDp(N-1, 0, 1, 1));
}
int main()
{
//freopen("input.txt", "r", stdin);
solve();
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define SZ(x) (int)(x.size())
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
const double eps = 1e-10;
const int MOD = 1000000007;
const int INF = 1000000000;
const ll LINF = 1ll<<50;
template<typename T>
void printv(const vector<T>& s) {
for(int i=0;i<(int)(s.size());++i) {
cout << s[i];
if(i == (int)(s.size())-1) cout << endl;
else cout << " ";
}
}
ll l, r;
ll dp[2][2][2][62];
ll f(int d, int msb, int lt, int rt) {
if(d == -1) return msb == 1;
if(dp[msb][lt][rt][d] >= 0) return dp[msb][lt][rt][d];
ll ret = 0;
for(int a=0;a<2;++a) {
for(int b=0;b<2;++b) {
if(a == 0 && lt && (l & (1ll<<d)) > 0) continue;
if(b == 1 && rt && (r & (1ll<<d)) == 0) continue;
if(a>b) continue;
if(msb == 0 && a != b) continue;
int nlt = lt && (((l>>d)&1)==a);
int nrt = rt && (((r>>d)&1)==b);
int nmsb = msb || (a == 1 && b == 1);
ret += f(d-1, nmsb, nlt, nrt);
}
}
return dp[msb][lt][rt][d] = ret % MOD;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(10);
cin >> l >> r;
for(int i=0;i<2;++i) {
for(int j=0;j<2;++j) {
for(int k=0;k<2;++k) {
for(int l=0;l<62;++l) {
dp[i][j][k][l] = -1;
}
}
}
}
cout << f(61, 0, 1, 1) << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int dp[60][2][2][2] = {};
long long L, R;
void add(int &x, int y) {
x = (x + y) % MOD;
}
bool bit(long long X, int i) {
return (X >> i & 1);
}
int f(int pos, int x, int y, int s) {
if(pos == -1) return 1;
if(dp[pos][x][y][s] != -1) return dp[pos][x][y][s];
int ret = 0;
if(x || !bit(L, pos)) add(ret, f(pos - 1, x, (bit(R, pos) ? 1 : y), s));
if((x || !bit(L, pos)) && (y || bit(R, pos)) && s) add(ret, f(pos - 1, x, y, s));
if(y || bit(R, pos)) add(ret, f(pos - 1, (!bit(L, pos) ? 1 : x), y, 1));
dp[pos][x][y][s] = ret;
return ret;
}
int main() {
cin >> L >> R;
memset(dp, -1, sizeof(dp));
cout << f(59, 0, 0, 0) << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define dbug(x) cout<<#x<<"="<<x<<endl
using namespace std;
template <typename T> void read(T &t) {
t=0; char ch=getchar(); int f=1;
while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll mod=(1e9)+7;
int mx=63;
int d[70],e[70];
ll dp[70][2][2][3];
void update(ll &x,ll y) {
x+=y; x%=mod;
}
ll solve(ll L,ll R) {
for (int i=0;i<mx;i++) {
if (R&(1LL<<i)) d[i]=1; else d[i]=0;
if (L&(1LL<<i)) e[i]=1; else e[i]=0;
}
//for (int i=0;i<mx;i++) printf("%d",d[i]); printf("\n");
//for (int i=0;i<mx;i++) printf("%d",e[i]); printf("\n");
memset(dp,0,sizeof(dp));
dp[mx][1][1][0]=1; int C;
for (int i=mx;i>=1;i--) {
for (int j=0;j<=1;j++)
for (int k=0;k<=1;k++)
for (int B=0;B<=2;B++) {
//printf("%d %d %d %d %lld\n",i,j,k,B,dp[i][j][k][B]);
for (int a=0;a<=1;a++)
for (int b=0;b<=1;b++) {
if (j&&a>d[i-1]) continue;
if (k&&b>e[i-1]) continue;
if (a<b) continue;
if (B==2) C=2;
else {
if (b||B) C=1;
else if (a-b) C=2;
else C=0;
}
update(dp[i-1][j&(a==d[i-1])][k&(b==e[i-1])][C],dp[i][j][k][B]);
}
}
}
ll res=0;
for (int j=0;j<=1;j++)
for (int k=0;k<=1;k++)
update(res,dp[0][j][k][1]);
//printf("%lld\n",res);
return res;
}
int main() {
//freopen("1.txt","r",stdin);
//freopen("1.out","w",stdout);
ll L,R; read(L); read(R);
ll ans=solve(R,R)-solve(L-1,R);
ans=(ans%mod+mod)%mod;
printf("%lld\n",ans);
/*ans=0;
for (int x=1;x<=R;x++)
for (int y=x;y<=R;y++)
if (y%x==(x^y)) ans++;
printf("%lld\n",ans);*/
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L, R = map(int, input().split())
MOD = 10 ** 9 + 7
l = '{:060b}'.format(L)[::-1] # 0左詰の二進数６０桁のstr を逆向きにスライス
r = '{:060b}'.format(R)[::-1] # (桁数60 は R<=10**18<2**60 より)
memo = [[[[-1 for l in range(2)] for k in range(2)] for j in range(2)] for i in range(60)]
# flagZは、既にx＝y＝1の位があったかチェックしている(MSB)
def f(pos, flagX, flagY, flagZ):
if pos == -1:
return 1
if memo[pos][flagX][flagY][flagZ] != -1:
return memo[pos][flagX][flagY][flagZ]
ret = 0
# x 0, y 0
if flagX or l[pos] == '0':
ret += f(pos - 1, flagX, 1 if r[pos] == '1' else flagY, flagZ)
# x 0, y 1
if (flagX or l[pos] == '0') and (flagY or r[pos] == '1') and flagZ:
ret += f(pos - 1, flagX, flagY, flagZ)
# x 1, y 1
if flagY or r[pos] == '1':
ret += f(pos - 1, 1 if l[pos] == '0' else flagX, flagY, 1)
ret %= MOD
memo[pos][flagX][flagY][flagZ] = ret
return ret
ans = f(59, 0, 0, 0) #最大桁からスタート
print(ans)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define Mo(a) (a)>=mo?(a)-=mo:(a)
long long l,r;
const int mo=1e9+7;
int Mem[70][2][2][2],len,R[100],cnt,a,ans,L[100];
long long le;
int DFS(int l,int mark,int h,int f) {//mark= <l h= >0 f= <r
// printf("%d %d %d %d %d %d\n",l,mark,h,f,f?R[l]:1,mark?L[l]:0);
if(l==0)return h;
if(~Mem[l][mark][h][f])return Mem[l][mark][h][f];
int re=0,ma=f?R[l]:1,mi=mark?L[l]:0;
if(ma) {
if(mi==0&&h) re+=DFS(l-1,mark&&mi==0,h,f&&ma==1);
re+=DFS(l-1,mark&&mi==1,1,f&&ma==1);
Mo(re);
}
if(mi==0) {
re+=DFS(l-1,mark&&mi==0,h,f&&ma==0);
Mo(re);
}
Mem[l][mark][h][f]=re;
return re;
}
int DO() {
len=0;
while(r)R[++len]=r&1,r>>=1;
le=1;
for(int i=1; i<=len; i++,le<<=1) L[i]=((l&le)>0);
return DFS(len,1,0,1);
}
int main() {
memset(Mem,-1,sizeof(Mem));
scanf("%lld %lld",&l,&r);
printf("%d",DO());
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
long long l,r,ans,dp[65][2][2];
long long solve(int idx,bool f1,bool f2)
{
if (idx<0)
return 1;
if (dp[idx][f1][f2]!=-1)
return dp[idx][f1][f2];
long long ret=0;
bool b1=(r&(1LL<<idx)),b2=(l&(1LL<<idx));
for (int i=0;i<2;i++)
{
if (f1 && i==1 && !b1)
continue;
for (int j=0;j<=i;j++)
{
if (f2 && !j && b2)
continue;
ret=(ret+solve(idx-1,(f1 && i==b1),(f2 && j==b2)))%mod;
}
}
return dp[idx][f1][f2]=ret;
}
int main()
{
cin >> l >> r;
int j=60;
for (;!(r&(1LL<<j));j--);
for (int i=j;;i--)
{
memset(dp,-1,sizeof(dp));
ans=(ans+solve(i-1,(i==j),(l&(1LL<<i))))%mod;
if (l&(1LL<<i))
break;
}
cout << ans;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
ll l,r;
ll bit[2][65];
ll dp[66][2][2];
ll ans;
void solve(ll l,ll r){
if(l>r)return;
//printf("%lld %lld\n",l,r);
int v=0;
ll tl=l,tr=r;
while(l>0){
bit[0][v]=l%2LL;
v++;
l/=2LL;
}
v=0;
while(r>0){
bit[1][v]=r%2LL;
v++;
r/=2LL;
}
memset(dp,0,sizeof(dp));
if(bit[0][v-1]==0){
dp[v-1][1][0]=1;
}else{
dp[v-1][0][0]=1;
}
for(int i=v-2;i>=0;i--){
for(int j=0;j<2;j++){
for(int k=0;k<2;k++){
for(int a=0;a<2;a++){
if(j==0 && a==0 && bit[0][i]==1)continue;
int nj=j;
if(a==1 && bit[0][i]==0)nj=1;
for(int b=0;b<2;b++){
if(k==0 && b==1 && bit[1][i]==0)continue;
if(a==1 && b==0)continue;
int nk=k;
if(b==0 && bit[1][i]==1)nk=1;
dp[i][nj][nk]+=dp[i+1][j][k];
dp[i][nj][nk]%=MOD;
}
}
}
}
}
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
ans+=dp[0][i][j];
ans%=MOD;
}
}
l=tl;
r=tr;
ll ig=1;
for(int i=0;i<v-1;i++){
ig*=2LL;
}
ig--;
solve(l,ig);
}
int main(void){
scanf("%lld%lld",&l,&r);
if(r==1LL){
printf("1\n");
return 0;
}
solve(l,r);
printf("%lld\n",ans);
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
// https://atcoder.jp/contests/abc138/submissions/7075855 を参考
#define mod 1000000007
// y と x は2進数で同じ桁数である必要がある
// このとき, y % x = y - x
// 繰り下がりが無いことが必要十分
ll L, R;
int dp[61][2][2][2];
// 現在の桁, Lの制約の有無, Rの制約の有無, 桁数が合ってるかどうか
int solve(int i, bool L_Free, bool R_Free, bool flag){
if(i == -1) return 1;
int &res = dp[i][L_Free][R_Free][flag];
if(res != -1) return res;
res = 0;
int lX = 0, rX = 1, lY = 0, rY = 1;
if (! L_Free) lX = (L >> i) & 1;
if (! R_Free) rY = (R >> i) & 1;
for (int x = lX; x <= rX; ++x) {
for (int y = lY; y <= rY; ++y) {
if (x <= y && (flag || x == y) ) {
res += solve(i-1, L_Free | (x > lX), R_Free | (y < rY), flag | (x == 1));
res %= mod;
}
}
}
return res;
}
int main() {
cin >> L >> R;
memset(dp, -1, sizeof dp);
cout << solve(60, 0, 0, 0);
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
#define fo(i,a,b) for((i)=(a);i<=(b);i++)
#define rfo(i,a,b) for((i)=(a);i>=(b);i--)
#define inrange(x,y,z) (((x)>=(y))&&((x)<=(z)))
#define ALL(vec) ((vec).begin(),(vec).end())
#define SOR(vec) sort(ALL(vec))
#define UNI(vec) (vec).erase(unique(ALL(vec)),(vec).end())
using namespace std;
#define tvple pair<int,pair<long long,long long> >
#define make_tvple(x,y,z) make_pair((x),make_pair((y),(z)))
#define mo 1000000007
map<tvple,int> mp;
int p3[70],aans;
long long l,r;
int solve(int d,long long l,long long r){
tvple tmp=make_tvple(d,l,r);
if(mp.count(tmp)) return mp[tmp];
long long s=(1ll<<d);
long long t=(1ll<<(d+1))-1;
if((r<s)||(l>t)) return mp[tmp]=0;
if(l<=s&&t<=r) return mp[tmp]=p3[d];
r=min(r,t);
l=max(l,s);
long long mid=(s+(1ll<<(d-1)));
long long dif=(1ll<<(d-1));
if(r<mid) return mp[tmp]=solve(d-1,l-dif,r-dif);
if(l>=mid) return mp[tmp]=solve(d-1,l-s,r-s);
int re=(solve(d-1,l-dif,r)+solve(d-1,dif,r-s))%mo;
if(l-dif<=r-s) re=(re+solve(d-1,l-dif,r-s))%mo;
return mp[tmp]=re;
}
int main(){
#ifdef FILIN
#ifndef DavidDesktop
freopen(FILIN,"r",stdin);
freopen(FILOUT,"w",stdout);
#endif
#endif
ios::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
p3[0]=1;
for(int i=1;i<=69;i++){
p3[i]=((p3[i-1]+p3[i-1])%mo+p3[i-1])%mo;
}
cin>>l>>r;
for(int i=0;i<=60;i++){
aans=(aans+solve(i,l,r))%mo;
}
cout<<aans<<endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | l,r = map(int,input().split())
lb = l.bit_length()
rb = r.bit_length()
ans = 0
mod = 10**9+7
def ignorel(r):
ret = 0
cntr = 0
rb = r.bit_length()
for i in range(rb-1)[::-1]:
if r&1<<i:
ret += 2**cntr*3**i
cntr += 1
ret += 2**cntr
return ret%mod
def ignorer(l):
cntl = 0
ret = 0
lb = l.bit_length()
for i in range(lb)[::-1]:
if l&1<<i == 0:
ret += 2**cntl*3**i
cntl += 1
ret += 2**cntl
return ret%mod
for i in range(lb+1,rb):
ans += 3**(i-1)
ans %= mod
if lb != rb:
ans += ignorer(l)
ans += ignorel(r)
ans %= mod
print(ans)
else:
def from0(x):
if x == 0:
return 1
elif x == 1:
return 3
xb = x.bit_length()
ret = 3**(xb-1)+2*from0(x-(1<<(xb-1)))
return ret
def calc(l,r):
ret = 0
lb = l.bit_length()
rb = r.bit_length()
if l > r:
return 0
if l == r:
return 1
if l == 2**(lb-1):
return ignorel(r)
elif r == 2**rb-1:
return ignorer(l)
for i in range(lb)[::-1]:
if l&1<<i == 0 and r&1<<i:
ret += from0((1<<i)-1-(l&(1<<i)-1))
ret += from0(r&((1<<i)-1))
ret += calc((l&((1<<i)-1))+(1<<i),(r&((1<<i)-1))+(1<<i))
break
return ret%mod
ans += calc(l,r)
print(ans%mod) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | mod = 10**9 + 7
K = 60
L, R = map(int, input().split())
Rb, Lb = [0], [0]
for i in range(K):
Rb.append((R>>i)%2)
Lb.append((L>>i)%2)
if Rb[-1]:
Rm = i+1
if Lb[-1]:
Lm = i+1
dp =[[1, 1, 1, 1] for _ in range(K+1)]
r = 0
for i in range(1, K+1):
dp[i][0] = dp[i-1][0] * 3 % mod
if Rb[i]:
dp[i][1] = dp[i-1][1] * 2 + dp[i-1][0] % mod
else:
dp[i][1] = dp[i-1][1]
if Lb[i]:
dp[i][2] = dp[i-1][2]
else:
dp[i][2] = dp[i-1][2] * 2 + dp[i-1][0] % mod
if Rb[i] and Lb[i]:
dp[i][3] = dp[i-1][3]
elif Rb[i]:
dp[i][3] = dp[i-1][1] + dp[i-1][2] + dp[i-1][3] % mod
elif Lb[i]:
dp[i][3] = 0
else:
dp[i][3] = dp[i-1][3]
if Rm > i and Lm < i:
r += dp[i-1][0]
elif Rm > i and Lm == i:
# このとき確実にLb[i] == 1
r += dp[i-1][2]
elif Rm == i and Lm < i:
r += dp[i-1][1]
elif Rm == i and Lm == i:
r += dp[i-1][3]
else:
r += 0
r = r % mod
print(r)
| PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | l,r=map(int,input().split())
M=len(bin(r))-2
L=format(l,"b").zfill(M)
R=format(r,"b")
mod=10**9+7
dp=[[0 for l in range(70)] for i in range(5)]
if L[0]=="1":
dp[2][0]=1
else:
dp[3][0]=1
dp[0][0]=1
for i in range(1,M):
if L[i]=="1" and R[i]=="1":
dp[1][i]=(dp[0][i-1]+dp[1][i-1])%mod
dp[2][i]=(dp[2][i-1])%mod
dp[3][i]=(2*dp[3][i-1])%mod
dp[4][i]=(dp[3][i-1]+3*dp[4][i-1])%mod
elif L[i]=="1" and R[i]=="0":
dp[1][i]=dp[1][i-1]+dp[0][i-1]
dp[3][i]=dp[3][i-1]
dp[4][i]=(dp[4][i-1]*3)%mod
elif L[i]=="0" and R[i]=="1":
dp[0][i]=dp[0][i-1]
dp[1][i]=(2*dp[1][i-1]+dp[2][i-1])%mod
dp[2][i]=dp[2][i-1]
dp[3][i]=(dp[2][i-1]+dp[3][i-1]*2)%mod
dp[4][i]=(dp[0][i-1]+dp[1][i-1]+dp[3][i-1]+dp[4][i-1]*3)%mod
else:
dp[0][i]=dp[0][i-1]
dp[1][i]=(dp[1][i-1]*2)%mod
dp[2][i]=dp[2][i-1]
dp[3][i]=(dp[3][i-1])%mod
dp[4][i]=(dp[0][i-1]+dp[1][i-1]+dp[4][i-1]*3)%mod
ans=0
for i in range(5):
ans+=dp[i][M-1]
ans%=mod
print(ans) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | MOD = 10**9 + 7
L, R = [int(item) for item in input().split()]
L_blen = L.bit_length()
R_blen = R.bit_length()
# dp[y loose/tight][x loose/tight][index]
dp = [[[0] * (R_blen+1) for _ in range(2)] for _ in range(2)]
for i in range(R_blen + 1):
L_bit = L & (1 << R_blen - i)
R_bit = R & (1 << R_blen - i)
# Form R's MSB to L's LSB can be the initial bit
curbit_idx = R_blen - i + 1
if curbit_idx <= R_blen and curbit_idx >= L_blen:
dp[R_blen == curbit_idx][L_blen == curbit_idx][i] += 1
# R=0, L=0
if not R_bit and not L_bit:
# y=1, x=1
dp[0][0][i] += dp[0][0][i-1]
dp[0][0][i] += dp[0][1][i-1]
# y=0, x=0
dp[1][1][i] += dp[1][1][i-1]
dp[1][0][i] += dp[1][0][i-1]
dp[0][0][i] += dp[0][0][i-1]
dp[0][1][i] += dp[0][1][i-1]
# y=1, x=0
dp[0][1][i] += dp[0][1][i-1]
dp[0][0][i] += dp[0][0][i-1]
# R=1, L=0
if R_bit and not L_bit:
# y=1, x=1
dp[0][0][i] += dp[0][0][i-1]
dp[1][0][i] += dp[1][0][i-1]
dp[0][0][i] += dp[0][1][i-1]
dp[1][0][i] += dp[1][1][i-1]
# y=0, x=0
dp[0][0][i] += dp[0][0][i-1]
dp[0][0][i] += dp[1][0][i-1]
dp[0][1][i] += dp[0][1][i-1]
dp[0][1][i] += dp[1][1][i-1]
# y=1, x=0
dp[0][0][i] += dp[0][0][i-1]
dp[1][0][i] += dp[1][0][i-1]
dp[0][1][i] += dp[0][1][i-1]
dp[1][1][i] += dp[1][1][i-1]
# R=0, L=1
if not R_bit and L_bit:
# y=1, x=1
dp[0][0][i] += dp[0][0][i-1]
dp[0][1][i] += dp[0][1][i-1]
# y=0, x=0
dp[0][0][i] += dp[0][0][i-1]
dp[1][0][i] += dp[1][0][i-1]
# y=1, x=0
dp[0][0][i] += dp[0][0][i-1]
# R=1, L=1
if R_bit and L_bit:
# y=1, x=1
dp[0][0][i] += dp[0][0][i-1]
dp[1][0][i] += dp[1][0][i-1]
dp[0][1][i] += dp[0][1][i-1]
dp[1][1][i] += dp[1][1][i-1]
# y=0, x=0
dp[0][0][i] += dp[1][0][i-1]
dp[0][0][i] += dp[0][0][i-1]
# y=1, x=0
dp[1][0][i] += dp[1][0][i-1]
dp[0][0][i] += dp[0][0][i-1]
# Take MOD
for i in range(2):
for j in range(2):
dp[i][j][i] %= MOD
ans = 0
for i in range(2):
for j in range(2):
ans += dp[i][j][-1]
ans %= MOD
print(ans) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define int long long
const int MAX = 1000000007;
vector<int> k[26];
signed main(){
int L, R;
cin>>L>>R;
int l=0,r=0,q=1;
while(L>=(q<<l))l++;
while(R>=(q<<r))r++;
int dpl[l],dpr[r],u;
dpl[0]=1;u=1;
for(int i=1;i<l;i++){
if(1&(L>>(i-1))){
dpl[i]=dpl[i-1];
}else{
dpl[i]=(2*dpl[i-1]+u)%MAX;
}
u=(u*3)%MAX;
}
// cerr<<'t';
u=1;dpr[0]=1;
for(int i=1;i<r;i++){
if(1&(R>>(i-1))){
dpr[i]=(2*dpr[i-1]+u)%MAX;
}else{
dpr[i]=dpr[i-1];
}
u=(u*3)%MAX;
}
//cerr<<'u';
if(l!=r){
u=1;
for(int i=0;i<l;i++)u=(u*3)%MAX;
int ans=(dpl[l-1]+dpr[r-1])%MAX;
for(int i=l+1;i<r;i++){
ans=(ans+u)%MAX;
u=(u*3)%MAX;
}
cout<<ans;
}else{
int ans=0;
for(int i=l-1;i>=0;i--){
if((1&(R>>i))&&(!(1&(L>>i)))){
ans=(ans+dpr[i]+dpl[i])%MAX;
}
if((1&(L>>i))&&(!(1&(R>>i)))){
ans-=1;
break;
}
}
ans++;
cout<<ans;
}
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
const int MD = 1e9+7;
int la, ra;
ll l, r, ans;
ll exp(ll b, ll n){
ll res = 1;
for(; n; n/=2,(b*=b)%=MD) if(n%2) (res *= b) %= MD;
return res;
}
int main(){
scanf("%lld%lld", &l, &r);
for(; l>>la; la++); la--;
for(; r>>ra; ra++); ra--; r = ~r;
if(ra != la) for(int i = la; i <= ra; i++){
if(la < i && i < ra) (ans += exp(3,i)) %= MD;
else{
ll dp1 = 1, dp2 = 0;
for(int j = i-1; j >= 0; j--){
if((la == i ? l : r)>>j&1) (dp2 *= 3) %= MD;
else{
dp2 = (dp2*3 + dp1) % MD;
(dp1 *= 2) %= MD;
}
}
(ans += dp2 + dp1) %= MD;
}
}else{
ll dp1 = 1, dp2 = 0, dp3 = 0, dp4 = 0;
r = ~r;
int j = ra-1;
for(; j >= 0; j--) if((r^l)>>j&1) break;
for(; j >= 0; j--){
int lb = l>>j&1, rb = r>>j&1;
if(lb && rb){
dp4 =(dp4*3 + dp2) % MD;
(dp2 *= 2) %= MD;
}else if(!lb && rb){
dp4 = (dp4*3 + dp2 + dp3) % MD;
dp2 = (dp2*2 + dp1) % MD;
dp3 = (dp3*2 + dp1) % MD;
}else if(lb && !rb){
(dp4 *= 3) %= MD;
dp1 = 0;
}else if(!lb && !rb){
dp4 =(dp4*3 + dp3) % MD;
(dp3 *= 2) %= MD;
}
}
ans = (dp1 + dp2 + dp3 +dp4) % MD;
}
printf("%lld\n", ans);
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
const int mod = 1e9 + 7;
LL dp[66][2][2],L, R;
LL dfs(int x, bool Left, bool Right, bool zero) {
if(x < 0) return 1;
if(dp[x][Left][Right] >= 0 && !zero)
return dp[x][Left][Right];
LL ans = 0;
int l = 0, r = 1;
if(Left) l = L >> x & 1;
if(Right) r = R >> x & 1;
for(int i = l ; i <= 1 ; i++) {
for (int j = i ; j <= r ; j++) {
if (j == 1 && zero) {
if(i == 1) ans += dfs(x-1, Left && i == l, Right && j == r, zero && j == 0);
}
else ans += dfs(x-1, Left && i == l, Right && j == r, zero && j == 0);
}
}
ans %= mod;
if(!zero) dp[x][Left][Right] = ans;
return ans;
}
int main() {
scanf("%lld%lld", & L, & R);
memset(dp, -1, sizeof(dp));
printf("%lld \n", dfs(60, 1, 1, 1));
//system("pause");
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define debug(x) cerr << #x << " " << x << '\n'
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pli = pair<ll,int>;
const int INF = 0x3f3f3f3f, N = 65;
const ll LINF = 1e18 + 5;
const int mod = 1e9 + 7;
ll dp[N][2][2][2], L, R;
ll dfs(int pos, int limx, int limy, int lead, ll x, ll y)
{
if(pos==-1) return 1;
if(dp[pos][limx][limy][lead]!=-1) return dp[pos][limx][limy][lead];
int upx = limx ? (x>>pos)&1 : 1;
int upy = limy ? (y>>pos)&1 : 1;
ll ans = 0;
for(int i=0; i<=upx; i++)
for(int j=0; j<=upy; j++)
{
if(lead && i==j) (ans += dfs(pos-1, limx&&i==upx, limy&&j==upy, lead&&i==0, x, y)) %= mod;
else if(!lead && j>=i) (ans += dfs(pos-1, limx&&i==upx, limy&&j==upy, lead, x, y)) %= mod;
}
return dp[pos][limx][limy][lead] = ans;
}
ll solve(ll l, ll r)
{
memset(dp, -1, sizeof(dp));
int len = 0;
ll x = max(l, r);
for(int i=0; i<63; i++)
if((x>>i)&1) len = i;
ll ans = dfs(len, 1, 1, 1, l, r);
return ans;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin >> L >> R;
ll ans = (solve(R, R) - solve(L-1, R) -solve(R, L-1) + solve(L-1, L-1))%mod;
ans = (ans + mod)%mod;
cout << ans;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);
#define l_ength size
void mul_mod(ll& a, ll b){
a *= b;
a %= MOD;
}
void add_mod(ll& a, ll b){
a = (a<MOD)?a:(a-MOD);
b = (b<MOD)?b:(b-MOD);
a += b;
a = (a<MOD)?a:(a-MOD);
}
ll dp[63][2][2][2][2];
int a[63],b[63],f[10];
int main(void){
int i,j,k;
ll l,r,ans=0ll;
cin >> l >> r;
for(i=0; i<60; ++i){
b[i] = l&1; l >>= 1;
a[i] = r&1; r >>= 1;
}
reverse(a,a+60);
reverse(b,b+60);
dp[0][1][1][0][1] = 1ll;
for(i=0; i<60; ++i){
for(j=0; j<64; ++j){
for(k=0; k<6; ++k){
f[k] = ((j&(1<<k))?1:0);
}
if(f[4] == 0 && f[5] == 1){
continue;
}
if(f[0] && f[4]>a[i]){
continue;
}
if(f[1] && f[5]<b[i]){
continue;
}
if(f[3] && f[2]>f[5]){
continue;
}
add_mod(dp[i+1][((f[0]&&(f[4]==a[i]))?1:0)][((f[1]&&(f[5]==b[i]))?1:0)][f[4]][((f[3] && (f[2]==f[5]))?1:0)],dp[i][f[0]][f[1]][f[2]][f[3]]);
}
}
for(j=0; j<16; ++j){
for(k=0; k<4; ++k){
f[k] = (j&(1<<k))?1:0;
}
if(f[3]){
continue;
}
add_mod(ans,dp[60][f[0]][f[1]][f[2]][f[3]]);
}
cout << ans << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int LGN = 60;
const int Z = 1e9+7;
void addm(int &a, int b) {
a = a + b >= Z ? a + b - Z : a + b;
}
int main() {
ios::sync_with_stdio(false);
ll l, r;
cin >> l >> r;
int dp[LGN+1][2][2][2];
fill(&dp[0][0][0][0], &dp[LGN+1][0][0][0], 0);
fill(&dp[0][0][0][0], &dp[1][0][0][0], 1);
for (int i = 1; i <= LGN; i++) {
bool bl = l >> (i - 1) & 1, br = r >> (i - 1) & 1;
for (int fl = 0; fl <= 1; fl++) {
for (int fr = 0; fr <= 1; fr++) {
for (int sb = 0; sb <= 1; sb++) {
if (!(bl && fl)) {
addm(dp[i][fl][fr][sb], dp[i-1][fl][br?0:fr][sb]);
}
if (!sb && !(bl && fl) && !(!br && fr)) {
addm(dp[i][fl][fr][sb], dp[i-1][fl][fr][sb]);
}
if (!(!br && fr)) {
addm(dp[i][fl][fr][sb], dp[i-1][!bl?0:fl][fr][0]);
}
}
}
}
}
cout << dp[LGN][1][1][1] << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define rep(i,s,e) for(i64 (i) = (s);(i) < (e);(i)++)
#define all(x) x.begin(),x.end()
#define let auto const
i64 dp[62][2][2][2];
i64 dp2[62][2][2][2];
int main() {
i64 L, R;
cin >> L >> R;
L--;
R++;
i64 mod = 1e9 + 7;
dp[61][0][0][0] = 1;
for(i64 d = 60; d >= 0; d--) {
for(i64 flag = 0;flag <= 1;flag++) {
i64 A = ((R >> d) & 1);
i64 B = ((L >> d) & 1);
if(flag == 1) A = 1;
for(i64 next = 0; next <= A; next++) {
i64 nf = !!(next < A || flag);
if(next == 0) {
if(B == next) {
dp[d][nf][0][0] += dp[d + 1][flag][0][0];
dp[d][nf][0][0] %= mod;
dp[d][nf][1][0] += dp[d + 1][flag][1][0];
dp[d][nf][0][0] %= mod;
}
dp[d][nf][1][1] += dp[d + 1][flag][1][1];
dp[d][nf][1][1] %= mod;
}
else {
if(B == next) {
dp[d][nf][1][0] += dp[d + 1][flag][0][0] + dp[d + 1][flag][1][0];
dp[d][nf][1][0] %= mod;
}
if(B < next) {
dp[d][nf][1][0] += dp[d + 1][flag][1][0];
dp[d][nf][1][0] %= mod;
dp[d][nf][1][1] += dp[d + 1][flag][0][0] + dp[d + 1][flag][1][0];
dp[d][nf][1][1] %= mod;
}
dp[d][nf][1][1] += dp[d + 1][flag][1][1] * 2;
dp[d][nf][1][1] %= mod;
}
}
}
}
cout << dp[0][1][1][1] << endl;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.util.Arrays;
public class Main {
private static void solve() {
long l = nl();
long r = nl();
long x = f(r, r);
long y = f(l - 1, r);
int mod = (int) 1e9 + 7;
System.out.println((x - y + mod) % mod);
}
private static long f(long a, long b) {
int mod = (int) 1e9 + 7;
// dp[n桁目][started][xがa上限][yがb上限]
long[][][] dp = new long[65][2][2];
int as = 1;
int bs = 1;
for (int i = 63; i >= 0; i--) {
int abit = (int) ((a >> i) & 1);
int bbit = (int) ((b >> i) & 1);
if ((abit == 1 || as == 0) && (bbit == 1 || bs == 0)) {
dp[i][as][bs]++;
}
// a && b
if (abit == 1 && bbit == 1) {
dp[i][1][1] += dp[i + 1][1][1]; // (1, 1)
dp[i][0][0] += dp[i + 1][1][1]; // (0, 0)
dp[i][0][1] += dp[i + 1][1][1]; // (0, 1)
} else if (abit == 0 && bbit == 1) {
dp[i][1][1] += dp[i + 1][1][1]; // (0, 1)
dp[i][1][0] += dp[i + 1][1][1]; // (0, 0)
} else if (abit == 0 && bbit == 0) {
dp[i][1][1] += dp[i + 1][1][1]; // (0, 0)
} else {
// abit == 1 && bbit == 0
dp[i][0][1] += dp[i + 1][1][1]; // (0, 0)
}
// !a && !b
dp[i][0][0] += dp[i + 1][0][0] * 3; // (0, 0) (0, 1), (1, 1);
// b only
if (bbit == 1) {
dp[i][0][1] += dp[i + 1][0][1] * 2; // (1, 1), (0, 1)
dp[i][0][0] += dp[i + 1][0][1]; // (0, 0);
} else {
dp[i][0][1] += dp[i + 1][0][1]; // (0, 0)
}
// a only
if (abit == 1) {
dp[i][1][0] += dp[i + 1][1][0]; // (1, 1)
dp[i][0][0] += dp[i + 1][1][0] * 2; // (0, 1), (0, 0);
} else {
dp[i][1][0] += dp[i + 1][1][0] * 2; // (0, 0), (0, 1);
}
if (abit == 1) {
as = 0;
}
if (bbit == 1) {
bs = 0;
}
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
dp[i][j][k] %= mod;
}
}
}
long ret = 0;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
ret += dp[0][i][j];
ret %= mod;
}
}
return ret;
}
public static void main(String[] args) {
new Thread(null, new Runnable() {
@Override
public void run() {
long start = System.currentTimeMillis();
String debug = args.length > 0 ? args[0] : null;
if (debug != null) {
try {
is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
} catch (Exception e) {
throw new RuntimeException(e);
}
}
reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
solve();
out.flush();
tr((System.currentTimeMillis() - start) + "ms");
}
}, "", 64000000).start();
}
private static java.io.InputStream is = System.in;
private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
private static java.util.StringTokenizer tokenizer = null;
private static java.io.BufferedReader reader;
public static String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new java.util.StringTokenizer(reader.readLine());
} catch (Exception e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
private static double nd() {
return Double.parseDouble(next());
}
private static long nl() {
return Long.parseLong(next());
}
private static int[] na(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = ni();
return a;
}
private static char[] ns() {
return next().toCharArray();
}
private static long[] nal(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nl();
return a;
}
private static int[][] ntable(int n, int m) {
int[][] table = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
table[i][j] = ni();
}
}
return table;
}
private static int[][] nlist(int n, int m) {
int[][] table = new int[m][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
table[j][i] = ni();
}
}
return table;
}
private static int ni() {
return Integer.parseInt(next());
}
private static void tr(Object... o) {
if (is != System.in)
System.out.println(java.util.Arrays.deepToString(o));
}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.util.*;
public class Main {
Scanner sc;
long L, R;
private static long MOD = 1000000007L;
private static final long[] MASK;
static {
MASK = new long[64];
MASK[0] = 0L;
for (int i = 1; i < 64; i++) MASK[i] = MASK[i-1] * 2 + 1;
}
private static final long[] POW3;
static {
POW3 = new long[64];
POW3[0] = 1L;
for (int i = 1; i < 64; i++) POW3[i] = (POW3[i-1] * 3L) % MOD;
}
public Main() {
sc = new Scanner(System.in);
}
private void calc() {
L = sc.nextLong();
R = sc.nextLong();
System.out.println(count(0, 0, 63));
System.out.flush();
}
private long count(long x, long y, int maskn) {
if ( (x|y) != 0 && (y>>>1) >= x) return 0L;
if (R < x || R < y) return 0L;
if ((y | MASK[maskn]) < L || (x | MASK[maskn]) < L) return 0L;
if (L <= x && (y | MASK[maskn]) <= R) return POW3[maskn];
maskn--;
long nextBit = MASK[maskn]+1L;
long su = count(x | nextBit, y | nextBit, maskn);
long sl = count(x, y, maskn);
if (x == y) return (su + sl + count(x, y | nextBit, maskn)) % MOD;
else if ( L < x ) return (2 * su + sl) % MOD;
else if ( (y | MASK[maskn+1]) < R) return (su + 2 * sl) % MOD;
return (su + sl + count(x, y | nextBit, maskn)) % MOD;
}
public static void main(String[] args) {
new Main().calc();
}
} | JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws IOException {
Main mainObj = new Main();
mainObj.solve();
}
public void solve() throws IOException {
FastScanner fs = new FastScanner();
long l = fs.nextLong();
long r = fs.nextLong();
int mod = 1_000_000_007;
long[][][][] dp = new long[61][2][2][2];
dp[60][0][0][0] = 1;
for(int i = 59; i >= 0; i--) {
for(int j = 0; j <2; j++) {
for(int k = 0; k < 2; k++) {
for(int s = 0; s < 2; s++) {
long pre = dp[i+1][j][k][s];
long lb = (l >> i) & 1;
long rb = (r >> i) & 1;
for(int x = 0; x < 2; x++) {
for(int y = 0; y < 2; y++) {
int nj = j;
int nk = k;
int ns = s;
if(x == 1 && y == 0) {
continue;
}
if(s == 0 && x == 0 && y == 1) {
continue;
}
if(x == 1 && y == 1) {
ns = 1;
}
if(j == 0 && lb == 1 && x == 0) {
continue;
}
if(k == 0 && rb == 0 && y == 1) {
continue;
}
if(lb == 0 && x == 1) {
nj = 1;
}
if(rb == 1 && y == 0) {
nk = 1;
}
dp[i][nj][nk][ns] = (dp[i][nj][nk][ns] + pre) % mod;
}
}
}
}
}
}
long ans = 0;
for(int j = 0; j < 2; j++) {
for(int k = 0; k < 2; k++) {
for(int s = 0; s < 2; s++) {
ans = (ans + dp[0][j][k][s])%mod;
}
}
}
System.out.println(ans);
}
public class FastScanner {
BufferedReader reader;
private StringTokenizer st;
public FastScanner() {
st = null;
reader = new BufferedReader(new InputStreamReader(System.in));
}
public String next() throws IOException {
if (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(reader.readLine());
}
return st.nextToken();
}
public String nextLine() throws IOException {
st = null;
String readLine = null;
readLine = reader.readLine();
return readLine;
}
public int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(next());
}
public long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(next());
}
public int[] nextIntArr(int n) throws NumberFormatException, IOException {
int[] retArr = new int[n];
for (int i = 0; i < n; i++) {
retArr[i] = nextInt();
}
return retArr;
}
public long[] nextLongArr(int n) throws NumberFormatException, IOException {
long[] retArr = new long[n];
for (int i = 0; i < n; i++) {
retArr[i] = nextLong();
}
return retArr;
}
public void close() throws IOException {
reader.close();
}
}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define int long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF emplace_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int la,lb,a[100],b[100];
const int mod=1e9+7;
int dp[100][2][2];
int dfs(int p,bool f1,bool f2){
if(p==0)RE 1;
if(dp[p][f1][f2]!=-1)RE dp[p][f1][f2];
int l=f1?a[p]:0,r=f2?b[p]:1,re=0;
FOR(i,l,r){
FOR(j,l,i)re=(re+dfs(p-1,f1&&j==l,f2&&i==r))%mod;
}
RE dp[p][f1][f2]=re;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int x,y;
cin>>x>>y;
while(x)a[++la]=x&1,x=x/2;
while(y)b[++lb]=y&1,y=y/2;
int ans=0;
FILL(dp,-1);
FOR(i,la,lb)ans=(ans+dfs(i-1,i==la,i==lb))%mod;
cout<<ans<<'\n';
RE 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <string>
#include <stdio.h>
#include <algorithm>
#include <vector>
#define llint long long
#define mod 1000000007
using namespace std;
llint L, R;
llint dp[63][2][2][2];
int main(void)
{
cin >> L >> R;
dp[62][0][0][0] = 1;
for(int i = 62; i > 0; i--){
for(int j = 0; j < 2; j++){
for(int k = 0; k < 2; k++){
for(int l = 0; l < 2; l++){
//cout << i << " " << j << " " << k << " " << l << " " << dp[i][j][k][l] << endl;
for(int u = 0; u < 2; u++){
for(int d = 0; d < 2; d++){
if(u == 0 && d == 1) continue;
if(l == 0 && u == 1 && d == 0) continue;
if(j == 0 && d == 0 && (L&(1LL<<(i-1)))) continue;
if(k == 0 && u == 1 && (R&(1LL<<(i-1))) == 0) continue;
int nj = j, nk = k, nl = l;
if(d == 1 && (L&(1LL<<(i-1))) == 0) nj = 1;
if(u == 0 && (R&(1LL<<(i-1)))) nk = 1;
if(u == 1 && d == 1) nl = 1;
(dp[i-1][nj][nk][nl] += dp[i][j][k][l]) %= mod;
}
}
}
}
}
}
llint ans = 0;
for(int j = 0; j < 2; j++){
for(int k = 0; k < 2; k++){
for(int l = 0; l < 2; l++){
ans += dp[0][j][k][l], ans %= mod;
}
}
}
cout << ans << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
#include <vector>
#include <chrono>
#include <random>
#include <cstring>
std::mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());
const int ms = 64;
int l[ms], r[ms];
void read(int a[ms]) {
long long x;
std::cin >> x;
for(int i = ms-1; i >= 0; x /= 2, i--) {
a[i] = (int)(x % 2);
}
}
const int MOD = 1e9 + 7;
int memo[ms][2][2][2][2][2];
int dp(int on, int L1, int R1, int L2, int R2, int got) {
if(on == ms) return 1;
int &ans = memo[on][L1][R1][L2][R2][got];
if(ans != -1) { return ans; }
ans = 0;
for(int i = 0; i < 2; i++) {
for(int j = 0; j < 2; j++) {
bool can = i >= j;
if(got == 0) {
can = can && i == j;
}
if(i == 0) can = can && (!L1 || l[on] == 0);
else can = can && (!R1 || r[on] == 1);
if(j == 0) can = can && (!L2 || l[on] == 0);
else can = can && (!R2 || r[on] == 1);
if(can) ans = (ans + dp(on+1, L1 && l[on] == i, R1 && r[on] == i, L2 && l[on] == j, R2 && r[on] == j, got || i)) % MOD;
}
}
return ans;
}
int main() {
std::ios_base::sync_with_stdio(false); std::cin.tie(NULL);
read(l);
read(r);
memset(memo, -1, sizeof memo);
std::cout << dp(0, 1, 1, 1, 1, 0) << '\n';
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 |
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.nio.charset.Charset;
import java.util.Arrays;
public class Main {
public static void main(String[] args) throws Exception {
boolean local = false;
boolean async = false;
Charset charset = Charset.forName("ascii");
FastIO io = local ? new FastIO(new FileInputStream("D:\\DATABASE\\TESTCASE\\Code.in"), System.out, charset) : new FastIO(System.in, System.out, charset);
Task task = new Task(io, new Debug(local));
if (async) {
Thread t = new Thread(null, task, "dalt", 1 << 27);
t.setPriority(Thread.MAX_PRIORITY);
t.start();
t.join();
} else {
task.run();
}
if (local) {
io.cache.append("\n\n--memory -- \n" + ((Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()) >> 20) + "M");
}
io.flush();
}
public static class Task implements Runnable {
final FastIO io;
final Debug debug;
int inf = (int) 1e8;
Modular mod = new Modular((int) 1e9 + 7);
Log2 log2 = new Log2();
Power power = new Power(mod);
public Task(FastIO io, Debug debug) {
this.io = io;
this.debug = debug;
}
@Override
public void run() {
solve();
// for (int i = 1; i < 100; i++) {
// for (int j = 1; j <= i; j++) {
// debug.debug("l", j);
// debug.debug("r", i);
// debug.assertTrue(bruteForce(j, i) == solve(j, i));
// }
// }
}
int[][][][] dp;
boolean[] lBits;
boolean[] rBits;
public int bruteForce(long l, long r) {
int sum = 0;
for (long i = l; i <= r; i++) {
for (long j = l; j <= i; j++) {
if (Long.highestOneBit(i) == Long.highestOneBit(j)
&& (i & j) == j) {
sum++;
}
}
}
return sum;
}
public void solve() {
long l = io.readLong();
long r = io.readLong();
io.cache.append(solve(l, r));
}
public int solve(long l, long r) {
lBits = new boolean[64];
rBits = new boolean[64];
toBits(lBits, 0, l);
toBits(rBits, 0, r);
reverse(lBits);
reverse(rBits);
//1st chosen, up bound, low bound
int[][][][] dp = new int[2][2][2][64];
dp[0][1][1][0] = 1;
for (int i = 1; i < 64; i++) {
//000
dp[1][0][0][i] = mod.plus(dp[1][0][0][i], dp[0][0][0][i - 1]);
dp[0][0][0][i] = mod.plus(dp[0][0][0][i], dp[0][0][0][i - 1]);
//001
if (lBits[i]) {
dp[1][0][1][i] = mod.plus(dp[1][0][1][i], dp[0][0][1][i - 1]);
} else {
dp[1][0][0][i] = mod.plus(dp[1][0][0][i], dp[0][0][1][i - 1]);
dp[0][0][1][i] = mod.plus(dp[0][0][1][i], dp[0][0][1][i - 1]);
}
//010
//011
if (rBits[i] && lBits[i]) {
dp[1][1][1][i] = mod.plus(dp[1][1][1][i],
dp[0][1][1][i - 1]);
} else if (rBits[i]) {
dp[1][1][0][i] = mod.plus(dp[1][1][0][i],
dp[0][1][1][i - 1]);
dp[0][0][1][i] = mod.plus(dp[0][0][1][i],
dp[0][1][1][i - 1]);
} else if (lBits[i]) {
} else {
dp[0][1][1][i] = mod.plus(dp[0][1][1][i], dp[0][1][1][i - 1]);
}
//100
dp[1][0][0][i] = mod.plus(dp[1][0][0][i], mod.mul(3, dp[1][0][0][i - 1]));
//101
if (lBits[i]) {
dp[1][0][1][i] = mod.plus(dp[1][0][1][i], dp[1][0][1][i - 1]);
} else {
dp[1][0][1][i] = mod.plus(dp[1][0][1][i], mod.mul(2, dp[1][0][1][i - 1]));
dp[1][0][0][i] = mod.plus(dp[1][0][0][i], dp[1][0][1][i - 1]);
}
//110
if (rBits[i]) {
dp[1][1][0][i] = mod.plus(dp[1][1][0][i], mod.mul(2, dp[1][1][0][i - 1]));
dp[1][0][0][i] = mod.plus(dp[1][0][0][i], dp[1][1][0][i - 1]);
} else {
dp[1][1][0][i] = mod.plus(dp[1][1][0][i], dp[1][1][0][i - 1]);
}
//111
if (lBits[i] && rBits[i]) {
dp[1][1][1][i] = mod.plus(dp[1][1][1][i], dp[1][1][1][i - 1]);
} else if (rBits[i]) {
dp[1][1][0][i] = mod.plus(dp[1][1][0][i], dp[1][1][1][i - 1]);
dp[1][1][1][i] = mod.plus(dp[1][1][1][i], dp[1][1][1][i - 1]);
dp[1][0][1][i] = mod.plus(dp[1][0][1][i], dp[1][1][1][i - 1]);
} else if (lBits[i]) {
} else {
dp[1][1][1][i] = mod.plus(dp[1][1][1][i], dp[1][1][1][i - 1]);
}
}
int sum = 0;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
sum = mod.plus(dp[1][i][j][63], sum);
}
}
return sum;
}
public int valueOf(boolean x) {
return x ? 1 : 0;
}
public void reverse(boolean[] data) {
int l = 0;
int r = data.length - 1;
while (l < r) {
boolean tmp = data[l];
data[l] = data[r];
data[r] = tmp;
l++;
r--;
}
}
public void toBits(boolean[] bits, int i, long x) {
if (x == 0) {
return;
}
bits[i] = (x & 1) == 1;
toBits(bits, i + 1, x >>> 1);
}
}
/**
* Mod operations
*/
public static class Modular {
int m;
public Modular(int m) {
this.m = m;
}
public int valueOf(int x) {
x %= m;
if (x < 0) {
x += m;
}
return x;
}
public int valueOf(long x) {
x %= m;
if (x < 0) {
x += m;
}
return (int) x;
}
public int mul(int x, int y) {
return valueOf((long) x * y);
}
public int mul(long x, long y) {
x = valueOf(x);
y = valueOf(y);
return valueOf(x * y);
}
public int plus(int x, int y) {
return valueOf(x + y);
}
public int plus(long x, long y) {
x = valueOf(x);
y = valueOf(y);
return valueOf(x + y);
}
@Override
public String toString() {
return "mod " + m;
}
}
/**
* Bit operations
*/
public static class BitOperator {
public int bitAt(int x, int i) {
return (x >> i) & 1;
}
public int bitAt(long x, int i) {
return (int) ((x >> i) & 1);
}
public int setBit(int x, int i, boolean v) {
if (v) {
x |= 1 << i;
} else {
x &= ~(1 << i);
}
return x;
}
public long setBit(long x, int i, boolean v) {
if (v) {
x |= 1L << i;
} else {
x &= ~(1L << i);
}
return x;
}
/**
* Determine whether x is subset of y
*/
public boolean subset(long x, long y) {
return intersect(x, y) == x;
}
/**
* Merge two set
*/
public long merge(long x, long y) {
return x | y;
}
public long intersect(long x, long y) {
return x & y;
}
public long differ(long x, long y) {
return x - intersect(x, y);
}
}
/**
* Power operations
*/
public static class Power {
final Modular modular;
public Power(Modular modular) {
this.modular = modular;
}
public int pow(int x, long n) {
if (n == 0) {
return 1;
}
long r = pow(x, n >> 1);
r = modular.valueOf(r * r);
if ((n & 1) == 1) {
r = modular.valueOf(r * x);
}
return (int) r;
}
public int inverse(int x) {
return pow(x, modular.m - 2);
}
public int pow2(int x) {
return x * x;
}
public long pow2(long x) {
return x * x;
}
public double pow2(double x) {
return x * x;
}
}
/**
* Log operations
*/
public static class Log2 {
public int ceilLog(int x) {
return 32 - Integer.numberOfLeadingZeros(x - 1);
}
public int floorLog(int x) {
return 31 - Integer.numberOfLeadingZeros(x);
}
public int ceilLog(long x) {
return 64 - Long.numberOfLeadingZeros(x - 1);
}
public int floorLog(long x) {
return 63 - Long.numberOfLeadingZeros(x);
}
}
public static class FastIO {
public final StringBuilder cache = new StringBuilder(1 << 13);
private final InputStream is;
private final OutputStream os;
private final Charset charset;
private StringBuilder defaultStringBuf = new StringBuilder(1 << 13);
private byte[] buf = new byte[1 << 13];
private int bufLen;
private int bufOffset;
private int next;
public FastIO(InputStream is, OutputStream os, Charset charset) {
this.is = is;
this.os = os;
this.charset = charset;
}
public FastIO(InputStream is, OutputStream os) {
this(is, os, Charset.forName("ascii"));
}
private int read() {
while (bufLen == bufOffset) {
bufOffset = 0;
try {
bufLen = is.read(buf);
} catch (IOException e) {
throw new RuntimeException(e);
}
if (bufLen == -1) {
return -1;
}
}
return buf[bufOffset++];
}
public void skipBlank() {
while (next >= 0 && next <= 32) {
next = read();
}
}
public int readInt() {
int sign = 1;
skipBlank();
if (next == '+' || next == '-') {
sign = next == '+' ? 1 : -1;
next = read();
}
int val = 0;
if (sign == 1) {
while (next >= '0' && next <= '9') {
val = val * 10 + next - '0';
next = read();
}
} else {
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
}
return val;
}
public long readLong() {
int sign = 1;
skipBlank();
if (next == '+' || next == '-') {
sign = next == '+' ? 1 : -1;
next = read();
}
long val = 0;
if (sign == 1) {
while (next >= '0' && next <= '9') {
val = val * 10 + next - '0';
next = read();
}
} else {
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
}
return val;
}
public double readDouble() {
boolean sign = true;
skipBlank();
if (next == '+' || next == '-') {
sign = next == '+';
next = read();
}
long val = 0;
while (next >= '0' && next <= '9') {
val = val * 10 + next - '0';
next = read();
}
if (next != '.') {
return sign ? val : -val;
}
next = read();
long radix = 1;
long point = 0;
while (next >= '0' && next <= '9') {
point = point * 10 + next - '0';
radix = radix * 10;
next = read();
}
double result = val + (double) point / radix;
return sign ? result : -result;
}
public String readString(StringBuilder builder) {
skipBlank();
while (next > 32) {
builder.append((char) next);
next = read();
}
return builder.toString();
}
public String readString() {
defaultStringBuf.setLength(0);
return readString(defaultStringBuf);
}
public int readLine(char[] data, int offset) {
int originalOffset = offset;
while (next != -1 && next != '\n') {
data[offset++] = (char) next;
next = read();
}
return offset - originalOffset;
}
public int readString(char[] data, int offset) {
skipBlank();
int originalOffset = offset;
while (next > 32) {
data[offset++] = (char) next;
next = read();
}
return offset - originalOffset;
}
public int readString(byte[] data, int offset) {
skipBlank();
int originalOffset = offset;
while (next > 32) {
data[offset++] = (byte) next;
next = read();
}
return offset - originalOffset;
}
public char readChar() {
skipBlank();
char c = (char) next;
next = read();
return c;
}
public void flush() throws IOException {
os.write(cache.toString().getBytes(charset));
os.flush();
cache.setLength(0);
}
public boolean hasMore() {
skipBlank();
return next != -1;
}
}
public static class Debug {
private boolean allowDebug;
public Debug(boolean allowDebug) {
this.allowDebug = allowDebug;
}
public void assertTrue(boolean flag) {
if (!allowDebug) {
return;
}
if (!flag) {
fail();
}
}
public void fail() {
throw new RuntimeException();
}
public void assertFalse(boolean flag) {
if (!allowDebug) {
return;
}
if (flag) {
fail();
}
}
private void outputName(String name) {
System.out.print(name + " = ");
}
public void debug(String name, int x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println("" + x);
}
public void debug(String name, long x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println("" + x);
}
public void debug(String name, double x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println("" + x);
}
public void debug(String name, int[] x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println(Arrays.toString(x));
}
public void debug(String name, long[] x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println(Arrays.toString(x));
}
public void debug(String name, double[] x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println(Arrays.toString(x));
}
public void debug(String name, Object x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println("" + x);
}
public void debug(String name, Object... x) {
if (!allowDebug) {
return;
}
outputName(name);
System.out.println(Arrays.deepToString(x));
}
}
}
| JAVA |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | L,R = map(int,input().split())
mod = 10**9+7
m = 64 +1
fac = [1]*m
ninv = [1]*m
finv = [1]*m
for i in range(2,m):
fac[i] = fac[i-1]*i%mod
ninv[i] = (-(mod//i)*ninv[mod%i])%mod
finv[i] = finv[i-1]*ninv[i]%mod
def comb(n,k):
return (fac[n]*finv[k]%mod)*finv[n-k]%mod
def f(L,R):
if L>R : return 0
R = bin(R)[2:]
N = len(R)
ret = f(L,int("0"+"1"*(N-1),2))
L = bin(L)[2:]
if len(L) != N : L = "1"+"0"*(N-1)
for i in range(N):
if R[i] == "0" : continue
R2 = R[:i] + "0" + "?"*(N-i-1)
if i==0: R2 = R
for j in range(N):
if L[j] == "1" and j!=0 : continue
L2 = L[:j] + "1" + "?"*(N-j-1)
if j==0 : L2 = L
if L2[0] == "0" : break
tmp = 1
for r,l in zip(R2[1:],L2[1:]):
if r=="0" and l=="1" : tmp *= 0
if r=="?" and l=="?" : tmp *= 3
if r=="?" and l=="0" : tmp *= 2
if r=="1" and l=="?" : tmp *= 2
ret += tmp
ret %= mod
return ret%mod
print(f(L,R)) | PYTHON3 |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
typedef long long ll;
typedef pair<ll, ll> P;
const ll MOD = 1e9+7;
const ll MOD2 = 998244353;
const ll INF = 1e18;
int main(){
ll L, R;
cin >> L >> R;
int l[61], r[61];
rep(i, 61){
l[i] = L%2, r[i] = R%2;
L /= 2, R /= 2;
}
ll dp[61][2][2][2];
rep(i, 61)rep(j, 2)rep(k, 2)rep(m, 2) dp[i][j][k][m] = 0;
dp[60][0][0][0] = 1;
rep3(i, 59, 0){
rep(j, 2)rep(k, 2)rep(m, 2){
ll pre = dp[i+1][j][k][m];
rep(x, 2)rep(y, 2){
if(x == 1 && y == 0) continue;
int nj = j, nk = k, nm = m;
if(x == 0 && l[i] == 1 && j == 0) continue;
if(x == 1 && l[i] == 0) nj = 1;
if(y == 1 && r[i] == 0 && k == 0) continue;
if(y == 0 && r[i] == 1) nk = 1;
if(x == 0 && y == 1 && m == 0) continue;
if(x == 1 && y == 1) nm = 1;
dp[i][nj][nk][nm] += pre;
dp[i][nj][nk][nm] %= MOD;
}
}
}
ll ans = 0;
rep(j, 2)rep(k, 2)rep(m, 2){
ans += dp[0][j][k][m];
ans %= MOD;
}
cout << ans << endl;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <iostream>
using namespace std;
typedef long long ll;
ll mod = 1e9 + 7;
ll dp[61][2][2][2];
int main(int argc, char *argv[])
{
ll L, R;
cin >> L >> R;
dp[60][0][0][0] = 1;
for (int i = 59; i >= 0; i--) {
int lb = L >> i & 1;
int rb = R >> i & 1;
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
for (int s = 0; s < 2; s++) {
ll pre = dp[i + 1][j][k][s] % mod;
for (int x = 0; x < 2; x++) {
for (int y = 0; y < 2; y++) {
if (x && !y) {
continue;
}
int nj = j, nk = k, ns = s;
if (!s && x != y) {
continue;
}
if (x && y) {
ns = 1;
}
if (!j && !x && lb) {
continue;
}
if (x && !lb) {
nj = 1;
}
if (!k && y && !rb) {
continue;
}
if (!y && rb) {
nk = 1;
}
dp[i][nj][nk][ns] += pre % mod;
dp[i][nj][nk][ns] %= mod;
}
}
}
}
}
}
ll ans = 0;
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
for (int s = 0; s < 2; s++) {
ans += dp[0][j][k][s] % mod;
ans %= mod;
}
}
}
cout << ans << endl;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//
ll dp[61][2][2][2];
ll L, R;
ll dfs(int pos, bool ismax, bool ismin, bool zero) {
if (pos == -1) return 1;
ll &res = dp[pos][ismax][ismin][zero];
if (~res) return res;
res = 0;
if (!ismax || ismax && R >> pos & 1) res += dfs(pos - 1, ismax, ismin & (L >> pos & 1), false);
if (!ismin || ismin && L >> pos & 1 ^ 1) {
if (!zero && (!ismax || ismax && R >> pos & 1)) res += dfs(pos - 1, ismax, ismin, zero);
res += dfs(pos - 1, ismax & (R >> pos & 1 ^ 1), ismin, zero);
}
res %= MOD;
return res;
}
int main() {
cin >> L >> R;
memset(dp, -1, sizeof(dp));
cout << dfs(60, true, true, true) << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
#define all(A) begin(A), end(A)
#define rall(A) rbegin(A), rend(A)
#define sz(A) int(A.size())
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
int main () {
ios::sync_with_stdio(false); cin.tie(0);
ll l, r;
cin >> l >> r;
const int B = 62;
const int MOD = 1e9 + 7;
int dp[B][2][2][2];
memset(dp, -1, sizeof dp);
function <int(int,int,int,int)> rec = [&] (int pos, int gt_x, int lt_y, int msb) {
if (pos == -1) return 1;
int& ret = dp[pos][gt_x][lt_y][msb];
if (ret != -1) return ret;
ret = 0;
int bit_l = (l >> pos) & 1;
int bit_r = (r >> pos) & 1;
// x = 0, y = 0
if (gt_x or (bit_l == 0)) {
ret += rec(pos - 1, gt_x, lt_y or (bit_r == 1), msb);
if (ret > MOD) ret -= MOD;
}
// x = 0, y = 1
if ((gt_x or (bit_l == 0)) and (lt_y or (bit_r == 1)) and msb) {
ret += rec(pos - 1, gt_x, lt_y, msb);
if (ret > MOD) ret -= MOD;
}
// x = 1, y = 1
if (lt_y or (bit_r == 1)) {
ret += rec(pos - 1, gt_x or (bit_l == 0), lt_y, 1);
if (ret > MOD) ret -= MOD;
}
return ret;
};
cout << rec(B - 1, 0, 0, 0) << '\n';
return (0);
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | /*
[abc138] F - Coincidence
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
#define ALL(c) (c).begin(), (c).end()
const int MAX_D = 64;
const ll MOD = 1e9 + 7;
ll L, R;
ll memo[MAX_D][2][2][2];
ll f(int d, int m, int l, int r) {
if (d == -1) {
return m == 1;
}
if (memo[d][m][l][r] >= 0) {
return memo[d][m][l][r];
}
ll ret = 0;
for (int a = 0; a < 2; a++) {
for (int b = 0; b < 2; b++) {
if ((l && a < ((L >> d) & 1)) || (r && b > ((R >> d) & 1)) ||
(a > b) || (!m && a != b)) {
continue;
}
int nm = m || (a & b);
int nl = l && (((L >> d) & 1) == a);
int nr = r && (((R >> d) & 1) == b);
(ret += f(d - 1, nm, nl, nr)) %= MOD;
}
}
return memo[d][m][l][r] = ret;
}
ll solve() {
memset(memo, -1, sizeof(memo));
return f(MAX_D - 1, 0, 1, 1);
}
int main() {
cin >> L >> R;
cout << solve() << endl;
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bitset>
#include <iostream>
using namespace std;
using lint = long long;
const int MOD = 1000000007;
int getCountOfPair(bitset<64> l, bitset<64> r) {
bool flag = false, diff = false, both = false;
int i = 63;
lint count[3] = {l != r, 0, 1};
do {
flag |= l[i];
diff |= l[i] != r[i];
} while (!r[i--]);
for (; i >= 0; i--) {
count[1] =
(count[1] * 3 + (count[0] * !l[i] + count[2] * r[i]) * diff) % MOD;
count[0] =
((count[0] << (!l[i] * flag * diff)) + both * !l[i] * r[i]) % MOD;
count[2] = ((count[2] << (r[i] * diff)) + both * !l[i] * r[i]) % MOD;
flag |= l[i];
bool check = diff;
diff |= l[i] != r[i];
check ^= diff;
both = (both * (!l[i] | r[i])) | check;
}
return int((count[0] + count[1] + count[2] + both) % MOD);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int answer;
lint l, r;
cin >> l >> r;
answer = getCountOfPair(l, r);
cout << answer;
return 0;
} | CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <cstdio>
long L, R, MOD=1e9+7;
long dp[66][2][2][2];
int main() {
scanf("%ld%ld", &L, &R);
dp[0][0][0][0] = 1;
bool x[3] = {1, 0, 0}, y[3] = {1, 1, 0};
for(int i = 0; i < 60; ++i) {
bool lb = L>>(59-i)&1, rb = R>>(59-i)&1;
for(int l = 0; l < 2; ++l)
for(int r = 0; r < 2; ++r)
for(int j = 0; j < 2; ++j)
for(int k = 0; k < 3; ++k) {
if(!l && !x[k] && lb) continue;
if(!r && y[k] && !rb) continue;
if(!j && x[k] ^ y[k]) continue;
int nl = (x[k] ^ lb ? 1 : l);
int nr = (y[k] ^ rb ? 1 : r);
int nj = (x[k] & y[k] ? 1 : j);
(dp[i+1][nl][nr][nj] += dp[i][l][r][j]) %= MOD;
}
}
long ans = dp[60][0][0][1] + dp[60][0][1][1] + dp[60][1][0][1] + dp[60][1][1][1];
printf("%ld\n", ans % MOD);
return 0;
}
| CPP |
p02938 AtCoder Beginner Contest 138 - Coincidence | Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.
What is \mbox{ XOR }?
The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:
* When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.
For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)
Constraints
* 1 \leq L \leq R \leq 10^{18}
Input
Input is given from Standard Input in the following format:
L R
Output
Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.
Examples
Input
2 3
Output
3
Input
10 100
Output
604
Input
1 1000000000000000000
Output
68038601 | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//
ll dp[61][2][2][2][2];
ll L, R;
ll dfs(int pos, bool ismax, bool ismin, bool ismaxy, bool zero) {
if (pos == -1) return 1;
ll &res = dp[pos][ismax][ismin][ismaxy][zero];
if (~res) return res;
res = 0;
if ((!ismax || ismax && R >> pos & 1) && (!ismaxy || ismaxy && R >> pos & 1)) res += dfs(pos - 1, ismax, ismin & (L >> pos & 1), ismaxy, false);
if (!ismin || ismin && L >> pos & 1 ^ 1) {
if (!zero && (!ismaxy || ismaxy && R >> pos & 1)) res += dfs(pos - 1, ismax & (R >> pos & 1 ^ 1), ismin, ismaxy, zero);
res += dfs(pos - 1, ismax & (R >> pos & 1 ^ 1), ismin, ismaxy & (R >> pos & 1 ^ 1), zero);
}
res %= MOD;
return res;
}
int main() {
cin >> L >> R;
memset(dp, -1, sizeof(dp));
cout << dfs(60, true, true, true, true) << endl;
return 0;
}
| CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] ant = new int[5];
int dis = 0;
for(int i = 0; i < 5; i++) {
ant[i] = sc.nextInt();
}
dis = sc.nextInt();
for(int i = 0; i < 4; i++) {
for(int j = i + 1; j < 5; j++) {
if(ant[j] - ant[i] > dis) {
System.out.println(":(");
System.exit(0);
}
}
}
System.out.println("Yay!");
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO 自動生成されたメソッド・スタブ
Scanner sc =new Scanner(System.in);
int a =sc.nextInt();
int b =sc.nextInt();
int c =sc.nextInt();
int d =sc.nextInt();
int e =sc.nextInt();
int k =sc.nextInt();
if(e-a>k) {
System.out.println(":(");
}
else {
System.out.println("Yay!");
}
}
}
| JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] a = new int[5];
for(int i=0;i<5;i++){
a[i]=sc.nextInt();
}
int k = sc.nextInt();
int b = a[4]-a[0];
if(b>k){
System.out.println(":(");
}else{
System.out.println("Yay!");
}
}
}
| JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int main()
{
int a[5],k;
for(int i=0;i<5;i++) cin>>a[i];
cin>>k;
int x=a[4]-a[0];
if(x<=k) cout<<"Yay!";
else cout<<":(";
return 0;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <stdio.h>
int main() {
int a, b, c, d, e, k;
scanf("%d %d %d %d %d %d", &a, &b, &c, &d, &e, &k);
if (e - a > k)
printf(":(\n");
else
printf("Yay!\n");
return 0;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int [] a = new int[4];
for(int i =0; i<4; i++)
{ a[i] = sc.nextInt();}
int e = sc.nextInt();
int k = sc.nextInt();
sc.close();
for(int i =0; i<4; i++)
{if(e-a[i]>k)
{System.out.println(":(");
System.exit(0);}
}
System.out.println("Yay!");
}
}
| JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int d = sc.nextInt();
int e = sc.nextInt();
int k = sc.nextInt();
if (e - a <= k) {
System.out.println("Yay!");
} else {
System.out.println(":(");
}
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
for (int i = 0; i < 3; i++) {
sc.next();
}
int e = sc.nextInt();
int k = sc.nextInt();
if (k >= e - a) {
System.out.println("Yay!");
} else {
System.out.println(":(");
}
}
}
| JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int main() {
int a,b,c,d,e,K;
cin >> a >> b >> c >> d >> e >> K;
if (e-a<=K) cout << "Yay!" << endl;
else cout << ":(" << endl;
}
| CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | a,*_,e,k=[int(input())for x in' '*6];print([':(','Yay!'][e-a<=k]) | PYTHON3 |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | A = [int(input()) for _ in range(6)]
ans = ":(" if A[4] - A[0] > A[5] else "Yay!"
print(ans) | PYTHON3 |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int a = scanner.nextInt();
int b = scanner.nextInt();
int c = scanner.nextInt();
int d = scanner.nextInt();
int e = scanner.nextInt();
int k = scanner.nextInt();
System.out.println(e - a > k ? ":(" : "Yay!");
scanner.close();
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include<iostream>
#include<string>
#include<cmath>
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,c,d,e,k;
cin>>a>>b>>c>>d>>e>>k;
cout<<(e-a <= k ?"Yay!" : ":(" );
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner in = new Scanner(System.in);
int min = 124;
int max = -1;
for (int i = 0; i < 5; i++) {
int n = in.nextInt();
if (n < min)
min = n;
if(n > max)
max = n;
}
int k = in.nextInt();
System.out.println((max - min <= k) ? "Yay!" : ":(");
in.close();
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include<iostream>
using namespace std;
int main(void){
int a,b,c,d,e,k;
cin>>a>>b>>c>>d>>e>>k;
if(e-a<=k){
cout<<"Yay!"<<endl;
}else{
cout<<":("<<endl;
}
return 0;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.*;
class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] a = new int[5];
int[] b = new int[5];
for(int i = 0; i<5; i++) {
a[i] = sc.nextInt();
b[i] = a[i];
}
int k = sc.nextInt();
for(int i = 0; i<5; i++) {
for(int j = 0; j<5; j++) {
if(Math.abs(a[i]-b[j]) > k) {
System.out.println(":(");
return;
}
}
}
System.out.println("Yay!");
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <iostream>
using namespace std;
int main(void){
int a,b,c,d,e,k;
cin >> a >> b >> c >> d >> e >> k;
if(e-a<=k) cout << "Yay!";
else cout << ":(";
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int main(){
int a,s,d,f,g,h;
cin>>a>>s>>d>>f>>g>>h;
if(g-a>h){
cout<<":(";
}else{
cout<<"Yay!";
}
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.*;
class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a[] = new int[5];
for (int i = 0; i < 5; i++) {
a[i] = sc.nextInt();
}
int k = sc.nextInt();
for (int i = 0; i < 4; i++) {
for (int j = i + 1; j < 5; j++) {
if (a[j] - a[i] > k) {
System.out.println(":(");
return;
}
}
}
System.out.println("Yay!");
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d, e, k;
cin >> a >> b >> c >> d >> e >> k;
cout << ((e - a <= k) ? "Yay!" : ":(") << endl;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <cstdio>
#include <iostream>
using namespace std;
int main() {
int a, e, k;
scanf("%d%*d%*d%*d%d%d", &a, &e, &k);
if (e - a > k) cout << ":(";
else cout << "Yay!";
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | a = [int(input()) for i in range(6)]
if a[4] - a[0] > a[5]:
print(':(')
else:
print('Yay!') | PYTHON3 |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <iostream>
using namespace std;
int main(int argc, char** argv) {
int a,b,c,d,e,k;
cin >> a >> b >> c >> d >> e >> k;
if(e-a<=k) puts("Yay!");
else puts(":(");
return 0;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int[] x = new int[5];
for(int i=0; i<5; i++){
x[i] = sc.nextInt();
}
int k = sc.nextInt();
if(x[4]-x[0]>k){
System.out.println(":(");
}
else if(x[4]-x[0]<=k){
System.out.println("Yay!");
}
}
} | JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include<iostream>
using namespace std;
int main(void)
{
int a,b,c,d,e,k;
cin >> a >> b >> c >> d >> e >> k;
if(e-a>k) cout << ":(";
else cout << "Yay!";
return 0;
}
| CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include<iostream>
int main(){
int a, b, c, d, e, k;
std::cin>>a>>b>>c>>d>>e>>k;
std::cout << ((e-a<=k) ? "Yay!" : ":(") << std::endl;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <iostream>
using namespace std;
int main(){
int A, B, C, D, E, k;
cin >> A >> B >> C >> D >> E >> k;
cout << (E-A <= k ? "Yay!" : ":(");
return 0;
}
| CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include<bits/stdc++.h>
using namespace std;
int a,b,c,d,e,k;
int main()
{
scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&k);
if(e-a>k)
puts(":(");
else
puts("Yay!");
return 0;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | l=[int(input()) for _ in range(6)]
print(':(' if l[4]-l[0]>l[5] else 'Yay!') | PYTHON3 |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <iostream>
using namespace std;
int main() {
int a, b, c, d, e, k;
cin >> a >> b >> c >> d >> e >> k;
cout << (e - a > k ? ":(" : "Yay!") << endl;
return 0;
}
| CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | #include <iostream>
using namespace std;
int main(){
int a,b,c,d,e,k;
cin >> a >> b >> c >> d >> e >> k;
if(e-a > k) cout << ":(" << endl;
else cout << "Yay!" << endl;
} | CPP |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] arr = new int[5];
for (int i = 0; i < 5; i++) {
arr[i] = sc.nextInt();
}
int k = sc.nextInt();
System.out.println(arr[4] - arr[0] > k ? ":(" : "Yay!");
}
}
| JAVA |
p03075 AtCoder Beginner Contest 123 - Five Antennas | In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.
Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.
Determine if there exists a pair of antennas that cannot communicate directly.
Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.
Constraints
* a, b, c, d, e and k are integers between 0 and 123 (inclusive).
* a < b < c < d < e
Input
Input is given from Standard Input in the following format:
a
b
c
d
e
k
Output
Print `:(` if there exists a pair of antennas that cannot communicate directly, and print `Yay!` if there is no such pair.
Examples
Input
1
2
4
8
9
15
Output
Yay!
Input
15
18
26
35
36
18
Output
:( | 6 | 0 | a,b,c,d,e,f=[int(input()) for i in range(6)]
print("Yay!" if abs(a-e)<=f else ":(") | PYTHON3 |
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