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| description
stringlengths 29
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| difficulty
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1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | n=int(input())
s=list(input())
c=s.copy()
k=[]
for i in range(n-1):
if s[i]=='W':
pass
else:
s[i]='W'
if s[i+1]=='W':
s[i+1]='B'
else:
s[i+1]='W'
k.append(i+1)
if s==['W']*n:
print(len(k))
print(*k)
else:
s=c.copy()
k=[]
for i in range(n-1):
if s[i]=='B':
pass
else:
s[i]='B'
if s[i+1]=='W':
s[i+1]='B'
else:
s[i+1]='W'
k.append(i+1)
if s==['B']*n:
print(len(k))
print(*k)
else:
print(-1)
| PYTHON3 |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | def sol():
n = int(input())
s = list(input())
b = s.count('B')
w = s.count('W')
if w % 2 == 1 and b % 2 == 1:
print(-1)
return
ans = []
s = list(s)
if w % 2 == 1:
odd = 'W'
even = 'B'
else:
odd = 'B'
even = 'W'
for i in range(n-1):
if s[i] != odd:
ans.append(i + 1)
s[i] = odd
s[i+1] = even if s[i+1] == odd else odd
print(len(ans))
print(*ans)
sol() | PYTHON3 |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 |
import java.io.BufferedReader;
import java.io.Console;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
public static int check(StringBuffer sf) {
int B = 0, W = 0;
for (int i = 0; i < sf.length(); i++)
if (sf.charAt(i) == 'B') B++;
else W++;
if (B == 0 || W == 0) return 1;
return 0;
}
public static void main(String[] args) throws IOException {
FastReader fs=new FastReader();
int l = fs.nextInt();
String s = fs.next();
char []ch=s.toCharArray();
StringBuffer sf=new StringBuffer(s);
int B = 0, W = 0;
for (int i = 0; i < l; i++) {
if (sf.charAt(i) == 'B') B++;
else W++;
}
if (W == 0 || B == 0) System.out.println(0);
else if (B % 2 != 0 && W % 2 != 0) System.out.println(-1);
else {
int cnt1 = 0;
List<Integer> list1 = new ArrayList<>();
for (int i = 0; i < l; i++) {
if (sf.charAt(i) == 'W' && i != l - 1) {
sf.setCharAt(i, 'B');
if (sf.charAt(i+1) == 'B') sf.setCharAt(i+1,'W');
else sf.setCharAt(i+1, 'B');
cnt1++;
list1.add(i + 1);
}
}
if(check(sf)==0){
for(int i=0;i<l-1;i+=2){
cnt1++;
list1.add(i+1);
}
}
System.out.println(cnt1);
for (int i = 0; i < cnt1; i++) {
if (i != 0) System.out.print(" ");
System.out.print(list1.get(i));
}
}
}
public static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | JAVA |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | n=int(input())
s=list(input())
ans=[]
if s[0]=="W":
ans.append(0)
s[0]="B"
if s[1]=="W":
s[1]="B"
else:
s[1]="W"
for i in range(1,n-1):
if s[i]=="W":
s[i]="B"
if s[i+1]=="W":
s[i+1]="B"
else:
s[i+1]="W"
ans.append(i)
if s!=["B" for i in range(n)] and n%2==0:
print(-1)
exit()
if s!=["B" for i in range(n)]:
for i in range(n-1):
if i%2==0:
ans.append(i)
print(len(ans))
print(" ".join(map(str,[i+1 for i in ans]))) | PYTHON3 |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | def solve():
n = int(input())
s = input()
ans = 0
res = []
w = 0
b = 0
for i in range(n):
if s[i]=="W":
w+=1
if s[i]=="B":
b+=1
delta = -1
if w==0 or b==0:
print(0)
return
if w%2==1:
if b%2==1:
print(-1)
return
else:
delta = "B"
else:
delta = "W"
s = list(s)
for i in range(n):
if s[i]==delta:
s[i]=0
else:
s[i]=1
for i in range(n-1):
if s[i]==0:
if s[i+1]==1:
s[i+1]=0
else:
s[i+1]=1
ans+=1
res.append(i+1)
print(ans)
for i in range(len(res)):
print(res[i]," ")
solve()
| PYTHON3 |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | n=int(input())
s=input()
l=[i for i in s]
s=list(l)
numofb=0
numofw=0
for i in s:
if(i=='B'):
numofb+=1
else:
numofw+=1
c=0
if(numofb%2==0):
c='W'
elif(numofw%2==0):
c='B'
if(c==0):
print(-1)
else:
numofs=0
l=[]
for i in range(n-1):
if(s[i]!=c):
if(s[i]=='W'):
s[i]='B'
else:
s[i]='W'
if(s[i+1]=='B'):
s[i+1]='W'
else:
s[i+1]='B'
l.append(i)
numofs+=1
print(numofs)
for i in l:
print(i+1,end=" ")
print()
| PYTHON3 |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | def change(a):
if a == 'W':
return 'B'
else:
return 'W'
n = int(input())
ss = input()
s = []
s[:0] = ss
ops = []
good = "w"
if s.count("B") == 0 or s.count("W") == 0:
print(0)
# print("uff")
exit()
elif s.count("B") % 2 == 0:
good = "W"
elif s.count("W") % 2 == 0:
good = "B"
else:
print(-1)
exit()
for i in range(n-1):
if s[i] != good:
s[i] = change(s[i])
s[i+1] = change(s[i+1])
ops.append(i+1)
print(len(ops))
for o in ops:
print(o, end=" ")
print("")
| PYTHON3 |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <class C>
void min_self(C &a, C b) {
a = min(a, b);
}
template <class C>
void max_self(C &a, C b) {
a = max(a, b);
}
long long mod(long long n, long long m = 1000000007) {
n %= m, n += m, n %= m;
return n;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM =
chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
unordered_map<long long, int, custom_hash> safe_map;
const int MAXN = 1e5 + 5;
const int LOGN = 21;
const long long INF = 1e14;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
string s;
cin >> n >> s;
vector<int> ans;
for (int i = 0; i < n; i++) {
if (s[i] != 'W' && i + 1 < n) {
ans.push_back(i);
s[i] = 'W' + 'B' - s[i];
s[i + 1] = 'W' + 'B' - s[i + 1];
}
}
int imp = 0;
for (int i = 0; i < n; i++) {
if (s[i] != 'W') {
imp = 1;
break;
}
}
if (imp) {
for (int i = 0; i < n; i++) {
if (s[i] != 'B' && i + 1 < n) {
ans.push_back(i);
s[i] = 'B' + 'W' - s[i];
s[i + 1] = 'B' + 'W' - s[i + 1];
}
}
int no = 0;
for (int i = 0; i < n; i++) {
if (s[i] != 'B') {
no = 1;
break;
}
}
if (no) {
cout << -1, cout << "\n";
} else {
cout << ans.size(), cout << "\n";
for (auto x : ans) cout << x + 1 << " ";
}
} else {
cout << ans.size(), cout << "\n";
for (auto x : ans) cout << x + 1 << " ";
}
cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
return 0;
}
| CPP |
1271_B. Blocks | There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.
You may perform the following operation zero or more times: choose two adjacent blocks and invert their colors (white block becomes black, and vice versa).
You want to find a sequence of operations, such that they make all the blocks having the same color. You don't have to minimize the number of operations, but it should not exceed 3 ⋅ n. If it is impossible to find such a sequence of operations, you need to report it.
Input
The first line contains one integer n (2 ≤ n ≤ 200) — the number of blocks.
The second line contains one string s consisting of n characters, each character is either "W" or "B". If the i-th character is "W", then the i-th block is white. If the i-th character is "B", then the i-th block is black.
Output
If it is impossible to make all the blocks having the same color, print -1.
Otherwise, print an integer k (0 ≤ k ≤ 3 ⋅ n) — the number of operations. Then print k integers p_1, p_2, ..., p_k (1 ≤ p_j ≤ n - 1), where p_j is the position of the left block in the pair of blocks that should be affected by the j-th operation.
If there are multiple answers, print any of them.
Examples
Input
8
BWWWWWWB
Output
3
6 2 4
Input
4
BWBB
Output
-1
Input
5
WWWWW
Output
0
Input
3
BWB
Output
2
2 1
Note
In the first example, it is possible to make all blocks black in 3 operations. Start with changing blocks 6 and 7, so the sequence is "BWWWWBBB". Then change blocks 2 and 3, so the sequence is "BBBWWBB". And finally, change blocks 4 and 5, so all blocks are black.
It is impossible to make all colors equal in the second example.
All blocks are already white in the third example.
In the fourth example it is possible to make all blocks white in two operations: first operation is to change blocks 2 and 3 (so the sequence is "BBW"), and then change blocks 1 and 2 (so all blocks are white). | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<int> v;
bool conv(string s, char w) {
v.clear();
for (int i = 0; i < s.size(); i++) {
if (s[i] == w)
continue;
else if (i != s.size() - 1) {
v.push_back(i + 1);
if (s[i + 1] == 'W')
s[i + 1] = 'B';
else
s[i + 1] = 'W';
}
if (i == s.size() - 1) return 0;
}
return 1;
}
int main() {
int n;
cin >> n;
string s;
cin >> s;
string tmp = "WB";
for (int i = 0; i < 2; i++)
if (conv(s, tmp[i])) {
cout << v.size() << endl;
for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
cout << endl;
return 0;
}
cout << -1 << endl;
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
import java.math.*;
public class Main{
public static void main(String[] args) throws Exception {
IO io = new IO();
PrintWriter out = new PrintWriter(System.out);
Solver sr = new Solver();
sr.solve(io,out);
out.flush();
out.close();
}
static class Solver
{
class Pair
{
int x, y;
public Pair(int a, int b)
{
x=a;
y=b;
}
}
void solve(IO io, PrintWriter out)
{
int i, j;
int t = io.nextInt();
while(t-->0)
{
int n = io.nextInt();
boolean flag=false;
int a=0, b=0, c=0;
for(i=2 ; i*i<=n ; i++)
{
if(n%i==0 && n/i!=i)
{
int m = n/i;
for(j=i+1 ; j*j<=m ; j++)
{
if(m%j==0 && m/j!=j)
{
a=i;
b=j;
c=m/j;
flag=true;
break;
}
}
}
if(flag)
break;
}
if(!flag)
out.println("NO");
else
out.println("YES\n"+a+" "+b+" "+c);
}
}
}
//Special thanks to Petr (on codeforces) who inspired me to implement IO class!
static class IO
{
BufferedReader reader;
StringTokenizer tokenizer;
public IO() {
reader = new BufferedReader(new InputStreamReader(System.in));
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public String nextLine() {
String s="";
try {
s=reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
return s;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
i=1
c=2
l=[]
y=n
while(i<=2):
if n%c==0:
n=n//c
i=i+1
l.append(c)
if c>=y**0.5:
break
c=c+1
if l==[] or len(l)==1:
print("NO")
elif y%(l[0]*l[1])==0:
x=y//(l[0]*l[1])
if x not in l and x!=1:
print("YES")
l.append(x)
for i in range(len(l)):
if i==len(l)-1:
print(l[i])
else:
print(l[i],end=" ")
else:
print("NO")
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | T = int(input().strip())
import math
for t in range(T):
N = int(input().strip())
status = True
for a in range(2, int(math.sqrt(N)+1)):
if N%a==0:
b=N//a
for x in range(2, int(math.sqrt(b)+1)):
if x!=a and b%x==0 and a!=b//x and x!=b//x:
print('YES')
print(a, x, b//x)
status=False
break
if status==False:
break
if status==True:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
int main(void) {
long long t;
scanf("%lld", &t);
long long num, sub_num, i, j;
while (t--) {
int cnt = 1;
scanf("%lld", &num);
for (i = 2; (i * i) <= num; ++i) {
if (!cnt) {
break;
}
if (!(num % i)) {
sub_num = (num / i);
for (j = (i + 1); (j * j) < sub_num; ++j) {
if (!(sub_num % j)) {
printf("YES\n%lld %lld %lld\n", i, j, (sub_num / j));
--cnt;
break;
}
}
}
}
if (cnt) {
printf("NO\n");
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def factors(n):
s=[]
i=2
while(i*i<=n):
if(n%i==0):
s.append(i)
n=n//i
if(len(s)==2):
if(n not in s):
s.append(n)
break
i=i+1
if(len(s)<3):
print("NO")
else:
print("YES")
print(*s)
for _ in range(int(input())):
n=int(input())
l1=factors(n) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input().strip())
import math
def primeFactors(n):
ans = []
# Print the number of two's that divide n
while n % 2 == 0:
ans.append(2)
n = n // 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
result = int(math.sqrt(n))
for i in range(3,result+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
ans.append(i)
n = n // i
# Condition if n is a prime
# number greater than 2
if n > 2:
ans.append(n)
return ans
for _ in range(t):
n = int(input().strip())
ans = primeFactors(n)
ts = set(ans)
if len(ts) == 3 and len(ts) == len(ans):
a = ans[0]
b = ans[1]
c = 1
for i in range(2,len(ans)):
c *= ans[i]
print('YES')
print(a,b,c)
elif (len(ts) >= 2 and len(ans) >= 4):
a = ts.pop()
b = ts.pop()
c = 1
for i in range(len(ans)):
if ans[i] == a:
ans[i] = 1
break
for i in range(len(ans)):
if ans[i] == b:
ans[i] = 1
break
for i in range(len(ans)):
c *= ans[i]
print('YES')
print(a,b,c)
elif (len(ts) >= 1 and len(ans) >= 6):
a = ans[0]
b = ans[1] * ans[2]
c = 1
for i in range(3,len(ans)):
c *= ans[i]
print('YES')
print(a,b,c)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n = int(input())
if n < 24:
print("NO")
continue
j = -1
a, b = 0, 0
for i in range(2, int(n**(1 / 3)) + 1):
if n % i == 0:
a = i
n //= i
j = i + 1
break
if j == -1:
print("NO")
continue
o = -1
for i in range(j, int(n**(1 / 2)) + 1):
if n % i == 0:
b = i
n //= i
o = 1
break
if o == -1 or n <= b:
print("NO")
continue
print("YES")
print(a, b, n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from collections import defaultdict
def primeFactors(n):
d=defaultdict(int)
while n % 2 == 0:
d[2]+=1
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
d[i]+=1
n = n / i
if n > 2:
d[n]+=1
return d
t=int(input())
for i in range(t):
n=int(input())
d= primeFactors(n)
# print(d)
if len( d.keys() )>=3:
print("YES")
s=[]
ww=1
for j in list(d.keys())[:2]:
s.append( int(j) )
ww*=j**(d[j]-1)
for j in list(d.keys())[2:]:
ww*= int(j**d[j])
s.append(ww)
print(*s)
elif len(list(d.keys()))==1:
w,w1 = int(list(d.keys())[0]), int(d[list(d.keys())[0]])
if w1>=6:
print("YES")
ans = "{} {} {}".format(w,w**2,w**(w1-3))
print(ans)
else:
print("NO")
elif len(list(d.keys()))==2:
keys= list(map(int,list(d.keys())))
value = list(map(int,list(d.values())))
if sum(value)>=4:
print("YES")
ans = "{} {} {}".format( keys[0], keys[1], keys[0]**(value[0]-1) * keys[1]**(value[1]-1) )
print(ans)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # from debug import debug
import sys
t_ = int(input())
while t_:
t_-=1
n = int(input())
if n<24:
print("NO")
else:
i = 2
a = b = c = 0
while i*i<n:
if n%i == 0:
a = i
n = n//i
break
i+=1
if a == 0:
print("NO")
else:
i = 3
while i*i<n:
if n%i == 0 and i != a:
b = i
c = n//i
break
i+=1
if b<2 or c<2 or b == c:
print("NO")
else:
print("YES")
print(a,b,c)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int tst;
long long n, a, b, c;
vector<long long> v;
bool flag;
cin >> tst;
while (tst--) {
flag = false;
cin >> n;
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
v.push_back(i);
n /= i;
}
if (v.size() == 2) {
v.push_back(n);
if (v[0] != v[1] && v[1] != v[2]) flag = true;
break;
}
}
if (flag) {
cout << "YES" << endl;
sort(v.begin(), v.end());
cout << v[0] << " " << v[1] << " " << v[2] << endl;
} else
cout << "NO" << endl;
v.clear();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long n;
cin >> n;
long long temp = n;
vector<long long> v;
for (long long i = 2; i <= sqrt(n); i++) {
while (n % i == 0) {
v.push_back(i);
n /= i;
}
}
if (n != 1) {
v.push_back(n);
}
if (v.size() < 3) {
cout << "No" << endl;
continue;
}
long long fi = v[0];
long long se = v[1];
if (se == fi) {
se = se * v[2];
}
long long th = temp / (fi * se);
if (th != fi && th != se && th != 1) {
cout << "Yes" << endl;
cout << fi << " " << se << " " << th << endl;
} else
cout << "No" << endl;
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import Counter
from collections import defaultdict
import math
import random
import heapq as hq
from math import sqrt
import sys
from functools import reduce
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def tinput():
return input().split()
def rinput():
return map(int, tinput())
def rlinput():
return list(rinput())
mod = int(1e9)+7
def solve(adj, colors, n):
pass
def root(parent, a):
while parent[a] != a:
parent[a] = parent[parent[a]]
a = parent[a]
return a
def union(parent, size, a, b):
roota, rootb = root(parent, a), root(parent, b)
if roota == rootb:
return
parent[rootb] = roota
size[roota] += size[rootb]
size[rootb] = 0
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
if __name__ == "__main__":
for i in range(iinput()):
n=iinput()
temp1=factors(n)
temp=list(factors(n))
temp.sort()
l=len(temp)
flag=True
for i in range(1,l-1):
for j in range(i+1,l-1):
if n%(temp[i]*temp[j])==0:
s=n/(temp[i]*temp[j])
if s!=temp[i] and s!=temp[j] and s!=1 and s!=n and s in temp1:
print('YES')
print(temp[i],temp[j],int(s))
flag=False
break
if not flag:break
if flag:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import ceil
def find(n,p,ans):
for i in range(2,ceil(n**(1/p))+1,1):
if n%i==0:
if p==3:
ans+=find(n//i,2,[i])
if len(set(ans))==3:
return ans
ans=[]
elif p==2 and n//i>1:
new=ans+[(n//i),i]
if len(set(new))==3:
return new
return ans
for i in range(int(input())):
n=int(input())
k=find(n,3,[])
if len(k)==3:
print("YES")
print(*k,sep=" ")
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())):
n=int(input())
g=0
for i in range(2,math.floor(math.sqrt(n))+1):
if n%i==0:
if n//i!=i:
p=n//i
for j in range(2,math.floor(math.sqrt(i))+1):
if i%j==0:
if i//j!=j and j!=p and p!=i//j:
g=1
l=[n//i,j,i//j]
break
for j in range(2,math.floor(math.sqrt(p))+1):
if p%j==0:
if p//j!=j and i!=j and i!=p//j:
g=1
l=[j,p//j,i]
break
if g==0:
print("NO")
else:
print("YES")
print(*l) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool isPrime(int x) {
for (long long int i = 2; i * i <= x; i++) {
if (x % i == 0) return false;
}
return true;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int aux;
cin >> aux;
while (aux--) {
int n;
cin >> n;
int num = n;
vector<int> fat;
for (int d : {2, 3, 5}) {
while (n % d == 0) {
fat.push_back(d);
n /= d;
}
}
int incremento[] = {4, 2, 4, 2, 4, 6, 2, 6};
int aux = 0;
for (int i = 7; i * i <= n; i += incremento[(aux++) % 8]) {
while (n % i == 0) {
fat.push_back(i);
n /= i;
}
}
if (n > 1) fat.push_back(n);
if (fat.size() < 3) {
cout << "NO\n";
continue;
}
int p = fat[0];
int s = fat[1];
if (s == p) {
s *= fat[2];
}
int t = num / (p * s);
if (p != s and p != t and s != t and s >= 2 and p >= 2 and t >= 2) {
cout << "YES\n";
cout << p << ' ' << s << ' ' << t << '\n';
continue;
}
cout << "NO\n";
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
struct di {
int div1, div2;
};
struct divv {
int divv1, divv2, divv3;
};
di a = {0, 0};
divv b = {0, 0, 0};
int n, t;
di f(int nr) {
for (int d = 2; d * d <= nr; d++)
if (nr % d == 0) {
di x = {d, nr / d};
return x;
}
return a;
}
divv esteprod(int nr) {
for (int d = 2; d * d <= nr; d++) {
if (nr % d == 0) {
di x = f(d);
if (x.div1 != 0 and x.div1 != x.div2 and x.div1 != nr / d and
x.div2 != nr / d) {
divv z = {x.div1, x.div2, nr / d};
return z;
}
x = f(nr / d);
if (x.div1 != 0 and x.div1 != x.div2 and x.div1 != d and x.div2 != d) {
divv z = {x.div1, x.div2, d};
return z;
}
}
}
return b;
}
int main() {
cin >> t;
for (int tt = 1; tt <= t; tt++) {
cin >> n;
divv z = esteprod(n);
if (z.divv1 != 0) {
cout << "YES" << endl;
cout << z.divv1 << ' ' << z.divv2 << ' ' << z.divv3 << endl;
} else
cout << "NO" << endl;
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | N = int(input())
for _ in range(N):
x = int(input())
d = 2
ans = []
for steps in range(2):
if d > x:
break
while d * d <= x:
if x % d == 0:
ans.append(d)
x //= d
d += 1
break
d += 1
if len(ans) == 2 and x >= d:
print('YES')
print(*(ans + [x]))
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | ### C. Product of Three Numbers
for _ in range(int(input())):
n=int(input())
c=[]
i=2
while len(c)<2 and i**2<n:
if n%i==0:
c.append(i)
n=n//i
i+=1
if len(c)==2 and n not in c:
print('YES')
print(*c,n)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for i in range(t):
n=int(input())
f=0
l=[]
l1=[]
for j in range(2,int(math.sqrt(n))+1):
if n%j==0:
l.append(j)
l.append(n//j)
l1=list(set(l))
l1.sort()
if len(l1)==0:
print("NO")
else:
a=l1[0]
n=n//a
for j in range(len(l1)):
if n%l1[j]==0 and l1[j]!=a:
b=l1[j]
n=n//b
f=1
break
if f==0:
print("NO")
else:
if n in l1 and n!=a and n!=b:
print("YES")
print(a,b,n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def solve(n,i):
a=-1
b=-1
while (i*i<=n):
if (n%i)==0:
a=i
b=n//i
if a==b:
a=-1
b=-1
break
i+=1
return a,b
for t in range(int(input())):
n =int(input())
a,b = solve(n,2)
if a==-1:
print("NO")
else:
b,c = solve(b,a+1)
if b==-1:
print("NO")
else:
print("YES")
print(a,b,c) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #####################################
import atexit, io, sys, collections, math, heapq, fractions
buffer = io.BytesIO()
sys.stdout = buffer
@atexit.register
def write(): sys.__stdout__.write(buffer.getvalue())
####################################
def f(n):
u = 2
on = n
r = []
limit = int(on **0.5)
while(u <= limit):
if n % u == 0:
c = 0
while(n % u == 0):
n /= u
c+=1
r.append((u,c))
u+= 1
if n > 1:
r.append((n,1))
r.sort(key = lambda x:-x[1])
if len(r) >= 3:
print 'YES'
print ' '.join(map(str,[r[0][0], r[1][0], on / (r[0][0] * r[1][0])]))
return
if len(r) == 2:
if r[0][1] >= 3:
e = r[0][0]
print 'YES'
print ' '.join(map(str,[e, e ** 2, on / (e ** 3)]))
return
if r[1][1] >= 3:
e = r[1][0]
print 'YES'
print ' '.join(map(str,[e, e ** 2, on / (e ** 3)]))
return
a = r[0][0]
b = r[1][0]
if on / (a*b) not in [a,b,1]:
print 'YES'
print ' '.join(map(str,[a, b, on / (a * b)]))
return
print 'NO'
return
if len(r) == 1:
if r[0][1] >= 6:
e = r[0][0]
print 'YES'
print ' '.join(map(str,[e, e** 2, on / (e ** 3)]))
return
print 'NO'
return
for i in range(int(raw_input())):
n = int(raw_input())
f(n) | PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
long long int t = 0;
cin >> t;
while (t--) {
long long int n = 0;
cin >> n;
if (n < 24) {
cout << "NO\n";
continue;
} else {
vector<long long int> f;
for (long long int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
f.push_back(i);
n /= i;
}
}
f.push_back(n);
sort(f.begin(), f.end());
if (f.size() > 3) {
for (long long int i = 3; i < f.size(); ++i) {
f[2] *= f[i];
}
}
if (f.size() < 3)
cout << "NO\n";
else if (f[0] != f[1] && f[0] != f[2] && f[1] != f[2]) {
cout << "YES\n";
cout << f[0] << " " << f[1] << " " << f[2] << "\n";
} else
cout << "NO\n";
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
t = int(stdin.readline())
for _ in xrange(t):
n = int(stdin.readline())
a = []
i = 2
while True:
if i * i > n:
break
if n%i==0:
a.append(i)
n/=i
i+=1
break
i+=1
while True:
if i*i > n:
break
if n%i==0:
a.append(i)
a.append(n/i)
break
i+=1
a = list(set(a))
if len(a)==3:
print "YES"
print a[0],a[1],a[2]
else:
print "NO" | PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n = int(input())
i = 2
while i * i <= n:
if n % i == 0:
k = n // i
j = i + 1
fl = 0
while j * j <= k:
if k % j == 0 and j != k // j:
print("YES")
print(i, j, k // j)
fl = 1
break
j += 1
if fl:
break
i += 1
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | """
author : dokueki
"""
import math
def divisor(x):
i = 1
div = []
while i <= math.sqrt(x):
if x%i == 0:
if x//i == i:
div.append(i)
else:
div.extend([i,x//i])
i += 1
div.remove(max(div))
div.remove(min(div))
return div
for _ in range(int(input())):
n = int(input())
div1 = divisor(n)
if len(div1) == 0:
a = False
else:
a = min(div1)
if a:
div2 = sorted(divisor(n // a))
if len(div2) == 0:
b = False
else:
for i in div2:
if i > a:
b = i
break
else:
b = False
#print(div2)
if a and b:
c = n//(a*b)
if c > a and c > b :
pass
else:
c = False
else:
c = False
#print((a,b,c))
if a and b and c:
print("YES")
print(a,b,c)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
#sys.stdin = open("division.in","r")
#sys.stdout = open("division.out","w")
t = int(input())
for i in range(t):
n = int(input())
i = 2
flag = 0
while(i ** 2 <= n):
if n % i == 0:
a = i
j = 2
while(j ** 2 <= (n // a)):
if (n // a) % j == 0:
b = (n // a) // j
c = (n // a) // b
if a != b and a != c and b != c:
flag = 1
break
j += 1
i += 1
if flag:
print("YES")
print(a,b,c)
break
if not flag:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools
from collections import deque,defaultdict,OrderedDict
import collections
def primeFactors(n):
pf=[]
# Print the number of two's that divide n
while n % 2 == 0:
pf.append(2)
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
pf.append(int(i))
n = n /i
# Condition if n is a prime
# number greater than 2
if n > 2:
pf.append(int(n))
return pf
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
for _ in range(ri()):
n=ri()
a=primeFactors(n)
amult=1
alen=len(a)
for i in range(1,alen-1):
amult=amult*a[i]
# print(a)
t=0
if alen<3:
ws("NO")
else:
z=len((set(a)))
if z>=3:
ws("YES")
ans=[a[0],a[-1],amult]
t=1
if z==2:
if alen>=4:
ws("YES")
ans=[a[0],a[-1],amult]
t=1
if t==0:
ws("NO")
if z==1:
if alen>=6:
ws("YES")
ans=[a[0],a[0]*2,n//(a[0]**3)]
t=1
if t==0:
ws("NO")
if t==1:
wia(ans)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())):
n = int(input())
a = b = c = -1;
ok = 0;
for j in range(2, int(math.sqrt(n))+1):
if n % j == 0:
if a == -1:
a = j;
n /= a
elif b == -1:
b = j;
n /= b;
if (n >= 2 and n != a and n != b):
c = n;
ok = 1;
else:
break
if ok:
print("YES")
print(a, b, int(c))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
k=set()
a=int(input())
for i in range(2,10000):
if len(k)<2:
if a%i==0:
a=a//i
k.add(i)
elif len(k)==2 and a>2:
k.add(a)
if len(k)==3:
print('YES')
for i in sorted(k):
print(i,end=' ')
print()
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.ArrayList;
import java.util.Scanner;
public class ProductofThreeNumbers {
public static ArrayList<Integer> factors;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
getFactors(n);
//brute force
boolean found = false;
for (int i=0;i<factors.size()-2;i++) {
for (int j=i+1;j<factors.size()-1;j++) {
for (int k=j+1;k<factors.size();k++) {
int mul = (factors.get(i) * factors.get(j) * factors.get(k));
if (mul == n) {
//print the numbers
System.out.println("YES");
System.out.println(factors.get(i) + " " + factors.get(j) + " " + factors.get(k));
found = true;
i = factors.size();
j = factors.size();
k = factors.size();
break;
}else if (mul > n) {
//no need to continue
break;
}
}
}
}
if (!found) {
System.out.println("NO");
}
}
in.close();
}
public static void getFactors(int n) {
factors = new ArrayList<>();
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0)
factors.add(i);
}
int size = factors.size() - 2;
if (factors.size() == 0) {
return;
}
if (factors.get(size + 1) * factors.get(size + 1) != n) {
factors.add(n / factors.get(size + 1));
}
for (int i=size;i>=0;i--) {
factors.add(n / factors.get(i));
}
}
public static void printFactors() {
for (int i=0;i<factors.size();i++) {
System.out.print(factors.get(i) + " ");
}
System.out.println();
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def primefactor(n):
l = []
while n%2==0:
l.append(2)
n = n//2
for i in range(3,int(n**0.5)+1,2):
if n % i == 0:
l.append(i)
n//=i
if n>2:
l.append(n)
return l
for _ in range(int(input())):
n = primefactor(int(input()))
l = []
if len(n)==3:
if len(set(n))==3:
print("YES")
print(*n)
else:
print("NO")
elif len(n)>3:
if n[0]==n[1]:
l.append(n[0])
l.append(n[1]*n[2])
c=1
for i in range(3,len(n)):
c *= n[i]
l.append(c)
if len(set(l))!=3:
print("NO")
else:
print("YES")
print(*l)
else:
l.append(n[0])
l.append(n[1])
c=1
for i in range(2,len(n)):
c *= n[i]
l.append(c)
if len(set(l))!=3:
print("NO")
else:
print("YES")
print(*l)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a[]=new int [(int)5e4+10];
ArrayList<Integer> al=new ArrayList<Integer>();
Arrays.fill(a,1);
for(int i=2;i<a.length;i++) {
if(a[i]==1) {
for(int j=2*i;j<a.length;j+=i) a[j]=0;
al.add(i);
}
}
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int g=0,ans=0;
for(int i=0;al.get(i)<=Math.sqrt(n);i++) {
if(n%al.get(i)==0) {ans=al.get(i);n/=ans;break;}
}
if(ans==0) {System.out.println("NO");continue;}
for(int i=2;i<=Math.sqrt(n);i++) {
if(i==ans) continue;
if(n%i==0&&n/i!=i&&n/i!=ans) {
System.out.println("YES\n"+ans+" "+i+" "+(n/i));
g=1;
break;
}
}
if(g==0)
System.out.println("NO");
}
sc.close();
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t=int(input())
for _ in range(t):
n=int(input())
n_root=int(math.sqrt(n))
ans=[]
count=0
for i in range(2,n_root+1):
if n%i==0:
ans.append(i)
count+=1
n//=i
if count==2 and n>ans[-1]:
count+=1
ans.append(n)
break
if count==3:
print("YES")
print(*ans)
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
for _ in range(int(input())):
n=(int(input()))
f=0
s=0
k=0
if(n<24):
f=1
else:
for i in range(2,int(math.sqrt(n))+1):
if(n%i==0):
f=0
k=i
break
if(k==int(math.sqrt(n))):
f=1
#print(f)
if(n>=24 and f==0):
f=1
for i in range(2,int(math.sqrt(n))+1):
if(n%i==0):
x=n//i
a=i
#print(x,i)
for j in range(2,int(math.sqrt(x))+1):
b=j
c=x//j
if(x%j==0):
if(a!=b and b!=c and c!=a):
f=0
#print(a,b,c)
break
if(f==0):
break
#print(f)
if(f==0):
print('YES')
print(a,b,c)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.PrintWriter;
import java.util.*;
public class TP {
public static PrintWriter out = new PrintWriter(System.out);
public static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
long s = System.currentTimeMillis();
int t = 1;
t = ni();
while (t-- > 0) solve();
out.flush();
tr(System.currentTimeMillis() - s + "ms");
}
private static void solve() {
int n = ni();
Set<Integer> set = new HashSet<>();
int i = 0;
int N = n;
for (int d = 2; d * d <= n; d++) {
if (n % d == 0) {
set.add(d);
n /= d;
if (set.size() >= 2) break;
}
}
set.add(n);
if (set.size() == 3) {
System.out.println("yes");
set.forEach((I) -> System.out.print(I + " "));
System.out.println();
} else
System.out.println("no");
}
static boolean isPrime(int n) {
for (int i = 2; i * i <= n; i++) if (n % i == 0) return false;
return true;
}
static boolean sorted(int[] a) {
for (int i = 0; i < a.length - 1; i++) {
if (a[i] > a[i + 1]) return false;
}
return true;
}
public static int gcd(int a, int b) {
while (b != 0) {
a %= b;
int t = a;
a = b;
b = t;
}
return a;
}
private static int ni() {
return in.nextInt();
}
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 long[] nal(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nl();
return a;
}
private static long nl() {
return in.nextLong();
}
private float nf() {
return in.nextFloat();
}
private static double nd() {
return in.nextDouble();
}
public static int[] facs(int n, int[] primes) {
int[] ret = new int[9];
int rp = 0;
for (int p : primes) {
if (p * p > n) break;
int i;
i = 0;
while (n % p == 0) {
n /= p;
i++;
}
if (i > 0) ret[rp++] = p;
}
if (n != 1) ret[rp++] = n;
return Arrays.copyOf(ret, rp);
}
public static int[] sieveEratosthenes(int n) {
if (n <= 32) {
int[] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
for (int i = 0; i < primes.length; i++) {
if (n < primes[i]) {
return Arrays.copyOf(primes, i);
}
}
return primes;
}
int u = n + 32;
double lu = Math.log(u);
int[] ret = new int[(int) (u / lu + u / lu / lu * 1.5)];
ret[0] = 2;
int pos = 1;
int[] isnp = new int[(n + 1) / 32 / 2 + 1];
int sup = (n + 1) / 32 / 2 + 1;
int[] tprimes = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
for (int tp : tprimes) {
ret[pos++] = tp;
int[] ptn = new int[tp];
for (int i = (tp - 3) / 2; i < tp << 5; i += tp) ptn[i >> 5] |= 1 << (i & 31);
for (int j = 0; j < sup; j += tp) {
for (int i = 0; i < tp && i + j < sup; i++) {
isnp[j + i] |= ptn[i];
}
}
}
// 3,5,7
// 2x+3=n
int[] magic = {
0, 1, 23, 2, 29, 24, 19, 3, 30, 27, 25, 11, 20, 8, 4, 13, 31, 22, 28, 18, 26, 10, 7, 12, 21,
17, 9, 6, 16, 5, 15, 14
};
int h = n / 2;
for (int i = 0; i < sup; i++) {
for (int j = ~isnp[i]; j != 0; j &= j - 1) {
int pp = i << 5 | magic[(j & -j) * 0x076be629 >>> 27];
int p = 2 * pp + 3;
if (p > n) break;
ret[pos++] = p;
if ((long) p * p > n) continue;
for (int q = (p * p - 3) / 2; q <= h; q += p) isnp[q >> 5] |= 1 << q;
}
}
return Arrays.copyOf(ret, pos);
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj) System.out.println(Arrays.deepToString(o));
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def pro(j):
for a in range(2,int(j**(1/3))+1):
if j%a==0:
for b in range(a+1,int(j**(1/2))):
if (j//a)%b==0 and (j//a)//b!=a and ((j//a)//b)!=int((j//a)**(1/2)):
print("YES")
print(a,b,(j//a)//b)
return
print("NO")
return
t=int(input())
l=[]
for i in range(t):
n=int(input())
l.append(n)
for j in l:
pro(j) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
public class ProductOfThreeNumbers{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++){
long n=sc.nextLong();
long arr[]=new long[3];
int c=0;
for(int j=2;j<=Math.sqrt(n);j++){
if(n%j==0){
arr[c++]=j;n/=j;
}
if(c==2){arr[c++]=n;break;}
}
if(c==3&&arr[0]!=arr[2]&&arr[1]!=arr[2])System.out.println("Yes\n"+arr[0]+" "+arr[1]+" "+arr[2]);
else System.out.println("NO");
}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def main():
n = int(input())
a = 2
while a * a * a < n:
if n % a == 0:
b = a + 1
while b * b < n // a:
if (n // a) % b == 0:
c = n // a // b
if a != c and b != c:
print("YES")
print(a, b, n // a // b)
return 0
b += 1
a += 1
print("NO")
p = int(input())
for i in range(p):
main() | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def factor(n):
ans=[]
d=2
while d*d<=n:
if n%d==0:
ans.append(d)
n=n//d
else:
d+=1
if n>1:
ans.append(n)
return ans
t=int(input())
for i in range(t):
n=int(input())
ans=factor(n)
if len(ans)<3:
print("NO")
else:
a=ans[0]
b=1
c=1
for i in range(1,len(ans)):
if b<=a:
b*=ans[i]
else:
c*=ans[i]
if c==1 or c==a or c==b:
print('NO')
else:
print("YES")
print(a,b,c) | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n = int(input())
c = 0
a = []
for i in range(2, int(n**0.5)+1):
if n % i == 0:
a.append(i)
n = n//i
if len(a) == 2:
break
if len(a) == 2 and n > a[1]:
print('YES')
print(a[0],a[1],n)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> v;
for (int i = 2; i <= sqrt(n); i++) {
while (n % i == 0) {
v.push_back(i);
n /= i;
}
}
if (n > 1) v.push_back(n);
set<int> s;
int cur = 1;
for (auto it : v) {
cur = cur * it;
if (s.size() <= 1 && s.find(cur) == s.end()) {
s.insert(cur);
cur = 1;
}
}
if (cur == 1 || s.find(cur) != s.end()) {
cout << "NO\n";
} else {
s.insert(cur);
cout << "YES\n";
for (auto it : s) cout << it << " ";
cout << "\n";
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def de(m):
ret = []
n = m
for i in range(2,int(math.sqrt(n))+1):
while n%i == 0:
ret.append(i)
n/= i
if n!= 1:
ret.append(n)
return ret
def solve(n):
lst = de(n)
args = list(set(lst))
if len(args) >= 3:
print "YES"
print args[0],args[1],n/args[0]/args[1]
elif len(args) == 2:
if len(lst)<=3:
print "NO"
else:
print "YES"
print args[0],args[1],n/args[0]/args[1]
elif len(args) == 1:
if len(lst) >= 6:
print "YES"
print args[0],args[0]*args[0],n/args[0]/args[0]/args[0]
else:
print 'NO'
t = input()
for _ in range(t):
solve(input()) | PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | n=int(input())
for i in range(n):
num=int(input())
list1=[]
j=2
while(j*j<num and len(list1)<2):
if(num%j==0):
num=num//j
list1.append(j)
j+=1
if(len(list1)==2 and num not in list1):
print("YES")
for k in list1:
print(k,end=" ")
print(num)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | ''' Hey stalker :) '''
INF = 10**10
def main():
#print = out.append
''' Cook your dish here! '''
n = get_int()
factors = prime_factors(n)
keys = list(factors.keys())
if len(factors)>=2:
a, b = keys[:2]
c = n // (a * b)
if c==a or c==b or c<2:
print("NO")
return
else:
if factors[keys[0]]>5:
a = keys[0]
b = keys[0]**2
else:
print("NO")
return
c = n // (a * b)
print("YES")
print(a,b,c)
def prime_factors(n): # n**0.5 complex
factors = dict()
for i in range(2, math.ceil(math.sqrt(n)) + 1):
while n % i == 0:
if i in factors:
factors[i] += 1
else:
factors[i] = 1
n = n // i
if n > 2:
factors[n] = 1
return (factors)
''' Pythonista fLite 1.1 '''
import sys
from collections import defaultdict, Counter
from bisect import bisect_left, bisect_right
#from functools import reduce
import math
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
out = []
get_int = lambda: int(input())
get_list = lambda: list(map(int, input().split()))
#main()
[main() for _ in range(int(input()))]
print(*out, sep='\n')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void fastIo() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
const long long N = 100000;
bitset<N + 5> p;
vector<long long> primes;
void spf() {
primes.push_back(2);
for (int i = 4; i <= N; i += 2) p[i] = 1;
for (long long i = 3; i <= N; i += 2) {
if (p[i] == 0) {
primes.push_back(i);
for (long long j = i * i; j <= N; j += i) p[j] = 1;
}
}
}
vector<long long> factorise(long long n) {
vector<long long> ans;
for (long long i = 0; primes[i] * primes[i] <= n; i++) {
if (n % primes[i] == 0) {
while (n % primes[i] == 0) {
ans.push_back(primes[i]);
n /= primes[i];
}
}
}
if (n != 1) ans.push_back(n);
return ans;
}
void solve() {
long long n;
cin >> n;
vector<long long> ans = factorise(n);
if (ans.size() < 3) {
cout << "NO\n";
return;
}
long long a = ans[0], b = ans[1], c = 1, s = 1;
if (a == b && ans.size() >= 4)
b *= ans[2], s = 2;
else if (a == b) {
cout << "NO\n";
return;
}
for (int i = s + 1; i < ans.size(); i++) c *= ans[i];
if (c == 1 || c == a || c == b) {
cout << "NO\n";
return;
}
cout << "YES\n" << a << " " << b << " " << c << endl;
}
int main() {
spf();
fastIo();
int t;
cin >> t;
while (t--) solve();
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | '''
5
64
32
97
2
12345
'''
def mi():
return map(int, input().split())
for _ in range(int(input())):
n = int(input())
i = 2
found=0
while i*i<=n:
if (n % i == 0):
n1 = n//i
j=2
comp=0
while j*j<=n1:
if n1%j==0:
comp=1
if i==n1//j or i==j or (n1//j==j) or n//j==1 or j==1:
j+=1
continue
else:
if found or n1//j==1 or j==1:
break
found = 1
print('YES')
print(i,n1//j,j)
j+=1
i+=1
if not found:
print ('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for _ in range(int(input())):
n=int(input())
ans=[]
for i in range(2,n):
if i*i>=n: break
if n%i==0:
ans.append(i)
n//=i
if len(ans)==2:
ans.append(n)
break
if len(ans)==3:
print('YES')
print(*ans)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.PrintWriter;
import java.util.Scanner;
public class ProductOfThreeNumbers {
public static Scanner sc = new Scanner(System.in);
public static PrintWriter out = new PrintWriter(System.out, true);
public static int n;
public static void Input() {
n = sc.nextInt();
}
public static int smallestDivisor(int x, int min)
{
int ans=0;
for (int i=min; i*i<=x; i++)
{
if (x % i == 0)
{
ans=i;
break;
}
}
return ans;
}
public static void Solve() {
int a=smallestDivisor(n, 2);
if (a>=n || a==0) {System.out.println("NO"); return;}
int b=0;
int min=2;
while (true)
{
b=smallestDivisor(n/a, min);
if (b>=n/a || b==0) {System.out.println("NO"); return;}
if (b>0 && b!=a) {break;}
min=b+1;
}
int c = n/a/b;
if (c==a || c==b) {System.out.println("NO"); return;}
System.out.println("YES");
System.out.println(a + " " + b + " " + c);
}
public static void main(String[] args) {
int t = sc.nextInt();
while (t-- > 0) {Input(); Solve();}
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
def Prime(d, n):
if n%d == 0:
return d
while d * d <= n:
if n % d == 0:
return d
d= d + 1
return 0
for _ in range(int(stdin.readline())):
n = int(stdin.readline())
if n < 24 :
print("NO")
continue
ans = []
count = 0
cond = True
d = 2
while(count<2):
m = Prime(d, n)
if m == 0:
cond = False
break
else:
ans.append(m)
count += 1
n = n//m
d = m+1
if cond == True:
if n in ans:
print("NO")
else:
ans.append(n)
print("YES")
print(*ans)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
I=sys.stdin.readline
ans=""
for _ in range(int(I())):
n=int(input())
fac=[]
for i in range(2,int(n**.5)):
if n%i==0:
fac.append((i,n//i))
break
#print(fac)
if len(fac)!=0:
x=fac[0][1]
flag=1
for i in range(2,int(x**.5)+1):
if x%i==0 and i!=fac[0][0]:
if i!=x//i:
ans+="YES\n"
ans+="{} {} {}\n".format(fac[0][0],i,x//i)
flag=0
break
if flag:
ans+="NO\n"
else:
ans+="NO\n"
print(ans)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
for (int g = 0; g < t; ++g) {
int n;
cin >> n;
vector<int> v;
for (int i = 2; i <= int(sqrt(n)); ++i) {
if (n % i == 0) {
v.push_back(i);
v.push_back(n / i);
}
}
bool T = false;
for (int i = 0; i < v.size(); ++i) {
for (int j = i + 1; j < v.size(); ++j) {
for (int k = j + 1; k < v.size(); ++k) {
if (v[i] * v[j] * v[k] == n) {
cout << "YES" << '\n' << v[i] << ' ' << v[j] << ' ' << v[k] << '\n';
T = true;
break;
}
}
if (T) break;
}
if (T) break;
}
if (!T) cout << "NO" << '\n';
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | /*
javac c.java && java c
*/
import java.io.*;
import java.util.*;
public class c {
public static void main(String[] args) { new c(); }
FS in = new FS();
PrintWriter out = new PrintWriter(System.out);
int t;
int n;
c() {
t = in.nextInt();
while (t-- > 0) {
n = in.nextInt();
int[] ans = new int[3];
int lp = 2;
boolean found = false;
for (; lp * lp <= n; lp++) {
if (n %lp == 0) {
found = true;
n /= lp;
ans[0] = lp;
break;
}
}
if (!found) {
out.println("NO");
}
else {
int nlp = ans[0] + 1;
found = false;
for (; nlp * nlp <= n; nlp++)
if (n %nlp == 0) {
found = true;
n /= nlp;
ans[1] = nlp;
break;
}
if (!found) out.println("NO");
else {
if (n < 2 || n == nlp) out.println("NO");
else {
out.println("YES");
out.println(ans[0] + " " + ans[1] + " " + n);
}
}
}
}
out.close();
}
int abs(int x) { if (x < 0) return -x; return x; }
long abs(long x) { if (x < 0) return -x; return x; }
int min(int a, int b) { if (a < b) return a; return b; }
int max(int a, int b) { if (a > b) return a; return b; }
long min(long a, long b) { if (a < b) return a; return b; }
long max(long a, long b) { if (a > b) return a; return b; }
int gcd(int a, int b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
long gcd(long a, long b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
long lcm(int a, int b) { return (((long) a) * b) / gcd(a, b); }
long lcm(long a, long b) { return (a * b) / gcd(a, b); }
void sort(int[] arr) {
int sz = arr.length, j;
Random r = new Random();
for (int i = 0; i < sz; i++) {
j = r.nextInt(i + 1);
arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
} Arrays.sort(arr);
}
class FS {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (Exception e) {}
} return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() { return Double.parseDouble(next()); }
}
}
| JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def isPrime(n):
if n == 2 or n == 3: return True
if n % 2 == 0 or n < 2: return False
for i in range(3, int(n ** 0.5) + 1, 2):
if n % i == 0:
return False
return True
t = int(input())
for _ in range(t):
f = 0
n = int(input())
if isPrime(n):
print("NO")
continue
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
if not isPrime(n // i):
m = n // i
for j in range(2, int(m ** 0.5) + 1):
if m % j == 0 and i != j != (m // j) != i:
print("YES")
print(i, j, m // j)
f = 1
break
if f:
break
if not f:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
for q in range(t):
n=int(input())
c=[]
i=2
while len(c)<2 and i*i<n:
if n%i==0:
c.append(i)
n=n/i
i=i+1
if len(c)==2 and n not in c:
print("YES")
print(*c,int(n))
else:
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long ans[3];
int main() {
int t;
cin >> t;
long long n;
for (int o = 0; o < t; o++) {
cin >> n;
vector<long long> z;
long long r = sqrt(n) + 1;
for (long long i = 2; i <= r; i++) {
{
while (n % i == 0) {
n /= i;
z.push_back(i);
}
}
}
if (n != 1) {
z.push_back(n);
}
long long prev = 1;
set<long long> anses;
for (auto x : z) {
prev *= x;
if (anses.size() >= 3) {
ans[2] *= x;
continue;
}
if (anses.find(prev) == anses.end()) {
anses.insert(prev);
ans[anses.size() - 1] = prev;
prev = 1;
}
}
if (anses.size() >= 3) {
cout << "YES" << endl;
cout << ans[0] << " " << ans[1] << " " << ans[2] << endl;
} else {
cout << "NO" << endl;
}
z.clear();
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
t = int(input())
for i in range(t):
n = int(input())
q = int( math.sqrt(n))
l = set()
for j in range(2, q + 1):
if n % j == 0:
l.add(j)
if len(l) < 2:
print('NO')
else:
l = sorted(list(l))
res = False
for k in range(len(l) - 1):
for p in range(k + 1, len(l)):
a = l[k]
b = l[p]
c = (n // a )// b
if a * b * c == n and b != c and a != c:
res = True
print("YES")
print(str(a) + ' ' + str(b) + ' ' + str(c))
break
if res:
break
if not res:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import defaultdict
for _ in range(int(input())):
n = int(input())
dr = n
i = 2
primes = defaultdict(int)
while i * i <= n:
while dr % i == 0:
primes[i] += 1
dr //= i
i += 1
if dr != 1:
primes[dr] += 1
if len(primes) >= 3:
print("YES")
i = 0
for par in primes:
print(par, end=" ")
n //= par
i += 1
if i == 2:
break
print(n)
elif len(primes) == 2:
primes = [(k, v) for k, v in primes.items()]
if primes[0][1] == 1:
if primes[1][1] <= 2:
print("NO")
continue
else:
print("YES")
print(primes[0][0], primes[1][0], n // (primes[0][0] * primes[1][0]))
elif primes[1][1] == 1:
if primes[0][1] <= 2:
print("NO")
continue
else:
print("YES")
print(primes[0][0], primes[1][0], n // (primes[0][0] * primes[1][0]))
else:
print("YES")
print(primes[0][0], primes[1][0], n // (primes[0][0] * primes[1][0]))
else:
primes = [(k, v) for k, v in primes.items()]
primes = primes[0]
if primes[1] >= 6:
print("YES")
print(primes[0], primes[0] ** 2, n // (primes[0] ** 3))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
def primes(n):
while n % 2 == 0:
prime.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
prime.append(i)
n = n / i
if n > 2:
prime.append(int(n))
t=int(input())
for _ in range(t):
n=int(input())
prime=[]
primes(n)
p={}
for i in prime:
if i in p.keys():
p[i]+=1
else:
p[i]=1
g=list(p.keys())
if len(g)==1:
if p[g[0]]>=6:
print("YES")
p1=p[g[0]]-2
print(g[0],g[0]**2,g[0]**(p[g[0]]-3))
else:
print("NO")
continue
if len(g)==2:
if p[g[0]]+p[g[1]]>3:
print("YES")
n1 = g[0]
n2 = g[1]
p1 = p[g[0]]
p2 = p[g[1]]
print(n1, n2, (pow(n1, p1 - 1) * pow(n2, p2 - 1)))
else:
print("NO")
else:
print("YES")
ans=pow(g[0],p[g[0]]-1)*pow(g[1],p[g[1]]-1)
for i in range(2,len(g)):
ans*=pow(g[i],p[g[i]])
print(g[0],g[1],ans)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
public class CodeForces1294C{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
for(int j = 0;j<t;j++){
int n = input.nextInt();
ArrayList<Integer> arr = new ArrayList<>();
int m = n;
for(int i = 2;i<=Math.sqrt(m);i++){
if(n%i == 0){
if(arr.size() == 2){
break;
}
arr.add(i);
n/= i;
}
}
if(arr.size() < 2 || arr.contains(n)){
System.out.println("NO");
}
else{
System.out.println("YES");
System.out.println(arr.get(0) + " " + arr.get(1) + " " + n);
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import floor, sqrt, ceil
def findPrimeFac(num):
ans = []
lim = ceil(sqrt(num))
while num % 2 == 0:
ans.append(2)
num //= 2
for p in range(3, lim, 2):
while num % p == 0:
ans.append(p)
num //= p
if num != 1:
ans.append(num)
return ans
t = int(input())
for _ in range(t):
n = int(input())
primeFacs = findPrimeFac(n)
#print(primeFacs)
freq = {}
for p in primeFacs:
if p not in freq:
freq[p] = 0
freq[p] += 1
numDist = 0
ans = []
l = list(freq.keys())
if len(l) >= 2:
a = l[0]
b = l[1]
c = n // (a * b)
if c != a and c != b and c != 1:
print("YES")
assert n == a*b*c
print(a, b, c)
else:
print("NO")
elif len(l) == 0 or freq[l[0]] < 6:
print("NO")
else:
print("YES")
p = l[0]
print(p, p*p, n//(p ** 3))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
def szuk_podz(liczba, a = 1):
for x in range(2, int(sqrt(liczba)) + 1):
if liczba % x == 0 and x != a:
return x
return -1
for _ in range(int(input())):
n = int(input())
a, b, c = szuk_podz(n), 0, 0
x = n // a
if a == -1:
print("NO")
continue
b = szuk_podz(x, a)
c = x // b
if c == a or c == b:
print("NO")
continue
if c > 0 and b > 0:
print("YES")
print(a, b, c)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for kek in range(t):
n = int(input())
i = 2
s = list()
while i < n ** (1 / 2) + 1:
if n % i == 0:
s.append(i)
n //= i
else:
i += 1
if n != 1:
s.append(n)
a = list()
a.append(s[0])
x = 1
for i in range(1, len(s)):
x *= s[i]
if x not in a:
a.append(x)
x = 1
else:
continue
a[-1] *= x
if len(a) >= 3:
x = 1
for i in range(2, len(a)):
x *= a[i]
print('YES')
print(a[0], a[1], x)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
out = []
n = int(input())
d = 2
while d * d <= n:
if n % d == 0:
out.append(d)
n //= d
if len(out) == 2:
break
d += 1
if len(out) == 2 and n > out[1]:
print("YES")
print(out[0], out[1], n)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
for xyz in range(t):
n=int(input())
a=[]
fa=1
p=n
i=2
while(i*i<=p):
if p%i==0 and i not in a:
a.append(i)
p=p/i
break
i=i+1
while(i*i<=p):
if p%i==0 and i not in a:
a.append(i)
p=p/i
break
i=i+1
l=len(a)
if l==0:
print('NO')
if l==1:
f=a[0]
s=a[0]*a[0]
t=n//(f*s)
if f*s*t==n and f!=s and s!=t and t!=f and t>1:
print('YES')
print(f,s,t)
else:
print('NO')
if l==2:
f=min(a)
c=a.copy()
for i in c:
if i==f:
c.remove(i)
s=min(c)
t=n//(f*s)
if f*s*t==n and f!=s and s!=t and t!=f and t>1:
print('YES')
print(f,s,t)
else:
print('NO')
if l>=3:
f=min(a)
c=a.copy()
for i in c:
if i==f:
c.remove(i)
s=min(c)
t=n//(f*s)
print('YES')
print(f,s,t)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import os,sys,math,random
def dep(a):
a0 = a
res = []
for b in range(2, int(a**(1/3))+1):
if a % b == 0:
res.append(b)
break
if len(res) == 0:
return "NO"
a /= res[0]
for b in range(2, int(a**(1/2))+1):
if a % b == 0 and b != res[0]:
res.append(b)
break
if len(res) == 1:
return "NO"
c = int(a0 / res[0] / res[1])
if c == 1 or c == res[0] or c == res[1]:
return "NO"
res.append(c)
return "YES\n" + " ".join(map(str, res))
T = int(input())
for _ in range(T):
a = int(input())
print(dep(a))
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import defaultdict
import math
def prime(n):
ans=[]
al=defaultdict(int)
while(n%2==0):
al[2]=al[2]+1
ans.append(2)
n=n//2
for i in range(3,int(math.sqrt(n))+1,2):
while(n%i==0):
al[i]+=1
ans.append(i)
n=n//i
if(n>2):
al[n]=1
ans.append(n)
ans=list(set(ans))
return ans,al;
t=int(input())
for i in range(t):
n=int(input())
ans,al=prime(n)
if(len(ans)>=3):
print('YES')
a=ans[0]**al[ans[0]]
b=ans[1]**al[ans[1]]
c=n//(a*b)
print(a,b,c)
elif(len(ans)==2):
if(al[ans[0]]+al[ans[1]]>=4):
a=ans[0]
b=ans[1]
c=n//(a*b)
print('YES')
print(a,b,c)
else:
print('NO')
else:
if(al[ans[0]]>5):
print('YES')
a=ans[0]*ans[0]
b=ans[0]
c=n//(a*b)
print(a,b,c)
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
class MergeFind:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
# self.lista = [[_] for _ in range(n)]
def find(self, a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self, a, b):
a = self.find(a)
b = self.find(b)
if a == b:
return
if self.size[a] < self.size[b]:
a, b = b, a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
# self.lista[a] += self.lista[b]
# self.lista[b] = []
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
# #
# to find factorial and ncr
# tot = 100005
# mod = 10**9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, tot + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
def comb(n, r):
if n < r:
return 0
else:
return fac[n] * (finv[r] * finv[n - r] % mod) % mod
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def arr1d(n, v):
return [v] * n
def arr2d(n, m, v):
return [[v] * m for _ in range(n)]
def arr3d(n, m, p, v):
return [[[v] * p for _ in range(m)] for i in range(n)]
def ceil(a, b):
return (a + b - 1) // b
# co-ordinate compression
# ma={s:idx for idx,s in enumerate(sorted(set(l+r)))}
# mxn=100005
# lrg=[0]*mxn
# for i in range(2,mxn-3):
# if (lrg[i]==0):
# for j in range(i,mxn-3,i):
# lrg[j]=i
def solve():
m = N()
c = 2
p = []
while len(p) < 2 and c * c < m:
if m % c == 0:
m = m // c
p.append(c)
c += 1
if len(p) == 2 and m not in p:
print("YES")
print(*(p+[m]))
else:
print("NO")
# solve()
testcase(int(inp()))
| PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | # -*- coding: utf-8 -*-
"""
Created on Wed Jan 22 18:45:17 2020
@author: npuja
"""
"""
t=int(input())
for i in range(0,t):
a,b,c,n=(int(k) for k in input().split())
mac=0
if a>=b and a>=c:
mac=a
elif b>=a and b>=c:
mac=b
else:
mac=c
test=n-3*(mac)+a+b+c
if test<0 or test%3!=0:
print("NO")
else:
print("YES")
"""
# Product of three no
t=int(input())
for i in range(0,t):
n=int(input())
l=int(n**(1/3))
a=1
for k in range(2,l+1):
if n%k==0:
a=k
break
b=1
c=1
if a==1:
print("NO")
else:
m=n/a
l=int(m**(1/2))
for k in range(a+1,l+1):
if m%k==0:
b=k
c=int(m/b)
break
if a!=1 and b!=1 and c!=1 and c!=b:
print("YES")
print(a,b,c)
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
public class Solution{
BufferedReader br;
PrintWriter out;
static String input[];
static int id = 0;
public String next() throws Exception{
if(input == null || input.length == id){
input = br.readLine().split(" ");
id=0;
}
return input[id++];
}
public String ns() throws Exception{
return next();
}
public int ni() throws Exception{
return Integer.parseInt(next());
}
public long nl() throws Exception{
return Long.parseLong(next());
}
public double nd() throws Exception{
return Double.parseDouble(next());
}
public int[] na(int n) throws Exception{
int[] arr = new int[n];
for(int i=0;i<n;i++){
arr[i] = ni();
}
return arr;
}
public void run() throws Exception{
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
int T=1;
for(int t=1;t<=T;t++){
solve();
}
out.flush();
}
public static void main(String[] args) throws Exception{new Solution().run();}
public void solve() throws Exception{
int[] spf = SPF(100000);
ArrayList<Integer> primes = getPrimes(spf);
int x = primes.size();
int T = ni();
for(int t=0;t<T;t++){
int n = ni();
int sqrt = (int)Math.sqrt(n);
int a=1;
int i=0;
while(i<x && primes.get(i) <= sqrt){
if(n%primes.get(i) == 0){
a = primes.get(i);
break;
}
i++;
}
if(a == 1){
out.println("NO");
continue;
}
n = n/a;
i = a + 1;
int b = 1;
while(i <= sqrt){
if(n%i== 0){
b = i;
if(n/b == a || b*b == n){
b=1;
}
break;
}
i++;
}
if(b==1){
out.println("NO");
continue;
}
out.println("YES");
out.println(a+" "+b+" "+n/b);
}
}
ArrayList<Integer> getPrimes(int[] spf){
int n = spf.length;
ArrayList<Integer> res = new ArrayList<Integer>();
res.add(2);
for(int i=3; i<n; i+=2){
if(spf[i]==i) res.add(i);
}
return res;
}
public int[] SPF(int n){
int[] arr = new int[n+1];
for(int i=2;i<=n;i+=2){
arr[i] = 2;
arr[i-1] = i-1;
}
for(int i=3;i*i <= n; i++){
if(arr[i]==i){
for(int j=i*i; j<=n; j+=i){
if(arr[j] == j){
arr[j]=i;
}
}
}
}
return arr;
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def product_of_3_numbers(n):
for i in range(2, int(n ** (1 / 2)) + 1):
if n % i == 0:
d = n // i
for j in range(i + 1, int(d ** (1 / 2)) + 1):
if d % j == 0 and j != d // j:
print("YES")
print(i, j, d // j)
return
print("NO")
t = int(input())
for _ in range(t):
n = int(input())
product_of_3_numbers(n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def lsi():
x=list(stdin.readline())
x.pop()
return x
def si(): return stdin.readline()
def sieve(x):
a=[True]*(x+1)
sq=floor(sqrt(x))
for i in range(3, sq+1, 2):
if a[i]:
for j in range(i*i, x+1, i):
a[j]=False
if x>1:
p=[2]
else:
p=[]
for i in range(3, x+1, 2):
if a[i]:
p.append(i)
return p
#vowel={'a', 'e', 'i', 'o', 'u', 'y', 'A', 'E', 'I', 'O', 'U', 'Y'}
#pow=[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592, 17179869184, 34359738368, 68719476736, 137438953472, 274877906944, 549755813888, 1099511627776, 2199023255552, 4398046511104, 8796093022208, 17592186044416, 35184372088832, 70368744177664, 140737488355328, 281474976710656, 562949953421312, 1125899906842624, 2251799813685248, 4503599627370496, 9007199254740992, 18014398509481984, 36028797018963968, 72057594037927936, 144115188075855872, 288230376151711744, 576460752303423488, 1152921504606846976, 2305843009213693952, 4611686018427387904, 9223372036854775808]
############# CODE STARTS HERE #############
p=sieve(32000)
for _ in range(ii()):
z=ii()
n=z
a=[]
i=0
while n>1:
if not n%p[i]:
n//=p[i]
a.append(p[i])
else:
i+=1
if p[i]**2>z:
break
if n>1:
a.append(n)
f=0
#print(a)
if len(a)>2:
ans=[a[0]]
c=a[1]
x=2
if c==ans[0]:
c*=a[2]
x=3
ans.append(c)
c=1
if x<len(a):
for i in a[x:]:
c*=i
if c!=ans[0] and c!=ans[1]:
ans.append(c)
ans.sort()
f=1
if f:
print('YES')
print(*ans)
else:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from collections import defaultdict as dd
from sys import stdin
input=stdin.readline
for _ in range(int(input())):
n=int(input())
br=0
for i in range(2,int(n**.5)+1):
fl=0
if n%i==0:
fl=1
x=i
for j in range(2,int(i**.5)+1):
if i%j==0 and len(set([j,i//j,n//i]))==3:
br=1
print('YES')
print(j,i//j,n//i)
break
if br:
break
for j in range(2,int((n//i)**.5)+1):
if (n//i)%j==0 and len(set([i,(n//i)//j,j]))==3:
br=1
print('YES')
print(i,(n//i)//j,j)
break
if br:
break
if br:
continue
else:
print('NO')
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | for i in range(int(input())):
n = int(input())
ans = []
o = 0
for i in range(2,int(n**0.5)+1):
if n % i == 0:
if n//i == i:
continue
ans.append(i)
o = n//i
break
for i in range(2,int(o**0.5)+1):
if o % i == 0:
if i not in ans:
if i == o//i:
continue
ans.append(i)
ans.append(o//i)
break
if len(ans) == 3:
print("YES")
print(*ans)
continue
print("NO") | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
def yes(factors):
print('YES')
print(' '.join(map(str, factors)))
def no():
print('NO')
for _ in range(t):
temp = n = int(input())
factors = []
i = 2
found = False
while(i <= pow(n, 1/2)):
while(n % i == 0):
if len(factors) == 0 or i not in factors[-1]:
factors.append({i:1})
else:
factors[-1][i] += 1
n //= i
i += 1
if len(factors) >= 3:
rv = [list(factors[0].keys())[0], list(factors[1].keys())[0]]
temp //= rv[0]
temp //= rv[1]
rv.append(temp)
yes(rv)
elif len(factors) == 2:
rv = [list(factors[0].keys())[0], list(factors[1].keys())[0]]
temp //= rv[0]
temp //= rv[1]
if temp == rv[0] or temp == rv[1] or temp == 1:
no()
else:
rv.append(temp)
yes(rv)
elif len(factors) == 1:
factor = list(factors[0].keys())[0]
if factors[0][factor] < 6:
if(factors[0][factor] < 3):
no()
else:
rv = [factor, factor*factor]
temp //= rv[0]
temp //= rv[1]
if temp == rv[0] or temp == rv[1] or temp == 1:
no()
else:
rv.append(temp)
yes(rv)
else:
rv = [factor, factor*factor]
temp //= rv[0]
temp //= rv[1]
rv.append(temp)
yes(rv)
else:
no()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.ArrayList;
import java.util.Scanner;
/**
*
* @author DELL
*/
public class Codechef {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++)
{
int n=sc.nextInt();
ArrayList<Integer> a=new ArrayList<>();
for(int j=2;j<=Math.sqrt(n);j++)
{
if(n%j==0)
{
if(n/j==j)
{
a.add(j);
}
else
{
a.add(j);
a.add(n/j);
}
}
}
int flag=0;
if(a.size()>=3)
{
for(int j=0;j<a.size()-2;j++)
{
for(int r=j+1;r<a.size()-1;r++)
{
for(int y=j+2;y<a.size();y++)
{
if(a.get(j)*a.get(r)*a.get(y)==n&&(a.get(j)!=(a.get(r)))&&(a.get(j)!=(a.get(y)))&&(a.get(y)!=a.get(r)))
{
System.out.println("YES");
System.out.println(a.get(j)+" "+a.get(r)+" "+a.get(y));
flag=1;
break;
}
}
if(flag==1)
break;
}
if(flag==1)
break;
}
}
if(flag==0)
System.out.println("NO");
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.util.*;
import java.io.*;
public class ProductofThreeNumbers
{
public static void main(String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int T=Integer.parseInt(br.readLine());
while(T-->0)
{
int n=Integer.parseInt(br.readLine());
Set<Integer> st=new HashSet<>();
for(int i=2;i*i<=n;i++)
{
if(n%i==0 && st.add(i))
{
n/=i;
break;
}
}
for(int i=2;i*i<=n;i++)
{
if(n%i==0 && st.add(i))
{
n/=i;
break;
}
}
if(st.size()<2 || st.contains(n) || n==1)
{
System.out.println("NO");
}
else
{
System.out.println("YES");
st.add(n);
for(int i:st)
System.out.print(i+" ");
System.out.println();
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const long long inf = 5e18;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
;
int t;
cin >> t;
while (t--) {
long long n;
cin >> n;
vector<long long> a;
for (long long i = 2; i <= min(100000ll, n); i++) {
if (n % i == 0) {
a.push_back(i);
n /= i;
if (a.size() == 2) {
break;
}
}
}
if (a.size() < 2) {
cout << "NO\n";
} else {
vector<long long> ans;
ans = a;
ans.push_back(n);
sort(ans.begin(), ans.end());
if (ans[1] == ans[2] or ans[0] == ans[1] or ans[0] < 2) {
cout << "NO\n";
} else {
cout << "YES\n";
for (long long i : ans) {
cout << i << ' ';
}
cout << '\n';
}
}
}
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys
input = sys.stdin.readline
def find(n,fa,fb):
f = 2
while f*f <= n:
if n % f == 0 and f != fa and f != fb:
return f
f += 1
return n
for _ in xrange(int(input())):
n = int(input())
nn = n
fa = find(n,None,None)
n = n // fa
fb = find(n,fa, None)
fc = nn // fa // fb
if fa*fb*fc == nn and fa >= 2 and fb >= 2 and fc >= 2 and fa != fb and fa != fc and fb != fc:
print("YES")
print fa,fb,fc
else:
print("NO")
| PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def foo(num):
isSimple = True
k = 0
while k < arr.__len__() and arr[k] <= math.sqrt(num):
if num % arr[k] == 0:
isSimple = False
num = num / arr[k]
break
k += 1
return num
import math
n = int(input())
arr = []
root = math.sqrt(1000000000)
for num in range(2, int(root) + 1):
isSimple = True
for delimiter in range(2, int(math.sqrt(num)) + 1):
if num % delimiter == 0:
isSimple = False
break
if isSimple:
arr.append(num)
for i in range(n):
num = int(input())
res = foo(num)
if res == num:
print("NO")
continue
else:
d1 = num / res
num = res
res = foo(num)
if res == num:
print("NO")
continue
else:
d2 = num / res
num = res
if d1 == d2:
res = foo(num)
if res == num:
print("NO")
continue
d2 *= num / res
num = res
if d1 != num and d2 != num and d1 != d2:
print("YES")
print("%d %d %d" % (d1, d2, res))
else:
print("NO")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void solve(long long N) {
vector<pair<long long, long long>> factors;
long long n = N;
for (long long i = 2; i * i <= n; i++) {
long long cnt = 0;
if (n % i == 0) {
while (n % i == 0) {
cnt++;
n /= i;
}
factors.push_back({i, cnt});
}
if (factors.size() == 3) break;
}
if (n != 1) {
factors.push_back({n, 1});
}
if (factors.size() >= 2) {
long long a = factors[0].first;
long long b = factors[1].first;
long long c = N / (a * b);
if (c != a and c != b and c > 1) {
cout << "YES" << '\n';
cout << a << " " << b << " " << c << '\n';
} else
cout << "NO" << '\n';
} else if (factors.size() == 1 and factors[0].second >= 6) {
cout << "YES" << '\n';
long long a = factors[0].first;
long long b = a * a;
long long c = N / (a * b);
cout << a << " " << b << " " << c << '\n';
} else
cout << "NO" << '\n';
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
;
long long t;
cin >> t;
while (t--) {
long long n;
cin >> n;
solve(n);
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(in.readLine());
for(int rc=0; rc<t; rc++) {
int n = Integer.parseInt(in.readLine());
int a = -1, b = -1, c = -1;
int cur = 2;
int div = 1;
while(cur <= Math.sqrt(n)) {
if(n%cur==0) {
div *= cur;
n /= cur;
if(a == -1) {
a = div;
div = 1;
} else if(b == -1 && div != a) {
b = div;
c = n;
break;
}
} else cur++;
}
if(a != -1 && b != -1 && c != -1 && c != b && c != a) {
System.out.println("YES\n"+a+" "+b+" "+c);
} else {
System.out.println("NO");
}
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools
from collections import deque,defaultdict,OrderedDict
import collections
def primeFactors(n):
pf=[]
# Print the number of two's that divide n
while n % 2 == 0:
pf.append(2)
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
pf.append(int(i))
n = n /i
# Condition if n is a prime
# number greater than 2
if n > 2:
pf.append(int(n))
return pf
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
for _ in range(ri()):
n=ri()
a=primeFactors(n)
amult=1
for i in range(1,len(a)-1):
amult=amult*a[i]
# print(a)
t=0
if len(a)<3:
print("NO")
else:
z=len((set(a)))
if z>=3:
print("YES")
ans=[a[0],a[-1],amult]
t=1
if z==2:
if len(a)>=4:
print("YES")
ans=[a[0],a[-1],amult]
t=1
if t==0:
print("NO")
if z==1:
if len(a)>=6:
print("YES")
ans=[a[0],a[0]*2,n//(a[0]**3)]
t=1
if t==0:
print("NO")
if t==1:
print(*ans)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.net.Inet4Address;
import java.util.*;
import java.lang.*;
import java.util.HashMap;
import java.util.PriorityQueue;
public class Solution implements Runnable
{
static class pair implements Comparable
{
int f;
int s;
pair(int fi, int se)
{
f=fi;
s=se;
}
public int compareTo(Object o)//desc order
{
pair pr=(pair)o;
if(s>pr.s)
return -1;
if(s==pr.s)
{
if(f>pr.f)
return 1;
else
return -1;
}
else
return 1;
}
public boolean equals(Object o)
{
pair ob=(pair)o;
if(o!=null)
{
if((ob.f==this.f)&&(ob.s==this.s))
return true;
}
return false;
}
public int hashCode()
{
return (this.f+" "+this.s).hashCode();
}
}
public class triplet implements Comparable
{
int f;
int s;
double t;
triplet(int f,int s,double t)
{
this.f=f;
this.s=s;
this.t=t;
}
public boolean equals(Object o)
{
triplet ob=(triplet)o;
int ff;
int ss;
double tt;
if(o!=null)
{
ff=ob.f;
ss=ob.s;
tt=ob.t;
if((ff==this.f)&&(ss==this.s)&&(tt==this.t))
return true;
}
return false;
}
public int hashCode()
{
return (this.f+" "+this.s+" "+this.t).hashCode();
}
public int compareTo(Object o)//asc order
{
triplet tr=(triplet)o;
if(t>tr.t)
return 1;
else
return -1;
}
}
void merge1(int arr[], int l, int m, int r)
{
int n1 = m - l + 1;
int n2 = r - m;
int L[] = new int [n1];
int R[] = new int [n2];
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
int i = 0, j = 0;
int k = l;
while (i < n1 && j < n2)
{
if (L[i]<=R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void sort1(int arr[], int l, int r)
{
if (l < r)
{
int m = (l+r)/2;
sort1(arr, l, m);
sort1(arr , m+1, r);
merge1(arr, l, m, r);
}
}
public static void main(String args[])throws Exception
{
new Thread(null,new Solution(),"Solution",1<<27).start();
}
public void run()
{
try
{
InputReader in = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
int t=in.ni();
while(t--!=0)
{
int n=in.ni();
int a=0,b=0,c=0;
for(int i=2;i<=Math.sqrt(n);i++)
{
if(n%i==0)
{
a=i;
n=n/i;
break;
}
}
for(int i=a+1;i<Math.sqrt(n);i++)
{
if(n%i==0)
{
b=i;
c=n/i;
break;
}
}
if(a==0 || b==0 || c==0)
out.println("NO");
else
{
out.println("YES");
out.println(a+" "+b+" "+c);
}
}
out.close();
}
catch(Exception e){
System.out.println(e);
}
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int ni() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nl() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = ni();
}
return a;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
}
} | JAVA |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import ceil, sqrt
def main():
t = int(input())
while t:
n = int(input())
findProd(n)
t -= 1
def findProd(n):
a = 2
found = 0
temp = n
while a <= ceil(sqrt(n)):
if n%a == 0:
n = n // a
b = a + 1
while b <= ceil(sqrt(n)):
c = n//b
if n%b == 0 and c > 1 and c != b and c != a :
found = 1
print("YES")
print(a),
print(b),
print(c)
break
else:
b += 1
if found:
break
else:
a += 1
if not found:
print("NO")
main() | PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool isPrime(int n) {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) return false;
}
return true;
}
int main() {
int tc;
cin >> tc;
while (tc--) {
int n;
cin >> n;
if (isPrime(n)) {
cout << "NO" << endl;
} else {
vector<int> factores;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
if (factores.size() == 0)
factores.push_back(i);
else if (i > factores.back())
factores.push_back(i);
n /= i;
}
}
if (n > 2) factores.push_back(n);
if (factores.size() < 3) {
cout << "NO" << endl;
} else {
int f = 1;
for (int i = 2; i < factores.size(); i++) f = f * factores[i];
if (f > factores[1]) {
cout << "YES" << endl;
cout << factores[0] << " " << factores[1] << " ";
cout << f << endl;
} else {
cout << "NO" << endl;
}
}
}
}
return 0;
}
| CPP |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
f = 0
for i in range(2, int(n**0.5)+1):
if n % i == 0:
a = i
k = n // a
for j in range(2, int(k ** 0.5)+1):
if k % j == 0 and j != a and k // j != a and j != k // j:
b = j
c = k // j
f = 1
break
if f == 1:
break
if f == 0:
print("NO")
else:
print("YES")
print(a, b, c, sep=" ")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | def factors(n):
factor = []
for i in range(2, int(n**0.5)+1):
if n % i == 0:
factor.append(i)
return factor
t = int(input())
for l in range(t):
n = int(input())
x = 0
factor = factors(n)
lenfactor = len(factor)
for i in range(lenfactor):
for j in range(i+1, lenfactor):
k = n/(factor[i]*factor[j])
if k%1 == 0 and k != factor[i] and k != factor[j]:
print('YES')
print(str(factor[i]) + ' ' + str(factor[j]) + ' ' + str(int(k)) )
x = 1
break
if x == 1:
break
if x == 0:
print('NO') | PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from sys import stdin,stdout
from math import sqrt
from collections import Counter
primes,primes_list = set([2]),None
composites = set()
def gen_primes(n):
for i in range(3,n+1,2):
if i not in composites:
primes.add(i)
for j in range(i*2,n+1,i):
composites.add(j)
def get_fact(n):
fact = []
if n not in primes:
for p in primes_list:
if n%p==0:
c = 0
while n%p==0:
n//=p
c+=1
fact.append((p,c))
if n==1:
break
elif len(fact) > 1:
fact.append((n,1))
break
else:
fact.append((n,1))
return fact
t = int(stdin.readline().strip())
ns = [int(stdin.readline().strip()) for _ in range(t)]
gen_primes(int(sqrt(max(ns)))+1)
primes_list = list(primes)
primes_list.sort()
#print(primes_list)
for n in ns:
fact = get_fact(n)
#print(n,fact)
div = []
if len(fact) == 3 or (len(fact) == 2 and fact[0][1]+fact[1][1]-2 > 1):
a,b = fact[0][0],fact[1][0]
div = [a,b,(n//a)//b]
elif len(fact) == 1 and fact[0][1] >= 6:
a = fact[0][0]
b = a*a
div = [a,b,(n//a)//b]
if div:
stdout.write("YES\n{} {} {}\n".format(*div))
else:
stdout.write("NO\n")
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | t=int(input())
for i in range(t):
n=int(input())
def check(n):
use=set()
i=2
while i*i<=n:
if n%i==0 and i not in use:
use.add(i)
n/=i
break
i+=1
i=2
while i*i<=n:
if n%i==0 and i not in use:
use.add(i)
n/=i
break
i+=1
if n==1 or n in use or len(use)<2:
return "NO"
else:
print "YES"
use.add(n)
for i in use:
print i,
return ""
print check(n) | PYTHON |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | from math import sqrt
def divide(n, start):
for k in range(start, int(sqrt(n)) + 1):
if n % k == 0:
return k
return n
for _ in range(int(input())):
n = int(input())
a = divide(n, 2)
if a == n:
print('NO')
continue
n //= a
b = divide(n, a + 1)
if b == n:
print('NO')
continue
n //= b
if n <= b:
print('NO')
continue
print('YES')
print(a, b, n)
| PYTHON3 |
1294_C. Product of Three Numbers | You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | 2 | 9 | import math
from collections import defaultdict as dq
def primeFactors(n):
d=dq(int)
while n % 2 == 0:
d[2]+=1
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
d[i]+=1
n = n // i
if n>1:
d[n]+=1
return d
for _ in range(int(input())):
n=int(input())
s=primeFactors(n)
if len(s)>2:
c=0
li=[]
for i in s.keys():
li.append(i)
c+=1
if c==2:
break
print("YES")
print(li[0],li[1],n//(li[0]*li[1]))
if len(s)==1:
for i in s.keys():
c=i
if s[c]>=6:
print("YES")
print(c,c*c,n//(c*c*c))
else:
print("NO")
if len(s)==2:
li=[]
for i in s.keys():
li.append(i)
if n//(li[0]*li[1]) in s.keys() or n//(li[0]*li[1])==1 :
print("NO")
else:
print("YES")
print(li[0],li[1],n//(li[0]*li[1])) | PYTHON3 |
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