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| description
stringlengths 29
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| source
int64 1
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| difficulty
int64 0
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| solution
stringlengths 7
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| language
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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 printDivisors(n,a) :
i=2
f=0
while(i*i<=n):
if n%i==0:
j=i+1
n=n//i
while(j*j<=n):
if n%j==0:
if j!=n//j:
print('YES')
print(i,j,n//j)
f=1
break
j+=1
if f==1:
break
i+=1
if f==0:
print('NO')
for _ in range(t):
a=[]
n=int(input())
printDivisors(n,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 | def main():
def divisors(n):
div = []
for i in range(2, int(n**0.5)): #n//2 + 1):#int(n ** 0.5)):
if n % i == 0:
div.append(i)
n //= i
if len(div) == 2:
if n not in div and n != 1:
div.append(n)
break
return div
for _ in range(int(input())):
n = int(input())
d = divisors(n)
#print(d)
if len(d) < 3:
print('NO')
continue
print('YES')
print(*d)
'''
a = d[0]
n = n // a
b = d[1]
c = n // b
if c != 1 and c != b and c != a:
print('YES')
print(a, b, c)
else:
print('NO')'''
main()
'''
i = 2
while i < n and len(div) < 2:
if n % i == 0:
div.append(i)
#t = n // i
n //= i
i += 1
if n not in div:
div.append(n)
return div''' | 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 Main{
public static void main(String args[]){
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-->0){
int n = in.nextInt();
int count=0;
int a=0,b=0;
for(int i=2;i<=Math.sqrt(n);i++){
if(n%i==0){
n=n/i;
if(a==0){
count++;
a=i;
}else if(b==0 && i!=n){
b=i;
count++;
break;
}
}
}
if(count==2){
System.out.println("YES");
System.out.println(a+" "+b+" "+n);
}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 math
T = int(input())
def check(n,p):
ans = -1
for i in range(p+1, int(math.sqrt(n)) + 1):
if n % i == 0:
ans = i
break
return ans
for i in range(T):
l = []
n = int(input())
ans = 'YES'
hello = 1
for i in range(3):
hello = check(n,hello)
if hello == -1 and n < 2:
ans = 'NO'
break
else:
if hello == -1 or i==2:
hello = n
if len(l) > 0 and l[-1] == hello:
ans = 'NO'
break
n = n // hello
l.append(hello)
print(ans)
if ans == 'YES':
print(str(l[0]) + ' ' + str(l[1]) + ' ' + str(l[2]))
| 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())
a=0
b=0
c=0
for i in range(2,int(n**0.5)+2):
if n%i==0:
n//=i
a=i
break
if a==0:
print("NO")
continue
else:
for j in range(a+1,int(n**0.5)+2):
if n%j==0 and n//j!=j and n//j!=a:
b=j
c=n//j
break
if b==0:
print("NO")
continue
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 math
t = int(input())
while t>0:
t-=1
n = int(input())
sn = n
l = []
for i in range(2,int(math.sqrt(n))+1):
if n%i == 0:
l.append(i)
n = n//i
c = i
break
else:
print("NO")
continue
for i in range(c+1,int(math.sqrt(n))+1):
if n%i == 0:
l.append(i)
n = n//i
break
else:
print("NO")
continue
if (sn//(l[0]*l[1]))!=1 and (sn//(l[0]*l[1]) not in l):
l.append(sn//(l[0]*l[1]))
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 sys
def fastio():
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
def debug(*var, sep = ' ', end = '\n'):
print(*var, file=sys.stderr, end = end, sep = sep)
INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br
def ok(n):
# for i in range()
pass
def fac(n):
res = []
for i in range(2, int(n ** 0.5 + 1)):
if n % i:
continue
res.append(i)
if i ** 2 != n:
res.append(n // i)
return sorted(res)
t, = I()
while t:
t -= 1
n, = I()
ans = 'YES'
f = fac(n)
if len(f) < 3:
print('NO')
continue
m = len(f)
for i in range(m):
for j in range(i + 1, m):
k = n // (f[i] * f[j])
if k in f and k != f[i] and k != f[j]:
print('YES')
print(f[i], f[j], k)
break
else:
continue
break
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 fact(n):
ans = 1
for i in range(2, n+1):
ans*= i
return ans
def comb(n, c):
return fact(n)//(fact(n-c)*c)
for _ in range(int(input())):
n= int(input())
ans = []
for i in range(2, int(math.sqrt(n))+1):
if(n%i==0 and i not in ans):
n//=i
ans.append(i)
break
for i in range(2,int(math.sqrt(n))+1):
if(n%i==0 and i not in ans):
n//=i
ans.append(i)
break
if(len(ans)<2 or n==1 or n in ans):
print('NO')
else:
ans.append(n)
print('YES')
print(ans[0], ans[1], ans[2])
| 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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
public class Solution{
public static void main(String[] args){
try{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
for(int i=0;i<t;i++){
int n=Integer.parseInt(br.readLine());
int u=0,v=0;
for(int j=2;j<Math.sqrt(n);j++){
if(n%j==0){
u=j;
n=n/j;
break;
}
}
if(u!=0){
for(int j=2;j<Math.sqrt(n);j++){
if(n%j==0 && j!=u && n/j != u){
v=j;
n=n/j;
break;
}
}
}
if(v!=0){
System.out.println("YES");
System.out.println(u+" "+v+" "+n);
}else{
System.out.println("NO");
}
}
}catch(Exception e){
System.out.println("kkkk");
}
}
static int lcm(int a,int b){
return (a*b)/hcf(a,b);
}
static int hcf(int a,int b){
if(a==0){
return b;
}
if(b==0){
return a;
}
if(a>b)
return hcf(a%b,b);
return hcf(a,b%a);
}
static class pair{
int a,b;
public pair(int a,int b){
this.a=a;
this.b=b;
}
}
} | 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.ArrayList;
import java.util.BitSet;
import java.util.Scanner;
public class CF1294C {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tc = sc.nextInt();
BitSet sieve = new BitSet();
for(int i=2; i<=1000000; i++) sieve.set(i);
for(int i=2; i<=1000000; i++)
if(sieve.get(i))
for(int j=i+i; j<=1000000; j+=i) sieve.set(j, false);
ArrayList<Integer> primes = new ArrayList<>();
for(int i=2; i<=1000000; i++) if(sieve.get(i)) primes.add(i);
while(tc-->0){
int n = sc.nextInt();
int sqrt = (int) Math.sqrt(n)+1;
ArrayList<Integer> ans = new ArrayList<>();
for(int i=0; ans.size()<3 && i<primes.size() && primes.get(i)<sqrt; i++){
int prime = primes.get(i);
if(n%prime==0){
n/=prime;
ans.add(prime);
if(ans.size()==2 && !ans.contains(n) && n>1) ans.add(n);
if(ans.size()<3 && n%(prime*prime)==0){
n/=prime*prime;
ans.add(prime*prime);
if(ans.size()==2 && !ans.contains(n) && n>1) ans.add(n);
}
}
}
System.out.println(ans.size()!=3 ? "NO" : "YES\n"+ans.get(0)+" "+ans.get(1)+" "+ans.get(2));
}
}
}
| 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 getprod(n):
res = []
i = 1
while(i*i <= n):
if(n%i == 0):
res.append(i)
if(n//i != i):
res.append(n//i)
i += 1
return res
for _ in range(int(input())):
num = int(input())
res = getprod(num)
if(len(res) <= 2):
print('NO')
else:
res.sort()
a,b,c = -1,-1,-1
i = 0
n = len(res)
flag = 0
while(i < n):
j,k = i+1,n-1
while(j < k):
x = res[i]*res[j]*res[k]
if(x == num and res[i]>= 2 and res[j]>= 2 and res[k]>= 2):
a,b,c = res[i],res[j],res[k]
flag = 1
break
elif(x < num):
j += 1
else:
k -= 1
if(flag == 1):
break
i += 1
if(a == -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 | /*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
//package ieee;
import java.util.Scanner;
import java.util.ArrayList;
/**
*
* @author LAPTOP
*/
public class IEEE {
/**
* @param args the command line arguments
*/
public static boolean isprime (long n )
{
if (n==2) return true;
if(n<2 || n%2==0) return false ;
for (long i = 3 ; i*i<=n ; i+=2)
if(n%i==0) return false;
return true;
}
public static void main(String[] args) {
Scanner sc = new Scanner (System.in);
int t = sc.nextInt();
while(t>0)
{
t--;
int n = sc.nextInt();
ArrayList <Integer> arr = new ArrayList<Integer>();
for (int i = 2 ; i*i < n ; i++)
{
if (n%i==0)
{
arr.add(i);
n/=i;
}
if (arr.size()==2)
{
arr.add(n);
break;
}
}
if (arr.size()<3 ) System.out.println("NO");
else
{
System.out.println("YES");
for (int i = 0; i < 3; i++) {
System.out.print(arr.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 | import java.util.*;
import java.io.*;
public class Product_of_Three_Numbers {
static int mod = (int) (1e9 + 7);
public static void main(String[] args) throws java.lang.Exception {
/*
Let a be the minimum divisor of n greater than 1.
Then let b be the minimum divisor of na that isn't equal a and 1.
If n/ab isn't equal a, b and 1 then the answer is "YES",
otherwise the answer is "NO".
Time complexity: O(√n) per query.
*/
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
int n=Integer.parseInt(st.nextToken());
Set<Integer> h=new HashSet<>();
for(int i=2; i*i<=n; i++) {
if(n%i==0 && !h.contains(i)) {
h.add(i);
n/=i;
break;
}
}
for(int i=2; i*i<=n; i++) {
if(n%i==0 && !h.contains(i)) {
h.add(i);
n/=i;
break;
}
}
if(h.size()<2 ||h.contains(n)|| n==1) System.out.println("NO");
else {
System.out.println("YES");
h.add(n);
for(int x: h) System.out.print(x+" ");
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 | import sys
I=sys.stdin.readline
ans=""
for _ in range(int(I())):
n=int(input())
fac=[(i,n//i) for i in range(2,int(n**.5)+1) if n%i==0]
#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 | """T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
import math
T=int(input())
for _ in range(0,T):
N=int(input())
n=N
L=[]
while (n % 2 == 0):
L.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while (n % i== 0):
L.append(i)
n = n // i
if (n > 2):
L.append(n)
if(len(L)<3):
print('NO')
else:
t1=L[0]
t2=L[1]
t3=1
if(L[1]==L[0]):
t2=L[1]*L[2]
for j in range(3,len(L)):
t3*=L[j]
else:
for j in range(2,len(L)):
t3*=L[j]
if(t1!=t2 and t2!=t3 and t1!=t3 and t1>1 and t2>1 and t3>1):
print('YES')
print(t1,t2,t3)
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 q in range(t):
n=int(input())
m=n
fact=[]
while m%2==0:
fact.append(2)
m = m / 2
for i in range(3,int(math.sqrt(m))+1,2):
while m % i== 0:
fact.append(i)
m = m / i
if m > 2:
fact.append(m)
dist = list(set(fact))
if len(dist)==1:
if fact.count(dist[0])>4:
o=dist[0]
t=dist[0]*dist[0]
if (n/(o*t))>2 and int((n/(o*t)))!=o and int((n/(o*t)))!=t:
print('YES')
print(int(o),int(t),int((n/(o*t))))
else:
print('NO')
else:
print('NO')
elif len(dist)==2:
o=dist[0]
t=dist[1]
if (fact.count(o)+fact.count(t))>=4:
if (n/(o*t))>2 and int((n/(o*t)))!=o and int((n/(o*t)))!=t:
print('YES')
print(int(o),int(t),int((n/(o*t))))
else:
print('NO')
else:
print('NO')
if len(dist)>2:
o=dist[0]
t=dist[1]
if (n/(o*t))>2 and int((n/(o*t)))!=o and int((n/(o*t)))!=t:
print('YES')
print(int(o),int(t),int((n//(o*t))))
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.*;import java.io.*;import java.math.*;
public class Main
{
public static void process()throws IOException
{
int n=ni();
if(n==2){
pn("NO");
return;
}
ArrayList<Pair> li = new ArrayList<>();
int t=n;
long po=0l;
int c=0;
while (n % 2 == 0) {
//System.out.print(2 + " ");
c++;
n /= 2;
}
if(c!=0){
li.add(new Pair(2,c));
po+=c;
}
for (int i = 3; i <= Math.sqrt(n); i += 2) {
c=0;
while (n % i == 0) {
// System.out.print(i + " ");
c++;
n /= i;
}
if(c!=0){
li.add(new Pair(i,c));
po+=c;
}
}
if (n > 2) {
if(n==t){
pn("NO");
return;
}
li.add(new Pair(n,1));
}
int a=li.get(0).n,b=1;
li.get(0).p=li.get(0).p-1;
// System.out.println("pow"+li.get(0).p);
if(li.get(0).p>=2)
b=(li.get(0).n)*(li.get(0).n);
else if(li.get(0).p==1){
if(li.size()==1){
pn("NO");
return;
}
b=(li.get(0).n)*(li.get(1).n);
}
else{
if(li.size()==1){
pn("NO");
return;
}
b=li.get(1).n;
}
long ca=t/(a*b);
if(ca==a || ca==b || ca==1){
pn("NO");
return;
}
pn("YES");
pn(a+" "+b+" "+ca);
}
static class Pair{int n,p; public Pair(int n,int p){this.n=n;this.p=p;}}
static AnotherReader sc;
static PrintWriter out;
public static void main(String[]args)throws IOException
{
out = new PrintWriter(System.out);
sc=new AnotherReader();
boolean oj = true;
oj = System.getProperty("ONLINE_JUDGE") != null;
if(!oj) sc=new AnotherReader(100);
long s = System.currentTimeMillis();
int t=ni();
while(t-->0)
process();
out.flush();
if(!oj)
System.out.println(System.currentTimeMillis()-s+"ms");
System.out.close();
}
static void pn(Object o){out.println(o);}
static void p(Object o){out.print(o);}
static void pni(Object o){out.println(o);System.out.flush();}
static int ni()throws IOException{return sc.nextInt();}
static long nl()throws IOException{return sc.nextLong();}
static double nd()throws IOException{return sc.nextDouble();}
static String nln()throws IOException{return sc.nextLine();}
static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
static boolean multipleTC=false;
/////////////////////////////////////////////////////////////////////////////////////////////////////////
static class AnotherReader{BufferedReader br; StringTokenizer st;
AnotherReader()throws FileNotFoundException{
br=new BufferedReader(new InputStreamReader(System.in));}
AnotherReader(int a)throws FileNotFoundException{
br = new BufferedReader(new FileReader("input.txt"));}
String next()throws IOException{
while (st == null || !st.hasMoreElements()) {try{
st = new StringTokenizer(br.readLine());}
catch (IOException e){ e.printStackTrace(); }}
return st.nextToken(); } int nextInt() throws IOException{
return Integer.parseInt(next());}
long nextLong() throws IOException
{return Long.parseLong(next());}
double nextDouble()throws IOException { return Double.parseDouble(next()); }
String nextLine() throws IOException{ String str = ""; try{
str = br.readLine();} catch (IOException e){
e.printStackTrace();} return str;}}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}
| 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;
int main() {
int t;
cin >> t;
while (t--) {
long long n;
cin >> n;
set<int> used;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
if (used.size() < 2 || used.count(n) || n == 1) {
cout << "NO" << endl;
} else {
cout << "YES"
<< "\n";
used.insert(n);
for (auto it : used) cout << it << " ";
cout << 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 | #include <bits/stdc++.h>
using namespace std;
void gao(int n) {
for (int i = 2; i * i < n; i++) {
if (n % i == 0) {
for (int j = i + 1; j * j < n / i; j++) {
if (n / i % j == 0) {
cout << "YES" << endl;
cout << i << ' ' << j << ' ' << n / i / j << endl;
return;
}
}
}
}
cout << "NO" << endl;
return;
}
int main() {
int t, n;
cin >> t;
for (int tt = 0; tt < t; tt++) {
cin >> n;
gao(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 IOException {
FastReader scan = new FastReader();
//PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("marathon.out")));
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)),true);
Task solver = new Task();
int t = scan.nextInt();
//int t = 1;
for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
out.close();
}
static class Task {
public void solve(int testNumber, FastReader sc, PrintWriter pw) {
int q = sc.nextInt();
int n = 0;
int m = 0;
boolean a = false;
boolean b = false;
for(n=2;n<=Math.sqrt(q);n++){
if(q%n==0){
a=true;
m=q/n;
break;
}
}
int o=n+1;
for(;o<=Math.sqrt(m);o++){
if(m%o==0){
b=true;
m/=o;
break;
}
}
if(n==o||!a||!b||o==m||n==m){
pw.println("NO");
}
else{
pw.println("YES");
pw.println(n+" "+m+" "+o);
}
}
}
static class tup implements Comparable<tup>{
int a, b;
tup(int a, int b){
this.a=a;
this.b=b;
}
@Override
public int compareTo(tup o) {
return Integer.compare(Math.abs(b),Math.abs(o.b));
}
}
static void shuffle(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
}
static void shuffle(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws FileNotFoundException {
br = new BufferedReader(new FileReader(new File(s)));
}
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 |
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
import math
input = sys.stdin.readline
ins = lambda: input().rstrip()
ini = lambda: int(input().rstrip())
inm = lambda: map(int, input().split())
inl = lambda: list(map(int, input().split()))
t = ini()
ans = []
andsindex = [0] * t
for _ in range(t):
n = ini()
i = 2
tmp = []
while n >= i * i:
if i == 5 and n % 5 == 0 and 4 in tmp:
tmp.remove(4)
if n % i == 0:
tmp.append(i)
if len(tmp) == 3:
break
i += 1
if len(tmp) <= 1:
ans.append("NO")
elif len(tmp) >= 2:
if len(tmp) == 3:
y = (n / (tmp[0] * tmp[2])).is_integer()
if tmp[0] * tmp[1] * tmp[2] == n:
ans.append("YES")
andsindex[len(ans) - 1] = f"{tmp[0]} {tmp[1]} {tmp[2]}"
# print(tmp[0], tmp[1], tmp[2])
continue
elif y and n / (tmp[0] * tmp[2]) not in tmp:
ans.append("YES")
andsindex[len(ans) - 1] = f"{tmp[0]} {tmp[2]} {n // (tmp[0] * tmp[2])}"
# print(tmp[0], tmp[2], n // (tmp[0] * tmp[2]))
continue
x = (n / (tmp[0] * tmp[1])).is_integer()
if x and n / (tmp[0] * tmp[1]) not in tmp:
ans.append("YES")
andsindex[len(ans) - 1] = f"{tmp[0]} {tmp[1]} {n // (tmp[0] * tmp[1])}"
# print(tmp[0], tmp[1], n // (tmp[0] * tmp[1]))
else:
ans.append("NO")
for i in range(t):
print(ans[i])
if ans[i] == "YES":
print(andsindex[i]) | 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() {
int n, i, cnt = 0, d, ar[3];
cin >> n;
for (i = 2; i * i <= n; i++)
if (n % i == 0) {
ar[cnt++] = i;
n /= i;
if (cnt == 2) break;
}
if (cnt == 2) {
ar[2] = n;
sort(ar, ar + 3);
if (ar[0] != ar[1] && ar[1] != ar[2]) {
cout << "YES\n" << ar[0] << ' ' << ar[1] << ' ' << ar[2] << '\n';
return;
}
}
cout << "NO\n";
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
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 | I = input
pr = print
def main():
for _ in range(int(I())):
ar=[]
n,i=int(I()),2
while i*i<n and len(ar)<2:
if n%i==0:ar.append(i);n//=i
i+=1
if len(ar)==2:pr('YES');pr(*ar,n)
else:pr('NO')
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 | from sys import stdin, stdout
from math import sqrt
def solve():
n = int(input())
st = set()
for i in range(2, int(sqrt(n)) + 1):
if n%i == 0 and i not in st:
st.add(i)
n /= i
break
for i in range(2, int(sqrt(n)) + 1):
if n%i == 0 and i not in st:
st.add(i)
n /= i
break
if n == 1 or n in st or len(st) < 2:
print("NO")
return 0
else:
print("YES")
A = []
for i in st: A.append(int(i))
A.append(int(n))
print(A[0], A[1], A[2])
return 0
t = int(input())
for i in range(t):
solve() | 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.*;
import java.math.BigInteger;
import java.util.*;
public class Main{
static Scanner scanner=new Scanner(System.in);
public static void main(String[] args) {
int q=scanner.nextInt();
while(q-->0) {
int n=scanner.nextInt();
Set<Integer>set=new HashSet<Integer>();
for(int i=2;i<=Math.sqrt(n);i++) {
if(n%i==0) {
set.add(i);
n/=i;
if(set.size()==2) {
set.add(n);
break;
}
}
}
if(set.size()==3) {
System.out.println("YES");
for(int i:set) {
System.out.print(i+" ");
}
System.out.println();
}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 | from math import sqrt
def fact(k):
for i in range(2,(int(sqrt(k)+1))):
if k%i == 0:
if k//i != i:
f = k//i
return f,i
return 0,0
def main():
t = int(input())
for __ in range(t):
n = int(input())
flag = False
for i in range(2,(int(sqrt(n)+1))):
if n%i == 0:
if n//i != i:
f = n//i
a,b = fact(i)
c,d = fact(f)
if a != f and b != f and(a != 0):
print("YES")
print(a,b,f)
flag = True
break
elif c != i and d != i and(d != 0):
print("YES")
print(c,d,i)
flag = True
break
if not flag:
print("NO")
if __name__ == '__main__':
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 | #include <bits/stdc++.h>
using namespace std;
vector<int> adj[200001];
long long int vist[200001], maxx;
void dfs(long long int node) {
vist[node] = 1;
maxx = max(maxx, node);
for (long long int child : adj[node])
if (!vist[child]) dfs(child);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int x = sqrt(n);
bool ok = false;
int i, j, k;
for (i = 2; i <= x; i++) {
int n1 = n;
if (n1 % i == 0) {
n1 /= i;
for (j = i + 1; j <= x; j++) {
if (n1 % j == 0 && j != i) {
k = n1 /= j;
if (k != i && k != j) {
ok = true;
break;
}
}
}
if (ok) break;
}
}
if (ok)
cout << "YES\n" << i << " " << j << " " << k << endl;
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 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void printDivisors(long long n, set<long long> &s, vector<long long> &v) {
for (long long i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i) {
s.insert(i);
v.push_back(i);
} else {
s.insert(i);
s.insert(n / i);
v.push_back(i);
v.push_back(n / i);
}
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
long long n;
cin >> n;
set<long long> s;
bool flag = 0;
vector<long long> v;
printDivisors(n, s, v);
for (int i = 0; i < v.size(); i++) {
for (int j = i + 1; j < v.size(); j++) {
if (n % (v[i] * v[j]) == 0 && s.find(n / (v[i] * v[j])) != s.end() &&
v[i] != v[j] && v[i] != n / (v[i] * v[j]) &&
v[j] != n / (v[i] * v[j])) {
cout << "YES" << '\n';
cout << v[i] << " " << v[j] << " " << n / (v[i] * v[j]) << '\n';
flag = 1;
break;
}
}
if (flag == 1) break;
}
if (flag == 0) 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 | import math
t = int(input())
for i in range(t):
a = 0
b = 0
n = int(input())
for j in range (int(math.sqrt(n))-1):
if n%(j+2) == 0:
n = n//(j+2)
a = j+2
break
if a == 0:
print("NO")
elif a != 0:
for j in range (int(math.sqrt(n))-2):
if n%(j+3) == 0 and j+3 != a:
n = n//(j+3)
if n != j+3:
b = j+3
break
if b == 0:
print("NO")
else:
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
t = int(input())
for xx in range(t):
n = int(input())
i = 2
fl = 0
divs = []
while(i*i <= n):
if n%i == 0:
divs.append(i)
n//=i
if len(divs) == 2:
break
i += 1
if len(divs) < 2: fl = 1
if fl == 1:
print("NO")
continue
kr = n
if kr == 1 or kr == divs[0] or kr == divs[1]:
print("NO")
continue
else:
divs.append(kr)
print("YES")
print(" ".join([str(v) for v in divs])) | 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 | a, i = int(input()), 0
def pro(a):
p, i = 0, 2
while i * i <= a:
if a % i == 0:
p = i
break
i = i + 1
return p
def f1(k, ak, a):
i, t = 2, 0
while i * i <= ak:
if a % i == 0:
if ak // i != i and i != k and (ak // i) * i * k == a:
print("YES")
print(int(ak / i), int(i), int(k))
t = 1
break
i = i + 1
if t == 0:
print("NO")
return 0
def ff(a):
k = pro(a)
if k == 0:
print("NO")
else:
f1(k, a / k, a)
return 0
while i != a:
x = int(input())
ff(x)
i = i + 1 | 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())):
x = int(input())
li = []
for i in range(2,int(x**0.5) +1):
if x%i==0:
li.append(i)
x = x//i
break
for i in range(2,int(x**0.5) +1):
if x%i==0 and i not in li:
li.append(i)
x = x//i
break
if len(li) <2 or x==1 or x in li:
print('NO')
else:
print('YES')
print(*li, x)
| 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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author GYSHGX868
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
TaskC solver = new TaskC();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++) {
solver.solve(i, in, out);
}
out.close();
}
static class TaskC {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt();
int[] result = new int[3];
int count = 0;
for (int i = 2; i * i < n; i++) {
if (n % i == 0) {
result[count++] = i;
n /= i;
if (count == 2) {
result[count++] = n;
break;
}
}
}
if (count != 3) {
out.printLine("NO");
} else if (result[0] == result[1] || result[1] == result[2] || result[0] == result[2]) {
out.printLine("NO");
} else {
out.printLine("YES");
out.printLine(result);
}
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void print(int[] array) {
for (int i = 0; i < array.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(array[i]);
}
}
public void printLine(int[] array) {
print(array);
writer.println();
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return nextString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| 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.io.*;
import java.util.*;
public class ProductofThreeNumbers {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
int n = in.nextInt();
int a,b,c;
a=b=c=-1;
for(int i=2;i*i<n;i++) {
if(n%i==0) {
a = i;
break;
}
}
if(a==-1) out.println("NO");
else {
n /= a;
for(int i=a+1;i*i<n;i++) {
if(n%i==0&&n/i!=i) {
b = i;
break;
}
}
if(b==-1) out.println("NO");
else {
n /= b;
c = n;
if(c<2||c==a) out.println("NO");
else {
out.println("YES");
out.println(a+" "+b+" "+c);
}
}
}
}out.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 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:
d={}
for i in a:
if i in d:
d[i]+=1
else:
d[i]=1
b=list(set(a))
z=len(b)
if z>=3:
print("YES")
ans=[a[0],a[-1],amult]
t=1
if z==2:
c=0
for i in d:
if d[i]>=3:
print("YES")
ans=[a[0],a[-1],amult]
t=1
break
if d[i]>=2:
c+=1
if c>=2:
print("YES")
ans=[a[0],a[-1],amult]
t=1
if t==0:
print("NO")
if z==1:
t=0
for i in d:
if d[i]>=6:
print("YES")
ans=[a[0],a[0]*2,n//(a[0]**3)]
t=1
break
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 | def factor(n):
dict1={}
for i in range(2,int(pow(n,0.5))+1):
if n%i==0:
if i not in dict1:
dict1[i]=1
if n//i not in dict1:
dict1[n//i]=1
return dict1
for t in range(int(input())):
n=int(input())
dict1=factor(n)
dict1=sorted(dict1)
#print(dict1)
a=0
b=0
c=-1
for i in dict1:
if a==0:
a=i
else:
b=i
d=n//(a*b)
if d!=a and d!=b and d in dict1:
c=d
break
if c==-1:
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 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
int n;
int num[maxn], cnt;
int seq[maxn];
bool Find(int res, int now) {
for (int i = 1; i <= now; i++) {
if (seq[i] == res) return true;
}
return false;
}
void Init();
int main() {
int T;
scanf("%d", &T);
while (T--) Init();
return 0;
}
void Init() {
scanf("%d", &n);
int cun = n;
cnt = 0;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
while (n % i == 0) {
n /= i;
num[++cnt] = i;
}
}
}
if (n > 1) num[++cnt] = n;
int now = 0, res = 1, i;
for (i = 1; i <= cnt; i++) {
res *= num[i];
if (!Find(res, now)) {
seq[++now] = res;
seq[now] = res;
if (now == 3) {
seq[now] = cun / seq[1] / seq[2];
break;
}
res = 1;
}
}
if (now == 3) {
puts("YES");
for (int i = 1; i <= 3; i++) printf("%d ", seq[i]);
puts("");
} else
puts("NO");
}
| 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;
void solve() {
int n;
cin >> n;
vector<int> v;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
v.push_back(i);
if (v.size() == 2) {
v.push_back(n / i);
break;
}
n /= i;
}
}
v.erase(unique(v.begin(), v.end()), v.end());
if (v.size() != 3) {
cout << "NO"
<< "\n";
} else {
cout << "YES"
<< "\n";
for (const auto &e : v) {
cout << e << " ";
}
cout << "\n";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
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 | t=int(input())
for i in range(t):
n=int(input())
l=[0, 0, 0]
tmp=0
m=n
i=2
while i**2<=m and tmp<2:
if n%i==0:
n//=i
l[tmp]=i
tmp+=1
i+=1
if tmp==2 and n>l[1]:
print("YES")
l[2]=n
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.Scanner;
public class Product_Of_Three_Numbers {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
//System.out.println("Enter the number of test cases :");
int t=sc.nextInt();
for(int i=0;i<t;i++) {
int n=sc.nextInt();
find(n);
}
}
public static void find(int n) {
int k = 0;
int m=n;
int res[] = new int[3];
for(int i=2;i*i<=n;i++) {
if(k==1 && n%i==0) {
res[k] = i;
k++;
break;
}
if(k<1 && n%i==0) {
res[k]=i;
n=n/i;
k++;
}
}
int r=0;
if(res[0]!=0 && res[1]!=0) {
r=m/(res[0]*res[1]);
if(r !=res[0] && r!=res[1] && r!=1) {
System.out.println("YES");
System.out.println(res[0]+" "+res[1]+" "+r);
}
else
System.out.println("NO");
}
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 | for t in range(int(input())):
n=int(input())
flag=0
for i in range(2,10001):
if(n%i==0):
flag=1
break
if(flag==1):
a=i
n=n//i
for i in range(2,100001):
if(n%i==0 and i!=a):
flag=2
break
b=i
c=n//i
if(flag==2 and b!=c and c!=1 and c!=a):
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 math
def printDivisors(n) :
i = 2
c = []
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
c.append(n//i)
else :
c.extend([i,n//i])
i = i + 1
return c
def countTriplets(arr, n, m):
count = 0
arr.sort()
for end in range(n - 1, 1, -1):
start = 0
mid = end - 1
while (start < mid):
prod = (arr[end]*arr[start]*arr[mid])
if (prod > m):
mid -= 1
elif (prod < m):
start += 1
elif (prod == m):
print("YES")
print(arr[end],arr[start],arr[mid])
count += 1
mid -= 1
start += 1
return count
return count
for _ in range(int(input())):
n = int(input())
a = printDivisors(n)
#print(len(a),a)
if countTriplets(a,len(a),n) == 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 |
import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
import java.util.Map.*;
public class codeforces {
public static void main(String [] args) throws IOException, InterruptedException {
Scanner sc=new Scanner(System.in);
PrintWriter pw=new PrintWriter(System.out);
int t=sc.nextInt();
while(t-->0) {
long y=sc.nextLong();
TreeSet<Long> set=printDivisors(y);
set.pollFirst(); set.pollLast();
// pw.println(set);
if(set.size()<3)pw.println("NO");
else {
StringBuilder sb=new StringBuilder();
boolean f=true;
long ans=set.first();
sb.append(ans+" ");
long k=y/ans;
TreeSet<Long> set2=printDivisors(k);
set2.pollFirst(); set2.pollLast();
while(!set2.isEmpty() && f) {
if(set2.first()!=set2.last() && set2.first()!=ans) {pw.println("YES"+"\n"+ans+" "+set2.first()+" "+set2.last()); f=false; }
else if(set2.size()==1) break ;
set2.pollFirst(); set2.pollLast();
}
if(f)pw.println("NO");
}
}
pw.flush();
pw.close();
}
static TreeSet<Long> printDivisors(long n)
{
TreeSet<Long> set=new TreeSet<>();
for (long i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i)
set.add(i);
else {set.add(i); set.add(n/i);}
}
}
return set;
}
} | 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
import math
import heapq
import bisect
import re
from collections import deque
from decimal import *
from fractions import gcd
def main():
# [int(i) for i in sys.stdin.readline().split()]
n = int(sys.stdin.readline())
q = []
for i in range(int(n ** 0.5) + 1, 0, -1):
if n % i == 0:
q.append(i)
if n // i != i:
q.append(n // i)
q.sort()
if len(q) <= 3:
print("NO")
return 0
i = 1
while i < len(q) - 3:
i += 1
f = n / q[1] / q[i]
if f != q[1] and f != q[i] and f >= 2 and f == int(f) and int(f) * q[1] * q[i] == n:
print("YES")
print(q[1], q[i], int(f))
return 0
print("NO")
for i in range(int(input())):
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
from functools import reduce
t = int(input())
for cas in range(t):
n = int(input())
ans = []
m = n
for i in range(2, int(math.sqrt(m))):
cal = 0
if n == 1:
break
while n % i == 0 and n > 1:
n //= i
ans.append(i)
if n > 1:
ans.append(n)
if len(ans) < 3:
print("NO")
elif len(ans) == 3:
if len(set(ans)) < 3:
print("NO")
else:
print("YES")
print(ans[0], ans[1], ans[2])
elif len(ans) == 4:
if len(set(ans)) == 1:
print("NO")
else:
print("YES")
print(ans[0], ans[1] * ans[2], ans[3])
elif len(ans) == 5:
if len(set(ans)) == 1:
print("NO")
else:
print("YES")
print(ans[0], ans[1] * ans[2] * ans[3], ans[4])
else:
print("YES")
print(ans[0], ans[1] * ans[2], reduce(lambda x, y: x * y, ans[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 | import math
def fun(s):
for i in range(2, int(math.sqrt(s)) + 1):
if s % i == 0:
w = s // i
for j in range(i + 1, int(math.sqrt(w) + 1)):
d = w // j
if w % j == 0 and d != j and d != i:
return i, j, d
return -1
for t in range(int(input())):
n = int(input())
q = fun(n)
if q != -1:
print('YES')
print(q[0], q[1], q[2])
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 i in range(t):
n = int(input())
a, x, d = [], 0, 2
while d * d <= n and x < 2:
if n%d == 0:
a.append(d)
n = n // d
x += 1
d += 1
if n >= d and x == 2:
a.append(n)
print("YES")
res = " ".join(map(str, a))
print(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 | def fun(n):
for i in range(2,int(n**0.5)+1):
x=n//i
if n%i==0:
for j in range(2,int(x**0.5)+1):
if x%j==0 and x//j!=j and i!=j and x//j!=i:
return True,[i,j,x//j]
return False,[-1,-1,-1]
t=int(input())
for _ in range(t):
n=int(input())
res,lst = fun(n)
lst.sort()
if res:
print("YES")
print(lst[0],lst[1],lst[2])
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 | /* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int k=0;k<t;k++){
int n=sc.nextInt();
int n1=n;
int[] a1=new int[3];
int count=0;
for(int i=2;i*i<=n;i++)
{
if(n%i==0 && (i!=a1[0] || i!=a1[1]))
{
a1[count]=i;
count++;
n=n/i;
if(count==2 && n>1 && (n!=a1[0] && n!=a1[1]))
{
a1[2]=n;
count++;
}
}
if(count==3)
{
System.out.println("YES");
System.out.print(a1[0]+" "+a1[1]+" "+a1[2]);
System.out.println();
break;
}
}
if(count!=3)
{
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 math
def primenum(x):
count=0
for i in range(2,int(math.floor(math.sqrt(x)))+1):
if(x%i==0):
count=count+1
if(count==0):
return True
else:
return False
t=int(input())
for i in range(t):
n=int(input())
m=n
if(primenum(n)==True):
print('NO')
else:
l=[]
for j in range(2,int(math.sqrt(m))+1):
if(n%j==0):
l.append(j)
n=n//j
if(len(l)==2):
x=m//(l[0]*l[1])
if(x>l[0] and x>l[1]):
l.append(x)
if(len(l)==3):
if((l[0]*l[1]*l[2])==m):
print('YES')
print(l[0],l[1],l[2])
break
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
class Read:
@staticmethod
def string():
return input()
@staticmethod
def int():
return int(input())
@staticmethod
def list(sep=' '):
return input().split(sep)
@staticmethod
def list_int(sep=' '):
return list(map(int, input().split(sep)))
def solve():
n = Read.int()
for i in range(2, math.ceil(math.sqrt(n))):
if n % i == 0:
t = n / i
for j in range(2, math.ceil(math.sqrt(t))):
if t % j == 0:
res = math.ceil(t / j)
if res == i or i == j or j == res:
continue
print('YES')
print('{} {} {}'.format(i, j, res))
return
print('NO')
# query_count = 1
query_count = Read.int()
while query_count:
query_count -= 1
solve()
| 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--) {
long long i, j, k, n;
cin >> n;
map<long long, long long> mp;
long long x = n;
while (x % 2 == 0) {
mp[2]++;
x /= 2;
}
for (i = 3; i * i <= x; i += 2) {
while (x % i == 0) {
mp[i]++;
x /= i;
}
}
if (x > 1) mp[x]++;
long long a = 1, b = 1, c = 1;
i = 0;
auto it = mp.begin();
a = it->first;
if (it->second > 2) {
b = it->first * it->first;
c = n / (it->first * it->first * it->first);
} else if (it->second == 2) {
b = it->first;
it++;
if (it == mp.end()) {
cout << "NO\n";
continue;
} else {
b *= it->first;
c = n / (a * b);
}
} else {
it++;
if (it == mp.end()) {
cout << "NO\n";
continue;
} else {
b *= it->first;
c = n / (a * b);
}
}
if (a == b || b == c || c == a || a < 2 || b < 2 || c < 2) {
cout << "NO\n";
} else {
cout << "YES\n";
cout << a << " " << b << " " << c << "\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 | t=int(input())
for _ in range(t):
n=int(input())
a,b,c=0,0,0
flag=0
flg=0
cnt=0
mx=2
l=[]
if(n<24):
print("NO")
else:
for _ in range(3):
for i in range(mx, int(n**0.5) + 1):
if n % i == 0:
flg=1
break
if(flg==0):
print("NO")
flag=1
break
mx=i+1
n=n//i
flg=0
if(i<n):
l.append(i)
cnt+=1
else:
print("NO")
flag=1
break
if(cnt==2):
l.append(n)
break
if(flag==0):
print("YES")
print(l[0],l[1],l[2])
| 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
N = int(input())
for i in range(N):
n = int(input())
res = []
is_good = False
for i in range(2, n//6 + 1):
if i > sqrt(n):
break
if n % i == 0:
res.append(i)
n /= i
if len(res) == 2:
if n > i:
is_good = True
break
if is_good:
print(f'YES\n{res[0]} {res[1]} {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 | t = int(input())
from math import *
import bisect
for i in range(t):
n = int(input())
x = set()
for j in range(2,int(sqrt(n))+1):
if(n%j==0):
x.add(j)
if(n%(n//j)==0):
x.add(n//j)
x = list(x)
x.sort()
flag = 0
for j in range(len(x)):
for k in range(j+1,len(x)):
y = n/(x[j]*x[k])
if(y in x and y!=x[j] and y!=x[k]):
flag = 1
print('YES')
print(x[j],x[k],int(y))
break
if(flag==1):
break
if(flag == 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 | t = int(input())
for _ in range(t):
n = int(input())
ans = []
i = 2
top = 10**6
while len(ans) < 2 and i <= n and i <= top:
if n%i == 0:
n //= i
ans.append(i)
i += 1
if len(ans) < 2 or n == 1 or n in ans:
print('NO')
else:
print('YES')
ans.append(n)
for i in ans:
print(i, end=' ')
print() | 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() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> f, c;
int r = n;
int co{0};
while (r % 2 == 0) {
r = r / 2;
co++;
}
if (co > 0) {
f.push_back(2);
c.push_back(co);
}
for (int i = 3; i * i < n; i++) {
co = 0;
while (r % i == 0) {
r = r / i;
co++;
}
if (co > 0) {
f.push_back(i);
c.push_back(co);
}
}
if (r > 2) {
f.push_back(r);
c.push_back(1);
}
if (f.size() >= 3) {
cout << "YES\n";
cout << f[0] << ' ' << f[1] << ' ';
n = n / f[0];
n = n / f[1];
cout << n << '\n';
} else if (f.size() == 2) {
if (c[0] + c[1] >= 4) {
cout << "YES\n";
cout << f[0] << ' ' << f[1] << ' ';
n = n / f[0];
n = n / f[1];
cout << n << '\n';
} else {
cout << "NO\n";
}
} else {
if (c[0] > 5) {
cout << "YES\n";
cout << f[0] << ' ' << f[0] * f[0] << ' ';
n = n / f[0];
n = n / f[0];
n = n / f[0];
cout << n << '\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 | t=int(input())
for t in range (t):
n=int(input())
d=2
cate=0
vec=[0]*10
while d*d <= n:
if(n%d == 0 and cate<2):
vec[cate]=d
cate+=1
n/=d
if cate == 2:
break
d+=1
if n > 1 and cate == 2:
vec[cate]=n
cate+=1
if cate < 3 or vec[2] == vec[1]:
print("NO")
else:
print("YES")
print(int(vec[0]),int(vec[1]),int(vec[2])) | 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 primeFactors(n):
f = []
while n % 2 == 0:
f.append(2),
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
f.append(i),
n = n / i
if n > 2:
f.append(n)
return f
t = int(input())
while t>0:
n = int(input())
l = primeFactors(n)
s = list(set(l))
ss = len(s)
sl = len(l)
if sl<=2:
print('NO')
elif ss==1 and sl<6:
print('NO')
elif ss==2 and sl<4:
print('NO')
else:
if ss>=2:
l1 = [int(s[0]),int(s[1]),int(n/(s[0]*s[1]))]
print('YES')
print(*l1)
else:
l2 = [int(s[0]),int(s[0]*s[0]),int(n/(s[0]*s[0]*s[0]))]
print('YES')
print(*l2)
t-=1 | 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 | # -*- encoding: utf-8 -*-
import sys
import math
r_input = sys.stdin.readline
def get_divisor(num):
divisors = []
length = int(math.sqrt(num)) + 1
for i in range(1, length):
if num % i == 0:
divisors.append(i)
divisors.append(num // i)
return divisors
if __name__ == '__main__':
T = int(r_input())
for _ in range(T):
number = int(r_input())
f_list = get_divisor(number)
flag = 0
for n in f_list:
tp = number // n
s_list = get_divisor(tp)
for m in s_list:
tmp = sorted([n, m, tp // m])
if not 1 in tmp:
if tmp[0] != tmp[1] and tmp[1] != tmp[2]:
print('YES')
print(*tmp)
flag = 1
break
if flag:
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 | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int MAXN = (int)1e5 + 5;
const int MOD = (int)1e9 + 7;
const int INF = 0x3f3f3f3f;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
bool mul(int n, int& a, int& b, int& c) {
a = INF, b = INF, c = INF;
for (int i = 2; i * i * i <= n; i++) {
if (n % i == 0) {
a = i;
break;
}
}
if (n % a == 0)
n = n / a;
else
return false;
for (int i = a + 1; i * i <= n; i++) {
if (n % i == 0) {
b = i;
break;
}
}
if (n % b == 0)
n = n / b;
else
return false;
if (n > b) {
c = n;
return true;
} else
return false;
}
int main() {
int T, a, b, c, n;
cin >> T;
while (T--) {
cin >> n;
if (mul(n, a, b, c))
printf("YES\n%d %d %d\n", a, b, c);
else
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 | from sys import stdin,stdout
t = 1
t=int(input())
def factorgreaterthan(n,x):
i=x+1
while (i*i)<=n:
if n%i==0:
return i
i+=1
return n
for i in range(t):
n=int(input())
N=n
l=[]
d1=factorgreaterthan(n,1)
n//=d1
l.append(d1)
d2=factorgreaterthan(n,d1)
n//=d2
l.append(d2)
l.append(n)
l.sort()
m=1
flag=True
p=1
for i in l:
m*=i
if i==p:
flag=False
p=i
if m!=N:
flag=False
if flag:
print("YES")
print(*l)
else:
print("NO")
#solve
| 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 Divisors(number):
Div=set()
sqr=int(number**0.5)
for i in range(2,sqr+1):
if number%i==0:
Div=Div|{i}
Div=Div|{(number//i)}
return list(Div)
def getDiv(Number):
D=list(Divisors(Number))
D.sort()
for i in range(len(D)):
for j in range(i+1,len(D)):
m=D[i]*D[j]
if m in D and Number//m !=D[i] and Number//m !=D[j] :
r=str(D[i])+' '+str(D[j])+' '+str(Number//m)
r='YES\n'+r
return r
return 'NO'
Answers=[]
n=int(input())
for i in range(n):
a=int(input())
Answers.append(getDiv(a))
for i in Answers:
print(i) | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void err() { cout << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) {
cout << arg << ' ';
err(args...);
}
using ll = long long;
using db = double;
using pII = pair<int, int>;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int N = 2e2 + 10;
int n;
int a[N], b[N];
int vis[N];
void RUN() {
memset(vis, 0, sizeof vis);
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> b[i];
}
for (int i = 1, j = 1; i <= n; ++i, j += 2) {
a[j] = b[i];
vis[b[i]] = 1;
}
for (int i = 2; i <= 2 * n; i += 2) {
int lst = -1;
for (int j = a[i - 1] + 1; j <= 2 * n; ++j) {
if (!vis[j]) {
vis[j] = 1;
lst = j;
break;
}
}
if (lst == -1) {
cout << -1 << "\n";
return;
}
a[i] = lst;
}
for (int i = 1; i <= 2 * n; ++i) {
cout << a[i] << " \n"[i == 2 * n];
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
cout << fixed << setprecision(20);
int T;
cin >> T;
for (int cas = 1; cas <= T; ++cas) {
RUN();
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
public class Unusual_Competitions {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t1=Integer.parseInt(br.readLine());
for(int t=0;t<t1;++t)
{
int n=Integer.parseInt(br.readLine());
ArrayList<Integer>al=new ArrayList<>();
StringBuilder ans=new StringBuilder();
for(int i=1;i<=2*n;++i)
{
al.add(i);
}
String s[]=br.readLine().trim().split(" ");
for(String a:s)
{
Integer k=new Integer(a);
al.remove(k);
}
boolean flag=true;
for(String a:s)
{
Integer k=new Integer(a);
ans.append(a+" ");
int f=0;
for(Integer i:al)
{
if(i>k)
{
f=1;
ans.append(i+" ");
al.remove(i);
break;
}
}
if(f==0)
{
System.out.println(-1);
flag=false;
break;
}
}
if (flag) {
System.out.println(ans);
}
}
}
}
| JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | t = int(input())
from collections import defaultdict
for _ in range(t):
dic = defaultdict(int)
dic2 = defaultdict(int)
n = int(input())
bl = list(map(int,input().split()))
for idx, b in enumerate(bl):
dic[b] = idx + 1
bl_sort = sorted(bl)
s = 0
kouho = []
for b in range(1, 2*n+1):
if not dic[b]:
if len(kouho) == 0:
print(-1)
break
min_idx = float("inf")
pos = -1
for k in kouho:
if min_idx > dic[k]:
min_idx = dic[k]
pos = k
dic2[pos] = b
kouho.remove(pos)
else:
kouho.append(b)
s += 1
else:
ansl = []
for b in bl:
ansl.append(b)
ansl.append(dic2[b])
print(*ansl)
continue
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int modPow(long long int x, long long int n, long long int mod) {
long long int r = 1;
x = x % mod;
while (n) {
if (n & 1) r = (r * x) % mod;
n = n >> 1;
x = (x * x) % mod;
}
return r;
}
int main() {
long long int t, i, j, k, l, n, m, a, b, c, x, y, z;
cin >> t;
while (t--) {
cin >> n;
vector<long long int> V;
map<long long int, long long int> M;
bool ans = true;
for (i = 0; i < n; ++i) {
cin >> l;
M[l]++;
V.push_back(l);
if (M[l] > 1) ans = false;
}
vector<long long int> P(2 * n, 0);
for (i = 0; i < n; ++i) P[i * 2] = V[i];
set<pair<long long int, long long int> > S;
for (i = 0; i < n; ++i) {
bool milla = false;
for (j = V[i]; j <= 2 * n; ++j) {
if (M[j] == 0) {
M[j]++;
P[2 * i + 1] = j;
milla = true;
break;
}
}
if (milla == false) {
ans = false;
break;
}
}
if (ans) {
for (auto it : P) cout << it << " ";
cout << "\n";
} else {
cout << -1 << "\n";
}
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | t=int(input())
for q in range(t):
n=int(input())
arr=list(map(int,input().split()))
marked=[0]*(2*n+1)
for i in range(len(arr)):
marked[arr[i]]=1
#print(marked)
ans=[]
for i in range(n):
ans.append(arr[i])
for j in range(arr[i]+1,len(marked)):
if marked[j]==0:
ans.append(j)
flag=1
marked[j]=1
break
if len(ans)==2*n:
for i in range(len(ans)):
print(ans[i],end=" ")
print()
else:
print(-1)
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 105;
int T, n, tot;
int a[N], b[N << 1];
bool vis[N << 1];
int main() {
scanf("%d", &T);
while (T--) {
memset(vis, false, sizeof(vis));
scanf("%d", &n);
for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]), vis[a[i]] = true;
bool jay = true;
for (register int i = 1; i <= n; ++i)
if (a[i] == 2 * n) {
jay = false;
break;
}
if (!jay) {
puts("-1");
continue;
}
int tot = 0;
for (register int i = 1; i <= n; ++i) {
bool flag = false;
for (register int j = a[i] + 1; j <= 2 * n; ++j)
if (!vis[j]) {
b[++tot] = a[i];
b[++tot] = j;
vis[a[i]] = vis[j] = true;
flag = true;
break;
}
if (!flag) {
jay = false;
break;
}
}
if (!jay) {
puts("-1");
continue;
}
for (register int i = 1; i <= tot; ++i) printf("%d ", b[i]);
puts("");
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native,avx2")
using namespace std;
const double eps = 1E-9;
const double Exp = 2.718281828459045;
const double Pi = 3.1415926535897932;
const int Max_Bit = 63;
const int INF = 2000000000;
const long long LINF = 1000000000000000007ll;
const int MOD = 1000000007;
const int NMAX = 1000005;
const int MMAX = 505;
const int KMAX = 6;
const string CODE = "CODEFORCES";
int t[4 * NMAX];
void build(vector<int> a, int v, int tl, int tr) {
if (tl == tr)
t[v] = a[tl];
else {
int tm = (tl + tr) / 2;
build(a, v * 2, tl, tm);
build(a, v * 2 + 1, tm + 1, tr);
t[v] = t[v * 2] + t[v * 2 + 1];
}
}
int sum(int v, int tl, int tr, int l, int r) {
if (l > r) return 0;
if (l == tl && r == tr) return t[v];
int tm = (tl + tr) / 2;
return sum(v * 2, tl, tm, l, min(r, tm)) +
sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
long long gcd(long long a, long long b, long long& x, long long& y) {
if (a == 0) {
x = 0;
y = 1;
return b;
}
long long x1, y1;
long long d = gcd(b % a, a, x1, y1);
x = y1 - (b / a) * x1;
y = x1;
return d;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int T;
for (cin >> T; T; T--) {
int n, a[100], b[200];
int o = 0;
priority_queue<int> q;
cin >> n;
for (int i = 0; i < n; ++i) cin >> a[i];
for (int i = 1; i <= 2 * n; ++i) {
int fnd = 0;
for (int j = 0; j < n; ++j)
if (a[j] == i) fnd = j + 1;
if (fnd == 0) {
if (q.empty()) {
o = 1;
break;
}
int pos = -q.top();
q.pop();
b[pos] = i;
} else {
b[fnd * 2 - 2] = i;
q.push(-fnd * 2 + 1);
}
}
if (o) {
cout << -1 << endl;
continue;
}
for (int i = 0; i < 2 * n; ++i) cout << b[i] << " ";
cout << "\n";
;
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | t=int(input())
for i in range(t):
n=int(input())
arr=list(map(int,input().split()))
ans=[]
for i in range(1,len(arr)*2+1):
if i not in arr:
ans.append(i)
fina=[]
for i in range(len(arr)):
fina.append(arr[i])
for j in range(len(ans)):
if ans[j]>arr[i]:
fina.append(ans[j])
ans.remove(ans[j])
break
if len(fina)!=2*n:
print(-1)
else:
print(*fina,sep=" ") | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import java.util.*;
public class Solutions {
static Scanner scr=new Scanner(System.in);
public static void main(String[] args) {
int t=read();
while(t-->0) {
solve();
}
}
static void solve() {
int n=read();
int []a=new int[n];
int []count=new int[2*n+1];
for(int i=0;i<n;i++) {
a[i]=read();
count[a[i]]=1;
}
int []res=new int[2*n];
for(int i=0;i<n;i++) {
res[2*i]=a[i];
}
int index=1;
for(int i=0;i<n;i++) {
int num=a[i];
int curr=a[i]+1;
while(curr<=2*n) {
if(count[curr]==0) {
count[curr]=1;
res[index]=curr;
index+=2;
break;
}else {
curr++;
}
}
if(curr==2*n+1) {
println(-1);
return;
}
}
printArray(res);
println();
}
static int[] sort(int a[]) {
Arrays.sort(a);
return a;
}
static String reverse(String s) {
String res="";
for(int i=s.length()-1;i>=0;i--) {
res+=s.charAt(i);
}
return res;
}
static int gcd(int a,int b) {
if(b==0) {
return a;
}
return gcd(b,a%b);
}
static void swap(int a[],int i,int j) {
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
static boolean contains(HashMap<Integer,Integer>hm,int ele) {
return hm.containsKey(ele);
}static boolean contains(HashMap<Character,Integer>hm,char ele) {
return hm.containsKey(ele);
}static int get(HashMap<Character,Integer>hm,char ele) {
return hm.get(ele);
}
static char[] cha(String s) {
return s.toCharArray();
}
static int len(String s) {
return s.length();
}
static int len(int a[]) {
return a.length;
}
static int len(long a[]) {
return a.length;
}
static long abs(long a) {
return Math.abs(a);
}static int abs(int a) {
return Math.abs(a);
}
static long min(long a,long b) {
return Math.min(a, b);
}static long max(long a,long b) {
return Math.max(a, b);
}
static void print(int a) {
System.out.print(a+" ");
}static void print(long a) {
System.out.print(a+" ");
}static void print(String a) {
System.out.print(a);
}static void println(int a) {
System.out.println(a);
}static void println(long a) {
System.out.println(a);
}static void println(String a) {
System.out.println(a);
}
static void println() {
System.out.println();
}
static void println(char []c) {
for(int i=0;i<c.length;i++) {
System.out.print(c[i]);
}
println();
}
static int read() {
return scr.nextInt();
}
static String readS() {
return scr.next();
}
static long readL() {
return scr.nextLong();
}
static int[]readIntArray(int n){
// int n=read();
int []a=new int[n];
for(int i=0;i<n;i++) {
a[i]=scr.nextInt();
}
return a;
}static long[]readLongArray(int n){
long []a=new long[n];
for(int i=0;i<n;i++) {
a[i]=scr.nextLong();
}
return a;
}
static void printArray(int a[]) {
int n=a.length;
for(int i=0;i<n;i++) {
print(a[i]);
}
}
}
| JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | # import sys
# input = sys.stdin.readline
# from bisect import bisect_left, bisect_right
# INF = 1 << 60
t = int(input())
for i in range(t):
n = int(input())
b = list(map(int, input().split()))
c = [0 for i in range(2 * n)]
ans = []
possible = True
for j in b:
c[j - 1] = 1
for j in b:
possible2 = False
for k in range(j, 2 * n):
if c[k] == 0:
c[k] = 1
ans.append(j)
ans.append(k + 1)
possible2 = True
break
if not possible2:
possible = False
break
if not possible:
print(-1)
# print(ans)
else:
print(*ans)
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | t=int(input())
for _ in range(t):
n=int(input())
b=[int(s) for s in input().split()]
vmax=max(b)
vmin=min(b)
a=[]
for i in range(0,n):
a.append(b[i])
k=b[i]
while k in b or k in a:
k+=1
if k>2*n:
print(-1)
break
a.append(k)
else:
print(*a) | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 |
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
public class C {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
while (T-->0)
{
int N = sc.nextInt();
int[] list = new int[N+1];
HashSet<Integer> set = new HashSet<>();
for(int i =1 ; i <= N;i++)
{
list[i] = sc.nextInt();
set.add(list[i]);
}
int[] finalList = new int[2*(N+1)];
Arrays.fill(finalList , -1);
for(int i = 1 ; i <= 2*N ; i+=2)
{
int idx = (int)(Math.ceil(i/2d));
finalList[i] = list[idx];
}
boolean flag = true;
for(int i = 1 ; i <= N ; i++)
{
int bi = list[i];
int min = Math.min(finalList[2*i],finalList[(2*i)-1]);
if(min>bi)
{
flag = false;
break;
}
//find the number
for(int k = Math.max(Math.max(finalList[2*i],finalList[(2*i)-1]) , 0) ;k<=2*N;k++)
{
//
if(!set.contains(k))
{
//bzrsh too
finalList[2*i] = k;
set.add(k);
break;
}
}
}
if(!flag)
System.out.println(-1);
else
{
StringBuilder sb = new StringBuilder();
for(int i = 1; i <= 2*N ; i++)
{
if(finalList[i] == -1)
{
flag = false;
System.out.println(-1);
break;
}
sb.append(finalList[i]+" ");
}
if(flag)
{
sb.deleteCharAt(sb.length()-1);
System.out.println(sb.toString());
}
}
}
}
}
| JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import java.util.*;
import java.io.*;
public class Main {
static StringBuilder sb = new StringBuilder();
static BufferedReader br = null;
static StringTokenizer st = null;
static int n;
static int[] ba;
static boolean[] ck;
public static void main(String[] args) throws Exception {
br = new BufferedReader(new InputStreamReader(System.in));
st = null;
int tc = Integer.parseInt(br.readLine());
while(tc-- >0) {
n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
ba = new int[n+1];
ck = new boolean[2*n+2];
boolean flg = false;
for(int i=1; i<=n; i++) {
ba[i] = Integer.parseInt(st.nextToken());
if(ba[i]>=2 * n) {
flg = true;
}else {
ck[ba[i]]=true;
}
}
StringBuilder ans = new StringBuilder();
for(int i=1; i<=n; i++) {
ans.append(ba[i]).append(" ");
int tmp = ++ba[i];
while(ck[tmp] && tmp<=2*n) {
tmp++;
}
if(tmp<=2*n) {
ans.append(tmp).append(" ");
ck[tmp]=true;
}else {
flg = true;
break;
}
}
if(flg) sb.append("-1").append("\n");
else sb.append(ans).append("\n");
}
System.out.println(sb);
}
} | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | def solve(n,b):
if 1 not in b or 2*n in b:
return ([-1])
else:
s=set(b)
arr=[]
for i in range(n):
j=b[i]+1
while True:
if j not in s and j<=2*n:
arr.append(j)
s.add(j)
break
elif j>2*n:
return ([-1])
j+=1
return ([str(b[x])+" "+str(arr[x]) for x in range(n)])
for _ in range(int(input())):
n=int(input())
b=[int(x) for x in input().split()]
print(*solve(n,b))
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | from bisect import bisect
for _ in range(input()):
n = input()
a = map(int,raw_input().split())
if 1 not in a or 2*n in a:
print -1
continue
l = []
for i in range(1,2*n+1):
if i not in a:
l.append(i)
d = {}
f = 0
for i in range(n):
p = bisect(l,a[i])
if p==len(l):
f = 1
break
d[a[i]] = l[p]
l.remove(l[p])
if f==1:
print -1
continue
for i in range(n):
print a[i],d[a[i]],
print
| PYTHON |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import bisect
t=int(input())
def find_gt(a, x):
'Find leftmost value greater than x'
i = bisect.bisect_right(a, x)
if i != len(a):
return a[i]
raise ValueError
for _ in range(t):
b=int(input())
B=list(map(int,input().split()))
d={}
v=[0]
x=[]
y=[]
for i in range(1,2*b+1):
v.append(0)
for i in B:
v[i]=1
for i in range(1,2*b+1):
if(v[i]==0):
y.append(i)
if(v[i]):
x.append(i)
i=0
f=1
while(i<b):
if(x[i]<y[i]):
d[x[i]]=y[i]
else:
print(-1)
f=0
break
i+=1
if(f):
for i in B:
s=find_gt(y,i)
y.remove(s)
print(i,s,end=" ")
print() | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
public class C implements Runnable {
private void solve() throws IOException {
int t = nextInt();
outer2: while(t-- > 0) {
int n = nextInt();
int[] b = new int[n];
for (int i = 0; i < n; i++) {
b[i] = nextInt();
}
boolean[] vis = new boolean[2 * n + 1];
int[] ans = new int[2 * n];
for (int i = 0; i < b.length; i++) {
ans[2 * i] = b[i];
vis[b[i]] = true;
}
outer: for (int i = 1; i < ans.length; i += 2) {
for (int j = ans[i - 1] + 1; j <= 2 * n; j++) {
if (!vis[j]) {
ans[i] = j;
vis[j] = true;
continue outer;
}
}
pl(-1); continue outer2;
}
for (int o : ans) p(o + " ");
pl();
}
}
public static void main(String[] args) {
new C().run();
}
BufferedReader reader;
StringTokenizer tokenizer;
PrintWriter writer;
public void run() {
try {
reader = new BufferedReader(new BufferedReader(new InputStreamReader(System.in)));
writer = new PrintWriter(System.out);
tokenizer = null;
solve();
reader.close();
writer.close();
} catch (Exception e) {
e.printStackTrace();
System.exit(1);
}
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
BigInteger nextBigInteger() throws IOException {
return new BigInteger(nextToken());
}
String nextToken() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
void p(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
void pl(Object... objects) {
p(objects);
writer.println();
}
int cc;
void pf() {
writer.printf("Case #%d: ", ++cc);
}
} | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(2 * n + 1), b(n + 1), c(2 * n + 1, 0);
set<int> d;
for (int i = 1; i <= n; i++) {
cin >> b[i];
c[b[i]] = 1;
}
for (int i = 1; i <= 2 * n; i++)
if (!c[i]) {
d.insert(i);
}
bool ok = true;
for (int i = 1; i <= n; i++) {
auto it = lower_bound(d.begin(), d.end(), b[i] + 1);
if (it == d.end()) {
ok = false;
break;
} else {
a[2 * i - 1] = b[i];
a[2 * i] = *it;
d.erase(it);
}
}
if (ok) {
for (int i = 1; i <= 2 * n; i++) {
cout << a[i] << (i < 2 * n ? ' ' : '\n');
}
} else {
cout << -1 << endl;
}
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import sys,os,io
# input = sys.stdin.readline # for strings
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # for non-strings
PI = 3.141592653589793238460
INF = float('inf')
MOD = 1000000007
# MOD = 998244353
def bin32(num):
return '{0:032b}'.format(num)
def add(x,y):
return (x+y)%MOD
def sub(x,y):
return (x-y+MOD)%MOD
def mul(x,y):
return (x*y)%MOD
def gcd(x,y):
if y == 0:
return x
return gcd(y,x%y)
def lcm(x,y):
return (x*y)//gcd(x,y)
def power(x,y):
res = 1
x%=MOD
while y!=0:
if y&1 :
res = mul(res,x)
y>>=1
x = mul(x,x)
return res
def mod_inv(n):
return power(n,MOD-2)
def prob(p,q):
return mul(p,power(q,MOD-2))
def ii():
return int(input())
def li():
return [int(i) for i in input().split()]
def ls():
return [i for i in input().split()]
for t in range(ii()):
n = ii()
a = li()
b = [0 for i in range(2*n)]
for i in range(n):
b[2*i] = a[i]
# print(b)
inds = [-1 for i in range(2 * n + 1)]
for i in range(n):
inds[a[i]] = i
a.sort()
a.reverse()
lol = 1
for i in a:
ind = inds[i]
flag = 0
for j in range(2*n , i, -1):
if inds[j ] == -1:
flag = j
if flag == 0:
lol = 0
break
b[2*ind + 1] = flag
inds[flag] = 2*ind + 1
if lol:
# print(b)
for i in range(n):
for j in range(i + 1, n):
if b[2*i + 1] > b[2*j] and b[2*j + 1] > b[2*i] and b[2*i + 1] > b[2*j+1]:
b[2*i + 1] ,b[2*j + 1] = b[2*j + 1] , b[2*i + 1]
# print(i , j)
print(*b)
else:
print(-1)
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int arr[1005];
int counter[1005];
int main() {
int tc;
cin >> tc;
while (tc--) {
memset(arr, 0, 1005);
memset(counter, 0, 1005);
int flag = 0;
vector<int> v;
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
counter[arr[i]]++;
}
int lim = 2 * n;
for (int i = 0; i < n; i++) {
v.push_back(arr[i]);
for (int j = arr[i] + 1;; j++) {
if (j > lim) {
flag = 1;
break;
}
if (!counter[j]) {
counter[j]++;
v.push_back(j);
break;
}
}
}
if (flag == 1)
cout << "-1" << endl;
else {
for (auto it = v.begin(); it != v.end(); ++it) cout << *it << " ";
cout << endl;
}
}
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import sys
input = sys.stdin.readline
t=int(input())
for tests in range(t):
n=int(input())
B=list(map(int,input().split()))
USE=[0]*(2*n+1)
A=[-1]*(2*n)
for i in range(n):
A[2*i]=B[i]
USE[B[i]]=1
for i in range(2*n):
if A[i]!=-1:
continue
for j in range(A[i-1]+1,2*n+1):
if USE[j]==0:
A[i]=j
USE[j]=1
break
if -1 in A:
print(-1)
else:
print(*A)
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | for _ in range(int(input())):
n=int(input())
b=[int(i) for i in input().strip().split()]
taken=set(b)
lim=2*n
a=[]
for x in b:
done=False
pairx=None
for y in range(x+1,lim+1):
if y not in taken:
done=True
pairx=y
break
if done==False:
break
a+=[x,pairx]
taken.add(pairx)
if done==False:
print(-1)
else:
for x in a:
print(x," ",sep='',end='')
print()
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t;
cin >> t;
int vis[205], p = 0;
memset(vis, 0, sizeof(vis));
while (t--) {
++p;
int n, b[205];
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> b[i];
vis[b[i]] = p;
}
int index;
bool f = false;
vector<int> ans;
for (int i = 0; i < n; ++i) {
index = -1;
for (int j = b[i] + 1; j <= 2 * n; ++j) {
if (vis[j] != p) {
index = j;
break;
}
}
if (index == -1) {
f = true;
break;
}
vis[index] = p;
ans.push_back(b[i]);
ans.push_back(index);
}
if (!f) {
for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
} else
cout << -1;
cout << endl;
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | t=int(input())
while(t):
t-=1
n=int(input())
li=list(map(int,input().split()))
b_li=[0]*(2*n)
for i in range(0,n):
b_li[i*2]=li[i]
have=[]
for i in range(1,2*n+1):
if i not in li:
have.append(i)
for i in range(1,2*n+1,2):
pre=b_li[i-1]
flag=0
for j in range(len(have)):
if have[j]>pre:
flag=1
break
if flag==1:
b_li[i]=have[j]
del(have[j])
else:
print(-1)
break
if (flag==1):
for i in b_li:
print(i,end=" ")
print()
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 |
import java.util.*;
import java.lang.*;
import java.io.*;
public class cf {
static PrintWriter out;
static int MOD = 1000000007;
static FastReader scan;
/*-------- I/O using short named function ---------*/
public static String ns() {
return scan.next();
}
public static int ni() {
return scan.nextInt();
}
public static long nl() {
return scan.nextLong();
}
public static double nd() {
return scan.nextDouble();
}
public static String nln() {
return scan.nextLine();
}
public static void p(Object o) {
out.print(o);
}
public static void ps(Object o) {
out.print(o + " ");
}
public static void pn(Object o) {
out.println(o);
}
/*-------- for output of an array ---------------------*/
static void iPA(int arr[]) {
StringBuilder output = new StringBuilder();
for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
out.println(output);
}
static void lPA(long arr[]) {
StringBuilder output = new StringBuilder();
for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
out.println(output);
}
static void sPA(String arr[]) {
StringBuilder output = new StringBuilder();
for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
out.println(output);
}
static void dPA(double arr[]) {
StringBuilder output = new StringBuilder();
for (int i = 0; i < arr.length; i++) output.append(arr[i] + " ");
out.println(output);
}
/*-------------- for input in an array ---------------------*/
static void iIA(int arr[]) {
for (int i = 0; i < arr.length; i++) arr[i] = ni();
}
static void lIA(long arr[]) {
for (int i = 0; i < arr.length; i++) arr[i] = nl();
}
static void sIA(String arr[]) {
for (int i = 0; i < arr.length; i++) arr[i] = ns();
}
static void dIA(double arr[]) {
for (int i = 0; i < arr.length; i++) arr[i] = nd();
}
/*------------ for taking input faster ----------------*/
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;
}
}
public static ArrayList<Integer> sieveOfEratosthenes(int n)
{
// Create a boolean array
// "prime[0..n]" and
// initialize all entries
// it as true. A value in
// prime[i] will finally be
// false if i is Not a
// prime, else true.
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
ArrayList<Integer> arr=new ArrayList<>();
// Print all prime numbers
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
// Method to check if x is power of 2
static boolean isPowerOfTwo(int x) {
return x != 0 && ((x & (x - 1)) == 0);
}
//Method to return lcm of two numbers
static int gcd(int a, int b) {
return a == 0 ? b : gcd(b % a, a);
}
//Method to count digit of a number
static int countDigit(long n) {
String sex = Long.toString(n);
return sex.length();
}
static void reverse(int a[]) {
int i, k, t;
int n = a.length;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
//Method for sorting
static void ruffle_sort(int[] a) {
//shandom_ruffle
Random r = new Random();
int n = a.length;
for (int i = 0; i < n; i++) {
int oi = r.nextInt(n);
int temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
//sort
Arrays.sort(a);
}
//Method for checking if a number is prime or not
static boolean isPrime(int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static void main(String[] args) throws java.lang.Exception {
OutputStream outputStream = System.out;
out = new PrintWriter(outputStream);
scan = new FastReader();
//for fast output sometimes
StringBuilder sb = new StringBuilder();
int t = ni();
String s="1";
while(t-- != 0) {
int n=ni();
int[] a=new int[n];
iIA(a);
HashSet<Integer> hs=new HashSet<>();
int count=0;int f=0;int[] b=new int[n*2];
for(int i=0;i<n;i++){
if(a[i]>n)
count++;
if(count>n/2){
f=1;break;
}
}
if(f==1){
pn("-1");
}
else{
int j=0;
for(int i:a)
hs.add(i);
for(int i=0;i<n;i++){
b[j++]=a[i];
int tt=a[i]+1;
while(hs.contains(tt))
tt+=1;
hs.add(tt);
b[j++]=tt;
}
int max=0;
for(int i:b)
max=Math.max(max,i);
if(max>2*n)
pn(-1);
else{
iPA(b);
}
}
}
out.flush();
out.close();
}
} | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
b = [i for i in range(1,2*n+1) if i not in a]
sol = [0]*(2*n)
for i in range(len(a)):
tmp=[j for j in b if j>a[i]]
if len(tmp)==0:
sol=-1
break
else:
sol[2*i]=a[i]
sol[2*i+1]=tmp[0]
b.remove(tmp[0])
if sol==-1: print(sol)
else: print(' '.join([str(i) for i in sol])) | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int t, n;
int main() {
cin >> t;
while (t--) {
int f = 0;
cin >> n;
vector<int> a;
a.push_back(0);
map<int, int> m;
int b[2 * n + 5];
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
if (x >= 2 * n) f = 1;
a.push_back(x);
m[x]++;
}
if (f) {
cout << -1 << endl;
continue;
}
for (int i = 1; i <= n && !f; i++) {
int x = a[i];
while (m.find(x) != m.end()) x++;
if (x > 2 * n) {
f = 1;
} else {
m[x]++;
b[2 * i - 1] = a[i];
b[2 * i] = x;
}
}
if (f)
cout << "-1" << endl;
else {
for (int i = 1; i <= 2 * n; i++) {
cout << b[i] << " ";
}
cout << endl;
}
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007;
const int MAXN = (int)3e5 + 1;
void setIO(string name) {
ios_base::sync_with_stdio(0);
cin.tie(0);
freopen((name + ".in").c_str(), "r", stdin);
freopen((name + ".out").c_str(), "w", stdout);
}
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
bool used[2 * n];
for (int i = 0; i < (2 * n); i++) {
used[i] = false;
}
int seq[n];
for (int i = 0; i < n; i++) {
cin >> seq[i];
seq[i]--;
used[seq[i]] = true;
}
int ans[2 * n];
bool can = true;
for (int i = 0; i < n; i++) {
int val = seq[i];
while (val < 2 * n && used[val]) {
val++;
}
if (val == 2 * n) {
can = false;
break;
}
used[val] = true;
ans[(2 * i)] = seq[i];
ans[(2 * i + 1)] = val;
}
if (!can) {
cout << -1;
} else {
for (int i = 0; i < 2 * n; i++) {
cout << ans[i] + 1 << " ";
}
}
cout << "\n";
}
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double EPS = -1e-2;
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
const long long mod = 1e9 + 7;
const int Mx = INT_MAX;
const int Mn = INT_MIN;
const long long MX = LLONG_MAX;
const long long MN = LLONG_MIN;
const int N = 1e5 + 9;
void Open() {}
long long arr[100005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
Open();
int t;
cin >> t;
while (t--) {
set<int> st;
set<int>::iterator it;
;
int n;
cin >> n;
vector<int> v(n + 1), ans(2 * n + 1);
for (int i = 1; i <= 2 * n; i++) st.insert(i);
for (int i = 1; i <= n; i++) {
cin >> v[i];
st.erase(v[i]);
}
bool ok = 1;
for (int i = 1; i <= n; i++) {
it = st.upper_bound(v[i]);
if (it == st.end()) {
ok = 0;
break;
}
ans[2 * i - 1] = v[i];
ans[2 * i] = *it;
st.erase(it);
}
if (ok) {
for (int i = 1; i <= 2 * n; i++) cout << ans[i] << " ";
} else
cout << -1;
cout << "\n";
}
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #import sys
#import math
#sys.stdout=open("C:/Users/pipal/OneDrive/Desktop/VS code/python/output.txt","w")
#sys.stdin=open("C:/Users/pipal/OneDrive/Desktop/VS code/python/input.txt","r")
t=int(input())
for i in range(t):
n=int(input())
l=list(map(int,input().split()))
if 1 not in l:
print("-1")
elif 2*n in l:
print(-1)
else:
new=[]
for i in range(n):
chc=1
new.append(l[i])
for k in range(l[i],2*n+1):
if k not in l:
l.append(k)
new.append(k)
chc=0
break
if chc==1:
print(-1)
break
if chc==0:
for i in range(len(new)):
print(new[i],end=' ')
print()
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | I=input
for _ in[0]*int(I()):
n=2*int(I());a=[0]*n;b=a[::2]=*map(int,I().split()),;c={*range(1,n+1)}-{*b};i=1
try:
for x in b:y=a[i]=min(c&{*range(x,n+1)});c-={y};i+=2
except:a=-1,
print(*a) | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | def solve(N, B):
A = [0] * (2 * N)
check = [False] * (2 * N)
flag = True
for i in range(N):
A[i*2] = B[i]
check[B[i]-1] = True
for i in range(N):
tmp = A[i*2]
for j in range(tmp, 2*N):
if not check[j]:
A[i*2+1] = j+1
check[j] = True
break
for i in range(N):
if A[2*i] > A[2*i+1]:
flag = False
if flag:
print(" ".join(map(str, A)))
else:
print(-1)
def main():
T = int(input())
for i in range(T):
N = int(input())
B = list(map(int, input().split()))
solve(N, B)
if __name__ == "__main__":
main() | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | import math
import heapq
import sys
num = int(raw_input())
for i in range(num):
n = int(raw_input())
nums = [int(x) for x in raw_input().split(" ")]
if 1 not in nums:
print("-1")
continue
cur_set = set()
for nn in nums:
cur_set.add(nn)
ans = []
flag = 0
for nn in nums:
ans.append(str(nn))
c = nn+1
while c in cur_set and c<=2*n:
c+=1
if c>2*n:
flag = 1
break
ans.append(str(c))
cur_set.add(c)
if flag == 1:
print("-1")
continue
print(" ".join(ans))
| PYTHON |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1e9 + 7;
const char nl = '\n';
void solve() {
int t;
cin >> t;
vector<int> b, a;
unordered_map<int, int> all;
while (t--) {
int n;
cin >> n;
b.resize(n);
all.clear();
bool all_unique = true;
for (int i = 0; i < n; ++i) {
cin >> b[i];
if (all.find(b[i]) != all.end()) {
all_unique = false;
}
all[b[i]] = i;
}
if (!all_unique) {
cout << -1 << "\n";
continue;
}
a.assign(2 * n, -1);
bool good = true;
for (int m = 1; m <= 2 * n; ++m) {
if (all.find(m) != all.end()) {
int pos = all[m];
a[2 * pos] = m;
} else {
bool found = false;
int ind = 0;
for (int i = 0; i < n; ++i) {
if (a[2 * i] != -1 && a[2 * i + 1] == -1) {
found = true;
ind = i;
break;
}
}
if (found) {
a[2 * ind + 1] = m;
} else {
good = false;
break;
}
}
}
if (!good) {
cout << -1;
} else {
for (int i = 0; i < 2 * n; ++i) {
cout << a[i] << " ";
}
}
cout << "\n";
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
solve();
return 0;
}
| CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | t=int(input())
for i in range(0,t):
o=[]
n=int(input())
l=list(map(int,input().split()))
e=0
for j in range(0,len(l)):
o.append(l[j])
c=l[j]+1
if c>2*n:
e=1
break
else:
while c in l or c in o:
c+=1
if c>2*n:
e=1
break
if e==1:
break
o.append(c)
if e==1:
print(-1)
else:
for k in range(0,len(o)):
print(o[k],end=' ')
print()
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | for T in range(int(input())):
n = int(input())
b = list(map(int, input().split()))
Ans = []
Exist = [False] * (2 * n + 100)
Bad = False
for i in b : Exist[i] = True
for i in range(n):
Ans.append(b[i])
Number = b[i] + 1
while Exist[Number] == True and Number < 2 * n: Number += 1
if Number <= 2 * n and Exist[Number] == False:
Ans.append(Number)
Exist[Number] = True
else:
Bad = True
break
if Bad : print(-1)
else : print(*Ans)
| PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).
The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).
It is guaranteed that the sum of n by all test cases doesn't exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number -1.
Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
Input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
Output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
set<int> prime;
void SieveOfEratosthenes(int n) {
bool prme[n + 1];
memset(prme, true, sizeof(prme));
for (int p = 2; p * p <= n; p++) {
if (prme[p] == true) {
for (int i = p * p; i <= n; i += p) prme[i] = false;
}
}
for (int p = 2; p <= n; p++)
if (prme[p]) prime.insert(p);
}
void solve() {
int n;
cin >> n;
vector<int> in(n), v;
set<int> s, temp;
for (int i = 0; i < n; i++) {
cin >> in[i];
s.insert(in[i]);
}
for (auto i : in) {
v.push_back(i);
int x = i + 1;
while (s.count(x) || temp.count(x)) {
x++;
}
v.push_back(x);
temp.insert(x);
}
auto it = temp.end();
it--;
if ((*it) != (2 * n)) {
cout << -1;
} else {
for (auto i : v) cout << i << ' ';
}
cout << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) solve();
}
| CPP |
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