A Central Meridian (ACM) Number N is a positive integer satisfies that given two positive integers A and B, and among A, B and N, we have 
N | ((A^2)*B+1) Then N | (A^2+B) 
Now, here is a number x, you need to tell me if it is ACM number or not. 

Input

The first line there is a number T (0<T<5000), denoting the test case number. 
The following T lines for each line there is a positive number N (0<N<5000) you need to judge.

Output

For each case, output “YES” if the given number is Kitty Number, “NO” if it is not.

Sample Input

2
3
7

Sample Output

YES
NO

Hint

Hint
X | Y means X is a factor of Y, for example 3 | 9;
X^2 means X multiplies itself, for example 3^2 = 9;
X*Y means X multiplies Y, for example 3*3 = 9.

这个题很题意很微妙啊

他的意思是给你一个数n 如果存在a b满足 

若a，b为正整数，且n|a2b+1，则n|a2+b.

实际上的意思是如果存在 ab 满足第一个公式 但不满足第二个公式 那这个数就不满足性质p

这样推导出来的结果是 以下。。反正我数学差我看不懂

#include <bits/stdc++.h>
using namespace std;
#define INF 1e8
#define eps 1e-4
#define ll __int64
#define maxn 500010
#define mol 1000000007
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        if(240%n==0)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

第二种做法打表就可以了

#include <bits/stdc++.h>
using namespace std;
#define INF 1e8
#define eps 1e-4
#define ll __int64
#define maxn 50000+5
#define mol 1000000007
int flag[maxn];
void init(){
    memset(flag,1,sizeof(flag));
    for(int k=1;k<=5000;k++){
        for(int i=1;i<=1000;i++){
            for(int j=1;j<=1000;j++){
                int x=i*i*j+1,y=i*i+j;
                if(x%k==0&&y%k!=0){
                    flag[k]=0;
                    goto ed;
                }
            }
        }
        ed:{}
    }
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        if(flag[n])printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}