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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

237

Sample Output

YESNO

Hint

HintX | 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.

【思路】

打表：

#include 
   
    #include 
    
     using namespace std;int f[5001];void init()//打表找规律的函数{    for(int i=1; i<=5000; i++)    {        for(int j=1; j<=1000; j++)        {            for(int k=1; k<=1000; k++)            {                if((j*j*k+1)%i==0&&(j*j+k)%i!=0)                {                    f[i]=1;                    break;                }            }            if(f[i]==1)                break;        }    }    for(int i=1;i<=5000;i++)        if(f[i]==0)        cout<
     <
    
   

正式代码：

#include
   
    using namespace std;int t;int n;int f[5001];void init(){    f[1]=1;    f[2]=1;    f[3]=1;    f[4]=1;    f[5]=1;    f[6]=1;    f[8]=1;    f[10]=1;    f[12]=1;    f[15]=1;    f[16]=1;    f[20]=1;    f[24]=1;    f[30]=1;    f[40]=1;    f[48]=1;    f[60]=1;    f[80]=1;    f[120]=1;    f[240]=1;}int main(){    init();    ios::sync_with_stdio(false);    cin>>t;    while(t--)    {        cin>>n;        if(f[n]==1)            cout<<"YES"<
   

 

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