HDU/HDOJ 2854 2009 Multi-University Training Contest 5 - Host by NUDT 数论

 

Central Meridian Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 210    Accepted Submission(s): 117

Problem Description

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 The following T lines for each line there is a positive number N (0

 

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.

   
     
     
     
     
    
  
    
    
    
    

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

其实是标题党。。

这货就是预处理一下就可以了。。纯暴力

 

我的代码：

#include

bool flag[5005];

void init()
{
	int i,j,n,x,y;
	for(n=1;n<=5000;n++)
	{
		for(i=1;i<=1000;i++)
		{
			for(j=1;j<=1000;j++)
			{
				x=i*i*j+1;
				y=i*i+j;
				if(x%n==0)
				{
					if(y%n!=0)
					{
						flag[n]=true;
						goto loop;
					}
				}
			}
		}
		loop:;
	}
}

int main()
{
	int n,t;
	init();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		if(flag[n])
			printf("NO\n");
		else
			printf("YES\n");
	}
	return 0;
}