hdu 3835 简单概率dp

R(N)

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 1

Problem Description

We know that some positive integer x can be expressed as x=A^2+B^2(A,B are integers). Take x=10 for example, 
10=(-3)^2+1^2.
We define R(N) (N is positive) to be the total number of variable presentation of N. So R(1)=4, which consists of 1=1^2+0^2, 1=(-1)^2+0^2, 1=0^2+1^2, 1=0^2+(-1)^2.Given N, you are to calculate R(N).

 

Input

No more than 100 test cases. Each case contains only one integer N(N<=10^9).

 

Output

For each N, print R(N) in one line.

 

Sample Input

   
   
   
   

    
    
    
    2
6
10
25
65
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
0
8
12
16

    
    
    
    
 
     
 
      Hint 
     
For the fourth test case, (A,B) can be (0,5), (0,-5), (5,0), (-5,0), (3,4), (3,-4), (-3,4), (-3,-4), (4,3) , (4,-3), (-4,3), (-4,-3) 
    
   
   
   
   

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(!n)
        {
            printf("1\n");
            continue;
        }
        int i,j,ans=0,ave=(int)sqrt(double(n/2.0));
        for(i=ave; i>=0; i--)
        {
             j=(int)sqrt(1.0*(n-i*i));
                if(i*i+j*j==n)
                {
                    if(i==0||j==0||i==j) ans+=4;
                    else ans+=8;
                }
        }
        printf("%d\n",ans);
    }
    return  0;
}
/*
2
6
10
25
65
*/

Source

2011 Multi-University Training Contest 1 - Host by HNU