HDU3835 R(N)

R(N)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1412    Accepted Submission(s): 729

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)

 

//求给定整数n可以分解成多少种情况使得n=x^2+y^2(x与y均为整数)
#include<stdio.h>
#include<math.h>

int main()
{
	int k,i,j,n,ans;
	while(~scanf("%d",&n))
	{
		if(!n)
		{
			printf("1\n");
			continue;
		}
		ans=0;
		k=(int)sqrt(n/2.0);
		for(i=0;i<=k;++i)
		{
			j=(int)sqrt(n-i*i);
			if(i*i+j*j==n)
			{
				if(i==j||i==0||j==0)  //其中有一个为0或者x==y时候只有四种情况
					ans+=4;
				else
					ans+=8;    //x与y不为0且不相等的时候有8种情况
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}