H - R(N)

H - R(N)

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3835

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<cstdio>
#include<math.h>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;
        for(int i=1;i*i<=n;i++)//i是平方因子11         {
            double m=sqrt(n-i*i);
            if(floor(m+0.50)==m) ans++;//floor这个函数是用来判断m 是否是整数，加0.5是用来减少误差的。
        }
        printf("%d\n",4*ans);//每种情况都有四种情况
    }
    return 0;
}