H - R(N)

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H - R(N) Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64uSubmit 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

26102565  

Sample Output

4081216

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)

 1 #include<cstdio> 2 #include<math.h> 3 using namespace std; 4 int main() 5 { 6     int n; 7     while(scanf("%d",&n)!=EOF) 8     { 9         int ans=0;10         for(int i=1;i*i<=n;i++)//i是平方因子11         {12             double m=sqrt(n-i*i);13             if(floor(m+0.50)==m) ans++;//floor這個函數是用來判斷m 是否是整數，加0.5是用來減少誤差的。14         }15         printf("%d\n",4*ans);//每種情況都有四種情況16     }17     return 0;18 }