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)

 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。
14         }
15         printf("%d\n",4*ans);//每种情况都有四种情况
16     }
17     return 0;
18 }

H - R(N)