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  • H

    H - R(N)
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

    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
    ",4*ans);//每种情况都有四种情况
    16     }
    17     return 0;
    18 }
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  • 原文地址:https://www.cnblogs.com/angledamon/p/3894109.html
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