hdu 3835:R(N)（水题，数学题）

R(N)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1815    Accepted Submission(s): 929

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)

 

Source

2011 Multi-University Training Contest 1 - Host by HNU

 

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　　水题，数学题。

　　题意是求一个正整数N分解成 N=A^2+B^2 的形式有多少种。A、B可以从0开始。

　　思路：遍历即可，注意两层循环会超时，尽量用一层循环。用cin、cout不会超时。

　　代码：

 #include <iostream>
 #include <stdio.h>
 #include <cmath>
 using namespace std;

 int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF){
         int sum = ;
         for(int i=;i<=sqrt(n);i++){
             int j = n-i*i;
             double t = sqrt(j);
             if(t-int(t)==){ //开方成功
                 if(i!= && j!=)
                     sum+=;
                 else if(i== || j==)
                     sum+=;
             }
         }
         printf("%d\n",sum);
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013