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：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <cmath>
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     int n;
 9     while(scanf("%d",&n)!=EOF){
10         int sum = 0;
11         for(int i=0;i<=sqrt(n);i++){
12             int j = n-i*i;
13             double t = sqrt(j);
14             if(t-int(t)==0){ //开方成功
15                 if(i!=0 && j!=0)
16                     sum+=4;
17                 else if(i==0 || j==0)
18                     sum+=2;
19             }
20         }
21         printf("%d
",sum);
22     }
23     return 0;
24 }

Freecode : www.cnblogs.com/yym2013