HDU 3835 R(N)(枚举)

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)

题解：暴力枚举一下，要注意if(i>j)break;这一句，否则会重复计算WA掉。

#include <cstdio>

#include <iostream>

#include <string>

#include <cstring>

#include <stack>

#include <queue>

#include <algorithm>

#include <cmath>

#include <map>

using namespace std;

//#define LOCAL

int main()

{

#ifdef LOCAL

    freopen("in.txt", "r", stdin);

#endif // LOCAL

    //Start

    int n;

    while(cin>>n)

    {

        int ans=;

        for(int i=; i<sqrt(n); i++)

        {

            double j=sqrt(n-i*i);

            if(i>j)break;

            if((int)j==j)

            {

                if(i==||j==||i==j)ans+=;

                else ans+=;

            }

        }

        printf("%d\n",ans);

    }

    return ;

}

巴特西

HDU 3835 R(N)(枚举)

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