HUST1544-R(N)

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最新推荐文章于 2025-02-16 16:31:52 发布

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本文介绍了一道数学编程题目，目标是计算一个正整数N可以被表示为两个整数平方和的不同方式的数量。文章提供了完整的C++代码实现，并通过样例解释了输出结果。

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R(N)

HUST - 1544

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

Sample Output

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)

题意：给出一个数n，问有多少种方式使得n=a*a+b*b

解题思路：构造

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int main()
{
    int n,sum,k;
    while(~scanf("%d",&n))
    {
        sum=0;
        if(n==0) printf("1\n");
        else
        {
            k=sqrt(1.0*n/2);
            for(int i=0;i<=k;i++)
            {
                int j=sqrt(n-i*i);
                if(i*i+j*j==n)
                {
                    if(i==0) sum+=4;
                    else if(i==j) sum+=4;
                    else sum+=8;
                }
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}