HUST 1544 R(N)

最新推荐文章于 2021-01-15 16:09:48 发布

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最新推荐文章于 2021-01-15 16:09:48 发布

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分类专栏： ----枚举 HUST 文章标签： hust

本文链接：https://blog.youkuaiyun.com/jtjy568805874/article/details/50625193

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本文深入探讨了如何通过平方数的组合来表示正整数，并介绍了计算特定整数表示方式数量的方法。通过实例演示和代码实现，展示了数学原理与编程实践的巧妙融合。

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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)

map预处理+枚举

#include<string>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
const int maxn = 10 + 5;
typedef long long LL;
int T, n, m;
map<int, int> M;

int main()
{
	for (int i = 0; i*i <= 1e9; i++)
	{
		M[i*i] = 1;
	}
	while (scanf("%d", &n) != EOF)
	{
		LL ans = 0;
		for (int i = 0; i*i <= n; i++)
		{
			if (M[n - i*i])
			{
				int t = 4;
				if (i == 0) t /= 2;
				if (n == i*i) t /= 2;
				ans += t;
			}
		}
		cout << ans << endl;
	}
	//while (scanf("%d", &n) != EOF){}
	return 0;
}