1120. Sum of Sequential Numbers

最新推荐文章于 2023-08-30 20:05:37 发布

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最新推荐文章于 2023-08-30 20:05:37 发布

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分类专栏： Timus 文章标签： numbers output pair

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本文介绍了一种用于解决序列求和问题的高效算法，通过数学推导和编程实现，展示了如何在限定时间内找到最大可能的序列长度，使得序列之和等于给定数。该算法特别适用于需要快速计算大量数据场景。

1120. Sum of Sequential Numbers

Time Limit: 1.0 second
Memory Limit: 16 MB

There is no involute formulation concerning factitiously activity of SKB Kontur in this problem. Moreover, there is no formulation at all.

Input

There is the only number N, 1 ≤ N ≤ 10⁹.

Output

Your program is to output two positive integers A and P separated with a space such that:

N = A + (A + 1) + … + (A + P − 1).
You are to choose a pair with the maximal possible value of P.

Sample

input	output
14	2 4

剪一下枝就可以了~

#include <iostream>
#include <cmath>
using namespace std;

bool temp(int p, int n)
{
	if ((n*2) % p != 0)
		return 0;
	int a = (n*2) / p + 1 - p;
	if (a <= 0 || (a&1))
		return 0;
	return 1;
}

int main()
{
	int n, p;
	cin>>n;
	p = sqrt(n*2.0);
	while (!temp(p, n))
		p--;
	cout<<(n*2/p + 1-p)/2<<" "<<p<<endl;
}