poj 2140 Herd Sums 判断S由连续的自然数组成共有几种情况

计算连续整数和的方法

最新推荐文章于 2019-05-19 14:14:46 发布

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本文介绍了一种计算方法，用于找出所有可能的连续整数序列，这些序列的和等于给定的整数N。通过枚举和数学推导，程序能够高效地找到所有满足条件的整数序列。

Herd Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11551		Accepted: 6891

Description

The cows in farmer John's herd are numbered and branded with consecutive integers from 1 to N (1 <= N <= 10,000,000). When the cows come to the barn for milking, they always come in sequential order from 1 to N.

Farmer John, who majored in mathematics in college and loves numbers, often looks for patterns. He has noticed that when he has exactly 15 cows in his herd, there are precisely four ways that the numbers on any set of one or more consecutive cows can add up to 15 (the same as the total number of cows). They are: 15, 7+8, 4+5+6, and 1+2+3+4+5.

When the number of cows in the herd is 10, the number of ways he can sum consecutive cows and get 10 drops to 2: namely 1+2+3+4 and 10.

Write a program that will compute the number of ways farmer John can sum the numbers on consecutive cows to equal N. Do not use precomputation to solve this problem.

Input

* Line 1: A single integer: N

Output

* Line 1: A single integer that is the number of ways consecutive cow brands can sum to N.

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
//2*S=n*(2*a1+n-1)
//S=n*(a1+an)/2 ,又因为是连续的,故n和a1+an必然是一奇一偶;
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        int cnt=0;
        double k=sqrt(2*n+0.5);
        for(int i=1;i<=k;i++)//枚举n
        {
            if(2*n%i==0&&((i&1)||(2*n/i)&1))
            {
                int a1=2*n/i-i+1;//计算a1 判断是否符合条件
                if(a1>=0) cnt++;
            }
        }
        printf("%d/n",cnt);
    }
    return 0;
}