POJ2140 HDU2715 Herd Sums【数学】

Herd Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18804		Accepted: 10991

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

15

Sample Output

4

Source

USACO 2003 March Orange

问题链接：POJ2140 HDU2715 Herd Sums

问题描述：（略）

问题分析：

  求n=a+a+1+a+2+...+a+k 的组合数。

  n=(k+1)*a+(k+1)*k/2=(k+1)(a+k/2)

  其中，n为整数，k+1为整数，(a+k/2)为整数，k为偶数，k+1为奇数

  当n和k+1确定时，a为定值

  故：解为n的奇因子个数。

程序说明：（略）

参考链接：（略）

题记：（略）

AC的C++语言程序如下：

/* POJ2140 HDU2715 Herd Sums */

#include <iostream>

using namespace std;

int main()
{
    int n;
    while(cin >> n) {
        int cnt = 0;
        for(int i = 1; i <= n; i += 2)
            if(n % i == 0)
                cnt++;
        cout << cnt << endl;
    }

    return 0;
}