（POJ2140）Herd Sums

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

题意是：求a + a + 1 + a +2 + a + 3……a + k = n，k有几个不同的解；

显然等差数列 (k + 1)(2 * a + k) / 2 = n;

-> k+1为偶数 或者 2 *a + k为偶数 才能把2约去；

当k+1为偶数时，2 *a + k一定是奇数 , n % (2 *a + k) == 0;

当2 *a + k为偶数时，k+1一定是奇数 , n % (k+1) == 0;

然后题目就退化到 n的奇因子有多少个。

#include <iostream>#include <string>#include <algorithm>#include <cmath>using namespace std;int cal(int n){int s = (int)sqrt((double)n);int i;int sum = 0;for(i = 1; i <= s; i++){if(n % i == 0){if(i % 2 != 0)sum++;if(n / i != i && n / i % 2 != 0)sum++;}}return sum;}int main(){int n;cin >> n;cout << cal(n) << endl;return 0;}