杭电2715 Herd Sums

Herd Sums

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 771    Accepted Submission(s): 374

Problem 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

看注释：

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;//2*s=(2*a1+n-1)*n;int i,j,k,l,m,n;double help;int main(){while(scanf("%d",&n)!=EOF){int ans=0;help=sqrt(2*n+1);for(i=1;i<=help+1;i++){m=(n*2+i-i*i)/2/i;//高斯求和反解出来的 if(i*i-i+2*m*i==2*n&&m>0&&(i%2==0||(2*m+i-1)%2==0))//i和 (2*m+i-1)有一个为偶数 ans++;//并且m要大于0； }printf("%d\n",ans);}}