2140 Herd Sums下载

weixin_39821746 2019-05-13 01:00:20

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
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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.

hutc-Herd Sums 参考代码hutc-Herd Sums 参考代码hutc-Herd Sums 参考代码

#include using namespace std; int main() { int n, i, j, sum, cnt = 0; cin >> n; //经分析可知，其i的取值不可能大于n/2+1，那样范围就缩小了！ for (i = 1; i <= n/2+1; i++){ sum = 0; for (j = i;

http://acm.pku.edu.cn/JudgeOnline/problem?id=2140
题意：给出一个数n，求n可以被分解成多少组连续的书相加。/*假设a,a+1,a+2...a+k ，为一组符合的答案有(k+1)a+0.5*k*(k＋1)=n成立划成(a + 0.5*k)(k+1)=n观察等式发现必须满足1. 0.5*k为整数. 所以k是偶数2. (k+1) 为n的因数，并且 (k+1) 为奇数所以综上n有多少个奇因数，就是本题的答案*/#include

#include <iostream>//这种在几个数里面任意去几个连续的数求和就用两个循环套一下就可以了 using namespace std; int main() { int n, i, j, sum, cnt = 0; cin >> n; //经分析可知，其i的取值不可能大于n/2+1，那样范围就缩小了！ for ...

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