首页 > 代码库 > POJ1844 Sum
POJ1844 Sum
Sum
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 10160 | Accepted: 6674 |
Description
Consider the natural numbers from 1 to N. By associating to each number a sign (+ or -) and calculating the value of this expression we obtain a sum S. The problem is to determine for a given sum S the minimum number N for which we can obtain S by associating signs for all numbers between 1 to N.
For a given S, find out the minimum value N in order to obtain S according to the conditions of the problem.
For a given S, find out the minimum value N in order to obtain S according to the conditions of the problem.
Input
The only line contains in the first line a positive integer S (0< S <= 100000) which represents the sum to be obtained.
Output
The output will contain the minimum number N for which the sum S can be obtained.
Sample Input
12
Sample Output
7
Hint
The sum 12 can be obtained from at least 7 terms in the following way: 12 = -1+2+3+4+5+6-7.
#include <stdio.h> int main() { int s, sum, i; while(scanf("%d", &s) == 1){ sum = i = 0; while(sum < s || (sum - s) & 1) sum += ++i; printf("%d\n", i); } return 0; }
POJ1844 Sum
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。