NYOJ-Sum

Sum

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

描述

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. 

输入

The input consists N test cases.
The only line of every test cases contains a positive integer S (0< S <= 100000) which represents the sum to be obtained.
A zero terminate the input.
The number of test cases is less than 100000.

输出

The output will contain the minimum number N for which the sum S can be obtained.

样例输入

3120

样例输出

27

来源

POJ

代码:

#include<stdio.h>#include<math.h>int main(){int n,i,sum;while(scanf("%d",&n),n){sum=0;for(i=1;;++i){sum+=i;if(sum>=n&&(sum-n)%2==0)break;}printf("%d\n",i);}return 0;}

解题思路:

 当累加时多加了 sum-n ,只需要找到 累加和为 (sum-n)/2 的若干数 的符号变为负号【解题重点！！！！！！！！】