NYOJ sum 215

Sum時間限制：1000 ms  |  記憶體限制：65535 KB難度：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. 

輸入

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.

輸出

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

範例輸入

3120

範例輸出

27

 

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