NYOJ 215 Sum

标签：nyoj 215 sum

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

3
12
0

样例输出

2
7

AC码：

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

NYOJ 215 Sum

标签：nyoj 215 sum

原文地址：http://blog.csdn.net/u012804490/article/details/24668311