杭电oj-sum && ACM数学题

Problem 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.

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

思路：
sum：从1到n的和
sum1：负数的和
T：给定的值
sum-2*sum1=T；===》（sum-T）%2 =0

代码：

```c
#include
int main()
{
    int T;
    scanf("%d",&T);
    int i=0,sum=0;
    while(1)
    {
        sum+=++i;
        if((sum-T)%2==0&&sum>T)
            break;
    }
    printf("%d\n",i);
    return 0;
}