UVA The ? 1 ? 2 ? ... ? n = k problem

 The ? 1 ? 2 ? ... ? n = k problem 

The problem

Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k
? 1 ? 2 ? ... ? n = k

For example: to obtain k = 12 , the expression to be used will be:
- 1 + 2 + 3 + 4 + 5 + 6 - 7 = 12 
with n = 7

The Input

The first line is the number of test cases, followed by a blank line.

Each test case of the input contains integer k (0<=|k|<=1000000000).

Each test case will be separated by a single line.

The Output

For each test case, your program should print the minimal possible n (1<=n) to obtain k with the above formula.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

2

12

-3646397

Sample Output

7

2701

 题意：给定任意一个值k，使k=(-)1+(-)2+(-)3+(-)4+(-)5++++(-)n，求最小的n

思路：S1=1+2+3+.....+n>=k，S2=1+2+3+...-x+...+n==k

       所以S1-S2=2x，所以只要有一个数导致S1和S2差为偶数就符合条件

       输出有空格，再次错了。

      数学真强大，这里完全体现

   

#include<stdio.h>
#include<stdlib.h>

int main()
{
    long long k,t;
    int n;
    int T,i;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&k);
        k=abs(k);
        for(i=1;;i++)
        {
            t=i*(i+1)/2;
            if(t>=k) break;
        }
        n=i;
        while(1)
        {
           int xx=n*(n+1)/2-k;
           if(xx%2==0) break;
           n++;
        }
        printf("%d\n",n);
        if(T) printf("\n");
    }
    return 0;
}