POJ 1844 Sum（数学）

题目链接：Click me！

                                                         Sum

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.Description

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

Hint

The sum 12 can be obtained from at least 7 terms in the following way: 12 = -1+2+3+4+5+6-7.

题意解析：

这道题呢，意思就是，给出一个S，你需要去寻找到一个最小的N，然后需要得出从1~N的和（可正可负）为S，输出这个N即可，比如12 = -1 + 2 + 3 + 4 + 5 + 6 - 7，那么结果就是7；

其实就两行代码，第一需要知道，最小的N，也就是在同样能得到S的情况下，符号绝大多数为正的话，才能尽可能得到最小的N，比如12 = -1 + 2 + 3 + 4 + 5 + 6 - 7，12 =  -1 + 2 + 3 - 4 + 5 + 6 - 7 + 8，故我们先令1~N中全为正，然后求和，如果当前的前N项和大于S，且（SUM-S）% 2 = 0，那么久可以得到这个S，而且得到的N是最小的N。

代码篇：

#include 
#include 
#include 
#include 
using namespace std;
int main()
{
    int sum, i, j, k;
    scanf("%d",&k);
    sum = 0;
    i = 1;
    while(1)
    {
        sum += i++;
        if(sum >= k && (sum - k) % 2 == 0)
        {
            printf("%d\n",i - 1);
            return 0;
        }
    }
    return 0;
}

 

OVER！