POJ1844 Sum【水题+数学题】

Sum

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10839		Accepted: 7107

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

Hint

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

Source

Romania OI 2002

问题链接：POJ1844 Sum。

题意简述：（略）

问题分析：对于n而言，Sn=(n+1)*n/2，需要满足Sn>=S。若存在负的S'(其值为1-n中的若干项之和，即取负值部分）使得Sn-2S'=S，则需要满足(Sn-S)%2=0。

程序说明：（略）

AC的C++语言程序如下：

/* POJ1844 Sum */

#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    int s;

    while(cin >> s) {
        for(int i=sqrt(s); ;i++) {
            int sn = (i + 1) * i / 2;
            if(sn >= s && (sn - s) % 2 == 0) {
                cout << i << endl;
                break;
            }
        }
    }

    return 0;
}