POJ 1844 Sum

Sum

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9543		Accepted: 6248

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
解题思路：简单数学，对于任意输入的数，先找到大于等于它的1+2+3+...+n的和的最接近值是第几个数，比如11大于等于它的是1+2+3+4+5，第五个数。接着考虑到数列中一个数i由正号变负号相当于减去2i， 分奇偶两种情况讨论，若输入的数是奇数，则应令大于等于它的数是奇数，因为只有奇数减偶数才是奇数，不是就继续加直至变成奇数，若输入的是偶数，同理应令大于等于它的数是偶数，操作同上。最后找出满足条件的是第几个数就是答案了。

#include<iostream>
using namespace std;
int main()
{
	long i,s,sum,ans;
	while(cin>>s)
	{
		i=1;
		sum=0;
		if(s%2)
		{
			while(sum<s)
		    {
			    sum+=i;
			    i++;
		    }
			while(sum%2==0)
			{
				sum+=i;
			    i++;
			}
		}
		else
		{
			while(sum<s)
		    {
			    sum+=i;
			    i++;
		    }
			while(sum%2)
			{
				sum+=i;
			    i++;
			}
		}
		ans=i-1;
		cout<<ans<<endl;
	}
	return 0;
}