POJ 1844：Sum ”滚动“数组

Sum

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10494		Accepted: 6895

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.

给一个数sum，问从1到n，各个数可以取正取负，想找到最小的n，从1到n加起来是sum。

这个题记一下在于数组的使用，做的时候发现开不了那么大的数组，于是就得利用当前的数只和前面一个数的状态有关，所以2个数组就好用了，&1这里决定要记一下。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int a[3][200005];

int main()
{
	int i, j, x;

	while (cin >> x)
	{
		memset(a[0], 0, sizeof(a[0]));
		memset(a[1], 0, sizeof(a[1]));
		a[0][100000] = 1;
		for (i = 1;; i++)
		{
			memset(a[i&1], 0, sizeof(a[i&1]));
			for (j = 0; j <= 200000; j++)
			{
				if (a[(i - 1) & 1][j] == 1)
				{
					a[i & 1][j + i] = 1;
					a[i & 1][j - i] = 1;
				}
			}
			if (a[i & 1][100000 + x] == 1 || a[i & 1][100000 - x] == 1)
			{
				cout << i << endl;
				break;
			}
		}
	}
	return 0;
}

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