POJ 1844：Sum ”滚动“数组

Sum

Time Limit: 1000MS Memory Limit: 30000KTotal 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;}