poj 2593

Description

Give you N integers a1, a2 ... aN (|ai| <=1000, 1 <= i <= N). 

You should output S. 

Input

The input will consist of several test cases. For each test case, one integer N (2 <= N <= 100000) is given in the first line. Second line contains N integers. The input is terminated by a single line with N = 0.

Output

For each test of the input, print a line containing S.

Sample Input

5 

-5 9 -5 11 20 

0

Sample Output

40

额， 算法弱菜写这些真不容易

一道很简单的动态规划，枚举出所有情况前后各部分最大子串的和，取最大值即可

#include <iostream>
#include <cstdio>
#include <climits>

int main()
{
    int m, i, s, max, a[100001], pre[100001];

    while (std::cin >> m, m) {
        s = 0;
        max = INT_MIN;
        for (i = 0; i != m; ++i) {
            scanf("%d", &a[i]);
            s += a[i];
            if (s > max) {
                max = s;
            }
            pre[i] = max;
            if (s < 0) {
                s = 0;
            }
        }
        s = 0;
        max = INT_MIN;
        for (i = m - 1; i != 0; --i) {
            s += a[i];
            if (s + pre[i - 1] > max) {
                max = s + pre[i - 1];
            }
            if (s < 0) {
                s = 0;
            }
        }
        std::cout << max << std::endl;
    }

    return 0;
}