poj 2593

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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 9 -5 11 20

Sample Output

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;}