poj2593

dp，分别从左侧和右侧求最大子段和，然后把fleft[i]变成0～i区间中最大子段和，fright以此类推。

然后枚举中间分割点。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 100003

int f[maxn], fleft[maxn], fright[maxn];
int n;

void input()
{
    for (int i = 0; i < n; i++)
        scanf("%d", &f[i]);
}

void work1()
{
    fleft[0] =f[0];
    for (int i = 1; i < n; i++)
        if (f[i] > fleft[i - 1] + f[i])
            fleft[i] = f[i];
        else
            fleft[i] = fleft[i - 1] + f[i];
    for (int i = 1; i < n; i++)
        fleft[i] = max(fleft[i - 1], fleft[i]);
}

void work2()
{
    fright[n - 1] =f[n - 1];
    for (int i = n - 2; i >= 0; i--)
        if (f[i] > fright[i + 1] + f[i])
            fright[i] = f[i];
        else
            fright[i] = fright[i + 1] + f[i];
    for (int i = n - 2; i >= 0; i--)
        fright[i] = max(fright[i + 1], fright[i]);
}

void work3()
{
    int ans = -1000000000;

    for (int i = 0; i < n - 1; i++)
        ans = max(ans, fleft[i] + fright[i + 1]);
    printf("%d\n", ans);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    while (scanf("%d", &n), n != 0)
    {
        input();
        work1();
        work2();
        work3();
    }
    return 0;
}