Max Sequence

题目链接：http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=2891

Max Sequence

Time Limit(Common/Java):3000MS/9000MS     Memory Limit:65536KByte
Total Submit: 75            Accepted: 20

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
分析：最大m子段问题，此题中m=2。方法还是DP，基本上还是用我们熟悉的最大子段和的方法。读入数据是时做一次DP，得到从前往后到第i（0=<i<n）个元素处时
的最大子段和，然后再从后往前反向做一次DP，加上前面求得的正向的最大字段和即可求出一个最大值。
代码：
#include<stdio.h>
const int MAXN=100005;
const int INF=1<<28;
int main()
{
    int n;
    while (scanf("%d", &n), n)
    {
        int lmax[MAXN], a[MAXN];
        int i;
        for (i=0; i<n; i++)
        {
            scanf("%d", &a[i]);
        }
        int nStart=a[0];
        int nAll=a[0];
        lmax[0]=a[0];
        for (i=1; i<n; i++)
        {
            if (nStart<0) nStart=0;
            nStart+=a[i];
            if (nStart>nAll) nAll=nStart;
            lmax[i]=nAll;
        }
        int max=-INF;
        nStart=-INF;
        nAll=-INF;
        for (i=n-1; i>=1; i--)
        {
            if(nStart<0) nStart=0;
            nStart+=a[i];
            if (nStart>nAll) nAll=nStart;
            if (nAll+lmax[i-1]>max) max=nAll+lmax[i-1];
        }
        printf("%d\n", max);
    }
    return 0;
}