Max Sequence （DP）

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

最大m子段问题，此题中m=2。主要方法还是DP，其实理解题意以后代码的编写不是很困难，就是这道题目很难理解，数学弱鸡瑟瑟发抖。

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。

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int inf=-0x3fffffff;int num[100001], dp[100001];int main(){    int n,max,sum,s;    while(scanf("%d",&n))    {        if(n==0)        break;sum=0;        max=inf;        for(int i=1;i<=n;i++)        {            scanf("%d", &num[i]);            sum +=num[i];            if(sum>max)                max=sum;            dp[i]=max;            if(sum<0)                sum=0;        }        dp[0]=s=max=inf;        sum=0;        for(int i=n;i>0;i--)         {            sum+= num[i];            if(sum>max)                max=sum;            if(s<max+dp[i-1])                s=max+dp[i-1];            if(sum<0)                sum=0;        }        printf("%d\n",s);    }}