POJ-2593 Max Sequence（不连续的最大子段和）

Max Sequence

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16001		Accepted: 6715

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

大致题意：给出一个数列，求出数列中不相交的两个子段和，要求子段和最大。

#include <stdio.h>
int main()
{
    int left[100005];
    int right[100005];
    int a[100005];
    while(1)
    {
        int n,i;
        scanf("%d",&n);
        if(n==0)
            break;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        int sumL=0;
        int max=-99999999;
        for(i=0;i<n;i++)//从左向右求到i结尾的最大连续子串的最大值    //如果这个循环看不懂的话，可以先做下hdu1003
        {
            sumL=sumL+a[i];
            if(sumL>max)
                max=sumL;
            if(sumL<0)
                sumL=0;
            left[i]=max;
        }
        max=-99999999;
        int tmp=-99999999;
        int sumR=0;
        for(i=n-1;i>=1;i--)
        {
            sumR=sumR+a[i];
            if(sumR>max)
                max=sumR;
            if(sumR<0)
                sumR=0;
            if(tmp<max+left[i-1])
            tmp=max+left[i-1];
         }
         printf("%d
",tmp);
     }
     return 0;
}