航电HDOJ 1003

温故而知新
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 339728 Accepted Submission(s): 80646

Problem Description
Given a sequence a[1],a[2],a[3]…a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6
最长子序列呢，最简单的应该就是在线处理法了吧，算法复杂度是O(n).
下面解释一下：
我们在读入数据的过程中，使用thissum累加和，如果thissum的值大于maxsum，我们将执行
maxstart=thisstart;
maxend=i;
maxsum=thissum;
记录最大值的下标和最大值
如果thissum的值小于0，我们就认为目前的序列只会让后面的序列减小，所以
thissum=0;
thisstart=i+1;
清空，并且将thisstart赋于下一个值。
上代码！！

#include<iostream>
using namespace std;
int main(){
    int t,j=1;
    cin>>t;
    while(t--)
    {
        int n,maxsum,thissum,thisstart=0,maxstart=0,maxend=0;
        cin>>n;
        cin>>maxsum;//第一个数使为maxsum
        thissum=maxsum;
        if(thissum<0){thissum=0;thisstart=1;//第一个数小于0，舍弃
        }
        for(int i=1;i<n;i++)
        {
            int temp;
            cin>>temp;
            thissum+=temp;
            if(thissum>maxsum){
                maxstart=thisstart;
                maxend=i;
                maxsum=thissum;
            }
            if(thissum<0)
            {
                thissum=0;
                thisstart=i+1;
            }
        }
        cout<<"Case "<<j++<<":"<<endl<<maxsum<<" "<<maxstart+1<<" "<<maxend+1<<endl;//输出
        if(t>0)
        cout<<endl;
    }
}

希望大佬指正！