hdu1003最大子序 记录起始与终止点

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 205709    Accepted Submission(s): 48066

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
   
   
   
   

这一题的升级版

http://blog.csdn.net/summer__show_/article/details/50616608

调试后容易理解

#include<stdio.h>

const int Max=100005;

int tag[Max];

int main()
{
        int K,t=1;
        scanf("%d",&K);
        while(K--)
        {
            int n;
            scanf("%d",&n);
            for(int i=1;i<=n;i++)
                scanf("%d",&tag[i]);

            int start=1,enda=1,tmpStart=1,tmpEnd=1;
            int ans=-0xfffffff;
            int sum=0;
            for(int i=1;i<=n;i++,tmpEnd=i){
                sum+=tag[i];
                if(sum>ans){
                    ans=sum;
                    start=tmpStart;
                    enda=tmpEnd;
                }
                if(sum<0){
                    tmpStart=tmpEnd+1;
                    sum=0;
                }
            }
            printf("Case %d:\n%d %d %d\n",t++,ans,start,enda);
            if(K){
                printf("\n");
            }
        }
        return 0;
}