HDU--1003 -- Max Sum [dp]

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 115384 Accepted Submission(s): 26761

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
  
  

Code:

简单的dp思想，可惜我还是没感觉，觉悟不够。。==

#include<stdio.h>
int a[100005];
int main()
{
    int t,i,c,n,maxsum,cursum,begin,end,pos;
    scanf("%d",&t);
    for(c=1;c<=t;c++)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        begin = end = pos = 0;
        cursum = maxsum = a[0];
        for(i=1;i<n;i++)
        {
            if(cursum+a[i]<a[i]){
                pos = i;
                cursum = a[i];
            }
            else    cursum = cursum + a[i];

            if(cursum>maxsum){
                end = i;
                begin = pos;
                maxsum = cursum;
            }
        }    
        printf("Case %d:\n%d %d %d\n",c,maxsum,begin+1,end+1);
        if(c<t) printf("\n");
    }
    return 0;
}