hdu 1003 Max Sum (DP)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 134690    Accepted Submission(s): 31181

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

 

 

Author

Ignatius.L

 

 

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简单dp思想：

//15 MS    236 KB    Visual C++    
#include<stdio.h>
int main(void)
{
    int t,n,k=1;
    int a;
    scanf("%d",&t);
    while(t--)
    {
        if(k>1) printf("\n");
        scanf("%d",&n);
        int l=1,r=1,sum=-0x7ffffff,tmp=0,tl=1;
        for(int i=1;i<=n;i++){
            scanf("%d",&a);
            tmp+=a;
            if(tmp>sum){
                sum=tmp;
                l=tl;
                r=i;
            }
            if(tmp<0){
                tmp=0;
                tl=i+1;          
            }
        }
        printf("Case %d:\n",k++);
        printf("%d %d %d\n",sum,l,r);
    }
    return 0;
}
/*

30
3 -1 0 -1
3 -1 0 1
3 -1 -2 -3

*/