Max Sum（dp水题）

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http://acm.hdu.edu.cn/showproblem.php?pid=1003

还是子序列求和

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 186386 Accepted Submission(s): 43487

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

#include <cstdio>int main(){    int t;    scanf("%d",&t);    for(int Case=1;Case<=t;Case++){        int a,max_sum=-1001,max_first,max_last,tot=-1,last,first;        int d;        scanf("%d",&d);        for(int i=1;i<=d;i++){            scanf("%d",&a);            if(tot<0){                first=last=i;                tot=a;            }            else {                tot+=a;last=i;            }            if(tot>max_sum){                max_sum=tot;                max_first=first;                max_last=last;            }        }    printf("Case %d:\n%d %d %d\n",Case,max_sum,max_first,max_last);    if(Case!=t)printf("\n");    }    return 0;}

AC之路，我选择坚持~~

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