Max Sum
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 114294    Accepted Submission(s): 26471

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<iostream>#include<algorithm>#include <cstdio>using namespace std;int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++){int m,temp;scanf("%d",&m);if(m) scanf("%d",&temp);int dp=temp;int max=temp;int max_a=0;int min_a=0;int head=0;for(int j=1;j<m;j++){scanf("%d",&temp);if(dp<0){head=j;}dp=(dp>=0?(dp+temp):temp);if(dp>max){min_a=head;max=dp;max_a=j;}}if(i!=0){printf("\n");}printf("Case %d:\n",i+1);printf("%d %d %d\n",max,min_a+1,max_a+1);}}return 0;}