Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 241450    Accepted Submission(s): 56990

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

 

/*分类：DP 来源：hdu Max Sum 思路：We are giants.create by Lee_SD on 2017/4/*/#include<queue>#include<iostream>#include<algorithm>#include<cmath>#include<stack>#include<string.h>#include<stdio.h>using namespace std;int  main(){int t,a[100003];int m;scanf("%d",&t);int kase=1;while(t--){scanf("%d",&m);for(int i=1;i<=m;i++)scanf("%d",&a[i]);int max=-9999;int sum=0;int start=1,end=1,k=1;for(int i=1;i<=m;i++){sum+=a[i];if(sum>max){start=k;max=sum;end=i;}if(sum<0){sum=0;k=i+1;}}printf("Case %d:\n",kase++);printf("%d %d %d\n",max,start,end);if(t>0)printf("\n");}}