1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 151503    Accepted Submission(s): 35379

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

 

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
  int test,count=0;
  scanf("%d",&test);
  int N = test;
  while(test--){
      int n,a,MAX=-1001,sum=0,first=0,k=1,last=0;
      scanf("%d",&n);
      for(int i = 0; i < n;i++){
              scanf("%d",&a);
              sum += a;
              if(sum > MAX){
                 MAX = sum; 
                 first = k; 
                 last = i+1;
              }
              if(sum < 0){sum=0;k=i+2;}
      }     
      printf("Case %d:\n",++count);         
      printf("%d %d %d\n",MAX,first,last);         
      if(count != N)printf("\n");                         
  }
  return 0;   
}