hdu1003 MaxSum (DP)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 180281    Accepted Submission(s): 42103

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

 

Recommend

We have carefully selected several similar problems for you:   ​​1087​​​  ​​​1069​​​  ​​​2084​​​  ​​​1058​​​  ​​​1159​​ 

 

#include <stdio.h>
#define num 100000
int main()
{
  int n,t,i,j,flag,a[num],star,q,g;
  long sum,maxsum;
  scanf("%d",&t);
  for(i=1;i<=t;i++)
  {
        scanf("%d",&n);
      for(j=0;j<n;j++)
        scanf("%d",&a[j]);
      maxsum=a[0];
      sum=0;
      q=flag=0;
      j=star=0;
      for(q=0;q<n;q++)
      {
        if(sum<0)
          sum=a[q],j=q;
        else
          sum=sum+a[q];
        if(sum>maxsum)
        {
          maxsum=sum;
          flag=q;
          star=j;
        }
      }
      printf("Case %d:\n",i);
      printf("%ld %d %d",maxsum,star+1,flag+1);
      if(i==t)
        printf("\n");
      else 
        printf("\n\n");
    
  }
  return 0;

}