HDUOJ--------1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 132258    Accepted Submission(s): 30652

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

 

最大连续和sum....运用动态规划思想

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
    int n,j,i,num,a,ans,maxc,posst,posen,temp;
      scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&num);
        maxc=0;
        posst=posen=1;
        ans=-0x3f3f3f3f ;
        temp=0;
     for(j=1;j<=num;j++)
     {
       scanf("%d",&a);
        maxc+=a;
       if(ans<maxc)
        {
          ans=maxc;
          posen=j;
          posst=temp+1;
        }
       if(maxc<0)
       {
        maxc=0;
        temp=j;
       }
    }
     printf("Case %d:
%d %d %d
",i,ans,posst,posen);
     if(i!=n)  putchar(10);
    }
    return 0;
}