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=;i<=n;i++)
     {
         scanf("%d",&num);
         maxc=;
         posst=posen=;
         ans=-0x3f3f3f3f ;
         temp=;
      for(j=;j<=num;j++)
      {
        scanf("%d",&a);
         maxc+=a;
        if(ans<maxc)
         {
           ans=maxc;
           posen=j;
           posst=temp+;
         }
        if(maxc<)
        {
         maxc=;
         temp=j;
        }
     }
      printf("Case %d:\n%d %d %d\n",i,ans,posst,posen);
      if(i!=n)  putchar();
     }
     return ;
 }