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 ;

 }

巴特西

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

Max Sum

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