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

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 int main()

 5 {

 6     int n,j,i,num,a,ans,maxc,posst,posen,temp;

 7       scanf("%d",&n);

 8     for(i=1;i<=n;i++)

 9     {

10         scanf("%d",&num);

11         maxc=0;

12         posst=posen=1;

13         ans=-0x3f3f3f3f ;

14         temp=0;

15      for(j=1;j<=num;j++)

16      {

17        scanf("%d",&a);

18         maxc+=a;

19        if(ans<maxc)

20         {

21           ans=maxc;

22           posen=j;

23           posst=temp+1;

24         }

25        if(maxc<0)

26        {

27         maxc=0;

28         temp=j;

29        }

30     }

31      printf("Case %d:\n%d %d %d\n",i,ans,posst,posen);

32      if(i!=n)  putchar(10);

33     }

34     return 0;

35 }

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