简单dp hdu1003

Max Sum

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

 

这题蛮悲剧的 总共交了13次 可能理解的还是不够深吧 有几个地方一直没注意

View Code

 1 #include<stdio.h>

 2 int a[100001];

 3 int main()

 4 {

 5     long t,n,nowmax,premax,x,y,k = 0,i,start;

 6     scanf("%ld", &t);

 7     while(t--)

 8     {

 9         k++;

10         scanf("%ld", &n);

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

12             scanf("%ld", &a[i]);          

13         premax = -999;//因为输入数据大于-1000，premax比-1000大就行了

14         nowmax = 0;

15         start = 1;

16         x = 1;

17         y = 1;

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

19         {

20             nowmax+=a[i];

21             if(nowmax>premax)

22             {

23                 y = i;

24                 start = x;//之前没有定义start 想着用下面的x就可以找出初始位置了 如果之前已经找到最大的premax 而后面又有负的nowmax 那x就不为初始位置 所以这里要跟着变动

25                 premax = nowmax;

26             }           

27             if(nowmax<0)//这里本来写的else if 如果nowmax是负的并且比premax大 就不对了

28             {

29                 x = i+1;

30                 nowmax = 0;

31             }

32         }

33         printf("Case %ld:\n",k);

34         printf("%ld %ld %d\n", premax,start,y);

35         if(t!=0)

36             printf("\n");

37     }

38     return 0;

39 }