HDU 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 105228    Accepted Submission(s): 24216

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

View Code

//第二组错误，输出了 7 7

#include <cstdio>

#include <cstring>

using namespace std;



const int maxn = 100010;



int ans[maxn] = {0};

int b[maxn][2] = {0};

int a[maxn];



int main()

{

    int i,j,k;

    int T;

    scanf("%d",&T);

    int n,m;

    int num;

    int flag = 1;

    int temp = T;

    while(T--)

    {

        scanf("%d",&num);

        memset(ans,0,sizeof(ans));

        memset(b,0,sizeof(b));

        memset(a,0,sizeof(a));





        for(i=1; i<=num; i++)

            scanf("%d",a+i);

        ans[1] = a[1];

        b[1][0] = b[1][1] = 1;

        int max = -1;

        int next;

        for(i=2; i<=num; i++)

        {

            if(ans[i-1]>0)

            {

                ans[i] = ans[i-1] + a[i];

                //if(a[i]>0)

                {

                    b[i][0] = b[i-1][0];

                    b[i][1] = i;

                }

               // else

               // {

               //     b[i][0] = b[i-1][0];

               //     b[i][1] = i-1;

               // }

            }

            else

            {

                ans[i] = a[i];

                b[i][0] = b[i][1] = i;

            }

            if(ans[i]>max)

            {

                max = ans[i];

                next = i;

            }



        }

        printf("Case %d:\n",flag);

        printf("%d %d %d\n",max,b[next][0],b[next][1]);

        if(flag<temp)

            printf("\n");

        flag++;

    }

    return 0;

}

 1 #include <cstdio>

 2 #include <cstring>

 3 using namespace std;

 4 

 5 const int maxn = 100010;

 6 int a[maxn] = {0};

 7 

 8 int main()

 9 {

10     int i,j,k;

11     int T;

12     scanf("%d",&T);

13     int num;

14     int flag = 1;

15     int sum = 0;

16     int start,end,last;

17     int temp = T;

18     while(T--)

19     {

20         

21         int max = (1<<31);

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

23         memset(a,0,sizeof(a));

24         sum = 0;

25         last = 1;

26         for(i=1; i<=num; i++)

27         {

28             scanf("%d",a+i);

29             sum += a[i];

30             if(sum>max)

31             {

32                 start = last;

33                 max= sum;

34                 end = i;

35             }

36             if(sum<0)

37             {

38                 sum = 0;

39                 last = i+1;

40             }

41 

42         }

43         printf("Case %d:\n",flag);

44         printf("%d %d %d\n",max,start,end);

45         if(flag<temp)

46             printf("\n");

47         flag++;

48     }

49     while(1);

50     return 0;

51 }