Max Sum hdu 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 125801    Accepted Submission(s): 29090

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

 

  

#include<stdio.h>
int main()
{
    int i,j,k,c,n,temp,max,t;
    int a[120055];
    scanf("%d",&c);
    k=0;
    for(k=1;k<=c;k++)
    {
        //k++;
        scanf("%d",&n);
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        int start=1,end=1;
        temp=a[1];
        max=a[1];
        int t=0;
        int p=1;
        for(i=2; i<=n; i++)
        {

            if(temp<0)
            {
                temp=0;
                p=i;
            }
            temp=temp+a[i];

            if(temp>max)
            {
                max=temp;
                end=i;
                start=p;
            }
        }
        if(k==c)
        {
            printf("Case %d:\n",k);
        printf("%d %d %d\n",max,start,end);
        }
        else
        {

        printf("Case %d:\n",k);
        printf("%d %d %d\n\n",max,start,end);


        }
    }
    return 0;
}