1003---Max Sum

http://acm.hdu.edu.cn/showproblem.php?pid=1003

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

 

 

Recommend

We have carefully selected several similar problems for you:  1176 1087 1069 2084 1058 

#include<stdio.h>
int main()
{
    int t,n,a,i,j,start,end,temp;
    long max,sum;
    scanf("%d",&t);
    for(i=1;i<=t;i++)
    {
        sum=0;
        temp=1;
        max=-1001;//由于存在全是负数的情况，将最大值赋值为负数
        scanf("%d",&n);
        for(j=1;j<=n;j++)
        {
            scanf("%d",&a);
            sum=sum+a;
            if(sum>max)
            {
                max=sum;
                start=temp;//将起始的下标标记
                end=j;//标记终止的下标
            }
            if(sum<0)
            {
                sum=0;
                temp=j+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",i,max,start,end);
        if(i<t)
            printf("\n");
    }
    return 0;
}