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

 

#include <stdio.h>

int main(){
    int T;
    int n;
    int number[100001];
    int i;
    int sum;
    int max;
    int start;
    int end;
    int time;
    int temp;

    
    scanf("%d",&T);
    time=1;

    while(T--){
        sum=0;
        max=-1000;
        start=0;
        end=0;
        temp=0;

        scanf("%d",&n);

        for(i=0;i<n;i++)
            scanf("%d",&number[i]);

        for(i=0;i<n;i++){
            sum+=number[i];

            if(sum>max){
                max=sum;
                start=temp;
                end=i;
            }

            if(sum<0){   //关键是这里，当求和小于0时，便把下一个数值作为开头再找最大值
                sum=0;
                temp=i+1;
            }
        }

        printf("Case %d:
",time);
        time++;
        printf("%d %d %d
",max,start+1,end+1);
        if(T!=0)
            printf("
");
    }

    return 0;
}