Max Sum

/*Max Sum


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147728    Accepted Submission(s): 34527




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
*/
//思路：i=0,sum=0;如果sum+=s[i+1]>s[i],max=sum記下下標起始點d1，繼續s+=s[i+2]與s[i+2]比較，如果s[i+2]為正數，max=s，末標識點d2=i；反之max=max；
//當sum<0時，起始標記點下標加1，sum重新置0； 
#include<stdio.h>
int s[100003];
int main()
{
    int i,T,n,kase;
    scanf("%d",&T);
    for(kase=1;kase<=T;kase++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        scanf("%d",&s[i]);
        int sum=0,f1=1,max=s[1];
        int d1=1,d2=1;//注意d1=d2=1； 
        for(i=1;i<=n;i++){//該題關鍵！！！ 
            sum+=s[i];
            if(sum>max)
            {  
               max=sum;
               d1=f1;
               d2=i;
            }
            if(sum<0)
            {
                sum=0;
                f1=i+1;
            }
        }
        printf("Case %d: ",kase);
        printf("%d %d %d ",max,d1,d2);
        if(kase!=T)
        printf(" ");
    }
    return 0;
}