Max Sum

                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                           Total Submission(s): 148609    Accepted Submission(s): 34731

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6
//思路：i=0,sum=0;如果sum+=s[i+1]>s[i],max=sum记下下标起始点d1，继续s+=s[i+2]与s[i+2]比较，如果s[i+2]为正数，max=s，末标识点d2=i；反之max=max；//当sum<0时，起始标记点下标加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++){//该题关键！！！             sum+=s[i];            if(sum>max)            {                 max=sum;               d1=f1;               d2=i;            }            if(sum<0)            {                sum=0;                f1=i+1;            }        }        printf("Case %d:\n",kase);        printf("%d %d %d\n",max,d1,d2);        if(kase!=T)        printf("\n");    }    return 0;}