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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

#include<stdio.h>int main(){    int t,n,tt=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        //尺取法         int maxx=-1001,sum=0,l=1,left,right,x;        for(int i=1;i<=n;i++)        {            scanf("%d",&x);            sum+=x;            if(sum>maxx)            {                maxx=sum;                left=l;//更新起点                 right=i;//更新终点             }            if(sum<0)//顺序不能反,考虑如果一组数全是负数             {                sum=0;//如果当前的累加和已经小于0则抛弃这一段重新开始                 l=i+1;            }        }        printf("Case %d:\n",tt++);        printf("%d %d %d\n",maxx,left,right);        if(t) printf("\n");    }    return 0;}