最大连续元素之和

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<iostream>

#include<cstdio>

using namespace std;

int main()

{

    int n,t,i,c;

    int a[100002];

    scanf("%d",&t);     //数据的组数

    for(c=1;c<=t;c++)   //用c来记录数据的组数和循环次数，还有控制换行的输出

    {

        int k=1,st=0,en=0,summax=-1000,sum=0;

        scanf("%d",&n);

        for(i=0;i<n;i++)  //输入数据

        {

            scanf("%d",&a[i]);

        }

        for(i=0;i<n;i++)  //此循环控制连续元素相加，sum与summax的比较，sum<0的情况

        {

            sum+=a[i];  //连续元素一个个加起来

            if(sum>summax) //如果sum>summax，把sum赋值给summax（把最大的连续元素之和用summax来存）

            {

                summax=sum;

                st=k; //记录头位置

                en=i+1;  //记录尾位置

            }

            if(sum<0) //若sum<0，则sum=0，且sum<0的这一连续元素不要了，从下一个元素开始

            {

                sum=0;

                k=i+2;  //从下一个元素开始，k是记录头位置，k=i+2即从尾位置的下一个元素开始算

            }

        }

        printf("Case %d:\n%d %d %d\n",c,summax,st,en);

        if(c!=t) //换行的控制
　　　　　　cout<<endl;  

    }

    return 0;

}