【dp】Max Sum

G - Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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 
     
     


     
     


     
     

      保存最佳答案，对可能的答案都进行相加。
     
     


     
     


     
     

      注意会取负值 所以设置时dp[0]应为一个非常小的负数。
     
     


     
     

      #include<iostream>
#include<cstdio>
#include <climits>
using namespace std;


int q[105000],pre[105000];
int main()
{
    int t;
    cin>>t;
    int num=1;


    while(t--)
    {
        if(num!=1)
            puts("");


        int n;
         int ans=0;
         int temp;


        cin>>n;
        q[0]=INT_MIN;
        pre[0]=1;
        for(int i=1;i<=n;i++)
            {
                cin>>temp;


                if(q[i-1]>=0)
                    {
                        pre[i]=pre[i-1];
                        q[i]=temp+q[i-1];
                    }
                else
                {
                    pre[i]=i;
                    q[i]=temp;
                }


                if(q[ans]<q[i])
                {
                    ans=i;
                }


            }
            cout<<"Case "<<num<<":"<<endl;
            num++;
            cout<<q[ans]<<" "<<pre[ans]<<" "<<ans<<endl;










    }
    return 0;
}