HDU Max Sum

2017-11-15 904

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简介：

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 191 Accepted Submission(s): 98

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

#include <iostream>
using namespace std;
int main()
{
    int T,N,max,sum,start,end,temp,t;
    cin>>T;
    int h = T;
    while(T--)
    {
        sum = 0;
        start = 1;
        end = 1;
        temp = 1;
        max = -1001;
        cin>>N;
       for(int i = 1; i <= N; i++)
       {
            cin>>t;
            sum +=t;
            if(sum>max)
            {
                max = sum;
                end = i;
                start = temp;
            }
            if(sum<0)
            {
                sum = 0;
                temp = i+1;
            }
       }
       if(T!=0)
       cout<<"Case "<<h-T<<":"<<endl<<max<<" "
       <<start<<" "<<end<<endl<<endl;
       else cout<<"Case "<<h-T<<":"<<endl<<max
       <<" "<<start<<" "<<end<<endl;
    }
    return 0;
}

本文转自NewPanderKing51CTO博客，原文链接：http://www.cnblogs.com/newpanderking/archive/2011/08/03/2125824.html，如需转载请自行联系原作者

文章标签：

Java

HDU Max Sum

Max Sum

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