杭电1003-Max Sum

Max Sum

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 <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char *argv[])
{
    int n,m,a,temp,start,end;
    long sum,max;
    cin>>n;
    for(int j=1;j<=n;j++)
    {
       cin>>m;
       sum=0;
       max=-1001;////由于存在全是负数，将最大值赋值为负数(更大也行)
       temp=1;
       for(int i=1;i<=m;i++)
       {
               cin>>a;
               sum+=a;
               if(sum>max)
               {
                  max=sum;
                  start=temp;//更新起始点
                  end=i;  //更新末点
               }
               if(sum<0)
               {
                  sum=0;
                  temp=i+1;//执行下一个i（也就是读取下一个数字）
               }
       }
       cout<<"Case "<<j<<":"<<endl;
       cout<<max<<" "<<start<<" "<<end<<endl;
       if(j<n)
       cout<<endl;
    }
    system("PAUSE");
    return EXIT_SUCCESS;
}