1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28629    Accepted Submission(s): 6070

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

 

自己写的代码：

 

#include<iostream>
using namespace std;
#define Max_size 100000
int main()
{
 int n,T,max,start,end,sum,a[Max_size],i=1;
 cin>>T;
 while(T--)
 {
  max=-1001;
  cin>>n;
  for(int t=0;t<n;t++)
   cin>>a[t];
  for(int j=0;j<n;j++)
  {
   sum=a[j];
   if(sum>max)
   {
    max=sum;
    start=end=j+1;
   }
   for(int k=j+1;k<n;k++)
   {
    sum+=a[k];
    if(sum>max)
    {
     max=sum;
     start=j+1;
     end=k+1;
    }

   }
  }
  cout<<"Case "<<i<<":"<<endl;
  cout<<max<<" "<<start<<" "<<end<<endl<<endl;
  i++;
 }
 return 0;
}
 

 

虽然结果是正确的，但是由于超时而不能AC，于是上网找了一下资料，发现一份牛人写的DP算法：

 

#include <iostream>
using namespace std;
int main()
{
         int T,N,num,startP,endP;
         cin>>T;
         for(int k=0;k<T;k++)
         {
            cin>>N;
                 int max=-1001,sum=0,temp=1;
                 for(int i=0;i<N;i++)
                 {
                                 cin>>num;
                                 sum+=num;
                                 if(sum>max)
                                 {
                                           max=sum;
                                           startP=temp;
                                           endP=i+1;
                                 }
                                 if(sum<0)
                                 {
                                         sum=0;
                                         temp=i+2;
                                 }
                 }
                 cout<<"Case "<<k+1<<":"<<endl<<max<<" "<<startP<<" "<<endP<<endl;
                 if(k!=T-1) cout<<endl;
        
         }
}