2013 - ECJTU 暑期12级训练赛第一场-Problem-K

K -K

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

SubmitStatus Practice HDU 1003

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

 25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 

 

Sample Output

 Case 1:14 1 4Case 2:7 1 6 
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AC代码：
#include<iostream>  using namespace std;  #define Min -10000  int main()  {      int DT[100000],qidian,zhongdian,i;      int m;      int p=1;      cin>>m;          while(m--)      {          int n;          cin>>n;          for (int i=1; i<=n;i++)              cin>>DT[i];          int max = Min;          int k=1;          int sum = 0;          for (i=1; i<=n; i++)          {              sum = sum + DT[i];              if (sum > max)              {                  max = sum;                  qidian=k;                  zhongdian=i;              }              if(sum<0)              {                  sum=0;                  k=i+1;              }          }          if(p!=1)              cout<<endl;          cout<<"Case "<<p<<":"<<endl;          cout<<max<<" "<<qidian<<" "<<zhongdian<<endl;          p++;      }      return 0;  } 

Hint

INPUT DETAILS:

Five cows with milk outputs of 1..5

OUTPUT DETAILS: