Hangzhou Electric oj 1003

Hangzhou University of Electronic Science and Technology online judge 1003 title

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.

Input2 the Sample
. 5. 4. 5. 6 -1 -7
. 7. 6 -1 0 -6. 7. 1 -5
the Sample the Output
Case. 1:
14. 4. 1
// Note here spaces
Case 2:
. 7. 6. 1
the Author
Ignatius.L

Personal ideas:
1. Determine the input case, the variable t is expressed by
2 to determine the length of the sequence, represented by n-
3. flexibility in the allocation memory, and sequentially input sequence
4. sequences determined starting position p1: p1> = 1;
determining the end position P2 sequences: sequence plus the maximum and j
5. output: output multi-row spaces Note

Code

#include<iostream>
using namespace std;
int main()
{
 int t,n;//执行情况(case)次数： t  每种情况输入数据个数：n   
 cin>>t;
 for(int i = 1;i<=t;i++){  
     int i1 = 0, sum = 0;//max为最大子序列和 
     int p1 =0,p2 = 0;//p1:起始位置 p2：末尾位置 
     cin>>n;
     int* seq = new  int[n];//灵活分配内存 
  for(int j = 0;j<n;j++){
      cin>>seq[j];
  }
     int max = seq[0];
  for(int j = 0;j<n;j++){
     sum = sum + seq[j];//顺序加和，逐次判断 
    if(sum>max){
    max = sum;
    p1 = i1;
    p2 = j;  
   }
   if(sum<0)
   { 
     sum = 0;
     i1 = j + 1;
   }
  }
  cout << "Case " << i << ":" << endl;
  if(i<t) cout<<max<<" "<<p1+1<<" "<<p2+1<<endl<<endl;
  else  cout<<max<<" "<<p1+1<<" "<<p2+1<<endl; 
  delete[] seq;//释放内存 
 } 
 return 0;
}