HDU-1003 Max Sum（动态规划）

题目回顾（HDU-1003）：

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

 

源码与解析

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 int main(){
 5     int T,N;
 6     int a[100010];
 7     cin>>T;
 8     for(int k=1;k<=T;k++){
 9         cin>>N;
10         for(int i=0;i<N;i++){
11             cin>>a[i];
12         }
13         int sum=0;
14         int ans=-10000;            //如果设置的数字不够小，出错。例如sum=-500，
15                                 　 //初始ans=-10，sum<ans无法标记下一步位置。 
16         int start,end;             //标记起点和终点。 
17         int temp=0;                //temp要初始化起点位置。 
18         for(int i=0;i<N;i++){
19             if(sum>=0){
20                 sum+=a[i];
21             }else{
22                 sum=a[i];
23                 temp=i;            //暂时保存新起点位置。 
24             }
25             
26             if(sum>ans){           //如果发现更大的sum，则保存起点和终点。 
27                 ans=sum;
28                 start=temp;
29                 end=i;
30             }
31         }
32         cout<<"Case "<<k<<":"<<endl<<ans<<" "<<start+1<<" "<<end+1<<endl;
33         if(k<T){                  //输出格式的要求。 
34             cout<<endl;
35         } 
36     }
37     return 0;
38 }