hdu1003最大连续子序列

Posted 2020-09-18 不积跬步无以至千里，不积小流无以成江海

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

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     int n,t,i,j,k,first,end,sum,x,max;
 8     cin>>t;
 9     for(j=1;j<=t;j++)
10     {
11         cin>>n;
12         sum=0;
13         max=-1001;
14         first=end=k=1;
15         for(i=1;i<=n;i++)
16         {
17             cin>>x;
18             sum+=x;
19             if(sum>max)
20             {
21                 max=sum;
22                 first=k;
23                 end=i;
24             }
25             if(sum<0)
26             {
27                 sum=0;
28                 k=i+1;
29             }
30         }
31          if(j!=1)
32          printf("\n");
33          printf("Case %d:\n",j);
34          printf("%d %d %d\n",max,first,end);
35 
36     }
37     return 0;
38 }