choir

                                                                                                              choir

 

Topic description

There are n students standing in a row, each student has an ability value, Niuniu wants to select k students in order from these n students, and requires that the difference between the position numbers of two adjacent students does not exceed d, so that the k students are The product of the student's ability values ​​is the largest, can you return the largest product?

Enter description:

Each input contains 1 test case. The first line of each test data contains an 
integer n (1 <= n <= 50), which represents the number of students, and the next line contains n integers, which represent the ability value ai of each student in order (- 50 <= ai <= 50). The next line contains two integers, k and d (1 <= k <= 10, 1 <= d <= 50).

Output description:

One line of output represents the largest product.

Example 1

enter

3
7 4 7
2 50

output

49

 

Analysis: This problem is implemented using dynamic programming because of the optimal substructure property.

Problem decomposition: Decomposition is the key, let f(n, k) mean that the last person is selected from n people first, and then k-1 people are selected from the remaining n-1 people, and let this one and the previous k -1 meets the constraints

Mathematical description [find the recursive formula]: a[end]>0 f(end,k)=max{f(then,k-1)}*a[end]} (max(k-1,end-d)= <then<=end-1)      

　　　　　　　　　　　   a[end]<0 f(end,k)=max{g(then,k-1)}*a[end]} (max(k-1,end-d)=<then<=end-1) //Because the ability value has a negative value, an additional array g[][] is added to store the minimum value of the product

 

Recursive transformation:

f(end,k)=max{f(then,k-1)*a[end],g(then,k-1)*a[end]    (max(k-1,end-d)=<then<=end-1)

g(end,k)=min{f(then,k-1)*a[end],g(then,k-1)*a[end]    (max(k-1,end-d)=<then<=end-1)

 

 

Note: The value of this question should be stored in the long type.

My code implementation:

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<limits.h>
 4 #define N 100
 5 int a[N];
 6 long f[N][N],g[N][N];
 7 
 8 int max1(int a,int b){
 9     return a>b?a:b;
10 }
11 
12 long max(long a,long b){
13     return a>b?a:b;
14 }
15 
16 long min(long a,long b){
17     return a<b?a:b;
18 }
19 void ability(int n,int kk,int d){
20     for(int i=1;i<=n;i++){
21         f[i][1]=a[i];
22         g[i][1]=a[i];  
23     }
24     for(long k=2;k<=kk;k++){
25         for(int end=k;end<=n;end++){
26             long tempmax=INT_MIN;
27             long tempmin=INT_MAX;
28             for(int then=max1(k-1,end-d);then<=end-1;then++){
29                 if(tempmax<max(f[then][k-1]*a[end],g[then][k-1]*a[end]))
30                     tempmax=max(f[then][k-1]*a[end],g[then][k-1]*a[end]);
31                 if(tempmin>min(f[then][k-1]*a[end],g[then][k-1]*a[end]))
32                     tempmin=min(f[then][k-1]*a[end],g[then][k-1]*a[end]);
33             }
34             f[end][k]=tempmax;
35             g[end][k]=tempmin;
36         }
37     }
38 }
39         
40 int main(){
41     int n,k,d;
42     scanf("%d",&n);
43     for(int i=1;i<=n;i++)
44         scanf("%d",&a[i]);
45     scanf("%d%d",&k,&d);
46     ability(n,k,d);
47     long result=INT_MIN;
48     for(int i=k;i<=n;i++){
49         if(result<f[i][k])result=f[i][k];
50     }
51     printf("%ld",result);
52 
53     return 0;
54 }