Series 1474 operation to find the law

Title Description

Given a positive integer of n columns, the number of columns in one action: the removal of two of a, b, and then add an a × b + 1. Each operation a reduced number of a row, n-1 by the number of operations after only a few columns. Find the maximum value of n-1 in the final thanks to a number of operations.

Enter a description

Input a plurality of sets, each set of two lines, the first line of the input integer n, a second input row n positive integer n <20

Output Description

For each input and output of a last remaining

Sample input

6

8 9 3 6 5 4

Sample Output

29493

Idea: Every time looking for the smallest of multiplying two numbers plus 1, the maximum final result

 1 #include<iostream>
 2 using namespace std;
 3 int main(){
 4     int n;
 5     while(cin>>n){
 6         int i,j;
 7         unsigned long long m;
 8         unsigned long long num[n];
 9         for(i=0;i<n;i++){
10             cin>>num[i];
11         }
12         while(n!=1){
13             for(i=0;i<n-1;i++){
14                 for(j=0;j<n-i-1;j++){
15                     if(num[j]>num[j+1]){
16                         m=num[j];
17                         num[j]=num[j+1];
18                         num[j+1]=m;
19                     }
20                 }
21             }
22             m=num[0]*num[1]+1;
23             num[0]=m;
24             n--;
25             for(i=1;i<n;i++){
26                 num[i]=num[i+1];
27             }
28         }
29         cout<<num[0]<<endl;
30     }
31     return 0;
32 }