Codeforces Round #446 (Div. 2) C. Pride

ACM_e發表於2017-11-18
IDE



C. Pride
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say xand y, and replace one of them with gcd(x, y), where gcd denotes the greatest common divisor.

What is the minimum number of operations you need to make all of the elements equal to 1?

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements in the array.

The second line contains n space separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the elements of the array.

Output

Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.

Examples
input
5
2 2 3 4 6
output
5
input
4
2 4 6 8
output
-1
input
3
2 6 9
output
4
Note

In the first sample you can turn all numbers to 1 using the following 5 moves:

  • [2, 2, 3, 4, 6].
  • [2, 1, 3, 4, 6]
  • [2, 1, 3, 1, 6]
  • [2, 1, 1, 1, 6]
  • [1, 1, 1, 1, 6]
  • [1, 1, 1, 1, 1]

We can prove that in this case it is not possible to make all numbers one using less than 5 moves.


#include<bits/stdc++.h>
using namespace std;
#define maxn 2000+100
int  a[maxn];
int  b[maxn];
int gcd(int a,int b){
    int temp;
    if(a<b){
        temp=a;
        a=b;
        b=temp;
    }
    while(b!=0){
        temp=a%b;
        a=b;
        b=temp;
    }
    return a;
}
int main(){
   int n;
   cin>>n;
   int i=0;
   for(int j=1;j<=n;j++){
      scanf("%d",&a[j]);
      if(a[j]==1) i++;
   }
   if(i){
      cout<<n-i<<endl;
      return 0;
   }
   int ans=20001;
   for(int j=1;j<n;j++){
      int t=a[j];
      for(int k=j+1;k<=n;k++){
         t=gcd(t,a[k]);
         if(t==1){
            ans=min(ans,k-j);
         }
      }
   }
   if(ans==20001) cout<<"-1"<<endl;
   else cout<<ans+n-1<<endl;

   return 0;
}





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