CodeForces 891A Pride （数学）

You have an array a with length n, you can perform operations. Each operation is like this: choose two adjacent elements from a, say x and 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.

Example
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.

题意：

相邻的两个数的GCD（最大公约数）可以替换其中的一个数，问最少需要多少步能将数列中的所有数全部都换成1。

题解：

首先，当数列中存在1的时候，答案明显就是N减去1的个数。不存在1的时候，想办法构造1出来，假设区间[L,R]的gcd为gcd(L, R) ，那么GCD(L, R + 1) = gcd(gcd(L,R), A(R + 1))。那么只要找到最短的区间使得gcd(L,R)==1，答案就是R-L+n-1了。

#include<iostream>
#include<algorithm>

using namespace std;
int a[2005];
const int INF=0x3f3f3f3f;
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int n;
    while(cin>>n)
    {
        int cnt=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(a[i]==1)
                cnt++;
        }
        if(cnt)
        {
            cout<<n-cnt<<endl;
        }
        else
        {
            int t=INF;
            for(int i=0;i<n;i++)
                for(int j=i,cur=0;j<n;j++)
                {
                    if((cur=gcd(cur,a[j]))==1)
                        t=min(t,j-i),j=n;
                }
            if(t==INF)
                cout<<"-1"<<endl;
            else
                cout<<t+n-1<<endl;
        }
    }
    return 0;
}