CodeForces - 891A-Pride (思维＋暴力)

CodeForces - 891A-Pride

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 t
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.

题目大意：

给定一个数列，我们定义一个操作是取相邻的两个数计算其 gcd ，然后替换掉其中的某一个数，问最少多少步可以将数列全部变为 1 。

解题思路：

只要能够出现1 那只要再进行n-1次就可以都变成1了,因为1跟任何数求公约数都是1，所有只要再进行n-1次就能全变成1了。
思路很简单，如果原数列里有1 最好了，没有的话就枚举什么时候能出现第一个1.

• AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e6+10;
int a[maxn];
int gcd(int x,int y)
{
	if(y==0)
		return x;
	else 
		return gcd(y,x%y);
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int num=0;
		for(int i=0;i<n;i++)
		{
			cin>>a[i];
			if(a[i]==1)
				num++;
		}
		if(num!=0)
		{
			cout<<n-num<<endl;
			continue;
		}
		int ans=0;
	
		for(int i=1;i<n;i++)
		{
			for(int j=0;j<n-i;j++)
			{
				a[j]=gcd(a[j],a[j+1]);
				if(a[j]==1)
				{
					ans=i+n-1;
					break;
				}
			}
			if(ans!=0)
				break;
		}
		if(ans!=0)
			cout<<ans<<endl;
		else
			cout<<"-1"<<endl;
		
	}
	return 0;
}```