G - Mike and gcd problem

Mike has a sequence A = [a₁, a₂, ..., a_n] of length n. He considers the sequence B = [b₁, b₂, ..., b_n] beautiful if the gcd of all its elements is bigger than 1, i.e. .

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers a_i, a_i + 1 and put numbers a_i - a_i + 1, a_i + a_i + 1in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

is the biggest non-negative number d such that d divides b_i for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence Abeautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequenceA beautiful.

Example

Input

21 1

Output

YES1

Input

36 2 4

Output

YES0

Input

21 3

Output

YES1题意：输入n个数 (2 ≤ n ≤ 100 000），操作:把a[i],a[i+1] 替换成 a[i]-a[i+1],a[i]+a[i+1]，问最少多少次操作使所有元素的gcd>1。

题解：假设两个数a,b。操作一次a-b,a+b. 操作两次 -2b，2a。gcd=2；所以任意两个数两次操作后gcd一定>1；当两个数是偶数时，需要0次操作当两个是奇数时，需要1次操作当一奇一偶时，需要2次操作先循环一遍两个都是奇数的，然后把这两个数更改为偶数，操作次数+1；在循环一遍一奇一偶成对的，然后把这两个数更改为偶数，操作次数+2；

代码：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cstdio>
long long gcd(long long a,long long b)
{if(b==0)return a;elsegcd(b,a%b);
}
using namespace std;
int main()
{long long a[100005],ans;int n,i,num=0;cin>>n;for(i=1;i<=n;i++)cin>>a[i];ans=gcd(a[1],a[2]);for(i=3;i<=n;i++)ans=gcd(ans,a[i]);if(ans>1)cout<<"YES"<<endl<<0<<endl;else{for(i=1;i<n;i++){if(a[i]%2&&a[i+1]%2){a[i]=0; a[i+1]=0; num++;}}for(i=1;i<n;i++){if(a[i]%2==0&&a[i+1]%2==1){a[i]=0; a[i+1]=0; num+=2;}else if(a[i]%2==1&&a[i+1]%2==0){a[i]=0; a[i+1]=0; num+=2;}elsecontinue;}cout<<"YES"<<endl<<num<<endl;}
}

转载于:https://www.cnblogs.com/GXXX/p/6814992.html

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