C.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

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 A

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

Example

2
1 1

YES
1

3
6 2 4

YES
0

2
1 3

YES
1

note

In the first example you can simply make one move to obtain sequence [0,?2] with .

In the second example the gcd of the sequence is already greater than 1.

解題思路：

題意為：輸入給定數字，求他們的大於1的最大公約數，如果沒有則進行操作

對數字a,b進行操作: a,b a-b,a+b -2b,2a

由上可知，不論兩個數字為什麽，最多操作兩次就有公約數2

那麽有三種情況：1.都為偶數，不需要進行操作。2.一個為奇數一個為偶數，需要操作兩次，3.都為奇數，只需操作一次；

然後隨便打下就行了

實現代碼：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

int main()
{
    ll m,a[100005],i,ans,num;
   cin>>m;
   num = 0;
   for(i=1;i<=m;i++)
    cin>>a[i];
   ans = gcd(abs(a[1]),abs(a[2]));
   for(i=3;i<=m;i++)
    ans = gcd(ans,abs(a[i]));
   if(ans>1)
    cout<<"YES"<<endl<<"0"<<endl;
   else
   {
       for(i=1;i<m;i++)

           if(a[i]%2&&a[i+1]%2)
           {
               a[i]=0;a[i+1]=0;num++;
           }
        for(i=1;i<m;i++)
           if((a[i]%2==0&&a[i+1]%2)||(a[i]%2&&a[i+1]%2==0))
            {
               a[i]=0;a[i+1]=0;num+=2;
            }

       cout<<"YES"<<endl<<num<<endl;
   }
   return 0;
}

codeforces 798C Mike and gcd problem