Mike has a sequence A = [a1, a2, …, an] of length n. He considers the sequence B = [b1, b2, …, bn] 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 ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in 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 bi 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, …, an (1 ≤ ai ≤ 109) — elements of sequence A.
Output
Output on the first line “YES” (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and “NO” (without quotes) otherwise.
If the answer was “YES”, output the minimal number of moves needed to make sequence A beautiful.
Examples
inputCopy
2
1 1
outputCopy
YES
1
inputCopy
3
6 2 4
outputCopy
YES
0
inputCopy
2
1 3
outputCopy
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.
题意：给定一个数组，求最少几次操作可以使 gcd( a1, a2,… ,an )>1
对于 第i个元素， 操作是将 ai, ai+1 —> ai-ai+1, ai+ ai+1;
考虑这三种情况：
1. 对于两个奇数，一次操作即可全变成偶数；
2. 对于两个偶数，0次操纵；
3. 对于一奇一偶，两次操作全变为偶数；
所以，遍历模拟即可：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define maxn 200005
const int mod=1e9+7;
#define eps 1e-6
#define pi acos(-1.0)
ll quickpow(ll a,ll b,ll m)
{
ll ans=1;
while(b){
if(b&1){
ans=ans*a%m;
}
a=a*a%m;
b>>=1;
}
return ans;
}
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
int a[maxn];
int main()
{
ios::sync_with_stdio(false);
int n;
cin>>n;
int i,j;
cin>>a[1]>>a[2];
ll d=gcd(a[1],a[2]);
for(i=3;i<=n;i++){
cin>>a[i];
d=gcd(d,a[i]);
}
if(d>1){
cout<<"YES"<<endl;
cout<<0<<endl;
}
else {
int cnt=0;
for(i=1;i<n;i++){
if(a[i]%2&&a[i+1]%2){
a[i]=0;a[i+1]=0;cnt++;
}
}
for(i=1;i<n;i++){
if(a[i]%2&&a[i+1]%2==0||a[i+1]%2&&a[i]%2==0){
a[i]=0;
a[i+1]=0;
cnt+=2;
}
}
cout<<"YES"<<endl;
cout<<cnt<<endl;
}
}