C. Mike and gcd problem
time limit per test:

2 seconds

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

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
input
2
1 1
output
YES
1
input
3
6 2 4
output
YES
0
input
2
1 3
output
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.

题目链接:http://codeforces.com/contest/798/problem/C

题意:一次操作中可以选择i (1 ≤ i < n),删除ai, ai + 1添加 ai - ai + 1, ai + ai + 1到相同的位子。尽可能的操作少的数量使得a数组的最大公约数大于1。注意

思路:ai - ai + 1,ai + ai + 1两个数相差2*ai + 1,2*ai + 1为它们之间最大公约数的倍数,即它们之间的最大公约数为2*ai + 1的约数。并且它们的最大公约数不可能为ai + 1的约数,否则ai,ai+1之间的最大公约数大于1,不需要进行操作。那么操作之后的a数组,之间的最大公约数为2。奇偶之间需要2次操作,奇奇之间需要1次操作,偶数直接跳过。

代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll MAXN=1e13+;
int n,a[maxn];
int main()
{
scanf("%d",&n);
for(int i=; i<=n; i++) scanf("%d",&a[i]);
int sign=a[];
for(int i=; i<=n; i++) sign=__gcd(sign,a[i]);
if(sign>)
{
cout<<"YES"<<endl<<<<endl;
return ;
}
int ans=;
for(int i=; i<=n; i++) a[i]=a[i]%;
for(int i=; i<=n; i++)
{
if(a[i]==) continue;
if(i==n) ans+=;
if(i+<=n)
{
if(a[i+]) ans+=;
else ans+=;
a[i+]=;
}
}
cout<<"YES"<<endl<<ans<<endl;
return ;
}

gcd

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