CF#798 C. Mike and gcd problem（思维）

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 numbersai, 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.

题意：给出一个序列 A ，我们可以将相邻两元素 a [ i ] , a [ i + 1 ] 改为 a [ i ] - a [ i + 1 ] 和 a [ i ] + a [ i + 1 ] ，问至少多少次操作后能使 gcd ( A1 ~ An) 大于1，若无法完成，则输出 NO

思路：显然我们把所有奇数修改成偶数即可满足条件，每次修改时若  a [ i ] 和 a [ i + 1 ] 同为奇数则修改一次，若一奇一偶则修改两次

#include<bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;

int n;
int a
,c
;

int gcd(int n,int m)
{
int t = 1;
while(t)
{
t = n % m;
n = m;
m = t;
}
return n;
}

int main()
{
while(scanf("%d", &n) == 1)
{
memset(c,0,sizeof(c));
for(int i=0; i < n; i++)
{
scanf("%d",& a[i]);
}
int ans = 0,tmp = gcd(a[0], a[1]);
for(int i = 0; i < n; i++)
{
tmp = gcd(tmp, a[i]);
if(a[i] & 1 && !c[i])
{
if(a[i + 1] & 1) ans++;
else ans += 2;
c[i] = 1;
c[i + 1] = 1;
}
}
if(tmp != 1) printf("YES\n0\n");
else printf("YES\n%d\n", ans);
}
}