CodeForces - 798C Mike and gcd problem (贪心+思维）

题目链接：http://codeforces.com/problemset/problem/798/C点击打开链接

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.

题意是 如果一个序列满足所有序列里数的gcd不为1 为好序列

现在可以使用操作把 ai ai+1变成 ai-ai+1 ai+ai+1 

问最少需要多少次

稍微算两个数就可以发现

在对同一队数操作两次之后结果为 -2ai+1 2ai

而此时必有gcd>1

因此这样每个序列都一定有解

再来考虑操作次数的问题

显然 当两个数中同样是偶数时 无需操作

两个数一奇一偶 则需操作两次

两个都是奇数 操作一次

这样我们贪心尽量让两个相邻的奇数凑在一起操作即可

注意需要判断原序列是否本身就是好序列

#include <bits/stdc++.h>
using namespace std;
int gcd(int x,int y)
{
if(x<y)
swap(x,y);
if(y==0)
return x;
return gcd(y,x%y);
}
int main()
{
int n;
cin >> n;
vector<int > s;
for(int i=0;i<n;i++)
{
int mid;
cin >>mid;
s.push_back(mid);

}
int ggcd=s[0];
for(int i=0;i<n;i++)
{
ggcd=gcd(ggcd,s[i]);
}
int ans=0;
for(int i=0;i<n;i++)
{
if(s[i]&1)
{
if(i!=n-1)
{
if(s[i+1]&1)
{
ans++;
i++;
}
else
{
ans+=2;
i++;
}
}
else
ans+=2;
}
}
//cout << ggcd << endl;
cout << "YES" << endl;
if(ggcd!=1)
ans=0;
cout << ans << endl;
}