CF#798 C. Mike and gcd problem（思维）

最新推荐文章于 2021-03-18 19:15:49 发布

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探讨通过特定操作使序列的各元素最大公约数大于1的方法，及其实现思路。

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. $\text{[math]}$ .

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.

$\text{[math]}$ 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 $\text{[math]}$ .

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[N],c[N];

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);
    }
}