CodeForces - 798C

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.

Example
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.

首先题干那种操作只能保证两个奇数出来一定是偶数
单个奇数变偶数只要两次没错就只能保证这个….
其他的啥都保证不了..除了先天出现的gcd还是别想了….
想到这个就暗全变偶数猜一发能过就过过不了算了

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
using namespace std;
int tu[100001];
int n;
typedef long long ll;
int gcd(int a, int b)
{
    if (b == 0) return a;
    return gcd(b, a%b);
}
int main()
{
    cin >> n;
    for(int a=1;a<=n;a++)scanf("%d",&tu[a]);
    int gg = tu[1];
    for (int a = 2; a <= n; a++)gg = gcd(gg, tu[a]);
    if (gg > 1)
    {
        cout << "YES" << endl;
        cout << 0;
        return 0;
    }
    int cz = 0;
    for (int a = 1; a <= n; a++)
    {
        int z = a - 1, y = a + 1;
        if (tu[a] % 2 == 0)continue;
        if (y <= n)
        {
            if (tu[y] % 2)tu[y] = tu[a] = 2,cz++;
            else
            {
                cz += 2;
                tu[a] = 2;
            }
            continue;
        }
        if (z > 0)
        {
            cz += 2;
            tu[a] = 2;
        }
    }
    cout << "YES" << endl;
    cout << cz;
}