Codeforces Round #410 (Div. 2)C. Mike and gcd problem（数论）

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Description

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.

Sample Input

2
1 1

2
1 3

3
6 2 4

Sample Output

YES
1

YES
1

YES
0

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.

思路

题解:

= =想了很久，虽然快接近思路，但是。。就差那么一点啊。。哎。。

First of all, the answer is always YES.

If  then the answer is 0.

Now suppose that the gcd of the sequence is 1. After we perform one operation on ai and ai + 1, the new gcd d must satisfy d|ai - ai + 1and d|ai + ai + 1  d|2ai and d|2ai + 1. Similarly, because d is the gcd of the new sequence, it must satisfy d|aj, j ≠ i, i + 1.

Using the above observations we can conclude that , so the gcd of the sequence can become at most 2 times bigger after an operation. This means that in order to make the gcd of the sequence bigger than 1 we need to make all numbers even. Now the problem is reduced to the following problem:

Given a sequence v1, v2, ... , vn of zero or one,in one move we can change numbers vi, vi + 1 with 2 numbers equal to . Find the minimal number of moves to make the whole sequence equal to 0.

It can be proved that it is optimal to solve the task for consecutive ones independently so we divide the array into the minimal number of subarrays full of ones, if their lengths are s1, s2, ... , st,the answer is .

Complexity is .

 

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#include<bits/stdc++.h> using namespace std;   int gcd(int a,int b) {     return b?gcd(b,a%b):a; }   int main() {     int n,res = 0,resgcd = 0,cnt = 0,tmp;     scanf("%d",&n);     while (n--)     {         scanf("%d",&tmp);         resgcd = gcd(resgcd,tmp);         if (tmp&1)  cnt++;         else    res += (cnt/2) + 2*(cnt&1),cnt = 0;     }     res += (cnt/2) +2*(cnt&1);     printf("YES\n");     if (resgcd > 1)  printf("0\n");     else    printf("%d\n",res);     return 0; }