Codeforces 798C Mike and gcd problem

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

21 1

output

YES1

input

36 2 4

output

YES0

input

21 3

output

YES1

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： 给定一个数组，问最少变幻多少次可以把他变成一个漂亮数组。变幻方式说明：每个数组只能和与它相邻的数组一起进行变幻。例如：a[i],a[i+1]  ------> a[i] - a[i+1],a[i+1] + a[i]，对漂亮数组的定义：整个数组的最大公约数大于1。

思路： 首先这个这个问题是不可能输出NO的，都是YES。然后，就是一些基本知识：1.任何数与0的最大公约数都是这个数本身。（确切的说是gcd值。）。2.而如果是负数， 那么可以把它转化成相反数来看。如果这个数组刚开始就是漂亮数组那么结果就是“0”，如果这个数组刚开始的是不是漂亮数组。然后，如果这个数组 初始中的最大公约数是1，那么最终变幻之后就只的最大公约数必然是2。然后，相邻的两个数是奇数那么变幻后就是 “奇数 - 奇数，奇数 + 奇数”。其中“奇数 - 奇数”是一个偶数，“奇数 + 奇数” 是偶数最大公约数中有2。如果是奇数与偶数相邻，则第一次变幻之后  “奇数 - 偶数，偶数 + 奇数”，（这两个数都是奇数，还需要在变幻一次才能得到两个偶数。）

基于这个思路接下来是我自己的AC代码：（个人感觉不是很好。）

#include<iostream>#define MAX 100005using namespace std;int GCD(int a,int b){return b == 0 ? a : GCD(b,a%b);}int num[MAX];int main( ){int n;while(cin >> n){for(int j = 0; j < n; j++)cin >> num[j];int gcd = GCD(num[0],num[1]);for(int j = 2; j < n; j++)gcd = GCD(gcd,num[j]);//cout << gcd << endl;if(gcd != 1){cout << "YES" << endl;cout << "0" << endl;continue;}int ans = 0;bool odd_ok;if((num[0]&1) == 0)odd_ok = false;elseodd_ok = true;for(int j = 1; j < n; j++){if((num[j]&1) == 1){if(odd_ok)ans++,odd_ok = false;elseodd_ok = true;}else{if(odd_ok)ans += 2,odd_ok = false;elseodd_ok = false;}}if(odd_ok)ans += 2;cout << "YES" << endl;cout << ans << endl;}}

然后，是另一份比较好的AC代码：

#include <bits/stdc++.h>using namespace std;int main ( ) {int n, ai, d = 0, ones = 0, ans = 0;scanf ( "%d", &n );for ( int i = 0; i <= n; ++i ) {if ( i == n ) ai = 0;else scanf ( "%d", &ai );d = __gcd(d,ai);if ( ai&1 ) ones++;else ans += ones/2 + 2*(ones%2), ones = 0;}printf ( "YES\n%d\n", d < 2 ? ans : 0 );return 0;}

在这份代码的思路完全一样，只是相同的思路不一样的水平写出来的差距就是有那么大。

这个题刚开始的时候把题目理解错了，我以为这个最大公约数是只是相邻两个数之间的最大公约数。结果在我好不容易看懂题意之后感觉太后悔了。QAQ.......