2 seconds
256 megabytes
standard input
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).
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 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.
2
1 1
YES
1
3
6 2 4
YES
0
2
1 3
YES
1
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.
题解:我们发现一个位置经过两次操作a[i]变成-2a[i+1],a[i+1]变成2a[i],所以当gcd为1时我们可以把他们都变为偶数,所以我们把所有的数都变为偶数
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+;
int a[maxn],n;
int main()
{
scanf("%d",&n);
for(int i=;i<n;i++)
{
scanf("%d",&a[i]);
}
int tmp=a[];
for(int i=;i<n;i++)
{
tmp=__gcd(tmp,a[i]);
}
if(tmp!=)
{
puts("YES\n0");
}
else
{
int ans=;
for(int i=;i<n;i++)
{
if(a[i]%==)continue;
else if(i==n-)
{
ans+=;
}
else
{
if(a[i+]%!=)ans++;
else ans+=;
i++;
}
}
printf("YES\n%d\n",ans);
} }