#410div2C. Mike and gcd problem
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.
题解:当$\gcd (x,y)! = 1$时,直接得出答案。
当$\gcd (x,y) = = 1$,令$d = \gcd (x - y,x + y)$,
所以,$d|(x - y),d|(x + y)$
由信安数基课本P4,$a|b,a|c \to a|tb + sc$,
$d|2x,d|2y$$\to d|\gcd (2x,2y) \to d|2\gcd (x,y) \to d|2$,$d = = 1or2$因为若再继续下去,必须满足此等式,故不必继续。
可以看出最后的d一定整除偶数,所以n个数必须都为偶数.
所以此题即变为,把n个数变为偶数的最小步数。
当$a[i]\% 2 = = 1 ,a[i + 1]\% 2 = = 1$,步数增加1,
当a[i]和a[i+1]有一个为偶数时,步数增加2.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
using namespace std;
ll n,a[],t=;
ll gcd(ll a,ll b){
while(b){
ll temp=b;
b=a%b;
a=temp;
}
return a;
}
int main(){
cin>>n;
for(int i=;i<n;i++){
cin>>a[i];
t=gcd(t,a[i]);
}
for(int i=;i<n;i++){
a[i]%=;
}
if(t!=){
cout<<"YES\n0\n";
return ;
}
ll ans=;
for(int i=;i<n;i++){
if(a[i]){
ans++;
if(!a[i+]){
ans++;
}
a[i]=a[i+]=;
}
}
cout<<"YES\n"<<ans<<endl; }
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