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time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputMike 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
2 output
YES input
3 output
YES input
2 output
YES 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. 题目大意:能否能找到最小的操作次数,使得gcd(b1,b2,...,bn)>1 。如果能,输出YES和最小操作次数,否则输出NO。 思路提示:首先,肯定能。 证明:假设b[i]和b[i+1]都为奇数,经过一次操作变换为b[i]-b[i+1]和b[i]+b[i+1]都是偶数(2的倍数); 假设b[i]和b[i+1]为一奇数一偶数,经过两次操作变换为2b[i]和-2b[i+1]都是偶数; 所以,可以经过有限次操作,使得gcd(b1,b2,...,bn)=2。 方法: 先计算gcd(b1,b2,...,bn),如果大于1,输出YES和0; 否则: 先把所有b[i]和b[i+1]都为奇数的经过一次变换为偶数,记录操作次数,再把b[i]和b[i+1]为一奇数一偶数的经过两次变换为偶数,记录操作次数。 最后,输出总的操作次数。 代码: #include<iostream> #include<cstdio> using namespace std; #define ll long long const int N=1e5+5; ll gcd(ll x,ll y) { while(y^=x^=y^=x%=y); return x; } int a[N]; int main() { ios::sync_with_stdio(false); cin.tie(0); int n; cin>>n; int cnt=0; int gd; for(int i=0;i<n;i++) { cin>>a[i]; if(i!=0)gd=gcd(gd,a[i]); else gd=a[i]; } if(gd>1) { cout<<"YES"<<endl,cout<<0<<endl; return 0; } for(int i=0;i<n-1;i++) { if(a[i]&1&&a[i+1]&1) cnt++,a[i]=a[i+1]=2; } for(int i=0;i<n-1;i++) { if((a[i]&1&&a[i+1]%2==0)||(a[i]%2==0&&a[i+1]&1)) cnt+=2,a[i]=a[i+1]=2; } cout<<"YES"<<endl; cout<<cnt<<endl; return 0; }
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