Codeforces 798C Mike and GCD problem

Mike and gcd problem

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.

Example
Input
2
1 1
Output
YES
1
Input
3
6 2 4
Output
YES
0
Input
2
1 3
Output
YES
1
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：一个数组，数组内的元素之间不互质，换句话说，就是所有元素的最大公约数大于1。

分析：起初，我以为要把素数处理了就行，后来发现这主要是考虑奇数与偶数的事。当有使公因数是1的时候，我们可以想到只要把奇数都变成偶数，这样最大公因数就大于1了。当两个数都是奇数时，只需要一步就可以；当两个数是一奇一偶时，我们要操作两步才能达到目的。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int a[100005];int gcd(int a,int b){    if(b==0)  return a;    return(gcd(b,a%b));}int main(){   int n;   while(~scanf("%d",&n))   {       int i;       for(i=0; i<n; i++)          scanf("%d",&a[i]);       int x=a[0];       for(i=1; i<n; i++)        {            x=gcd(x,a[i]);        }       if(x>1)          printf("YES\n0\n");        else        {            int sum=0;            for(i=0; i<n-1; i++)                if(a[i]%2 && a[i+1]%2)               {                   int t1=a[i]+a[i+1],t2=a[i]-a[i+1];                   a[i]=t1;                   a[i+1]=t2;                   sum++;               }            for(int i=0; i<n; i++)                if(a[i]%2)               {                  if(i==0)                  {                      int t1=-2*a[i+1],t2=2*a[i];                      a[i]=t1;                      a[i+1]=t2;                      sum+=2;                  }                  else                  {                      int t1=-2*a[i],t2=2*a[i-1];                      a[i-1]=t1;                      a[i]=t2;                      sum+=2;                  }               }               printf("YES\n%d\n",sum);        }   }}