798C - Mike and gcd problem

题意

从而新序列的gcd一定为2，所以先求出所有数字的gcd（因为我太菜的原因不知道算gcd的复杂度其实不高，log（max（a,b））这样，如果不等于1，直接输出0。

然后再贪心扫两遍就行。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define MAX_NUM   1000003
int num[MAX_NUM];
int dp[MAX_NUM][2];
int exchang[MAX_NUM];
int gcd(int a, int b){
    if( b == 0 )
        return a;
    return gcd(b,a%b);
}
int main(int argc, char const *argv[])
{
    int n;
    scanf("%d",&n);
    for (int i = 1; i <= n; ++i)
        scanf("%d",&num[i]);
    int pre = gcd(num[1],num[2]);
    for (int i = 3; i <= n ; ++i)
        pre =  gcd(pre,num[i]);
    if(pre!=1){
        printf("YES\n");
        printf("0\n");
        return 0;
    }
    int ans = 0;
    for (int i = 2; i <= n; ++i)
    {
        if(num[i-1]&1&&num[i]&1){
            ans++;
            num[i-1] = 0;
            num[i] = 0;
        }
    }
    for (int i = 2; i <= n; ++i)
    {
        if(num[i-1]&1||num[i]&1){
            ans+=2;
            num[i-1] = 0;
            num[i] = 0;
        }
    }
    printf("YES\n%d\n",ans );
    return 0;
}