洛谷 CF798C Mike and gcd problem

 1 #include<iostream>
 2 #include<cstdio>
 3 
 4 using namespace std;
 5 
 6 int n, a[100005], ans;
 7 
 8 int gcd(int a, int b){
 9     if(b == 0)
10         return a;
11     return gcd(b, a % b);
12 }
13 
14 int main(){
15     scanf("%d", &n);
16     for(int i = 1; i <= n; i++)
17         scanf("%d", &a[i]);
18     if(n == 1){
19         printf("YES\n0\n");
20         return 0;
21     }//特判 
22     int now = gcd(a[1], a[2]);
23     for(int i = 3; i <= n; i++)
24         now = gcd(a[i], now);//gcd 
25     if(now != 1){
26         printf("YES\n0\n");
27         return 0;
28     }//特判 
29     else{
30         a[n + 1] = 0;
31         for(int i = 1; i <= n; i++){
32             if(a[i] % 2 == 1 && a[i + 1] % 2 == 1){
33                 ans++;
34                 a[i + 1] = 0;//已经操作成偶数，所以赋值成任何一个偶数都可 
35             }
36             else if(a[i] % 2 == 1 && a[i + 1] % 2 == 0)//一奇一偶 
37                 ans += 2;
38         }
39         printf("YES\n%d\n", ans);
40     }
41     return 0;
42 }