codeforces798C - Mike and gcd problem (数论+思维)

原题链接：http://codeforces.com/contest/798/problem/C

题意：有一个数列A，gcd(a₁,a₂,a₃...,a_n)>1 时称这个数列是“漂亮”的。存在这样的操作，使a_i,a_i+1变为(a_i-a_i+1), (a_i+a_i+1)。问最少进行这样的操作使数列是“漂亮”的。

思路：考虑gcd(a₁,a₂,a₃...,a_n)>1 的情况：

我们对a_i,a_i+1进行两次操作可以得到2a_i,2a_i+1，也就是说一对相邻的数字最多操作2次使它们的gcd=2>1。而对于一对奇数来说，操作一次就能使它们成为偶数。

现在就是要把数列中所有的数变为偶数。先对相邻奇数进行操作（每次+1），再对单个奇数进行操作（每次+2）即可。

AC代码：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 const int MAXN=2e6+5;
 6 long long num[MAXN];
 7 long long gcd(long long a, long long b){
 8     return (b==0)?a:gcd(b, a%b);
 9 }
10 int main()
11 {
12     int n;
13     while(cin>>n){
14         memset(num, 0, sizeof(num));
15         for(int i=0;i<n;i++){
16             cin>>num[i];
17         }
18         long long g=gcd(num[0], num[1]);
19         for(int i=2;i<n;i++){
20             g=gcd(g, num[i]);
21             if(g==1)
22                 break;
23         }
24         if(g>1){
25             cout<<"YES"<<endl;
26             cout<<0<<endl;
27             continue;
28         }
29         int res=0;
30         for(int i=1;i<n;i++){
31             if(num[i-1]%2&&num[i]%2){
32                 num[i-1]++;
33                 num[i]++;
34                 res++;
35             }
36         }
37         for(int i=1;i<n;i++){
38             if(num[i-1]%2&&num[i]%2==0||(num[i]%2&&num[i-1]%2==0))    {
39                 num[i-1]=0;
40                 num[i]=0;
41                 res+=2;
42             }
43         }
44         cout<<"YES"<<endl;
45         cout<<res<<endl;
46     }
47
48     return 0;
49 }

转载于:https://www.cnblogs.com/MasterSpark/p/7436987.html

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