POJ3641 Pseudoprime numbers (幂取模板子)

给你两个数字p,a。如果p是素数，并且a^p mod p = a，输出“yes”，否则输出“no”。

很简单的板子题。核心算法是幂取模（算法详见《算法竞赛入门经典》315页）。

幂取模板子：

 int pow_mod(int a,int n,int m)
 {
     if(n==) return ;
     int x = pow_mod(a, n / , m);
     long long ans = (long long)x * x % m;
     if(n%) ans = ans * a % m;
     return (int)ans;
 }

题目代码也比较简单，有一个坑点是如果用筛素数打表，数组开不了这么大。

报错：error: total size of array must not exceed 0x7fffffff bytes

报错代码：

 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 #include <cstdio>
 #include <cstring>
 #include <stack>
 #include <functional>
 #include <queue>
 #include <cmath>
 using namespace std;
 typedef long long ll;
 ll pow_mod(ll a, ll n, ll m)
 {
     if (n == )
         return ;
     ll x = pow_mod(a, n / , m);
     ll ans = x * x % m;
     if (n % )
         ans = ans * a % m;
     return ans;
 }
 const long long maxn =  + ;
 int *vis = new int[maxn];
 void prime()
 {
     memset(vis, , sizeof(vis));
     int len = sqrt(maxn * 1.0);
     for (int i = ; i <= len; i++)
         if (!vis[i])
             for (int j = i * ; j <= maxn; j += i)
                 vis[j] = ;
 }
 int main()
 {
     int p, a;
     prime();
     while (cin >> p >> a)
     {
         if (!vis[p])
             cout << "no\n";
         else
         {
             if (pow_mod(a, p, p) == a)
                 cout << "yes\n";
             else
                 cout << "no\n";
         }
     }
     delete[] vis;
     return ;
 }

AC代码：

 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 #include <cstdio>
 #include <cstring>
 #include <stack>
 #include <functional>
 #include <queue>
 #include <cmath>
 using namespace std;
 typedef long long ll;
 ll pow_mod(ll a, ll n, ll m)
 {
     if (n == )
         return ;
     ll x = pow_mod(a, n / , m);
     ll ans = x * x % m;
     if (n %  == )
         ans = ans * a % m;
     return ans;
 }
 bool prime(ll a)
 {
     if (a == )
         return ;
     if (a == )
         return ;
     for (int i = ; i * i <= a; i++)
         if (a % i == )
             return ;
     return ;
 }
 int main()
 {
     ll a, p;
     while (cin >> p >> a)
     {
         if (a ==  && p == )
             break;
         else
         {
             if (prime(p))
             {
                 cout << "no\n";
                 continue;
             }
             if (pow_mod(a, p, p) == a)
                 cout << "yes\n";
             else
                 cout << "no\n";
         }
     }
 }