给你两个数字p,a。如果p是素数，并且ap 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：
1 int pow_mod(int a,int n,int m) 2 { 3 if(n==0) return 1; 4 int x = pow_mod(a, n / 2, m); 5 long long ans = (long long)x * x % m; 6 if(n%2) ans = ans * a % m; 7 return (int)ans; 8 }
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：
1 #include <iostream> 2 #include <algorithm> 3 #include <iomanip> 4 #include <cstdio> 5 #include <cstring> 6 #include <stack> 7 #include <functional> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 typedef long long ll; 12 ll pow_mod(ll a, ll n, ll m) 13 { 14 if (n == 0) 15 return 1; 16 ll x = pow_mod(a, n / 2, m); 17 ll ans = x * x % m; 18 if (n % 2) 19 ans = ans * a % m; 20 return ans; 21 } 22 const long long maxn = 1000000000 + 1; 23 int *vis = new int[maxn]; 24 void prime() 25 { 26 memset(vis, 0, sizeof(vis)); 27 int len = sqrt(maxn * 1.0); 28 for (int i = 2; i <= len; i++) 29 if (!vis[i]) 30 for (int j = i * 2; j <= maxn; j += i) 31 vis[j] = 1; 32 } 33 int main() 34 { 35 int p, a; 36 prime(); 37 while (cin >> p >> a) 38 { 39 if (!vis[p]) 40 cout << "no "; 41 else 42 { 43 if (pow_mod(a, p, p) == a) 44 cout << "yes "; 45 else 46 cout << "no "; 47 } 48 } 49 delete[] vis; 50 return 0; 51 }
AC代码：
1 #include <iostream> 2 #include <algorithm> 3 #include <iomanip> 4 #include <cstdio> 5 #include <cstring> 6 #include <stack> 7 #include <functional> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 typedef long long ll; 12 ll pow_mod(ll a, ll n, ll m) 13 { 14 if (n == 0) 15 return 1; 16 ll x = pow_mod(a, n / 2, m); 17 ll ans = x * x % m; 18 if (n % 2 == 1) 19 ans = ans * a % m; 20 return ans; 21 } 22 bool prime(ll a) 23 { 24 if (a == 1) 25 return 1; 26 if (a == 2) 27 return 1; 28 for (int i = 2; i * i <= a; i++) 29 if (a % i == 0) 30 return 0; 31 return 1; 32 } 33 int main() 34 { 35 ll a, p; 36 while (cin >> p >> a) 37 { 38 if (a == 0 && p == 0) 39 break; 40 else 41 { 42 if (prime(p)) 43 { 44 cout << "no "; 45 continue; 46 } 47 if (pow_mod(a, p, p) == a) 48 cout << "yes "; 49 else 50 cout << "no "; 51 } 52 } 53 }