判定1-2^31以内的数是否是素数，比一个一个判定要高很多。
普通筛法：
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cmath> 5 #include <cstring> 6 #include <map> 7 using namespace std; 8 9 const int maxn = 1e6; 10 int vis[maxn + 1] = { 1,1 }; 11 int prime[maxn + 1]; 12 int cnt = 0; 13 14 void create_prime() 15 { 16 int i, j; 17 int m = sqrt(maxn + 0.5); 18 for (i = 2; i <= m; i++) 19 if (!vis[i]) { 20 for (j = i*i; j <= maxn; j += i) 21 vis[j] = 1; 22 } 23 for (i = 2; i <= maxn; i++) 24 if (!vis[i]) prime[cnt++] = i; 25 } 26 27 int judge(int n) 28 { 29 if (n == 0 || n == 1) return 0; 30 for (int i = 0; prime[i] * prime[i] <=n&&i<cnt; i++) { 31 if (n%prime[i] == 0) return 0; 32 } 33 return 1; 34 } 35 36 int main() 37 { 38 int n; 39 create_prime(); 40 while (scanf("%d",&n)==1) 41 { 42 if (n == -1) break; 43 printf("%d ", judge(n)); 44 } 45 return 0; 46 }
更高效的筛：
1 #include <cstdio> 2 #include <iostream> 3 #include <cmath> 4 using namespace std; 5 6 #define N 50001 7 8 bool isprime[N]; 9 int prime[N], nprime; 10 11 void doprime() 12 { 13 long long i, j; 14 nprime = 0; 15 for (i = 0; i<N; i++) 16 isprime[i] = true; 17 isprime[1] = 0; 18 for (i = 2; i<N; i++) 19 { 20 if (isprime[i]) 21 { 22 prime[++nprime] = i; 23 for (j = i*i; j<N; j += i) 24 isprime[j] = false; 25 } 26 } 27 } 28 29 int isp(int n) 30 { 31 if (n == 0 || n == 1) return 0; 32 int i, k = (int)sqrt(double(n)); 33 for (i = 1; prime[i] <= k; i++) 34 if (n%prime[i] == 0) 35 return 0; 36 return 1; 37 } 38 39 int main() 40 { 41 freopen("E:\1.txt", "w", stdout); 42 doprime(); 43 int n; 44 while (scanf("%d",&n)==1) 45 { 46 if (n == -1) break; 47 printf("%d ", isp(n)); 48 } 49 return 0; 50 }
优化的
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 using namespace std; 6 const int maxn = 65537; 7 int vis[maxn + 1] = { 1,1 }; 8 int prime[maxn + 1]; 9 int cnt = 0; 10 11 void eratosthenes() 12 { 13 int i, j; 14 int m = sqrt(maxn + 0.5); 15 for (i = 2; i <= m; i++) { 16 if (!vis[i]) { 17 for (j = i*i; j <= maxn; j+=i) vis[j] = 1; 18 } 19 } 20 for (i = 2; i <= maxn; i++) 21 if (!vis[i]) prime[cnt++] = i; 22 } 23 24 int isp(int n) 25 { 26 if (n == 0 || n == 1) return 0; 27 int k = sqrt(n + 0.5); 28 for (int i = 0; prime[i]<=k; i++) 29 if (n%prime[i] == 0) return 0; 30 return 1; 31 } 32 33 int main() 34 { 35 int n; 36 eratosthenes(); 37 while (scanf("%d",&n)==1) { 38 if (n == -1) break; 39 printf("%d ", isp(n)); 40 } 41 return 0; 42 }