埃氏筛法
#define exp 2.7182818285
#define Y 0.57721566490153286060651209
#define ln2 0.69314718055995
#define ln3 1.0986122886681
#define lg2 0.301029995663981
#define lg5 0.698970004336018
#define ln5 1.6094379124341
int prime[N];
void pre()
{
memset(prime,1,sizeof(prime));prime[1]=0;
for(int i=2;i { if(prime[i]) for(int j=2*i;j } } 线性筛 int p[MAX],cnt; bool isprime[MAX]; void prime() { cnt=1; memset(isprime,1,sizeof(isprime));//初始化认为所有数都为素数 isprime[0]=isprime[1]=0;//0和1不是素数 for(int i=2;i { if(isprime[i]) p[cnt++]=i;//保存素数i for(int j=1;j { isprime[p[j]*i]=0; If(i%isprimep[j]=0)break; //筛掉小于等于i的素数和i的积构成的合数 } } } Meisell-Lehmer //求n以内的素数个数 #include using namespace std; typedef long long LL; const int N = 5*1e6 + 2; bool np[N]; int prime[N], pi[N]; int getprime() { int cnt = 0; np[0] = np[1] = true; pi[0] = pi[1] = 0; for(int i = 2; i < N; ++i) { if(!np[i]) prime[++cnt] = i; pi[i] = cnt; for(int j = 1; j <= cnt && i * prime[j] < N; ++j) { np[i * prime[j]] = true; if(i % prime[j] == 0) break; } } return cnt; } const int M = 7; const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17; int phi[PM + 1][M + 1], sz[M + 1]; void init() { getprime(); sz[0] = 1; for(int i = 0; i <= PM; ++i) phi[i][0] = i; for(int i = 1; i <= M; ++i) { sz[i] = prime[i] * sz[i - 1]; for(int j = 1; j <= PM; ++j) { phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1]; } } } int sqrt2(LL x) { LL r = (LL)sqrt(x - 0.1); while(r * r <= x) ++r; return int(r - 1); } int sqrt3(LL x) { LL r = (LL)cbrt(x - 0.1); while(r * r * r <= x) ++r; return int(r - 1); } LL getphi(LL x, int s) { if(s == 0) return x; if(s <= M) return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s]; if(x <= prime[s]*prime[s]) return pi[x] - s + 1; if(x <= prime[s]*prime[s]*prime[s] && x < N) { int s2x = pi[sqrt2(x)]; LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2; for(int i = s + 1; i <= s2x; ++i) { ans += pi[x / prime[i]]; } return ans; } return getphi(x, s - 1) - getphi(x / prime[s], s - 1); } LL getpi(LL x) { if(x < N) return pi[x]; LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1; for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) { ans -= getpi(x / prime[i]) - i + 1; } return ans; } LL lehmer_pi(LL x) { if(x < N) return pi[x]; int a = (int)lehmer_pi(sqrt2(sqrt2(x))); int b = (int)lehmer_pi(sqrt2(x)); int c = (int)lehmer_pi(sqrt3(x)); LL sum = getphi(x, a) + LL(b + a - 2) * (b - a + 1) / 2; for (int i = a + 1; i <= b; i++) { LL w = x / prime[i]; sum -= lehmer_pi(w); if (i > c) continue; LL lim = lehmer_pi(sqrt2(w)); for (int j = i; j <= lim; j++) { sum -= lehmer_pi(w / prime[j]) - (j - 1); } } return sum; } int main() { init(); LL n; while(cin >> n) { cout << lehmer_pi(n) << endl; } } Min_25 prime #include #include #include #include #define M 5333333 using namespace std; typedef long long ll; ll A[M], B[M]; double inv[M]; int main() { ll n; scanf("%lld", &n); int lim = sqrt(n) + 1; if(n <= 10) return printf("%dn", (n >= 2) + (n >= 3) + (n >= 5) + (n >= 7)), 0; for(int i = 1; i <= lim; i++) inv[i] = 1.0/ (i * 2 - 1), A[i] = (n / (i * 2 - 1) + 1) / 2, B[i] = (i + 1) / 2 - (i == 1); for(int i = 3; i <= lim; i += 2)if(B[i] != B[i - 1]) { unsigned l = B[i - 1]; for(int j = 1; i * (j * 2 - 1) <= lim; j++) A[j] -= A[i * j - (i - 1) / 2] - l; double tmp = n / i; for(int j = (lim / i + 1) / 2 + 1, s = min(n / i / i, (ll) lim); j * 2 - 1 <= s; j++) A[j] -= B[(int) (tmp * inv[j] + 1e-7)] - l; for(int j = lim / i; j >= i; j--) for(int k = i - 1; k >= 0; k--) B[j * i + k] -= B[j] - l; } printf("%lldn", A[1]); return 0; } Miller-rabin //判断n是否为素数 #include #include #include #include #include #include using namespace std; const int times = 20; int number = 0; //map long long Random( long long n ) //生成[ 0 , n ]的随机数 { return ((double)rand( ) / RAND_MAX*n + 0.5); } long long q_mul( long long a, long long b, long long mod ) //快速计算 (a*b) % mod { long long ans = 0; while(b) { if(b & 1) { b--; ans =(ans+ a)%mod; } b /= 2; a = (a + a) % mod; } return ans; } long long q_pow( long long a, long long b, long long mod ) //快速计算 (a^b) % mod { long long ans = 1; while(b) { if(b & 1) { ans = q_mul( ans, a, mod ); } b /= 2; a = q_mul( a, a, mod ); } return ans; } bool witness( long long a, long long n )//miller_rabin算法的精华 {//用检验算子a来检验n是不是素数 long long tem = n - 1; int j = 0; while(tem % 2 == 0) { tem /= 2; j++; } //将n-1拆分为a^r * s long long x = q_pow( a, tem, n ); //得到a^r mod n if(x == 1 || x == n - 1) return true; //余数为1则为素数 while(j--) //否则试验条件2看是否有满足的 j { x = q_mul( x, x, n ); if(x == n - 1) return true; } return false; } bool miller_rabin( long long n ) //检验n是否是素数 { if(n == 2) return true; if(n < 2 || n % 2 == 0) return false; //如果是2则是素数，如果<2或者是>2的偶数则不是素数 for(int i = 1; i <= times; i++) //做times次随机检验 { long long a = Random( n - 2 ) + 1; //得到随机检验算子 a if(!witness( a, n )) //用a检验n是否是素数 return false; } return true; } main( ) { int t;cin>>t; while(t--) { long long n;scanf("%lld",&n); if(miller_rabin(n)) { puts("Prime");continue; } else puts("NO"); } } ©著作权归作者所有：来自51CTO博客作者qinXpeng的原创作品，如需转载，请注明出处，否则将追究法律责任
