poj 2689 Prime Distance （素数二次筛法）
没怎么研究过数论，还是今天才知道有素数二次筛法这样的东西。
题意是，要求求出给定区间内相邻两个素数的最大和最小差。
二次筛法的意思其实就是先将1~sqrt(b)内的素数先筛出来，然后再对[a,b]区间用那些筛出来的素数再次线性筛。
代码如下：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 6 using namespace std; 7 8 const int N = 66666; 9 int prm[N >> 3], pn; 10 bool np[N]; 11 12 void getbase() { 13 np[pn = 0] = np[1] = true; 14 prm[pn++] = 2; 15 for (int i = 3; i < N; i++, i++) { 16 if (!np[i]) prm[pn++] = i; 17 for (int j = 0, tmp; j < pn; j++) { 18 tmp = i * prm[j]; 19 if (tmp >= N) break; 20 np[tmp] = true; 21 if (i % prm[j] == 0) break; 22 } 23 } 24 // cout << pn << endl; 25 // for (int i = 0; i < 10; i++) cout << prm[i] << endl; 26 } 27 28 bool mk[1111111]; 29 30 int main() { 31 // freopen("in", "r", stdin); 32 getbase(); 33 int T; 34 long long a, b; 35 while (~scanf("%lld%lld", &a, &b)) { 36 a = max(2ll, a); 37 memset(mk, 0, sizeof(mk)); 38 for (int i = 0; i < pn && prm[i] <= b; i++) { 39 long long t = a / prm[i] * prm[i]; 40 if (t != a) t += prm[i]; 41 t = max(t, (long long) prm[i] << 1); 42 for ( ; t <= b; t += prm[i]) mk[t - a] = true; 43 } 44 long long p = a; 45 int mini = 1111111, maxi = -1111111, minp, maxp; 46 while (p <= b && mk[p - a]) p++; 47 for (long long i = p + 1; i <= b; i++) { 48 if (mk[i - a]) continue; 49 // cout << i << endl; 50 if (maxi < i - p) maxi = i - p, maxp = p; 51 if (mini > i - p) mini = i - p, minp = p; 52 p = i; 53 } 54 if (maxi < 0) puts("There are no adjacent primes."); 55 else printf("%d,%d are closest, %d,%d are most distant.\n", minp, minp + mini, maxp, maxp + maxi); 56 } 57 return 0; 58 }
——written by Lyon
