POJ-2689 Prime Distance(区间素数筛--经典题)

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题意：

给出一个区间[L, R]，求该区间内的所有素数(1 <= L,R <= 2.1E9, R-L <= 1e6)

思路：

区间素数筛，枚举不超过sqrt(R)的所有质数p，然后再去枚举[L, R]区间内p的倍数，将之划去，最后区间内剩下的就都是素数了。注意一点，1需要特判不是素数。

代码：

#include <algorithm>#include <string.h>#include <cstdio>#define LL long longusing namespace std;const int N = 1e5;const int maxn = 1e6+5;int isprime[maxn], prime[maxn], len;LL L, R;int ans[maxn];void prime_init(){len = 0;memset(isprime, 0, sizeof isprime);for(int i = 2; i <= N; ++i)if(!isprime[i]){prime[++len] = i;if(1ll*i*i > N) continue;for(int j = i*i; j <= N; j+=i)isprime[j] = 1;}}void work(){memset(ans, 0, sizeof ans);for(int i = 1; i <= len; ++i){if(1ll*prime[i]*prime[i] > R) break;LL star = ((L-1)/prime[i]+1)*prime[i];for(LL j = star; j <= R; j+=prime[i])if(j != prime[i]) ans[j-L+1] = 1;}if(L == 1) ans[1] = 1;}int main(){prime_init();while(~scanf("%lld %lld", &L, &R)){work();LL i, mins, maxx, ans1, ans2, ans3, ans4;ans1 = ans2 = -1, i = 0;while(i < R-L+1 && (ans1 == -1 || ans2 == -1)){++i;if(ans[i]) continue;if(ans1 == -1) ans1 = i+L-1;else if(ans2 == -1) ans2 = i+L-1;}if(ans1 == -1 || ans2 == -1) puts("There are no adjacent primes.");else{mins = maxx = ans2-ans1;ans3 = ans1, ans4 = ans2;int pre = ans2;for(++i; i <= R-L+1; ++i){if(ans[i]) continue;if(i+L-1-pre < mins){mins = i+L-1-pre;ans1 = pre, ans2 = i+L-1;}if(i+L-1-pre > maxx){maxx = i+L-1-pre;ans3 = pre, ans4 = i+L-1;}pre = i+L-1;}printf("%lld,%lld are closest, %lld,%lld are most distant.\n", ans1, ans2, ans3, ans4);}}return 0;}

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