大区间素数筛选（POJ 2689）

/*
 *POJ 2689 Prime Distance
 *给出一个区间[L,U],找出区间内容、相邻的距离最近的两个素数和距离最远的两个素数
 *1<=L<U<=2147483647 区间长度不超过1000000、就是要筛选出[L,U]之间的素数 
*/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <stack>
#include <algorithm>
using namespace std;

const int MAXN = 100010;
int prime[MAXN+1];
void getPrime() {
	memset(prime, 0, sizeof(prime));
	for (int i = 2; i<=MAXN; i++) {
		if (!prime[i])
			prime[++prime[0]] = i;
		for (int j = 1; j<=prime[0] && prime[j] <= MAXN/i; j++) {
			prime[prime[j]*i] = 1;
			if (i%prime[j] == 0)
				break;
		}
	}
}

bool notprime[1000010];
int prime2[1000010];

void getPrime2(int L, int R) {
	memset(notprime, false, sizeof(notprime));
	if (L < 2)
		L = 2;
	for (int i = 1; i<=prime[0]&& (long long)prime[i]*prime[i] <= R; i++) {
		int s = L /prime[i] + (L%prime[i] > 0);
		if (s == 1)
			s = 2;
		for (int j = s; (long long)j*prime[i] <= R; j++) {
			if ((long long)j*prime[i] >= L)
				notprime[j*prime[i]-L] = true;
		}
	}
	prime2[0] = 0;
	for (int i = 0; i<=R-L; i++) {
		if (!notprime[i])
			prime2[++prime2[0]] = i+L;
	}
}

int main() {
	getPrime();
	int L, U;
	while (scanf("%d%d",&L,&U) == 2) {
		getPrime2(L, U);
		if (prime2[0] < 2)
			printf("There are no adjacent primes.
");
		else {
			int x1 = 0, x2 = 100000000, y1 = 0, y2 = 0;
			for (int i = 1; i<prime2[0]; i++) {
				if (prime2[i+1]-prime2[i] < x2-x1) {
					x1 = prime2[i];
					x2 = prime2[i+1];
				}
				if (prime2[i+1] - prime2[i] > y2-y1) {
					y1 = prime2[i];
					y2 = prime2[i+1];
				}
			}
			printf("%d,%d are closest, %d,%d are most distant.
",x1,x2,y1,y2);
		}
	}
	
	return 0;
}