POJ2689 Prime distance - 筛法

这是一道非常典型的筛法，利用区间长度比较小，以及质数比较少，用少量的质数，只筛区间内部的合数，复杂度就不会很高
建议多开long long，很多时候你难以注意到哪里会爆int
还有就是可以自己估摸着数量级提前把素数表打完，别每次都重打一遍素数表

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 1000000 + 10;
typedef long long ll;
ll l,r,max_ans,min_ans,l1,r1,l2,r2,tot,n,cnt;
ll prime[MAXN],vis[MAXN];
void primes(int n) {
	memset(vis, 0, sizeof(vis));
	memset(prime, 0, sizeof(prime));
	tot = 0;
	for(int i=2; i<=n; i++) {
		if(!vis[i]) prime[++tot] = i;
		for(int j=1; j <= tot && prime[j]*i <= n; j++) {
			vis[i*prime[j]] = 1;
			if(i % prime[j] == 0) break;
		}
	}
}

int main() {
	primes(1e5);
	while(scanf("%lld%lld", &l, &r) != EOF) {
		if(l == 1) l = 2;
		memset(vis, 0, sizeof(vis));
		max_ans = 0, min_ans = 1 << 30;
		for(int i=1; i<=tot; i++) {
			int now = prime[i];
			for(int j = l/now; j <= r/now; j++) {
				ll val = j * now;
				if(val < l || j <= 1) continue;
				int pos = val - l;
				vis[pos] = 1;
			}
		}
		cnt = 0;
		ll last = -1, max_ans = 0, min_ans = 1<<30;
		for(int i=0; i<=r-l; i++) {
			if(!vis[i]) {
				if(last != -1) {
					if(max_ans < i - last) {
						max_ans = i - last;
						l1 = last, r1 = i;
					}
					if(min_ans > i - last) {
						min_ans = i - last;
						l2 = last, r2 = i;
					}
				}
				last = i;
			}
		}
		if(!max_ans) 
			printf("There are no adjacent primes.\n");
		else 
			printf("%lld,%lld are closest, %lld,%lld are most distant.\n", l2 + l, r2 + l, l1 + l, r1 + l);
	}
	return 0;
}