POJ2689 Prime distance - 筛法

Posted 2021-01-16 zolrk

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.
");
        else 
            printf("%lld,%lld are closest, %lld,%lld are most distant.
", l2 + l, r2 + l, l1 + l, r1 + l);
    }
    return 0;
}