POJ2689 Prime distance - 篩法
阿新 • • 發佈:2018-10-29
tin scan i++ lse 一道 const queue The define
這是一道非常典型的篩法,利用區間長度比較小,以及質數比較少,用少量的質數,只篩區間內部的合數,復雜度就不會很高
建議多開long long,很多時候你難以註意到哪裏會爆int
還有就是可以自己估摸著數量級提前把素數表打完,別每次都重打一遍素數表
#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; }
POJ2689 Prime distance - 篩法