Prime Distance POJ - 2689 線性篩

一個數 $n$ 必有一個不超過 $\sqrt n$ 的質因子。

打表處理出 $1$ 到 $\sqrt n$ 的質因子后去篩掉屬于 $L$ 到 $R$ 區間的素數即可。??

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int Range=50000;
const int N=1000000+233;
int f[N],vis[Range+233],prime[Range];
int cnt;
void get_prime(){for(int i=2;i<=Range;++i){if(!vis[i])prime[++cnt]=i;for(int j=1;j<=cnt&&prime[j]*i<=Range;++j){vis[prime[j]*i]=1;if(i%prime[j]==0)break;}}
}
int main(){get_prime();                      int L,U;while(scanf("%d%d",&L,&U)!=EOF){memset(f,0,sizeof(f));if(L==1)L=2;for(int i=1;i<=cnt;++i){     int a=L%prime[i]==0?L/prime[i]:L/prime[i]+1;     int b=U/prime[i];for(int j=a;j<=b;++j)if(j>1)f[j*prime[i]-L]=1;}int p=-1,x1,x2,maxans=-1,minans=N,y1,y2;for(int i=0;i<=U-L;++i)if(f[i]==0){if(p==-1){p=i;continue;};if(maxans<i-p){maxans=i-p,x1=p+L,x2=i+L;}if(minans>i-p){minans=i-p,y1=p+L,y2=i+L;}p=i;}if(maxans==-1)cout<<"There are no adjacent primes."<<endl;else cout<<y1<<","<<y2<<" are closest, "<<x1<<","<<x2<<" are most distant."<<endl;
}return 0;
}