POJ 2689 - Prime Distance 任意区间内筛素数

来源：互联网发布：cgwang插画网络编辑：程序博客网时间：2024/05/01 04:42

和1~n范围内筛素数差不多...只不过对于每个素数加了一个偏移量了..比如在5 10这个区间..筛掉2的倍数..那么从6开始剔除...由于题目所给的数据..最大的合数..其质因子不会大于10^5..所以先用传统的筛素数法把1~10^5内的素数找出..然后再来筛[L,U]这个区间的素数...

Program:

#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<queue>#include<stack>#include<set>#include<algorithm>#define ll long long #define oo 1000000007#define pi acos(-1.0)#define MAXN 1000005using namespace std;ll Prime[MAXN],ans[MAXN];int Pnum,anum;bool P[MAXN];void PreWork(){       int i,x;       memset(P,true,sizeof(P));        for (Pnum=0,i=2;i<=100000;i++)          if (P[i])            for (Prime[++Pnum]=i,x=i*2;x<=100000;x+=i)               P[x]=false; }int main(){         PreWork();        for (ll L,U;~scanf("%I64d%I64d",&L,&U);)       {                ll n=U-L;                memset(P,true,sizeof(P));                for (int i=1;i<=Pnum;i++)                {                       ll h=Prime[i],x;                       if (h*h>U) break;                       x=(h-(L%h))%h;                       if (L+x==h) x+=h;                       for (;x<=n;x+=h) P[x]=false;                }                if (L==1) P[0]=false;                anum=0;                for (int i=0;i<=n;i++)                   if (P[i]) ans[++anum]=i+L;                if (anum<2) printf("There are no adjacent primes.\n");                    else                 {                      int i,a,b;                      a=b=1;                      for (i=1;i<anum;i++)                      {                           if (ans[i+1]-ans[i]<ans[a+1]-ans[a]) a=i;                           if (ans[i+1]-ans[i]>ans[b+1]-ans[b]) b=i;                      }                      printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n",ans[a],ans[a+1],ans[b],ans[b+1]);                }       }       return 0;}