POJ 2689 区间素数筛选

题意：给区间[L,U]，要求该区间内，距离最小的素数对，和距离最大的素数对，输出素数对和对应距离。（距离的定义：2个相邻素数的差的绝对值）

分析：

1、筛选法

2、由于区间L,U (1<=L< U<=2,147,483,647)本身可能很大，直接筛选数组都开不下。

　  但是题目说“The difference between L and U will not exceed 1,000,000.”也就是说U-L<=1000 000,这就给我们启示只筛选区间内部的素数。

　　然后由于区间是一个连续区间，所以我们用x(L=<x<=U)%1000000就可以把x映射到1000 000内的数了。这样开一个1000 000的数组存素数就好（当然，区间内的素数肯定没这么多）

3、注意用long long因为筛选过程可能超int（嘛……如果用int的话，本地都应该跑不出来才对吧……前提是自己随便输一个大一点的数……我也是程序死循环了才改的……T T）

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<map>
#include<string>
using namespace std;
const int MAXN = 50001;
const int MOD = 1000001;
int prime[MAXN];
int isprime[MAXN*20];
int prim[MAXN*20];
int init(){
  int k = 0;
  memset(prime,0,sizeof prime);
  for(int i = 2;i<MAXN;i++){
    //cout<<"
"<<i<<endl;
    if(!prime[i]){
      prime[k++] = i;
      long long j = (long long )i*(long long )i;
      //cout<<j<<endl;
      for(;j<MAXN;j+=i)prime[j] = 1;
    }
  }
  return k;
}
int shaixuan(long long L,long long U,int pn){
  memset(isprime,0,sizeof isprime);
  //cout<<prime[0]<<"
U:"<<U<<endl;
  for(long long i = 0;prime[i]<U+1 && i<pn;i++){
    //cout<<"???
";
    long long tp = L/prime[i];
    if(tp < 2) tp = (long long )prime[i]*(long long )prime[i];
    else{
      tp = tp*(long long )prime[i];
      if(tp < L)tp+=(long long )prime[i];
    }
    //cout<<prime[i]<<" "<<tp<<"
";
    //if(!isprime[tp%MOD]){
      //cout<<tp%MOD<<endl;
      for(long long j = tp;j<U+1;j+=(long long )prime[i]){
        isprime[j%MOD] = 1;
      }
    //}
  }
  int k = 0;
  for(long long i = L;i<U+1;i++){
    if(i == 1)continue;
    if(!isprime[i%MOD]){
      prim[k++] = i;
      //cout<<i<<endl;
    }
  }
  return k;
}
int main()
{
#ifdef LOCALL
//    freopen("in.txt","r",stdin);
  //  freopen("out.txt","w",stdout);
#endif
  int pn = init();
  int L,U;
  while(cin>>L>>U){
    if(U == 1 || U == 2){cout<<"There are no adjacent primes."<<endl;continue;}
    int pn2 = shaixuan(L,U,pn);
    if(pn2 == 1){cout<<"There are no adjacent primes."<<endl;continue;}
    long long minn ,maxn = 0,minx,maxx;
    minx = maxx = 0;
    minn = 1<<30;
    //cout<<pn2<<endl;
    //cout<<minn<<endl;
    for(int i = 1;i<pn2;i++){
     // cout<<"prim["<<i<<"]:"<<prim[i]<<endl;
      //cout<<"sub:"<<prim[i]-prim[i-1]<<endl;
      if(prim[i]-prim[i-1] < minn) {minn = prim[i]-prim[i-1]; minx = i-1;}
      if(prim[i]-prim[i-1] > maxn) {maxn = prim[i]-prim[i-1]; maxx = i-1;}
    }

    //cout<<"min:"<<minn<<" maxx:"<<maxn<<"maxx:"<<maxx<<endl;
    cout<<prim[minx]<<","<<prim[minx+1]<<" are closest, "<<prim[maxx]<<","<<prim[maxx+1]<<" are most distant.
";

  }
  return 0;
}

总之，筛选法简直打素数表之万能神器……