poj 2689 巧妙地运用素数筛选

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long LL;
const int MAXN = 50000;
int primes[MAXN];
bool vst[MAXN];
int notPrimes[1000010];
int pos[1000010];
int top,pcnt;

void init(){
    top = 0;
    memset(vst,0,sizeof(vst));
    vst[0] = vst[1] = 1;
    for(int i = 2;i < MAXN;++i)if(!vst[i]){
        primes[top++] = i;
        for(int j = i + i;j < MAXN;j += i) vst[j] = 1;
    }
    //printf("top: %d\n",top);
}

void solve(int L,int R){
    memset(notPrimes,0,sizeof(notPrimes));

    if(L == 1) L = 2;              /// 防止筛掉全部该区间的素数本身!!!!!
    for(int i = 0;i < top&&(LL)primes[i]*primes[i] <= R;++i){  //筛选因子
        int s = L / primes[i] + (L % primes[i] > 0); //当前素数的最小倍数达到L
       
        s = (s == 1 ? 2 : s);    /// 防止筛掉全部该区间的素数本身!!!!!
       
        for(int j = s;(LL)j*primes[i] <= R;++j){
            if((LL)j*primes[i] >= L)   //合数
               notPrimes[j*primes[i] - L] = 1;   // 相当与离散化
        }
    }

    pcnt = 0;
    for(int i = 0;i <= R - L;++i){
        if(!notPrimes[i]){
            pos[pcnt++] = i + L;
            //printf("i -- > %d\n",i + L);
        }
    }
    
    
    if(pcnt < 2){
        puts("There are no adjacent primes.");
    } else {
        int minl,minr,maxl,maxr,minv = 999999,maxv = -1;
        for(int i = 1;i < pcnt;++i){
            if(pos[i] - pos[i-1] > maxv){
                maxv = pos[i] - pos[i-1];
                maxl = pos[i-1];
                maxr = pos[i];
            }
            if(pos[i] - pos[i-1] < minv){
                minv = pos[i] - pos[i-1];
                minl = pos[i-1];
                minr = pos[i];
            }
        }
        printf("%d,%d are closest, %d,%d are most distant.\n",minl,minr,maxl,maxr);
    }
}
int main()
{
    init();
    int L,R;
    while(~scanf("%d%d",&L,&R)){
        solve(L,R);
    }
    return 0;
}