POJ2689——区间筛法——Prime Distance

http://poj.org/problem?id=2689

/*
区间筛法适用于L, R范围比较大,但是区间长度较小时
套个模板然后取出最小最大就可以
*/
/************************************************
* Author        :Powatr
* Created Time  :2015-8-17 9:38:50
* File Name     :POJ2689.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e6 + 1e4;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 50010;

bool prime[MAXN];
bool sieve[N];
int len;
void segent_sieve(ll L, ll R)
{
    len = R - L + 1;
    for(int i = 0; i < len; i++) prime[i] = true;
    if(1 - L >= 0) prime[1-L] = false;//特判
    for(ll i = 2; i*i < R; i++){
        if(seive[i]){
            for(ll j = max(1ll*2, (L-1+i)/i)*i; j <= R; j+=i)//(L-1+i/i)得到最接近L的i的倍数
                prime[j-L] = false;
            //偏移量,因为单纯j会导致数组越界,把范围弄到0到len上
        }
    }
}
int main(){
    //艾氏筛法 O(nlognlogn)
    for(int i = 2; i < N; i++) sieve[i] = true;
    for(int i = 2; i*i < N; i++){
        if(sieve[i]){
            for(int j = i*2; j < N; j+=i){
                sieve[j] = false;
            }
        }
    }
    int L, R;
    while(~scanf("%d%d", &L, &R)){
        segent_seive(L, R);
    int mmax, mmin;
    int lmax, rmax, lmin, rmin;
    mmax = -1, mmin = 1 << 30;
    int t = -1;
    for(int i = 0; i < len; i++){
        if(prime[i]){
            if(t >= 0){
                if(mmax < i - t) mmax = i - t, lmax = L + t, rmax = L + i;
                if(mmin > i - t) mmin = i - t, lmin = L + t, rmin = L + i;
                t = i;
            }
            else t = i;
        }
    }
    if(mmax != -1 || mmin != 1 << 30) printf("%d,%d are closest, %d,%d are most distant.
", lmin, rmin, lmax, rmax);
    else printf("There are no adjacent primes.
");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/zero-begin/p/4735915.html