素数区间筛

素数区间筛

POj 2689
题意：给出区间[L,R]求出区间相邻最近的素数对，和相邻最远的素数对，没有输出没有，1<=L，R<= (2 ^ 31 - 1), 且保证 R - L <= 1e6.
思路：区间筛法，预处理出2 ~ sqrt®的所有质数p，再把(L,R)中能被p整除的数标记，即标记 j * p( [L / p ] <= j <= [ R / P] 为合数，注意(j >= 2, j == 1 就是素数p本身不要误筛)。

code：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const int N = 50000; // sqrt(2^31) 46340
bool isprime[maxn]; //2 ~ sqrt(R)
bool isprime_big[maxn];
//pre
void init(){
    for(int i = 2; i < N; ++i) isprime[i] = 1;
    for(int i = 2; i * i < N; ++i){
        if(isprime[i]){
            for(int j = 2 * i; j < N; j += i){
                isprime[j] = 0;
            }
        }
    }
}

void solve(ll L,ll R){
    int len = R - L + 1;
    for(int i = 0; i < len; ++i) isprime_big[i] = 1;
    if(L == 1) isprime_big[0] = 0;
    for(ll i = 2; i * i <= R; ++i){
        if(isprime[i]){
            for(ll j = max(2LL,(L - 1 + i) / i) * i; j <= R; j += i){
                isprime_big[j - L] = 0;
            }
        }
    }
}

int main(){
    init();
    ll L,R;
    while(~scanf("%lld%lld",&L,&R)){
        solve(L,R);
        int len = R - L + 1;
        int lmin,rmin,lmax,rmax,cur = -1;
        int maxx = -1, minn =  (1LL << 31) - 1;
        for(int i = 0; i < len; ++i){
            if(isprime_big[i]){
                if(cur >= 0){
                    if(maxx < i - cur){
                        maxx = i - cur, lmax = cur + L, rmax = i + L;
                    }
                    if(minn > i - cur){
                        minn = i - cur, lmin = cur + L, rmin = i + L;
                    }
                    cur = i;
                }
                else{
                    cur = i;
                }
            }
        }
        if(maxx >= 0){
            printf("%d,%d are closest, %d,%d are most distant.\n",lmin,rmin,lmax,rmax);
        }
        else{
            puts("There are no adjacent primes.");
        }
    }
    return 0;
}