AcWing 196.质数距离

题意

https://www.acwing.com/problem/content/198/

思路

题目旨在问对于一个较大的L、R区段如何对区段内数字判断isprime？

假设A是合数 则A必有一个因子< sqrt(A)

所以我们可以对sqrt(R)以内的数据范围线性筛出所有素数prime[]

对于现有这PNum个素数 将他们在[L,R]内的倍数(>=2倍)全部标记为合数（埃筛） 数据范围过大可以用相对位置标记

代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
const int MaxN = 1e6 + 5;

int L,R;
int prime[MaxN],pNum = 0;
bool p[MaxN] = {0};

int ansP[MaxN],cnt = 0;
bool ok[MaxN] = {0};

void Find_Prime(int N){
	for(int i = 2;i <= N; i++){
		if(p[i] == false) prime[++pNum] = i;
		for(int j = 1;j <= pNum && i * prime[j] < N;j++){
			p[i * prime[j]] = true;
			if(i % prime[j] == 0) break;
		}
	}
}

int main(){
	Find_Prime(5e4);//sqrt(2^31)
	while(~scanf("%d %d",&L,&R)){
		cnt = 0;
		memset(ansP,0,sizeof ansP);
		memset(ok,0,sizeof ok);
		//先处理出每个p的p0
		for(int i = 1;i <= pNum; i++){
			long long P = prime[i];
			long long P0 = max(2 * P,(L + P - 1) / P * P);
			for(long long j = P0;j <= R; j += P) ok[j - L] = 1;
		}
		for(int i = 0;i <= R - L; i++){
			if(!ok[i] && i + L >= 2) ansP[++cnt] = i + L;
		}
		if(cnt < 2) printf("There are no adjacent primes.\n");
		else{
			int Min = 1e6;
			int Max = 0;
			int Minx = 0,Maxx = 0;
			for(int i = 1;i < cnt; i++){
				int cur = ansP[i + 1] - ansP[i];
				if(cur > Max){
					Max = cur;
					Maxx = i;
				}
				if(cur < Min){
					Min = cur;
					Minx = i;
				}
			}
			printf("%d,%d are closest, %d,%d are most distant.\n",ansP[Minx],ansP[Minx + 1],ansP[Maxx],ansP[Maxx + 1]);
		}
	}
	return 0;
}