简简单单AC代码

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
const int AX = 5e4;
const int maxn = 1e6+666;
const int INF = 1e9;
int vis[AX];
int f[maxn];
int prime[AX];
int ans;
void prime1(){                        //筛选出50000以内的素数
	ans = 0;
	memset(vis,0,sizeof(vis));
	int q = (int)sqrt(AX+0.5);
	for(int i=2;i<=q;i++){
		if(!vis[i]){
			for(int j= i*i;j<AX;j+=i){
				vis[j] = 1;
			}
		}
	}
	for(int i=2;i<AX;i++){
		if(!vis[i]){
			prime[ans++] = i;
		}
	}
}
 
void prime_2(int l,int u){           //求l，u合数
	int a,b;
	for(int i=0;i<ans;i++){
		a = (l-1)/prime[i]+1;
		b = u/prime[i];
		for(int j=a;j<=b;j++){
			if(j>1)  f[j*prime[i]-l] = 1;        //偏移减去l，后面再加回来。
		}
	}
}
 
int main(){
	long long l,u;
	prime1();
	while(cin>>l>>u){
		memset(f,0,sizeof(f));
		if(l == 1) l = 2;
		prime_2(l,u);
		int mi = INF;
		int ma = -1;
		int p = -1;
		int x1,x2,y1,y2;
		for(int i=0;i<=u-l;i++){          //枚举50000以内素数
			if(!f[i]){
				if(p == -1) {p=i;continue;}
				if(mi > i-p) mi = i-p,x1=p+l,y1=i+l;     //偏移量加回来
				if(ma < i-p) ma = i-p,x2=p+l,y2=i+l;
				p = i;	
			}
		}
		if(ma == -1) printf("There are no adjacent primes.\n");
		else cout<<x1<<","<<y1<<" are closest, "<<x2<<","<<y2<<" are most distant."<<endl;
	}
	return 0;
}