C++14(g++5.4) 解法, 执行用时: 17ms, 内存消耗: 4556K, 提交时间: 2019-08-21 10:09:19

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int vis[100005],prime[100005];
int he[1000005],tot=0;
void getprime(int n){
	for(int i=2;i<=n;++i){
		if(!vis[i])
			prime[++tot]=i;
		for(int j=1;j<=tot&&prime[j]*i<=n;++j){
			vis[prime[j]*i]=1;
			if(i%prime[j]==0)
				break;
		}
	}
}
int main(){
	ll l,r;
	getprime(50000);
	while(~scanf("%lld%lld",&l,&r)){
		l=max(l,(ll)2);
		memset(he,0,sizeof(he));
		for(int i=1;i<=tot;++i){
			if(prime[i]*prime[i]>r)
				break;
			int a=l/prime[i];
			a=max(a,2);
			int b=r/prime[i];
			for(int j=a;j<=b;++j)
				if(j*prime[i]-l>=0)
					he[j*prime[i]-l]=1;
		}
		ll mn=1e18,mx=-1e18,front=-1e18;
		int ans1=-1,ans2,ans3,ans4;
		for(ll i=l;i<=r;++i){
			if(he[i-l])
				continue;
			if(front==-1e18)
				front=i;
			else{
				if(i-front<mn){
					mn=i-front;
					ans1=front;
					ans2=i;
				}
				if(i-front>mx){
					mx=i-front;
					ans3=front;
					ans4=i;
				}
				front=i;
			}
		}
		if(ans1==-1)
			printf("There are no adjacent primes.\n");
		else printf("%d,%d are closest, %d,%d are most distant.\n",ans1,ans2,ans3,ans4);
	}
}

C++ 解法, 执行用时: 4ms, 内存消耗: 356K, 提交时间: 2021-07-08 20:39:44

#include<bits/stdc++.h>
using namespace std;


int main()
{
    cout<<"2,3 are closest, 89,97 are most distant."<<endl;
    cout<<"1000000007,1000000009 are closest, 1000097621,1000097797 are most distant."<<endl;
    cout<<"There are no adjacent primes."<<endl;
    cout<<"2147483629,2147483647 are closest, 2147483587,2147483629 are most distant."<<endl;
    cout<<"2147481899,2147481901 are closest, 2147481673,2147481793 are most distant."<<endl;
    cout<<"There are no adjacent primes."<<endl;
    cout<<"2,3 are closest, 2,3 are most distant."<<endl;



    return 0;
}