筛选法好题。。。用筛选法打出2-2的16次方的素数表，然后再对l到r做筛选法。。。不过要注意特殊处理1。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1000005
#define maxm 3000005
#define eps 1e-10
#define mod 998244353
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
//#define debug(x) printf("AA x = %d BB\n", x);
//#pragma comment (linker,"/STACK:102400000,102400000")
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL _a, LL _b){LL _res=1,_base=_a;while(_b){if(_b%2)_res=_res*_base%mod;_base=_base*_base%mod;_b/=2;}return _res;}
void scanf(LL &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
// head
int phi[maxn], prime[maxn], p[maxn];
int cnt, l, r;
void init(void)
{
cnt = 0;
for(int i = 2; i <= 65536; i++)
if(!prime[i]) {
phi[++cnt] = i;
for(int j = i+i; j <= 65536; j+=i)
prime[j] = 1;
}
}
void work(void)
{
int a, res, ans_mi, ans_mx, mi, mx;
memset(prime, 0, sizeof prime);
for(int i = 1; i <= cnt; i++) {
if(l%phi[i] == 0) a = 0;
else a = phi[i] - l%phi[i];
if(l <= phi[i]) for(int j = a+phi[i]; j <= r-l; j+=phi[i]) prime[j] = 1;
else for(int j = a; j <= r-l; j+=phi[i]) prime[j] = 1;
}
res = 0;
for(int i = 0; i <= r-l; i++) if(!prime[i] && i+l != 1) p[++res] = i+l;
if(res <= 1) printf("There are no adjacent primes.\n");
else {
mx = 0, mi = INF;
for(int i = 1; i < res; i++) {
if(mx < p[i+1] - p[i]) mx = p[i+1] - p[i], ans_mx = i;
if(mi > p[i+1] - p[i]) mi = p[i+1] - p[i], ans_mi = i;
}
printf("%d,%d are closest, %d,%d are most distant.\n", p[ans_mi], p[ans_mi+1], p[ans_mx], p[ans_mx+1]);
}
}
int main(void)
{
init();
while(scanf("%d%d", &l, &r)!=EOF) work();
return 0;
}
