题意：

给一个区间[L,U],(1<=L< U<=2,147,483,647),U-L<=1000000,求出[L,U]内距离近期和距离最远的素数对。

因为L,U都小于2^32,所以区间内的合数的最小质因子必定小于2^16,所以先筛出2^16以内的素数，用筛出来的素数去筛[L,U]内的合数。然后把[L,U]内的素数保存下来，再搜索近期和最远的素数对就可以。注意两整数相乘可能溢出32位，注意对1的推断。

代码：

#include <cstdlib>

#include <cctype>

#include <cstring>

#include <cstdio>

#include <cmath>

#include<climits>

#include <algorithm>

#include <vector>

#include <string>

#include <iostream>

#include <sstream>

#include <map>

#include <set>

#include <queue>

#include <stack>

#include <fstream>

#include <numeric>

#include <iomanip>

#include <bitset>

#include <list>

#include <stdexcept>

#include <functional>

#include <utility>

#include <ctime>

using namespace std;

#define PB push_back

#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)

#define FOR(i,l,h) for(int i=(l);i<=(h);++i)

#define FORD(i,h,l) for(int i=(h);i>=(l);--i)

#define SZ(X) ((int)(X).size())

#define ALL(X) (X).begin(), (X).end()

#define RI(X) scanf("%d", &(X))

#define RII(X, Y) scanf("%d%d", &(X), &(Y))

#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))

#define DRI(X) int (X); scanf("%d", &X)

#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)

#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)

#define OI(X) printf("%d",X);

#define RS(X) scanf("%s", (X))

#define MS0(X) memset((X), 0, sizeof((X)))

#define MS1(X) memset((X), -1, sizeof((X)))

#define LEN(X) strlen(X)

#define F first

#define S second

#define Swap(a, b) (a ^= b, b ^= a, a ^= b)

#define Dpoint  strcut node{int x,y}

#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME

    freopen("in.txt","r",stdin);

    #endif*/

const int MOD = 1e9+7;

typedef vector<int> VI;

typedef vector<string> VS;

typedef vector<double> VD;

typedef long long LL;

typedef pair<int,int> PII;

//#define HOME

int Scan()

{

	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//推断正负

		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数

		res = ch - '0';

	while((ch = getchar()) >= '0' && ch <= '9' )

		res = res * 10 + ch - '0';

	return flag ? -res : res;

}

/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/

#define MAXN 100000

int prime[MAXN];

int vis[MAXN+5];

int cnt;

void getprime()

{cnt=0;

for(int i=2;i<=MAXN;i++)

    if(!vis[i])

{

    prime[cnt++]=i;

    for(int j=0;j<cnt&&prime[j]<=MAXN/i;j++)

    {

        vis[prime[j]*i]=1;

        if(i%prime[j]==0)

            break;

    }

}

}

int notprime[1000000+5];

int prime2[1000000+5];

int cnt2;

void getprime2(int L,int U)

{

    for(int i=0;i<cnt;i++)

    {   if(prime[i]>=U)

            break;

        int s=L/prime[i];

        if(s<=1)

            s=2;

        for(int j=s;(long long)prime[i]*j<=U;j++)

            if((long long )prime[i]*j>=L)

        {

            notprime[(long long )prime[i]*j-L]=1;

        }

    }

    cnt2=0;

    REP(i,0,U-L+1)

    {

        if(!notprime[i]&&(i+L)!=1&&(i+L)!=0)

            prime2[cnt2++]=i+L;

    }

}

int main()

{getprime();

int L,U;

while(RII(L,U)!=EOF)

{

    MS0(notprime);

    getprime2(L,U);

    int ans1=INT_MAX;

    int ans2=0;

    int n1,n2,f1,f2;

    if(cnt2<2)

    {

        printf("There are no adjacent primes.\n");

        continue;

    }

    REP(i,0,cnt2-1)

    {

       if(prime2[i+1]-prime2[i]<ans1)

       {

           ans1=prime2[i+1]-prime2[i];

           n1=prime2[i];

           n2=prime2[i+1];

       }

       if(prime2[i+1]-prime2[i]>ans2)

       {

           ans2=prime2[i+1]-prime2[i];

           f1=prime2[i];

           f2=prime2[i+1];

       }

    }

    printf("%d,%d are closest, %d,%d are most distant.\n",n1,n2,f1,f2);

}

        return 0;

}