大区间素数筛选 POJ2689

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;
}