大区间素数筛选 POJ2689

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题意：

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

由于L,U都小于2^32,所以区间内的合数的最小质因子必然小于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，再搜索最近和最远的素数对即可。注意两整数相乘可能溢出32位，注意对1的判断。

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 HOMEint Scan(){int res = 0, ch, flag = 0;if((ch = getchar()) == '-')//判断正负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 100000int 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;}

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