poj-2689-素数区间筛

Prime Distance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22420		Accepted: 5954

Description

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers. 
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).

Input

Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.

Output

For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.

Sample Input

2 17
14 17

Sample Output

2,3 are closest, 7,11 are most distant.
There are no adjacent primes.

Source

Waterloo local 1998.10.17

　　　　给出[L,R],求区间内的素数，R<=2147483647,R-L<=1000000, 注意到只用sqrt(R)以内的素数就可以筛出[L,R]里面的素数，

可以先对sqrt(MAX_INT)内的素数打一个表。对于[L,R]的询问，用小于等于sqrt(R)的素数筛一下然后统计一下就好了。注意L<2的时候

要特判一下否则容易把1也给打进去。

　　　　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
#define LL long long 
#define mp make_pair
#define pb push_back
#define inf 0x3f3f3f3f
int maxn=100005;
vector<int>prime;
vector<int>p;
bool is[1000100];
void init(){
    is[0]=is[1]=1;
    for(LL i=2;i<=maxn;++i){
        if(!is[i]) prime.push_back(i);
        for(LL j=0;j<prime.size()&&i*prime[j]<=maxn;j++){
            is[i*prime[j]]=1;
            if(i%prime[j]) break;
        }
    }
}
void solve(LL L,LL R){
    p.clear();
    memset(is,0,sizeof(is));
    for(LL i=0;i<prime.size()&&1LL*prime[i]*prime[i]<=R;i++){
        LL s=L/prime[i]+(L%prime[i]>0);
        if(s==1)s=2;
        for(LL j=s;j*prime[i]<=R;j++){
            if(j*prime[i]>=L) is[j*prime[i]-L]=1;
        }
    }
    for(int i=0;i<=R-L;i++){
        if(!is[i]&&i+L>=2) p.push_back(i+L);
    }
}
int main(){
    LL L,R;
    init();
    while(scanf("%lld%lld",&L,&R)!=EOF){
        solve(L,R);
    
        if(p.size()<2) puts("There are no adjacent primes.");
        else{
            int c1,c2,m1,m2;
            c1=m1=p[0];
            c2=m2=p[1];
            for(int i=1;i<p.size();++i){
                if(p[i]-p[i-1]<c2-c1){
                    c1=p[i-1];
                    c2=p[i];
                }
                if(p[i]-p[i-1]>m2-m1){
                    m1=p[i-1];
                    m2=p[i];
                }
            }
            printf("%d,%d are closest, %d,%d are most distant.
",c1,c2,m1,m2);
        }
    }
    return 0;
}