POJ2689:Prime Distance（大数区间素数筛）

Posted 2020-10-25 ---学习ing---

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.

 

题意：求一个长度不超过1e6的区间[L,R]（L和R很大）中，距离最近和最远的素数对。

思路：求出1到sqrt(R)间的素数，然后利用这些素数筛去[L,R]中的合数。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100000;
int p[maxn+10],vis[maxn+10],cnt,L,R;
int wl,wr,vl,vr,tag[maxn*10+10],times;
void prime()
{
    for(int i=2;i<=maxn;i++){
        if(!vis[i]) p[++cnt]=i;
        for(int j=1;j<=cnt&&p[j]*i<=maxn;j++){
            vis[i*p[j]]=1;
            if(i%p[j]==0) break;
        }
    }
}
void find()
{
    for(int i=1;i<=cnt&&p[i]<=R;i++){
        int Now=(max(L,maxn)/p[i])*p[i];
        while(Now<L) Now+=p[i];
        while(Now<=R){
            tag[Now-L+1]=times;
            Now+=p[i];
        }
    }
    int pre=0;
    for(int i=L;i<=R;i++){
        if((i<=maxn&&!vis[i])||(i>maxn&&tag[i-L+1]!=times)) {
            if(pre) {
                if(wl==0) wl=vl=pre,wr=vr=i;
                else {
                    if(i-pre>wr-wl) wl=pre,wr=i;
                    if(i-pre<vr-vl) vl=pre,vr=i;
                }
            }
            pre=i;
        }
    }
    if(wl==0) printf("There are no adjacent primes.\n");
    else printf("%d,%d are closest, %d,%d are most distant.\n",vl,vr,wl,wr);
    return ;
}
int main()
{
    prime();
    while(~scanf("%d%d",&L,&R)){
        times++;
        wl=wr=vl=vr=0;//da xiao
        find();
    }
    return 0;
}