poj 2689 Prime Distance 篩法+區間篩素數

poj 2689 Prime Distance 篩法+區間篩素數

Prime Distance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16488		Accepted: 4384

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.

L和U的範圍很大,所以不能直接使用線性篩,而L和U的距離不超過一百萬,所以可以先找2~sqrt(INF)的素數,再用篩法找L和U之間的素數,這種篩法是找一個區間內的素數,比較經典,具體看代碼:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
#define M 65538
#define LL long long
const int maxn=1000001;
bool visit[M];
bool is[maxn];
int ans[maxn];
LL p[M];
int cnt=0;
void iii()//找2~sqrt(INF)的素數,65537是素數且大於sqrt(2^31-1);
{
    memset(visit,true,sizeof(visit));
    visit[1]=0;
    for(long long i=2; i<M; ++i)
    {
        if(visit[i]==true)
        {
            p[cnt++]=i;
            for(long long  j=i*i; j<M; j+=i)
            {
                visit[j]=false;
            }
        }
    }
}
int main()
{
    int n,m;
    iii();
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=0; i<maxn; i++)
            is[i]=1;
        if (n == 1) n = 2;
        int d=sqrt(m+0.5);
        for (LL i = 0; i < cnt && p[i]<= d; i++)//區間篩法,找到大於等於n,小於等於m的素數
        {
            int s = n/ p[i] + (n% p[i] > 0);
            if (s == 1) s = 2;
            for (LL j = s; j*p[i] <= m; j++)
                if (j*p[i] >= n)
                    is[j*p[i]-n] = 0;
        }
        int c = 0;
        for (int i = 0; i <= m-n; i++)
            if (is[i])
                ans[c++] = i + n;
        if (c < 2)
        {
            printf("There are no adjacent primes.\n");
            continue;
        }
        else
        {
            int min1=ans[0],min2=ans[1],max1=ans[0],max2=ans[1];
            for(int i=1; i<c-1; i++)
            {
                if(ans[i+1]-ans[i]<min2-min1)
                {
                    min2=ans[i+1];
                    min1=ans[i];
                }
                if(ans[i+1]-ans[i]>max2-max1)
                {
                    max2=ans[i+1];
                    max1=ans[i];
                }
            }
            printf("%d,%d are closest, %d,%d are most distant.\n",min1,min2,max1,max2);
        }
    }
    return 0;
}