Prime Distance(數論)

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.

解題思路

找出給出的範圍內，相鄰最近和最遠的兩個素數，用到了篩素數法，因爲 1 <= L < U <= 2,147,483,647 所以只要先找出50000以內的素數，再對給出的範圍再次篩選就行了。

因爲範圍的數值較大，所以數組記錄素數時要用到映射。

AC代碼

#include <cstdio>
#include <cstring>
#define maxn 1000001
#define LL long long
LL p_list[50000], p_cnt;                                                //p_list[]存儲50000範圍內的素數，p_cnt計數
LL map[maxn], p_map[maxn], count;                                       //map[]存儲給定範圍內素數的情況p_map[]，count計數

void get_prime()
{
    LL p_tmp[50000];
    memset(p_tmp, 0, sizeof(p_tmp));
    for(int i = 2; i < 50000; i++)
    {
        if( p_tmp[i] == 0 )
        {
            p_list[p_cnt++] = i;
            for(int k = i * 2; k < 50000; k += i)
                p_tmp[k] = 1;
        }
    }
}

int main()
{
    LL a, b;
    get_prime();
    while(scanf("%lld%lld", &a, &b) != EOF)
    {
        if(a == 1)
            a++;
        memset(map, 0, sizeof(map));
        for(LL i = 0; i < p_cnt && p_list[i] * p_list[i] <= b; i++)                       //篩選[a, b]內的素數，map[]記錄篩選結果
        {
            LL tmp = a / p_list[i];
            if(tmp * p_list[i] != a)
                tmp++;
            if(tmp == 1)
                tmp++;
            for(LL j = tmp * p_list[i]; j <= b; j += p_list[i])
                map[j - a] = 1;                                                         //映射
        }
        count = 0;
        for(LL i = a; i <= b; i++)                                                        //p_map[]記錄篩選出來的素數
            if(map[i - a] == 0)
                p_map[count++] = i;

        if(count < 2)
                printf("There are no adjacent primes.\n");
        else
        {
            LL m1 = p_map[0], m2 = p_map[1], n1 = p_map[0], n2 = p_map[1];                    //找出相鄰最近和最遠的兩個素數

            for(int i = 2; i < count; i++)
            {
                if(p_map[i] - p_map[i - 1] < m2 - m1)
                    m2 = p_map[i], m1 = p_map[i - 1];
                if(p_map[i] - p_map[i - 1] > n2 - n1)
                    n2 = p_map[i], n1 = p_map[i - 1];

            }
            printf("%lld,%lld are closest, %lld,%lld are most distant.\n", m1, m2, n1, n2);

        }
    }
    return 0;
}






<pre name="code" class="csharp">