素数筛选（2次）：poj2689 Prime Distance

Prime Distance

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 1714 17

Sample Output

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

解题思路：

首先，不同于普通素数筛选，由于上限U很大，只进行一次从2-2147483647的筛选绝对会超时，跟不行不通，除非改进算法，否则就要换一下思路。

然后，可以考虑只对L-U内的素数进行筛选，筛选的方法是从1-sqrt（2147483647）内的素数筛选（素因子肯定不超过sqrt(2147483647)）,getprime()函数实现。

接着，用可能的素因子去筛选L-U区间的数，和筛法的思想相同（但要注意从哪开始，begin怎么确定），将所有的素数存到数组q[]中。

最后，for循环一次，找到相邻的q[i+1]、q[i]的最大值和最小值，输出。

思路并不难，最蹩脚的是数据的溢出问题。

①for（int i=l;i<=u;i++） 当i=u=2147483647时，i++会溢出，进入死循环，所以要处理一下。

②long long begin=l%p[i]?l+(p[i]-l%p[i]):l+l%p[i]；在确定begin时，有加法可能会溢出，可以用long long p[]和long long begin,也可以都换成long long或unsigned int

参考代码和部分注释：