POJ - 2689 Prime Distance(大區間素數篩選)

POJ - 2689 Prime Distance(大區間素數篩選)

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

題意：輸出區間[l, r]中，差最大和最小的兩對素數。

思路：因為范圍很大，區間相對較小，所以我們先篩選出2^16范圍內的素數，然後通過小的區間去刪除大區間的合數（每個合數都能通過素數相乘得到）。最後遍歷

#include 
#include 
#include 
#include 
#include 
typedef long long ll;
using namespace std;
const int INF = 1<<30;
const int inf = 60000;
const int maxn = 1000005;

int prime[maxn], vis[maxn], cnt;

void init() {
	cnt = 0;
	for (ll i = 2; i < inf; i++) 
		if (!vis[i]) {
			prime[cnt++] = i;
			for (ll j = i*i; j < inf; j += i)
				vis[j] = 1;
		}
}

int isprime[maxn], a[maxn], c;

int main() {
	int l, r;
	init();
	while (scanf("%d%d", &l, &r) != EOF) {
		memset(isprime, 1, sizeof(isprime));
		if (l == 1) l = 2;
		for (int i = 0; i < cnt && (ll)prime[i]*prime[i] <= r; i++) {
			int s = l / prime[i] + (l % prime[i] > 0);
			if (s == 1) s = 2;
			for (int j = s; (ll)j*prime[i] <= r; j++) 
				if ((ll)j*prime[i] >= l) 
					isprime[j*prime[i]-l] = 0;
		}
		c = 0;
		for (int i = 0; i <= r-l; i++) 
			if (isprime[i])
				a[c++] = i + l;
		if (c < 2) {
			printf("There are no adjacent primes.\n");
			continue;
		}
		else {
			int x1 = 0, x2 = 0, y1 = 0, y2 = INF;	
			for (int i = 1; i < c; i++) {
				if (a[i] - a[i-1] > x2 - x1) {
					x1 = a[i-1];
					x2 = a[i];
				}
				if (a[i] - a[i-1] < y2 - y1) {
					y1 = a[i-1];
					y2 = a[i];
				}
			}
			printf("%d,%d are closest, %d,%d are most distant.\n", y1, y2, x1, x2);
		}
	}
	return 0;
}