POJ - 2689 Prime Distance(大区间素数筛选)

标签：des   style   blog   http   io   color   ar   os   for   

2 17
14 17

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

Source

题意：输出区间[l, r]中，差最大和最小的两对素数。

思路：因为范围很大，区间相对较小，所以我们先筛选出2^16范围内的素数，然后通过小的区间去删除大区间的合数（每个合数都能通过素数相乘得到）。最后遍历

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
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;
}

POJ - 2689 Prime Distance(大区间素数筛选)

标签：des   style   blog   http   io   color   ar   os   for   

原文地址：http://blog.csdn.net/u011345136/article/details/40796987