POJ 2689 Prime Distance 素数筛选法应用

题目来源：POJ 2689 Prime Distance

题意：给出一个区间L R 区间内的距离最远和最近的2个素数 并且是相邻的 R-L <= 1000000 但是L和R会很大

思路：一般素数筛选法是拿一个素数 然后它的2倍3倍4倍...都不是 然后这题可以直接从2的L/2倍开始它的L/2+1倍L/2+2倍...都不是素数

首先筛选出一些素数 然后在以这些素数为基础 在L-R上在筛一次因为 R-L <= 1000000 可以左移开一个1百万的数组

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1000010;
typedef __int64 LL;
int vis[maxn];
int p[maxn];
int prime[maxn];
//筛素数 
void sieve(int n)
{
	int m = sqrt(n+0.5);
	memset(vis, 0, sizeof(vis));
	vis[0] = vis[1] = 1;
	for(int i = 2; i <= m; i++)
		if(!vis[i])
			for(int j = i*i; j <= n; j += i)
				vis[j] = 1;
}

int get_primes(int n)
{
	sieve(n);
	int c = 0;
	for(int i = 2; i <= n; i++)
		if(!vis[i])
			prime[c++] = i;
	return c;
}
int main()
{
	LL L, R;
	int c = get_primes(100000);
	while(scanf("%I64d %I64d", &L, &R) != EOF)
	{
		if(L <= 2)
			L = 2;
		memset(p, 0, sizeof(p));
		for(int i = 0; i < c; i++)
		{
			if(prime[i] > R)
				break;
			LL t = L / prime[i];
			if(t <= 1)
				t = 2;
			LL j = t*prime[i];
			while(j < L)
				j += prime[i];
			
			for(; j <= R; j += prime[i])
			{
				//printf("%d,,", j);
				p[j-L] = 1;
			}
		}
		LL ans1 = 999999999, ans2 = 0;
		LL x1, y1, x2, y2, x = -1, y = -1;
		for(LL i = L; i <= R; i++)
		{
			if(!p[i-L])
			{
				//printf("%d\n", i);
				x = y;
				y = i;
				if(x != -1 && y != -1)
				{
					if(ans1 > y-x)
					{
						ans1 = y-x;
						x1 = x;
						y1 = y;
					}
					if(ans2 < y-x)
					{
						ans2 = y-x;
						x2 = x;
						y2 = y;
					}
				}
			}
		}
		if(x == -1 || y == -1)
		{
			puts("There are no adjacent primes.");
			continue;
		}
		printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n", x1, y1, x2, y2);
	}
	return 0;
}

POJ 2689 Prime Distance 素数筛选法应用,布布扣,bubuko.com

POJ 2689 Prime Distance 素数筛选法应用

原文：http://blog.csdn.net/u011686226/article/details/25890819