poj 2689 Prime Distance(大区间筛素数)

http://poj.org/problem?id=2689

题意:给出一个大区间[L,U],分别求出该区间内连续的相差最小和相差最大的素数对。

因为L<U<=2147483647,直接筛素数是不行的,数组就开不了。但是可以根据素数筛的原理。我们先筛出sqrt(2147483647)以内的素数,然后拿这些素数去筛[L,U]之间的素数,即两次素数筛。但是L,U还是很大,但U-L<=1000000,所以进行区间平移,将[L,U]平移为[0,U-L],就能用数组放得下。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;

const int maxn = 47000;
int prime[maxn];
bool flag[maxn];
bool a[1000010];
LL L,U;

void get_prime()
{
	memset(flag,true,sizeof(flag));
	prime[0] = 0;
	flag[0] = flag[1] = false;
	for(int i = 2; i < maxn; i++)
	{
		if(flag[i])
			prime[++prime[0]] = i;
		for(int j = 1; j <= prime[0]&&prime[j]*i < maxn; j++)
		{
			flag[prime[j]*i] = false;
			if(i%prime[j] == 0)
				break;
		}
	}
}
//用已筛选的素数去筛[L,U]之间的素数,区间平移为[0,U-L];
void solve(LL L, LL u)
{
	memset(a,true,sizeof(a));
	if(L == 1)//特殊判断L为1的时候,映射为0,但1不是素数,所以a[0]为false。
		a[0] = false;
	for(int i = 1; i <= prime[0]; i++)
	{
		if(prime[i]*prime[i] > U)
			break;
		int l = L/prime[i];
		if(L%prime[i]!=0)
			l++;
		if(l == 1) l = 2; // l=1时,L为素数,不能把L本身筛掉
		int u = U/prime[i];

		for(int j = l; j <= u; j++)
		{
			LL t = (long long)j*prime[i];
			a[t-L] = false;
		}
	}
}

int main()
{
	get_prime();
	while(~scanf("%lld %lld",&L,&U))
	{
		solve(L,U);
		int Min = 1000010,Max = -1;
		int ans[4];
		int x,y;
		x = y = -1;
		for(int i = 0; i <= U-L; i++)
		{
			if(a[i] == true)
			{
				x = y;
				y = i;

				if(x == -1)
					continue;
				int t = y-x;
				if(t < Min)
				{
					ans[0] = x;
					ans[1] = y;
					Min = t;
				}
				if(t > Max)
				{
					ans[2] = x;
					ans[3] = y;
					Max = t;
				}
			}
		}
		if(x != -1 && y != -1)
		{
			printf("%lld,%lld are closest, %lld,%lld are most distant.\n",ans[0]+L,ans[1]+L,ans[2]+L,ans[3]+L);
		}
		else
			printf("There are no adjacent primes.\n");
	}
	return 0;
}
时间: 2024-10-11 13:37:34

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