poj2689

题意：求区间[l,u]内，距离最近的两个素数，和距离最远的两个素数。区间长度最大1000000，l和u都在int范围内。

分析：由于l,u在2^31内，所以先从1到2^16筛素数。这些素数足以用来判断2^31内的所有数字是否为素数，对于每对l,u，用已知的1～2^16内的素数圈定一个可以筛l,u之间的数字的范围，来筛l,u之间的数字。

View Code

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

using namespace std;



#define maxn 1 << 17

#define maxl 1000005



int l, u;

bool is[maxn];

long long prm[maxn];

int n;

bool ans[maxl];

int prime[maxl];

int num;



int getprm(int n)

{

    int i, j, k = 0;

    int s, e = (int) (sqrt(0.0 + n) + 1);

    memset(is, 1, sizeof(is));

    prm[k++] = 2;

    is[0] = is[1] = 0;

    for (i = 3; i < e; i += 2)

        if (is[i])

        {

            prm[k++] = i;

            for (s = i * 2, j = i * i; j < n; j += s)

                is[j] = 0;

        }

    for (; i < n; i += 2)

        if (is[i])

            prm[k++] = i;

    return k;

}



void work()

{

    for (int i = 0; i < u - l + 1; i++)

        ans[i] = true;

    if (l == 1)

        ans[0] = false;

    num = 0;

    for (int i = 0; i < n && prm[i] * prm[i] <= u; i++)

    {

        long long temp = (l + prm[i] - 1) / prm[i] * prm[i];

        if (temp == prm[i])

            temp += prm[i];

        while (temp <= u)

        {

            ans[temp - l] = false;

            temp += prm[i];

        }

    }

    for (int i = 0; i < u - l + 1; i++)

        if (ans[i])

            prime[num++] = l + i;

    if (num < 2)

    {

        printf("There are no adjacent primes.\n");

        return;

    }

    int ans1 = 0, ans2 = 0;

    for (int i = 0; i < num - 1; i++)

    {

        if (prime[ans1 + 1] - prime[ans1] > prime[i + 1] - prime[i])

            ans1 = i;

        if (prime[ans2 + 1] - prime[ans2] < prime[i + 1] - prime[i])

            ans2 = i;

    }

    printf("%d,%d are closest, %d,%d are most distant.\n", prime[ans1], prime[ans1 + 1], prime[ans2], prime[ans2 + 1]);

}



int main()

{

    //freopen("t.txt", "r", stdin);

    n = getprm(1 << 16);

    while (~scanf("%d%d", &l, &u))

    {

        work();

    }

    return 0;

}