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。

#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;
}