Prime Distance(质数距离)

Prime Distance

L,R的范围很大,任何已知算法都无法在规定时间内生成[1,R]中的所有质数

但R-L的值很小,并且任何一个合数n必定包含一个不超过sqrt(n)的质因子

解:用筛法求出2~sqrt®之间的所有质数.对于每个质数p,将范围内其倍数标记,未被标记的即为合数

(这里的标记也可以用下题中,选倍数的方法,可以进一步优化时间复杂度)

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int l, u;
    int v[1 << 16];
    bool jud[1000005];
    while (cin >> l >> u)
    {
        if (l == 1)
            l++;
        memset(jud, 0, sizeof(jud));
        vector<int> su;
        su.clear();
        memset(v, 0, sizeof(v));
        for (int i = 2; i <= sqrt(u); i++)
        {
            if (v[i])
                continue;
            // 素数运算,i为素数
            for (int j = ceil(l / i); j <= floor(u / i); j++)
            {
                if (j == 1)
                    continue;
                jud[i * j - l] = 1;
            }
            for (int j = i; j <= sqrt(u) / i; j++)
                v[i * j] = i;
        }
        int maxx = 0, leftt, r;
        for (int i = 0; i <= u-l; i++)
        {
            if (!jud[i])
            {
                su.push_back(i+l);
            }
        }
        if (su.size() < 2)
            cout << "There are no adjacent primes.\n";
        else
        {
            int ans_l, ans_r, ansl2, ansr2, l2, r2, minn;
            ansl2 = ans_l = leftt = l2 = su[0], ansr2 = ans_r = r = r2 = su[1];
            maxx = minn = r - leftt;
            for (int i = 2; i < su.size(); i++)
            {
                leftt = r;
                r = su[i];
                if (maxx < r - leftt)
                {
                    maxx = r - leftt;
                    ans_l = leftt;
                    ans_r = r;
                }
                if (minn > r - leftt)
                {
                    minn = r - leftt;
                    ansl2 = leftt;
                    ansr2 = r;
                }
            }
            printf("%d,%d are closest, %d,%d are most distant\n", ansl2, ansr2, ans_l, ans_r);
        }
    }
    return 0;
}

注意点:map不能轻用,其添加,存取,删除的时间复杂度为log(n),不是O(1)