质数

相邻质数距离 https://www.acwing.com/problem/content/198/

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1000010;

int primes[N], cnt;
bool st[N];

void init(int n)
{
    memset(st, 0, sizeof st);
    cnt = 0;
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    int l, r;
    while (cin >> l >> r)
    {
        init(50000);//sqrt(n)

        memset(st, 0, sizeof st);
        for (int i = 0; i < cnt; i ++ )
        {
            LL p = primes[i];
            for (LL j = max(p * 2, (l + p - 1) / p * p); j <= r; j += p)//找l-r内的质数
                st[j - l] = true;//这里防止爆内存
        }

        cnt = 0;
        for (int i = 0; i <= r - l; i ++ )
            if (!st[i] && i + l >= 2)
                primes[cnt ++ ] = i + l;

        if (cnt < 2) puts("There are no adjacent primes.");//
        else
        {
            int minp = 0, maxp = 0;
            for (int i = 0; i + 1 < cnt; i ++ )//相邻所以是 i i+1
            {
                int d = primes[i + 1] - primes[i];
                if (d < primes[minp + 1] - primes[minp]) minp = i;
                if (d > primes[maxp + 1] - primes[maxp]) maxp = i;
            }

            printf("%d,%d are closest, %d,%d are most distant.\n",
                primes[minp], primes[minp + 1],
                primes[maxp], primes[maxp + 1]);
        }
    }

    return 0;
}

阶乘分解 https://www.acwing.com/problem/content/199/

8!=1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 要求阶乘里面2的个数
1 1 1 1 2的倍数4个
1 1 4的个数2个
1 8的个数1个 (知道x的n次方大于n)

出现2的个数有7个 虽然没什么逻辑但是就是能这么求出来2的个数 横着求出来

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1000010;

int primes[N], cnt;
bool st[N];

void init(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    int n;
    cin >> n;
    init(n);

    for (int i = 0; i < cnt; i ++ )
    {
        int p = primes[i];
        int s = 0;
        for (int j = n; j; j /= p) s += j / p;
        printf("%d %d\n", p, s);
    }

    return 0;
}