【题解】SDOI2008莎拉公主的困惑

挺有趣的恩：洛谷P2155

在纸上打打草稿，写出n!个数，从先往后，遇到不互质的就筛掉——发现一个奇妙的性质！：筛掉的次数、顺序好像是周期性出现的呢~

而且更加妙妙的是，好像还是m!一轮..那么因为n!一定能被m!整除，所以问题转变为：（n!\m! - 有多少个循环节）*(φ(m))。

接下来，φ(m) = m!*(1 - 1/p1)*(1 - 1/p2)...任务就只剩下打出阶乘表&逆元啦。离线的处理会快很多。

#include <bits/stdc++.h>
using namespace std;
#define maxn 10000050
#define ll long long
#define int long long
int maxx, now = , P, T, tot, inv[maxn], ans[], pri[maxn],fac_a[maxn], fac_b[maxn], fac_c[maxn];
bool is_prime[maxn];
struct query
{
    int n, m, id, pri;
}Q[];

int read()
{
    int x = ;
    char c;
    c = getchar();
    while(c < '' || c > '') c = getchar();
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x;
}

bool cmp1(query a, query b)
{
    return a.m < b.m;
}

int Get_Pri(int n)
{
    for(int i = ; i <= n; i ++)
    {
        if(!is_prime[i])
        {
            pri[++ tot] = i;
            while(now <= T && pri[tot] > Q[now].m)
            {
                Q[now].pri = tot - ;
                now ++;
            }
        }
        while(now <= T && i == n)
            {
                Q[now].pri = tot;
                now ++;
            }
        for(int j = ; j <= tot; j ++)
        {
            if(i * pri[j] > n) break;
            is_prime[i * pri[j]] = ;
            if(!(i % pri[j])) break;
        }
    }
}

int Get_fac(int n)
{
    fac_a[] = fac_a[] = fac_b[] = fac_b[] = fac_c[] = fac_c[] = ;
    inv[] = inv[] = ;
     for(int i = ; i <= n; i ++)
     {
         fac_a[i] = (fac_a[i - ] * i) % P;
        inv[i] = ((P - P / i) * inv[P % i]) % P;
    }
    for(int i = ; i <= tot; i ++)
    {
        fac_b[i] = inv[pri[i]];
        fac_b[i] = (fac_b[i] * fac_b[i - ]) % P;
        fac_c[i] = pri[i] - ;
        fac_c[i] = (fac_c[i] * fac_c[i - ]) % P;
    }
}

signed main()
{
    T = read(), P = read();
    for(int i = ; i <= T; i ++)
    {
        Q[i].n = read(), Q[i].m = read(), Q[i].id = i;
        maxx = max(maxx, max(Q[i].n, Q[i].m));
    }
    sort(Q + , Q +  + T, cmp1);
    Get_Pri(maxx);
    Get_fac(maxx);
    for(int i = ; i <= T; i ++)
        ans[Q[i].id] = ((fac_a[Q[i].n] * fac_b[Q[i].pri]) % P * fac_c[Q[i].pri]) % P;
    for(int i = ; i <= T; i ++)
        printf("%lld\n", ans[i]);
    return ;
}