题目链接：uva 11440 - Help Tomisu

题目大意：给定n和m，求从2~n！中的数x，要求x的质因子均大于m，问说x有多少个，答案模上1e9+7。

解题思路：

(1)n!=k?m!(n≥m)

(2) 如果有gcd(x,T)=1，那么gcd(x+T,T)=gcd(x,T)=1

题目要求说x的质因子必须要大于m，也就是说x不能包含2~m的因子，那么gcd(x,m!)=1,于是我们求出?(m!),小于m！并且满足gcd(x,m!)=1的个数。

那么根据(2)可得从[m!+1, 2*m!]中的x个数也是?(m!)个；因为如果存在gcd(x,T)=a,那么gcd(x+T,T)=gcd(x,T)=a.

又因为(1),所以最后n!以内的x个数为：n!??(m!)m!

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e7;
const ll MOD = 100000007;

int np, pri[maxn+5], vis[maxn+5];
ll fact[maxn+5], phi[maxn+5];


void prime_table (ll n) {
    np = 0;
    for (ll i = 2; i <= n; i++) {
        if (vis[i])
            continue;

        pri[np++] = i;
        for (ll j = i * i; j <= n; j += i)
            vis[j] = 1;
    }
}

void gcd (ll a, ll b, ll& d, ll& x, ll& y) {
    if (b == 0) {
        d = a;
        x = 1;
        y = 0;
    } else {
        gcd(b, a%b, d, y, x);
        y -= (a/b) * x;
    }
}

inline ll inv_number (ll a, ll n) {
    ll d, x, y;
    gcd(a, n, d, x, y);
    return (x + n) % n;
}

void init (ll n) {
    fact[1] = phi[1] = 1;
    for (ll i = 2; i <= n; i++) {
        fact[i] = (fact[i-1] * i) % MOD;
        phi[i] = phi[i-1];

        if (vis[i] == 0) {
            phi[i] *= ((i-1) * inv_number(i, MOD)) % MOD;
            phi[i] %= MOD;
        }
    }
}

ll solve (int n, int m) {
    ll ans = fact[n] * phi[m] % MOD;
    return (ans - 1 + MOD) % MOD;
}

int main () {
    prime_table(maxn);
    init (maxn);

    int n, m;
    while (scanf("%d%d", &n, &m) == 2 && n + m) {
        printf("%lld\n", solve(n, m));
    }
    return 0;
}