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)
假设有gcd(x,T)=1，那么gcd(x+T,T)=gcd(x,T)=1

题目要求说x的质因子必需要大于m，也就是说x不能包括2~m的因子，那么gcd(x,m!)=1,于是我们求出ϕ(m!),小于m！

而且满足gcd(x,m!)=1的个数。

那么依据(2)可得从[m!+1, 2*m!]中的x个数也是ϕ(m!)个。由于假设存在gcd(x,T)=a,那么gcd(x+T,T)=gcd(x,T)=a.

又由于(1),所以最后n!以内的x个数为：n!∗ϕ(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;

}