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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;
}
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原文地址:http://blog.csdn.net/keshuai19940722/article/details/37654075