[SDOI2008]沙拉公主的困惑

题面

传送门

Sol

题目要求∑n!i=1[gcd(i,m!)==1]

设N=n!，M=m!，莫比乌斯反演一波

就变成了∑d|Mμ(d)Nd

因为M|N所以d|N

而有个定理∑d|Mμ(d)d=φ(M)M

那么就是求φ(M)M∗N

就是φ(m!)∗n!m!

而φ(m!)=φ(m)∗(m−1)!

化简

ans=n!∗ΠP|m(1−1P) (P为质数)=n!∗ΠP|mP−1P

那就变成SB题了

预处理就好了

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1);

IL ll Read(){
RG char c = getchar(); RG ll x = 0, z = 1;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}

int n, m, Zsy, prime[_], num, fac[_], inv[_], id[_];
bool isprime[_];

IL int Pow(RG ll x, RG ll y){
RG ll ret = 1;
for(; y; y >>= 1, x = x * x % Zsy) if(y & 1) ret = ret * x % Zsy;
return ret;
}

IL void Sieve(){
isprime[1] = 1; fac[1] = 1;
for(RG int i = 2; i < _; ++i){
if(!isprime[i]) prime[++num] = i , inv[num] = Pow(i, Zsy - 2);
for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
isprime[i * prime[j]] = 1;
if(!(i % prime[j])) break;
}
fac[i] = 1LL * fac[i - 1] * i % Zsy;
}
for(RG int i = 1; i < num; ++i)
for(RG int j = prime[i]; j < prime[i + 1]; ++j) id[j] = i;
inv[0] = prime[0] = 1;
for(RG int i = 1; i <= num; ++i){
prime[i] = 1LL * (prime[i] - 1) * prime[i - 1] % Zsy;
inv[i] = 1LL * inv[i] * inv[i - 1] % Zsy;
}
}

IL int Calc(){  return 1LL * fac
* prime[id[m]] % Zsy * inv[id[m]] % Zsy; }

int main(RG int argc, RG char* argv[]){
RG int T = Read(); Zsy = Read();
Sieve();
while(T--){
n = Read(); m = Read();
printf("%d\n", Calc());
}
return 0;
}