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发现好多人的做法并不对......

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【分析】

先化简一波式子：

\(\quad Ans\)
\(\displaystyle =\sum_{i=1}^{N!}[\gcd(i, M!)=1]\)
\(\displaystyle =\sum_{i=1}^{N!}\sum_{d\mid i\wedge d\mid (M!)}\boldsymbol \mu(d)\)
\(\displaystyle =\sum_{d\mid (M!)}\boldsymbol \mu(d)\sum_{i=1}^{N!}[d\mid i]\)

由于 \(M\leq N\) 故 \((M!)\mid (N!)\) 有 \(d\mid (N!)\)

\(\therefore Ans\)
\(\displaystyle =\sum_{d\mid (M!)}\boldsymbol \mu(d){N!\over d}\)
\(\displaystyle ={N!\over M!}\sum_{d\mid (M!)}\boldsymbol \mu(d){M!\over d}\)
\(\displaystyle ={N!\over M!}(\boldsymbol \mu*\boldsymbol {id})(M!)\)
\(\displaystyle ={N!\over M!}\boldsymbol \varphi(M!)\)
\(\displaystyle ={N!\over M!}\cdot M!\cdot \prod_{p\mid (M!)}{p-1\over p}\)
\(\displaystyle =N!\prod_{p\leq M}{p-1\over p}\)

答案要求对 \(R\) 取模，但并没有表明 \(R\) 是否严格大于 \(M\)

故类似于 exLucas 的想法：设 \(f_R(n)\) 表示 \(n\) 去除 \(R\) 质因子后，剩余质因子的乘积；\(g_R(n)\) 表示 \(n\) 中 \(R\) 质因子的次数

故 \(\displaystyle n=f_R(n)\cdot R^{\displaystyle g_R(n)}\)

代入所求式子得：\(\displaystyle Ans={f_R(N!)\cdot f_R(\prod_{p\leq M}(p-1))\over f_R(\prod_{p\leq M}p)}\cdot R^{\displaystyle g_R(N!)+g_R(\prod_{p\leq M}(p-1))-g_R(\prod_{p\leq M}p)}\)

当 \(\displaystyle g_R(N!)+g_R(\prod_{p\leq M}(p-1))-g_R(\prod_{p\leq M}p)>0\) 时，\(Ans\equiv 0\pmod R\)

当 \(\displaystyle g_R(N!)+g_R(\prod_{p\leq M}(p-1))-g_R(\prod_{p\leq M}p)=0\) 时，\(\displaystyle Ans\equiv f_R(N!)\cdot f_R(\prod_{p\leq M}(p-1)) \cdot f_R^{-1}(\prod_{p\leq M}p)\pmod R\)

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现考虑如何线性求解三个数字的 \(f_R\) 和 \(g_R\)

对于 \(N!\) 有 \(\begin{cases} f_R(N!)\equiv f_R((N-1)!)\cdot f_R(N)\pmod R \\ g_R(N!)\equiv g_R((N-1)!)+g_R(N)\pmod R \end{cases}\)

其中，\(f_R(N)\) 和 \(g_R(N)\) 可以通过对 \(N\) 暴力拆解得到，复杂度为 \(T(N)=O(N)+O(N)+O({N\over R})+O({N\over R^2})+O({N\over R^3})+\cdots =O(N)+O({N\over 1-{1\over R}})=O(N)\)

对于 \(\displaystyle \prod_{p\leq M} p\) 有 \(\begin{cases} \begin{cases} \displaystyle f_R(\prod_{p\leq M}p)\equiv f_R(\prod_{p\leq M-1}p)\pmod R \\ \displaystyle g_R(\prod_{p\leq M}p)\equiv g_R(\prod_{p\leq M-1}p)\pmod R \end{cases}, M\not\in Prime \\\ \\ \begin{cases} \displaystyle f_R(\prod_{p\leq M}p)\equiv f_R(\prod_{p\leq M-1}p)\cdot M^{[M\neq R]}\pmod R \\ \displaystyle g_R(\prod_{p\leq M}p)\equiv g_R(\prod_{p\leq M-1}p)+[M=R]\pmod R \end{cases}, M\in Prime \end{cases}\)

复杂度为 \(O(N)\)

对于 \(\displaystyle \prod_{p\leq M}(p-1)\) 有 \(\begin{cases} \begin{cases} \displaystyle f_R(\prod_{p\leq M}(p-1))\equiv f_R(\prod_{p\leq M-1}(p-1))\pmod R \\ \displaystyle g_R(\prod_{p\leq M}(p-1))\equiv g_R(\prod_{p\leq M-1}(p-1))\pmod R \end{cases}, M\not\in Prime \\\ \\ \begin{cases} \displaystyle f_R(\prod_{p\leq M}(p-1))\equiv f_R(\prod_{p\leq M-1}(p-1))\cdot f_R(M-1)\pmod R \\ \displaystyle g_R(\prod_{p\leq M}(p-1))\equiv g_R(\prod_{p\leq M-1}(p-1))+g_R(M-1)\pmod R \end{cases}, M\in Prime \end{cases}\)

对于 \(f_R(M-1)\) 与 \(g_R(M-1)\) 同样是暴力拆解，复杂度不大于 \(f_R(N!)\) 和 \(g_R(N!)\) 的求解，故也是 \(O(N)\) 的

因此预处理总复杂度为 \(O(N)+O(N)+O(N)=O(N)\)

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【代码】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=1e7+10;
int R, prime[MAXN], cntprime, frac[MAXN], cntR[MAXN], prodP[MAXN], cntRP[MAXN], prodP1[MAXN], cntRP1[MAXN];
bool nprime[MAXN];
inline ll fpow(ll a,ll x) { ll ans=1; for(;x;x>>=1,a=a*a%R) if(x&1) ans=ans*a%R; return ans; }
inline int ans(int N, int M){
    if( cntR[N]+cntRP1[M]-cntRP[M] ) return 0;
    return (ll)frac[N]*prodP1[M]%R*fpow(prodP[M], R-2)%R;
}
inline void pre(){
    frac[0]=frac[1]=1;
    prodP[1]=prodP1[1]=1;
    for(int i=2;i<=1e7;++i){
        if(!nprime[i]) prime[++cntprime]=i;
        for(int j=1;j<=cntprime;++j)
            if((ll)prime[j]*i>1e7) break;
            else{
                nprime[prime[j]*i]=1;
                if(i%prime[j]==0) break;
            }
        cntR[i]=cntR[i-1];
        int Val=i;
        while(Val%R==0) ++cntR[i], Val/=R;
        frac[i]=(ll)frac[i-1]*Val%R;

        prodP[i]=prodP[i-1];
        cntRP[i]=cntRP[i-1];
        prodP1[i]=prodP1[i-1];
        cntRP1[i]=cntRP1[i-1];
        if(prime[cntprime]!=i) continue;
        if(i==R) ++cntRP[i];
        else prodP[i]=(ll)prodP[i]*i%R;
        Val=i-1;
        while(Val%R==0) ++cntRP1[i], Val/=R;
        prodP1[i]=(ll)prodP1[i]*Val%R;
    }
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int T, N, M; cin>>T>>R;
    pre();
    while(T--&&cin>>N>>M) cout<<ans(N, M)<<"\n";
    cout.flush();
    return 0;
}