BZOJ 2186 [Sdoi2008]沙拉公主的困惑

数论重学篇.

首先需要看出答案是:

　　n!/m!*phi(m!)

　　小于m!的数中与m!互质的数有phi(m!)个,由辗转相除法可知,

　　gcd(x,m!)=1 <==>  gcd(x+m!,m!)=1

　　然后直接算就行了.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define FILE "dealing"
#define up(i,j,n) for(int i=j;i<=n;i++)
#define db long double 
#define pii pair<int,int>
#define pb push_back
#define mem(a,L) memset(a,0,sizeof(int)*(L+1))
template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
template<class T> inline T squ(T a){return a*a;}
const ll maxn=10000100+10,inf=1e9+10,limit=1e7;
ll read(){
    ll x=0,f=1,ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    return x*f;
}
int T,mod;
int fac[maxn],inv[maxn],f[maxn];
int prime[maxn],tail;
bool b[maxn];
int main(){
    freopen(FILE".in","r",stdin);
    freopen(FILE".out","w",stdout);
    T=read();mod=read();
    fac[0]=1;up(i,1,limit)fac[i]=(ll)fac[i-1]*i%mod;
    inv[0]=inv[1]=1;up(i,2,limit)inv[i]=(mod-(ll)inv[mod%i]*(mod/i)%mod)%mod;
    f[1]=1;
    for(int i=2;i<=limit;i++){
        if(!b[i])prime[++tail]=i,f[i]=(ll)f[i-1]*(i-1)%mod*inv[i]%mod;
        else f[i]=f[i-1];
        for(int j=1;j<=tail&&prime[j]*i<=limit;j++){
            b[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
    while(T--){
        int n=read(),m=read();
        printf("%lld\n",(ll)fac[n]*f[m]%mod);
    }
    return 0;
}