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.
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define FILE "dealing" 5 #define up(i,j,n) for(int i=j;i<=n;i++) 6 #define db long double 7 #define pii pair<int,int> 8 #define pb push_back 9 #define mem(a,L) memset(a,0,sizeof(int)*(L+1)) 10 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;} 11 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;} 12 template<class T> inline T squ(T a){return a*a;} 13 const ll maxn=10000100+10,inf=1e9+10,limit=1e7; 14 ll read(){ 15 ll x=0,f=1,ch=getchar(); 16 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 17 while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); 18 return x*f; 19 } 20 int T,mod; 21 int fac[maxn],inv[maxn],f[maxn]; 22 int prime[maxn],tail; 23 bool b[maxn]; 24 int main(){ 25 freopen(FILE".in","r",stdin); 26 freopen(FILE".out","w",stdout); 27 T=read();mod=read(); 28 fac[0]=1;up(i,1,limit)fac[i]=(ll)fac[i-1]*i%mod; 29 inv[0]=inv[1]=1;up(i,2,limit)inv[i]=(mod-(ll)inv[mod%i]*(mod/i)%mod)%mod; 30 f[1]=1; 31 for(int i=2;i<=limit;i++){ 32 if(!b[i])prime[++tail]=i,f[i]=(ll)f[i-1]*(i-1)%mod*inv[i]%mod; 33 else f[i]=f[i-1]; 34 for(int j=1;j<=tail&&prime[j]*i<=limit;j++){ 35 b[i*prime[j]]=1; 36 if(i%prime[j]==0)break; 37 } 38 } 39 while(T--){ 40 int n=read(),m=read(); 41 printf("%lld ",(ll)fac[n]*f[m]%mod); 42 } 43 return 0; 44 }