Light OJ 1095 Arrange the Numbers(容斥)

给定n,m,k,要求在n的全排列中，前m个数字中恰好有k个位置不变，有几种方案?
首先，前m个中k个不变，那就是C(m,k)，然后利用容斥原理可得

ans=ΣC(m,k)*(-1)^i*C(m-k,i)*(n-k-i)! (0<=i<=m-k)

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define ll long long
 const ll Mod=;
 int n,m,k;
 ll jc[],jcny[];
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void exgcd(ll a,ll b,ll &x,ll &y){
     if (b==){
         x=;
         y=;
         return;
     }
     exgcd(b,a%b,x,y);
     ll t=x;
     x=y;
     y=t-(a/b)*y;
 }
 void init(){
     jc[]=;jc[]=;jcny[]=;
     for (int i=;i<=;i++)
      jc[i]=(jc[i-]*i)%Mod;
     ll x,y;
     exgcd(jc[],Mod,x,y);
     jcny[]=x%Mod;
     for (int i=-;i>=;i--)
      jcny[i]=(jcny[i+]*(i+))%Mod;
 }
 ll powit(ll x,ll y){
     ll res=;
     while (y){
         if (y%) res=(res*x)%Mod;
         x=(x*x)%Mod;
         y/=;
     }
     return res;
 }
 ll C(int n,int m){
     if (m==) return ;
     return ((((jc[n]%Mod)*(jcny[n-m]%Mod))%Mod)*jcny[m])%Mod;
 }
 ll A(int x){
     if (x==) return ;
     return jc[x]%Mod;
 }
 int main(){
     init();
     int Tcase=;
     int T=read();
     while (T--){
         printf("Case %d: ",++Tcase);
         n=read(),m=read(),k=read();
         ll ans=%Mod;
         for (int i=;i<=m-k;i++)
          ans=(ans+powit((ll)-,(ll)i)*A(n-k-i)*C(m-k,i)+Mod)%Mod;
         ans=(ans*C(m,k))%Mod;
         printf("%lld\n",ans);
     }
 }