UVa Live 7040 (二项式反演+线性求逆元)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int MOD=1e9+7,N=1e6+10;
typedef long long ll;
int n,m,k,T,kase=0;
ll inv
,Ck
,Cm
;
void calinverse()
{
inv[1]=1;
for(int i=2;i<N;i++)
inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD;
}
void getC()
{
Ck[0]=Cm[0]=1;
fo(i,1,k) {Ck[i]=Ck[i-1]%MOD*(k-i+1)%MOD*inv[i]%MOD;
Cm[i]=Cm[i-1]%MOD*(m-i+1)%MOD*inv[i]%MOD;}
}
ll qpow(ll a,int b)
{
ll ret=1;
for(int i=b;i;i>>=1,a=a*a%MOD)
if(i&1) ret=ret*a%MOD;
return ret;
}
int main()
{
calinverse();
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
int sgn=1;
ll sum=0;
getC();
for(int i=k;i>=1;i--){
sum=(sum+Ck[i]*i%MOD*qpow(i-1,n-1)*sgn%MOD+MOD)%MOD;
sgn=-sgn;
}
sum=(sum*Cm[k])%MOD;
printf("Case #%d: %lld\n",++kase,sum);
}
return 0;
}