Full_of_Boys训练6总结

#include<bits/stdc++.h>
typedef long long ll;
const ll mod = 1e9 + 7;
using namespace std;
ll q_pow(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1)ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
ll n,m,k,inv[1000007];
ll C(ll n,ll m){ll t=1;
    for(ll i=n-m+1;i<=n;++i)t=(t*i)%mod;
    for(ll i=1;i<=m;++i)t=(t*inv[i])%mod;
    return t;
}
ll a(ll x){return ((x%mod)*q_pow(x-1,n-1)%mod)%mod;}
int T,K;
int main() {
    scanf("%d",&T);
    for(int i=1;i<=1e6;++i)inv[i]=q_pow(i,mod-2);
    while(T--){
        scanf("%lld%lld%lld",&n,&m,&k);
        ll ans=0,w=1,Cki=1;if(k%2)w=-1;
        for(int i=0;i<=k;++i,w=-w){
            ans = (ans%mod + (w*(Cki%mod*a(i)%mod)%mod+mod)%mod)%mod;
            Cki=(((Cki%mod)*(k-i)%mod)*inv[i+1]%mod)%mod;
        }
        ans = (ans*C(m,k))%mod;
        printf("Case #%d: %lld\n",++K,ans);
    }
}