Color

C - Color

Reference: CF2 GYM 100 548 Color (2014ACM Xi'an game field F. Problem) codeforces Gym 100548F Color (+ number of combinations of inclusion and exclusion)

Idea: You can refer to the idea of ​​a blog, it is easy to understand

Caveats: Because data is large so believe it will burst, it is best to have a long long

ll ans=0;
        int flag=1;
        for(int i=k;i>=1;i--)
            ans=(ans+(1ll*flag*i*qpow(i-1,n-1)%mod*c[i]%mod)+mod)%mod,flag=-flag;
            /*这个地方如果不乘1ll就会爆，如果不乘1ll的话，因为flag是int，所以乘起来就是按int来乘的*/

Code:

// Created by CAD on 2019/8/11.
#include <bits/stdc++.h>
using namespace std;
using pii=pair<int, int>;
using piii=pair<pair<int, int>, int>;
using ll=long long;
const int mod= 1e9+7;
const int maxn=1e6+5;
int c[1000005];
int inv[1000005];

ll qpow(ll x,ll n)
{
    ll re=1;
    while(n>0)
    {
        if(n&1) re=(re*x)%mod;
        n>>=1;
        x=x*x%mod;
    }
    return re;
}
void init_inv()
{
    inv[1]=1;
    for(int i=2;i<maxn;++i)
        inv[i]=ll(mod-mod/i)*inv[mod%i]%mod;
}
int k;
void init_C(ll n)
{
    c[0]=1;
    for(int i=1;i<=k;++i)
        c[i]=c[i-1]*(n-i+1)%mod*inv[i]%mod;

}
ll n,m;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    init_inv();
    int t; cin>>t;
    for(int Case=1;Case<=t;Case++)
    {
        cin>>n>>m>>k;
        init_C(k);
        ll ans=0;
        int flag=1;
        for(int i=k;i>=1;i--)
            ans=(ans+(1ll*flag*i*qpow(i-1,n-1)%mod*c[i]%mod)+mod)%mod,flag=-flag;
        init_C(m);
        cout<<"Case #"<<Case<<": "<<ans*c[k]%mod<<endl;
    }
    return 0;
}