UVALive 7040 Color 容斥原理 + 组合数学递推公式 +lucas

题意：
给你m种颜色，你要选出其中k种并全部用上去给那朵花涂颜色，相邻两朵花的颜色要不同有几种方案
思路：
设dp[k]为刚好用k种颜色的答案
如果k种颜色不用完则ans=k*(k-1)^（n-1）
则

dp[k]=ckk*f(k)-ckk-1*f(k-1)+ckk-2*f(k-2)-….

#include<bits/stdc++.h>using namespace std;#define ll long longconst int mod = 1e9 + 7;const int MAXN = 1000010;ll inv[MAXN], fac[MAXN];  //  fac[i] =i!%mod    inv[i] =1/(i!) %modvoid init(){    inv[0] = fac[0] = inv[1] = fac[1] = 1;    for(int i = 2; i < MAXN; i++)        fac[i] = fac[i - 1] * i % mod;    for(int i = 2; i < MAXN; i++)    inv[i] = (mod - (mod / i)) * inv[mod % i] % mod;    for(int i = 2; i < MAXN; i++)        inv[i] = inv[i - 1] * inv[i] % mod;}ll qpow(ll a, ll b){    ll ans = 1;    while(b)    {        if(b & 1) ans = a * ans % mod;        a = a * a % mod;        b >>= 1;    }    return ans;}ll CC(ll a, ll b){    if(a < b) return 0;    if(a < MAXN && b < MAXN)    return (fac[a] * inv[b] % mod) * inv[a - b] %mod;    if(b > a - b) b = a - b;    ll up = 1, down = 1;    for(ll i = 0; i < b; i++)    {        up = up * (a - i) % mod;        down = down * (i + 1) % mod;    }    return up * qpow(down, mod - 2) % mod;}ll C(ll a, ll b){    ll ans=fac[a];    ans=(ans*inv[b])%mod;    ans=(ans*inv[a-b])%mod;    return ans;}ll lucas(ll a, ll b){    if(b == 0) return 1;    return CC(a % mod, b % mod) * lucas(a / mod, b / mod) % mod;}ll n, m, k;ll f(ll x){    return (k-x)*qpow(k-x-1,n-1)%mod;}ll solve( ){       ll ans;      if(n==k) ans=fac[k];     else if(k==1) ans=0;     else     {      ans=0;      for(int i=0;i<k;i++)      if(i&1) ans=(ans-(f(i)*lucas(k,k-i))%mod+mod)%mod;      else   ans=(ans+(f(i)*lucas(k,k-i))%mod)%mod;      }     return (ans*lucas(m,k))%mod;}int main(){      init();    int T, kase = 1, i, j;    cin >> T;    while(T--)    {       scanf("%lld %lld %lld", &n, &m, &k);       printf("Case #%d: %lld\n", kase++, solve());    }}