Gym 100548F Color (数论容斥原理+组合数)

题意：给定 m 种颜色，把 n 盆花排成一直线的花涂色。要求相邻花的颜色不相同，且使用的颜色恰好是k种。问一共有几种涂色方法。

析：首先是先从 m 种颜色中选出 k 种颜色，然后下面用的容斥原理，当时没想出来，如果是只用一种颜色，那么肯定不行，如果用两种颜色，可以有这么方法，

2 * (2-1) ^ (n-1)种，如果是只用 i 种那么就是  i * (i-1) ^ (n-1)。然后依次求就好。再就是求组合数的时候，由于太大，不能用递推，所以要用逆元。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <ctime>
#include <cstdlib>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL f[maxn];
LL cm[maxn], ck[maxn];

void init(){
    f[1] = 1;
    for(int i = 2; i < maxn; ++i)
        f[i] = (mod - mod/i) * f[mod%i] % mod;
}

LL quick_pow(LL a, LL n){
    LL ans = 1LL;
    while(n){
        if(n & 1)  ans = ans * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ans;
}

LL solve(LL n, LL m, LL k){
    ck[0] = cm[0] = 1;
    for(int i = 1; i <= k; ++i){
        cm[i] = cm[i-1] * (m - i + 1) % mod * f[i] % mod;
        ck[i] = ck[i-1] * (k - i + 1) % mod * f[i] % mod;
    }
    LL ans = 0;
    LL cnt = 1;
    for(int i = k; i >= 1; --i, cnt = -cnt)
        ans = (mod + ans + ck[i] * i % mod * quick_pow(i-1, n-1) * cnt % mod) % mod;
    return ans * cm[k] % mod;
}

int main(){
    init();
    int T;  cin >> T;
    LL n, m, k;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%I64d %I64d %I64d", &n, &m, &k);
        printf("Case #%d: %I64d\n", kase, solve(n, m, k));
    }
    return 0;
}