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;

}