目录
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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 | #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; } |
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