Uva 10247 (组合计数)
2016-07-27 14:49
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题目:
有一颗 K 叉完全树, 深度 D,有 N 个节点, 给你 [1 - N] 给这颗树的节点标号。父节点 要比所有子节点的标号小。 问有多少种标号方法。
分析:
题目的标号是不允许重复的。所以根节点的表示是最小的。对于一颗 K 叉 D 深度的树, 其节点数为 Nk,d, 首先根节点的标号数是最小的。
然后就会剩下 N-1 个标号, 就相当于把 N-1个数分为 K 组, 每组 Nk,d−1个
对于这些分配方案就有 T=∏(Nk,d−1−k∗Nk,d−1Nk,d−1);
每颗子树的标号方案是互不影响的, 所以有答案: Ansk,d=T∗(Ansk,d−1)k
Code:
import java.math.*; import java.util.*; public class Main { public static int[][] N = new int[25][25]; public static BigInteger[][] Ans = new BigInteger[25][25]; public static BigInteger C(int m, int n) { if(n > m) return BigInteger.ZERO; if(n > (m-n)) return C(m,m-n); BigInteger ans = BigInteger.ONE; //System.out.println(n); for(int i = 0; i < n; ++i) { ans = ans.multiply(BigInteger.valueOf((long)(m-i))).divide(BigInteger.valueOf((long)(i+1))); } return ans; } public static void INIT() { for(int i = 1; i < 25; ++i) { int tmp = 1; N[i][0] = 1; for(int j = 1; j < 25; ++j) { tmp = tmp * i; N[i][j] = N[i][j-1] + tmp; } } } public static void GetAns() { for(int i = 1; i < 25; ++i) { Ans[i][0] = BigInteger.ONE; for(int j = 1; i*j < 25; ++j) { Ans[i][j] = BigInteger.ONE; int n = N[i][j-1]; int m = N[i][j]; for(int k = 0; k < i; ++k) { Ans[i][j] = Ans[i][j].multiply(Ans[i][j-1]); if(m-1-k*n >= n) { Ans[i][j] = Ans[i][j].multiply(C(m-1-k*n, n)); } //System.out.println(Ans[i][j]); } } } } public static void main(String[] args) { INIT(); GetAns(); Scanner cin = new Scanner(System.in); while(cin.hasNext()) { System.out.println(Ans[cin.nextInt()][cin.nextInt()]); //System.out.println(C(cin.nextInt(), cin.nextInt())); } } }
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