HDOJ 4359 Easy Tree DP? DP

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Easy Tree DP?

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1482 Accepted Submission(s): 565

Problem Description

A Bear tree is a binary tree with such properties : each node has a value of 2⁰,2¹…2^(N-1)(each number used only once),and for each node ,its left subtree’s elements’ sum<its right subtree’s elements’ sum(if the node hasn’t left/right subtree ,this limitation is invalid).
You need to calculate how many Bear trees with N nodes and exactly D deeps.

Input

First a integer T(T<=5000),then T lines follow ,every line has two positive integer N,D.(1<=D<=N<=360).

Output

For each test case, print "Case #t:" first, in which t is the number of the test case starting from 1 and the number of Bear tree.(mod 10⁹+7)

Sample Input

22 24 3

Sample Output

Case #1: 4Case #2: 72

Author

smxrwzdx@UESTC_Brightroar

Source

2012 Multi-University Training Contest 6

/* ***********************************************Author        :CKbossCreated Time  :2015年09月04日 星期五 16时40分18秒File Name     :1009.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;typedef long long int LL;const int maxn=400;const LL mod=1e9+7;LL dp[maxn][maxn];LL C[maxn][maxn];void init(){for(int i=1;i<maxn;i++) C[i][0]=C[i][i]=1LL;for(int i=1;i<maxn;i++){for(int j=1;j<i;j++){C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;}}for(int i=1;i<=360;i++) dp[1][i]=1;for(int i=2;i<=360;i++) dp[2][i]=4;for(int i=3;i<=360;i++){for(int j=1;j<=360;j++){dp[i][j]=(dp[i][j]+dp[i-1][j-1]*2*i)%mod;for(int left=1;left<=i-2;left++){int right=i-1-left;dp[i][j]=(dp[i][j]+dp[left][j-1]*dp[right][j-1]%mod*i%mod*C[i-2][left]%mod)%mod;}}}}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    init();int T_T,cas=1,n,d;scanf("%d",&T_T);while(T_T--){scanf("%d%d",&n,&d);printf("Case #%d: %lld\n",cas++,(dp[n][d]-dp[n][d-1]+mod)%mod);}return 0;}

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