Andrew Stankevich's Contest (21) J dp+组合数

坑爹的，，组合数模板，，，

6132

njczy2010

1412

Accepted

5572 MS

50620 KB

C++

1844 B

2014-10-02 21:41:15

J - 2-3 Trees

Time Limit: 12000/6000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

SubmitStatus

Problem Description

2-3 tree is an elegant data structure invented by John Hopcroft. It is designed to implement the same functionality as the binary search tree. 2-3 tree is an ordered rooted tree with the following properties:

• the root and each internal vertex have either 2 or 3 children;
• the distance from the root to any leaf of the tree is the same.

The only exception is the tree that contains exactly one vertex — in this case the root of the tree is the only vertex, and it is simultaneously a leaf, i.e. has no children. The main idea of the described properties is that the tree with l leaves has the height O(log l).       Given the number of leaves l there can be several valid 2-3 trees that have l leaves. For example, the picture below shows the two possible 2-3 trees with exactly 6 leaves.

Given l find the number of different 2-3 trees that have l leaves. Since this number can be quite large, output it modulo r.

Input

      Input file contains two integer numbers: l and r (1 ≤ l ≤ 5 000, 1 ≤ r ≤ 10⁹).

Output

      Output one number — the number of different 2-3 trees with exactly l leaves modulo r.

Sample Input

6 1000000000
7 1000000000

Sample Output

2
3

题解转自：http://acdream.info/topic?tid=3623

J题：问你一棵有l个叶子的2-3叉树有多少种拓扑结构，结果对r取余。。。 2-3叉树的定义为所有非叶子节点要么有2个儿子，要么有3个儿子，并且所有叶子到根的距离相等。。。 （关于那个O(logn)。。。其实这句话完全是废话，貌似好多人被这句废话给坑到了。。大O只是一个标记，表示渐进的意思，就是说这种树的高度和叶子数n成渐进对数关系）

J题：首先预处理出前2500行的组合数（杨辉三角递推即可，别忘了取余），然后初始化dp[1]=1，若只有根，也有一种情况
然后进行dp，注意到一个性质，2-3树的所有叶子都在同一层，于是就可以通过排列组合来计算叶子的方法数。
于是就转移到了上一层的情况，一层一层记忆化上去，或者递推下来= =。。反正是单组数据~~~

注意用 lld。。。

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<string>
 //#include<pair>

 #define N  5005
 #define M 15
 #define mod 10000007
 //#define p 10000007
 #define mod2 100000000
 #define ll long long
 #define LL long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 ll l,r;
 ll dp[N];
 ll C[N/][N/];

 void ini()
 {
    // memset(C,0,sizeof(C));
     int i,j;
     for(i=; i<=l/; ++i)
     {
         C[i][] = ;
         C[i][i] = ;
         for(j=; j<=l/; ++j){
             C[i][j] = (C[i-][j] + C[i-][j-]) % r;
         }

     }
 }

 void solve()
 {
     ll i;
     ll num;
     ll st;
     memset(dp,,sizeof(dp));
     dp[]=;
     dp[]=;
     dp[]=dp[]=;
     for(i=;i<=l;i++){
         st=;
         if(i%==){
             st=;
         }
         for(;*st<=i;st+=){
             num=st+(i-*st)/;
            dp[i]=(dp[i]+(C[num][st]*dp[num])%r)%r;
         }
     }
 }

 void out()
 {
     //for(int i=1;i<=l;i++) printf(" i=%d dp=%d\n",i,dp[i]);
     printf("%lld\n",dp[l]);
 }

 int main()
 {
    // freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     //scanf("%d",&T);
    // for(int ccnt=1;ccnt<=T;ccnt++)
    // while(T--)
     while(scanf("%lld%lld",&l,&r)!=EOF)
     {
         //if(n==0 && m==0 ) break;
         //printf("Case %d: ",ccnt);
         ini();
         solve();
         out();
     }

     return ;
 }