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/2][N/2];

void ini()
{
   // memset(C,0,sizeof(C));
    int i,j;
    for(i=0; i<=l/2; ++i)
    {
        C[i][0] = 1;
        C[i][i] = 1;
        for(j=1; j<=l/2; ++j){
            C[i][j] = (C[i-1][j] + C[i-1][j-1]) % r;
        }

    }
}

void solve()
{
    ll i;
    ll num;
    ll st;
    memset(dp,0,sizeof(dp));
    dp[0]=0;
    dp[1]=1;
    dp[2]=dp[3]=1;
    for(i=4;i<=l;i++){
        st=0;
        if(i%2==1){
            st=1;
        }
        for(;3*st<=i;st+=2){
            num=st+(i-3*st)/2;
           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 0;
}