HDU 3240 counting Binary Trees)

Counting Binary Trees

Problem descriptionthere are 5 distinct Binary Trees of 3 nodes:

Let T (n) be the number of distinct non-empty Binary Trees of no more NNodes, your task is to calculate T (n) mod M.
Inputthe input contains at most 10 test cases. each case contains two integers n and M (1 <= n <= 100,000, 1 <= m <= 109) on a single line. the input ends with N = m = 0.
Outputfor each test case, print T (n) mod m.
Sample Input

3 1004 100 0

 
Sample output

82

 
Source2009 "nit Cup" national invitational contest
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Question:

Ask you the number of Binary Tree solutions with no more than N nodes, and the result requires M remainder.

Solution:

The number of schemes can be divided between the left and right sides, that is, H (n) = H (0) * H (n-1) + H (1) * (H-2) + .................. + H (n-1) * H (0), which is the number of catlands,

The number of catlands must be about the efficiency of NLG ^ n.

Baidu:

H (0) = 1, H (1) = 1,

The Catalan number meets the following requirements:

Recursive Formula [1]: H (n) = H (0) * H (n-1) + H (1) * H (n-2) +... + H (n-1) H (0) (N> = 2)
Recursive Formula [2]: H (n) = H (n-1) * (4 * N-2)/(n + 1 );
Recursive Formula [3]: H (n) = C (2n, N)/(n + 1) (n =, 2 ,...)
Recursive Formula [4]: H (n) = C (2n, n)-C (2n, n + 1) (n =, 2 ,...)

We choose, recursive formula [2]: H (n) = H (n-1) * (4 * N-2)/(n + 1 );

However, this recursion requires the remainder, and the multiplication requires the remainder. Division requires the inverse element. The necessary and sufficient condition for the Division is gcd (A, MoD) = 1, that is, the mass of the divisor and the number to be modeled.

So, for MOD decomposition factors, for factors in denominator

(1) Calculate the inverse element directly for the denominator of each other.

(2) numbers without mutual quality cannot reverse the element, and only molecules can be used for elimination.

Solution code:

#include <iostream>#include <cstdio>#include <map>#include <vector>#include <cstring>using namespace std;typedef long long ll;const int maxn=50;int tol,prime[maxn],cnt[maxn],n,mod;void getPrime(){    tol=0;    int x=mod;    for(int i=2;i*i<=x;i++){        if(x%i==0){            prime[tol++]=i;            while(x%i==0) x/=i;        }    }    if(x>1) prime[tol++]=x;}void extend_gcd(int a,int b,int &x,int &y){     if(b==0){        x=1;        y=0;     }else{        extend_gcd(b,a%b,x,y);        int tmp=x;        x=y;        y=tmp-(a/b)*y;     }}int inv(int a,int mod){    int x,y;    extend_gcd(a,mod,x,y);    return (x%mod+mod)%mod;}void deal1(ll &ret,int x){    for(int i=0;i<tol;i++){        while(x%prime[i]==0){            x/=prime[i];            cnt[i]++;        }    }    ret=(ret*x)%mod;}void deal2(ll &ret,int x){    for(int i=0;i<tol;i++){        while(x%prime[i]==0){            x/=prime[i];            cnt[i]--;        }    }    if(x>1){        int tmp=inv(x,mod);        ret=(ret*tmp)%mod;    }}ll pow_mod(ll a,ll b,ll p){    ll sum=1;    while(b>0){        if(b&1) sum=(sum*a)%p;        a=(a*a)%p;        b/=2;    }    return sum%p;}void solve(){    ll ans=1,ret=1;    memset(cnt,0,sizeof(cnt));    getPrime();    for(int i=2;i<=n;i++){        deal1(ret,4*i-2);        deal2(ret,i+1);        ll tmp=ret;        for(int t=0;t<tol;t++){            tmp=( tmp*pow_mod(prime[t],cnt[t],mod) )%mod;        }        ans=(ans+tmp)%mod;    }    cout<<ans<<endl;}int main(){    while(cin>>n>>mod && (n||mod) ){        solve();    }    return 0;}

HDU 3240 counting Binary Trees)