HDOJ1133

点击打开链接

卡特兰数的运用
(C(m+n,n)-C(m+n,m+1))*m!*n!化简即(m+n)!*(m-n+1)/(m+1)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long  s[205][5002];
long long  tem[205][5002];
long long  temp[205][5002];
void f()
{
    int i , j;
    memset(s,0,sizeof(s));
    s[1][0]=1;s[2][0]=2;s[3][0]=6;
    for(i=4;i<205;i++)
    {
        for(j=0;j<5002;j++)
        {
            s[i][j]=s[i-1][j]*i;
        }
        for(j=0;j<5002;j++)
        {
            if(s[i][j]>=10)
            {
                
                s[i][j+1]+=(s[i][j]/10);
                s[i][j]%=10;
            }
        }
    }
}
int main()
{
    f();
    int i, j, d, t1, t2, count = 1;
    int m,n;
    while(cin>>m>>n)
    {
        if(m==0 && n==0)
            break;
        if(n>m)
        {
            cout<<"Test #"<<count<<":"<<endl;
            cout<<0<<endl;
            count++;
            continue;
        }
        t1 = (m-n+1);
        t2 = m+1;
        memset(tem,0,sizeof(tem));
        for(i=0;i<5002;i++)
        {
            tem[m+n][i] += s[m+n][i]*t1;
            tem[m+n][i+1] += tem[m+n][i]/10;
            tem[m+n][i] %= 10;
        }   
        memset(temp,0,sizeof(temp));
        d = 0;
        for(i=5001;i>=0;i--)
        {
            d=d*10+tem[m+n][i];
            temp[m+n][i]=d/t2;
            d%=t2;
        }
        
        cout<<"Test #"<<count<<":"<<endl;
        count++;
        i = 5001;
        while(i--)
        {
            if(temp[m+n][i]!=0)
                break;
        }
        for(;i>=0;i--)
        {
            cout<<temp[m+n][i];
        }
        cout<<endl;
    }
    return 0;
}