Poj 2084 Game of Connections（高精度卡特兰数）

Game of Connections

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7610 Accepted: 3847

Description

This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, . . . , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. 
And, no two segments are allowed to intersect. 
It's still a simple game, isn't it? But after you've written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs? Life is harder, right?

Input

Each line of the input file will be a single positive number n, except the last line, which is a number -1. 
You may assume that 1 <= n <= 100.

Output

For each n, print in a single line the number of ways to connect the 2n numbers into pairs.

Sample Input

23-1

Sample Output

25

高精度卡特兰数；自大从kuangbin手里买来模板后终于用到了，索性全写了当高精度用吧；加减乘除取余比较都有，简直6的飞起；

//高精度卡特兰数/* *By Kuangbin *输入 cin>>a; *输出 a.printf();*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define Maxn 9999#define MaxSize 1010#define Dlen 4class BigNum{private:    int a[500];    int len;public:    BigNum(){        len=1;        memset(a,0,sizeof(a));    }    BigNum(const int);    BigNum(const char*);    BigNum(const BigNum &);    BigNum &operator=(const BigNum &);    friend istream& operator>>(istream&,BigNum&);    friend ostream& operator<<(ostream&,BigNum&);    BigNum operator+(const BigNum &)const;    BigNum operator-(const BigNum &)const;    BigNum operator*(const BigNum &)const;    BigNum operator/(const int &)const;    BigNum operator^(const int &)const;    int operator%(const int &)const;    bool operator>(const BigNum &T)const;    bool operator>(const int &t)const;    void print();};//讲一个 int 类型的变量转化为大数BigNum::BigNum(const int b){    int c,d=b;    len=0;    memset(a,0,sizeof(a));    while(d>Maxn){        c=d-(d/(Maxn+1)*(Maxn+1));        d=d/(Maxn+1);        a[len++]=c;    }    a[len++]=d;}//将一个字符串类型的变量转化为大数BigNum::BigNum(const char *s){    int t,k,index,L,i;    memset(a,0,sizeof(a));    L=strlen(s);    len=L/Dlen;    if(L%Dlen)        len++;    index=0;    for(i=L-1;i>=0;i-=Dlen){        t=0;        k=i-Dlen+1;        if(k<0)            k=0;        for(int j=k;j<=i;j++)            t=t*10+s[j]-'0';        a[index++]=t;    }}//拷贝构造函数BigNum::BigNum(const BigNum &T):len(T.len){    int i;    memset(a,0,sizeof(a));    for(i=0;i<len;i++)        a[i]=T.a[i];}//重载赋值运算符，大数之间复制运算BigNum & BigNum::operator=(const BigNum &n){    int i;    len=n.len;    memset(a,0,sizeof(a));    for(i=0;i<len;i++)        a[i]=n.a[i];    return *this;}//重载输入运算符istream& operator>>(istream &in,BigNum &b){    char ch[MaxSize*4];    int i=-1;    in>>ch;    int L=strlen(ch);    int count=0,sum=0;    for(i=L-1;i>=0;){        sum=0;        int t=1;        for(int j=0;j<4&&i>=0;j++,i--,t*=10){            sum+=(ch[i]-'0')*t;        }        b.a[count]=sum;        count++;    }    b.len=count++;    return in;}//重载输出运算符ostream& operator<<(ostream& out,BigNum& b){    int i;    cout<<b.a[b.len-1];    for(i=b.len-2;i>=0;i--){        printf("%04d",b.a[i]);    }    return out;}//两个大数之间的相加运算BigNum BigNum::operator+(const BigNum &T)const{    BigNum t(*this);    int i,big;    big=T.len>len?T.len:len;    for(i=0;i<big;i++){        t.a[i]+=T.a[i];        if(t.a[i]>Maxn){            t.a[i+1]++;            t.a[i]-=Maxn+1;        }    }    if(t.a[big]!=0)        t.len=big+1;    else t.len=big;    return t;}//两个大数之间的相减运算BigNum BigNum::operator-(const BigNum &T)const{    int i,j,big;    bool flag;    BigNum t1,t2;    if(*this>T){        t1=*this;        t2=T;        flag=0;    }    else{        t1=T;        t2=*this;        flag=1;    }    big=t1.len;    for(i=0;i<big;i++){        if(t1.a[i]<t2.a[i]){            j=i+1;            while(t1.a[j]==0)                j++;            t1.a[j--]--;            while(j>i)                t1.a[j--]+=Maxn;            t1.a[i]+=Maxn+1-t2.a[i];        }        else t1.a[i]-=t2.a[i];    }    t1.len=big;    while(t1.a[t1.len-1]==0&&t1.len>1){        t1.len--;        big--;    }    if(flag)        t1.a[big-1]=0-t1.a[big-1];    return t1;}//两个大数之间的相乘BigNum BigNum::operator*(const BigNum &T)const{    BigNum ret;    int i,j,up;    int temp,temp1;    for(i=0;i<len;i++){        up=0;        for(j=0;j<T.len;j++){            temp=a[i]*T.a[j]+ret.a[i+j]+up;            if(temp>Maxn){                temp1=temp-temp/(Maxn+1)*(Maxn+1);                up=temp/(Maxn+1);                ret.a[i+j]=temp1;            }            else{                up=0;                ret.a[i+j]=temp;            }        }        if(up!=0)            ret.a[i+j]=up;    }    ret.len=i+j;    while(ret.a[ret.len-1]==0&&ret.len>1) ret.len--;    return ret;}//大数对一个整数进行相除运算BigNum BigNum::operator/(const int &b)const{    BigNum ret;    int i,down=0;    for(i=len-1;i>=0;i--){        ret.a[i]=(a[i]+down*(Maxn+1))/b;        down=a[i]+down*(Maxn+1)-ret.a[i]*b;    }    ret.len=len;    while(ret.a[ret.len-1]==0&&ret.len>1)        ret.len--;    return ret;}//大数对一个 int 类型的变量进行取模int BigNum::operator%(const int &b)const{    int i,d=0;    for(i=len-1;i>=0;i--)        d=((d*(Maxn+1))%b+a[i])%b;    return d;}//大数的n次方运算/*当n<0的时候有错误*/BigNum BigNum::operator^(const int &n)const{    BigNum t,ret(1);    int i;    //if(n<0) exit(-1);    if(n==0)return 1;    if(n==1)return *this;    int m=n;    while(m>1){        t=*this;        for(i=1;(i<<1)<=m;i<<1)            t=t*t;        m-=i;        ret=ret*t;        if(m==1)ret=ret*(*this);    }    return ret;}//大数和另一个大数的大小比较bool BigNum::operator>(const BigNum &T)const{    int ln;    if(len>T.len)return true;    else if(len==T.len){        ln=len-1;        while(a[ln]==T.a[ln]&&ln>=0)            ln--;        if(ln>=0&&a[ln]>T.a[ln])            return true;        else            return false;    }    else        return false;}//大数和一个 int 类型的变量的大小比较bool BigNum::operator>(const int &t)const{    BigNum b(t);    return *this>b;}//输出大数void BigNum::print(){    int i;    printf("%d",a[len-1]);    for(i=len-2;i>=0;i--)        printf("%04d",a[i]);    printf("\n");}BigNum f[110];int main(){    f[0]=1;    for(int i=1;i<=100;i++)        f[i]=f[i-1]*(4*i-2)/(i+1);    int n;    while(scanf("%d",&n)==1){        if(n==-1)            break;        f[n].print();    }    return 0;}