Game of ConnectionsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3327Accepted Submission(s): 1896
Problem 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
Source
#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>#include <math.h>#include <ctype.h>#include <time.h>#include <queue>#include <iterator>using namespace std;#define MAXN 9999#define MAXSIZE 1010#define DLEN 4class BigNum{private:int a[1000]; //可以控制大数的位数int len;public:BigNum(){ len = 1; memset(a, 0, sizeof(a)); } //构造函数BigNum(const int);//将一个int类型的变量转化成大数BigNum(const char*); //将一个字符串类型的变量转化为大数BigNum(const BigNum &); //拷贝构造函数BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算friend istream& operator>>(istream&, BigNum&); //重载输入运算符friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符BigNum operator+(const BigNum &)const; //重载加法运算符，两个大数之间的相加运算BigNum operator-(const BigNum &)const; //重载减法运算符，两个大数之间的相减运算BigNum operator*(const BigNum &)const; //重载乘法运算符，两个大数之间的相乘运算BigNum operator/(const int &)const;//重载除法运算符，，大数对一个整数进行相除运算BigNum operator^(const int &)const;//大数的n次方运算int operator%(const int &)const;//大数对一个int类型的变量进行取模运算bool operator>(const BigNum &T)const; //大数和另一个大数的大小比较bool operator>(const int &t)const;//大数和一个int类型的变量的大小比较void print();//输出大数};BigNum::BigNum(const int b) //将一个int类型的变量转化为大数{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;}BigNum::BigNum(const char *s) //将一个字符串类型的变量转化为大数{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;}}BigNum::BigNum(const BigNum &T) :len(T.len) //拷贝构造函数{int i;memset(a, 0, sizeof(a));for (i = 0; i<len; i++)a[i] = T.a[i];}BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符，大数之间赋值运算{int i;len = n.len;memset(a, 0, sizeof(a));for (i = 0; i<len; i++)a[i] = n.a[i];return *this;}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;}ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符{int i;cout << b.a[b.len – 1];for (i = b.len – 2; i >= 0; i–){printf("%04d", b.a[i]);}return out;}BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算{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;}BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算{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;}BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘{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;}BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算{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;}int BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模{int i, d = 0;for (i = len – 1; i >= 0; i–)d = ((d*(MAXN + 1)) % b + a[i]) % b;return d;}BigNum BigNum::operator^(const int &n)const //大数的n次方运算{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;}bool BigNum::operator>(const BigNum &T)const //大数和另一个大数的大小比较{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;elsereturn false;}elsereturn false;}bool BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较{BigNum b(t);return *this>b;}void BigNum::print() //输出大数{int i;printf("%d", a[len – 1]);for (i = len – 2; i >= 0; i–)printf("%04d", a[i]);printf("\n");}BigNum p[3010];int main(){int i, n;p[0] = 1;for (int i = 1; i<= 100; i++)p[i] = p[i – 1] * (4 * i – 2) / (i + 1);while (cin>>n && n!=-1){p[n].print();}return 0;}
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