PKU 2084 Game of Connections

Game of Connections

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7567 Accepted: 3832

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

2
5
我滴个乖乖，《算法竞赛入门经典》给的公式是错误的！！！我算了好久，但是找到了卡特兰数的模板。高精度不会，还是先学会这种吧。
#include <iostream>#include <algorithm>#include <string>#include <cstdio>#include <cstring>using namespace std;int a[105][1050],b[1000];void catalan() //求卡特兰数{    int i, j, len, carry, temp;    a[1][0] = b[1] = 1;    len = 1;    for(i = 2; i <= 100; i++)    {        for(j = 0; j < len; j++) //乘法        a[i][j] = a[i-1][j]*(4*(i-1)+2);        carry = 0;        for(j = 0; j < len; j++) //处理相乘结果        {            temp = a[i][j] + carry;            a[i][j] = temp % 10;            carry = temp / 10;        }        while(carry) //进位处理        {            a[i][len++] = carry % 10;            carry /= 10;        }        carry = 0;        for(j = len-1; j >= 0; j--) //除法        {            temp = carry*10 + a[i][j];            a[i][j] = temp/(i+1);            carry = temp%(i+1);        }        while(!a[i][len-1]) //高位零处理        len --;        b[i] = len;    }}int main(void){    memset(a,0,sizeof(a));  catalan();    int n;    while(cin>>n&&n!=-1)    {    //    for( j=999;j>=0;j--) if(a[n][j]) break;        for(int i=b[n]-1;i>=0;i--) cout<<a[n][i];        cout<<endl;       // cout<<a[n]<<endl;    }    return 0;}