POJ 2084 Game of Connections

 

Game of Connections

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 4839 Accepted: 2515

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

Shanghai 2004 Preliminary

 

 

/*
卡特兰数 + 大数运算，一开始自己推公式得到的结论是:
f(2 * n) = f(0) * f(2 * n - 2) + f(2) * f(2 * n - 4) + f(4) * f(2 * n - 6) + ... + f(2 * n - 2) * f(0);
然后纯粹基于递推来算, 发现结果是对的，但是速度太慢，超时了。后来想了想，这不就是卡特兰数吗？呵呵，问题解决了
F(n) = F(0) * F(n - 1) + F(1) * F(n - 2) + ... + F(n - 1) * F(0) = C(2 * n, n) / (n + 1), F(0) = 1
代码很长，大部分是自己写的大数运算工具模板.实在不想用JAVA水了，没意思，哈哈
*/
#include <iostream>#include <string>#define BIGINTEGER_LEN_TYPE int#define MAX_N 201using namespace std;string num[MAX_N];string cval[MAX_N][MAX_N];//大数加法string bigIntegerAdd(const string left, const string right){ BIGINTEGER_LEN_TYPE lenl = static_cast<BIGINTEGER_LEN_TYPE>(left.size()); BIGINTEGER_LEN_TYPE lenr = static_cast<BIGINTEGER_LEN_TYPE>(right.size()); std::string res = ""; //get the minimum length of the two string BIGINTEGER_LEN_TYPE minLen = (lenl <= lenr) ? lenl : lenr; BIGINTEGER_LEN_TYPE maxLen = (lenl <= lenr) ? lenr : lenl; std::string longerStr = (lenl <= lenr) ? right : left; int residue = 0, sum = 0, quotient = 0; //add operation, first part BIGINTEGER_LEN_TYPE pos1; for(pos1 = 0; pos1 <= minLen - 1; pos1++) { sum = (left[pos1] - '0') + (right[pos1] - '0') + quotient; quotient = sum / 10; residue = sum % 10; res += residue + '0'; } //add operation, second part for(BIGINTEGER_LEN_TYPE pos2 = pos1; pos2 <= maxLen - 1; pos2 ++) { sum = (longerStr[pos2] - '0') + quotient; quotient = sum / 10; residue = sum % 10; res += residue + '0'; } //add the extra carry while(quotient != 0) { residue = quotient % 10; quotient = quotient / 10; res += residue + '0'; } //std::cout<<"res size:"<<res.length()<<std::endl; return res;}//大数乘法string bigIntegerMult(const string left, const string right){ BIGINTEGER_LEN_TYPE minLen = static_cast<BIGINTEGER_LEN_TYPE>(left.length()); BIGINTEGER_LEN_TYPE maxLen = static_cast<BIGINTEGER_LEN_TYPE>(right.length()); std::string longerStr = ""; std::string shorterStr = ""; std::string res = ""; //get the longer and short strings and corresponded length if(left.length() <= right.length()) { minLen = static_cast<BIGINTEGER_LEN_TYPE>(left.length()); maxLen = static_cast<BIGINTEGER_LEN_TYPE>(right.length()); shorterStr = left; longerStr = right; } else { minLen = static_cast<BIGINTEGER_LEN_TYPE>(right.length()); maxLen = static_cast<BIGINTEGER_LEN_TYPE>(left.length()); shorterStr = right; longerStr = left; } BIGINTEGER_LEN_TYPE pos1 = 0; BIGINTEGER_LEN_TYPE pos2 = 0; //start from the first digit of the shorter string for(pos1 = 0; pos1 <= minLen - 1; pos1++) { //multiply each digit of the longer string int residue = 0, product = 0, quotient = 0; BIGINTEGER_LEN_TYPE curPos = 0; for(pos2 = 0; pos2 <= maxLen - 1; pos2++) { curPos = pos1 + pos2; //new space if(res.length() == 0 || curPos > int(res.length() - 1)) { product = (shorterStr[pos1] - '0') * (longerStr[pos2] - '0') + quotient; quotient = product / 10; residue = product % 10; res += residue + '0'; } //space already exists, just add the original value to product else { product = (shorterStr[pos1] - '0') * (longerStr[pos2] - '0') + quotient + (res[curPos] - '0'); quotient = product / 10; residue = product % 10; res[curPos] = residue + '0'; } } //process the extra quotient //start position curPos = maxLen + pos1; while(quotient != 0) { if(res.length() == 0 || curPos > int(res.length() - 1)) { residue = quotient % 10; quotient = quotient / 10; res += residue + '0'; } else { quotient = quotient + (res[curPos] - '0'); residue = quotient % 10; quotient = quotient / 10; res[curPos] = residue + '0'; } curPos++; } } //std::cout<<res.length()<<std::endl; return res; }//对字符串str进行翻转string getReverse(const string str){ std::string revStr = ""; BIGINTEGER_LEN_TYPE pos = static_cast<BIGINTEGER_LEN_TYPE>(str.length()) - 1; for(; pos >= 0; pos--) revStr += str[pos]; return revStr;}//比较两个大数的大小int bigIntegerCmp(const string left, const string right){ BIGINTEGER_LEN_TYPE lenl = static_cast<BIGINTEGER_LEN_TYPE>(left.length()); BIGINTEGER_LEN_TYPE lenr = static_cast<BIGINTEGER_LEN_TYPE>(right.length()); if(lenl < lenr) return -1; else if(lenl > lenr) return 1; return strcmp(getReverse(left).c_str(), getReverse(right).c_str());}//大数减法string bigIntegerSub(const string left, const string right){ std::string biggerStr = ""; std::string smallerStr = ""; std::string res = ""; BIGINTEGER_LEN_TYPE minLen = 0; BIGINTEGER_LEN_TYPE maxLen = 0; //get the bigger integer and smaller integer int cmpRes = bigIntegerCmp(left, right); if(cmpRes == -1 || cmpRes == 0) { smallerStr = left; biggerStr = right; } else { smallerStr = right; biggerStr = left; } //get related size minLen = static_cast<BIGINTEGER_LEN_TYPE>(smallerStr.length()); maxLen = static_cast<BIGINTEGER_LEN_TYPE>(biggerStr.length()); int extra = 0, temp = 0, subRes = 0; //subtraction operation, first part BIGINTEGER_LEN_TYPE pos1; for(pos1 = 0; pos1 <= minLen - 1; pos1++) { int curL = biggerStr[pos1] - '0'; int curR = smallerStr[pos1] - '0'; temp = 0; while((subRes = (curL - curR + 10 * temp - extra)) < 0) temp++; res += subRes + '0'; extra = temp; } //subtraction operation, second part for(BIGINTEGER_LEN_TYPE pos2 = pos1; pos2 <= maxLen - 1; pos2 ++) { int curL = biggerStr[pos2] - '0'; temp = 0; while((subRes = (curL + 10 * temp - extra)) < 0) temp++; res += subRes + '0'; extra = temp; } //remove the zero at head while(res[res.length() - 1] == '0' && res.length() >= 2) res = res.substr(0, res.length() - 1); //std::cout<<"res size:"<<res.length()<<std::endl; return res;}//把整数转换为字符串string getStr(int a){ string res = ""; if(a == 0) return "0"; else { while(a) { res = char(a % 10 + '0') + res; a = a / 10; } } return res;}//大数除法void bigIntegerDiv(const std::string left, const std::string right, std::string & quotient, std::string & residual){ int type = bigIntegerCmp(left, right); //left < right if(type == -1) { quotient = "0"; residual = left; return; } //left = right else if(type == 0) { quotient = "1"; residual = "0"; return; } //left > right //initialize the related data quotient = ""; residual = ""; std::string dividend = left; std::string divisor = right; BIGINTEGER_LEN_TYPE lenDend = static_cast<BIGINTEGER_LEN_TYPE>(left.length()); BIGINTEGER_LEN_TYPE lenDsor = static_cast<BIGINTEGER_LEN_TYPE>(right.length()); std::string temp = ""; std::string curStr = ""; std::string tryStr = ""; BIGINTEGER_LEN_TYPE curPos = 0; int choice = 0; bool visitedFirst = false; //start from the head of the dividend for(curPos = lenDend - 1; curPos >= 0; curPos--) { curStr = dividend[curPos] + curStr; //curStr > divisor, can be processed if(bigIntegerCmp(curStr, divisor) >= 0) { //try the quotient from 1 to 10 until the result of choice * divisor is the biggest less than curStr for(choice = 1; choice <= 10; choice++) { temp = ""; temp += char(choice + '0'); tryStr = bigIntegerMult(divisor, temp); if(bigIntegerCmp(tryStr, curStr) > 0) { choice--; break; } } //right now, choice is just the current quotient and should be appended ahead to quotient quotient = char(choice + '0') + quotient; temp = ""; temp += char(choice + '0'); //update curStr curStr = bigIntegerSub(curStr, bigIntegerMult(divisor, temp)); if(curPos == 0) residual = curStr; //set visitedFirst, visitedFirst is used to prevent producing '0' quotient visitedFirst = true; } //curStr is not enough to be divided by divisor, so the current quotient is '0' and should be appended to 'quotient' else if(visitedFirst) quotient = '0' + quotient; if(curPos == 0) { residual = curStr; break; } if(curStr == "0") curStr = ""; } residual = curStr; return;}string getCval(int p, int q){ if(p == q || q == 0) return "1"; if(q == 1) return getReverse(getStr(p)); if(cval[p][q] != "0") return cval[p][q]; return cval[p][q] = bigIntegerAdd(getCval(p - 1, q - 1), getCval(p - 1, q));} //计算k的卡特兰数，下标从0开始string solve1(int k){ if(num[k] != "0") return num[k]; string quotient, residual; bigIntegerDiv(getCval(2 * k, k), getReverse(getStr(k + 1)), quotient, residual); return quotient;}int main(){ int i, j, p; for(i = 1; i < MAX_N; i++) { num[i] = "0"; for(j = 1; j < MAX_N; j++) cval[i][j] = "0"; } num[1] = "1"; while(scanf("%I64d", &p) && p != -1) cout<<getReverse(solve1(p))<<endl; return 0;}