acm 杭电 train problem ii 1023

#include <iostream>
#include <algorithm>
#include <string>
#include <stdlib.h>
using namespace std;
class big_int : protected string
{
//实际数字123456在内存中以字符串"654321"存放
friend ostream &operator <<(ostream &os, const big_int &right);
friend big_int operator +(const big_int &left, const big_int &right);
friend bool operator ==(const big_int &left, int right);
public:
big_int &operator =(int right);
big_int(int r) {    operator =(r);  }
big_int(const string &r):
string(r) {}
big_int():
string("0") {}
};
ostream &operator <<(ostream &os, const big_int &right)
{
int i = right.size();
while(i > 0)
{
--i;
os << right[i];
}
return os;
}
big_int operator +(const big_int &left, const big_int &right)
{
int iter_l = 0;//左操作数下标
int iter_r = 0;//右操作数下标
char mod = '0';//MOD 10
char carry = '0';//满10进位
string sum;//和
while(iter_l < left.size() && iter_r < right.size())
//当两个操作数都参与运算
{
mod = (left[iter_l] - '0' + right[iter_r] - '0' + carry - '0') % 10 + '0';
carry = (left[iter_l] - '0' + right[iter_r] - '0' + carry - '0') / 10 + '0';
sum += mod;
++iter_l;
++iter_r;
}
const big_int &rest = left.size() > right.size() ? left : right;//应用位数长的那一个操作数
int iter_rest = iter_l;//位数更长的那个操作数的下标
while(iter_rest < rest.size())
//剩下一个操作数
{
mod = (rest[iter_rest] - '0' + carry - '0') % 10 + '0';
carry = (rest[iter_rest] - '0' + carry - '0') / 10 + '0';
sum += mod;
++iter_rest;
}
if(carry != '0')
{
sum += carry;
}
return sum;
}
bool operator ==(const big_int &left, int right)
{
return dynamic_cast<const string&>(left) ==
static_cast<string>(big_int(right));
}
big_int &big_int::operator =(int right)
{
char str[20];
itoa(right, str, 10);
reverse(str, str + strlen(str));
string::operator =(str);
return *this;
}
big_int mat[101][101];
big_int &t(int n, int m)
//计算train排列数目的公式:
//t(n, m) = m,                      (n == 1)
//t(n, m) = t(n-1, m),              (n == m)
//t(n, m) = t(n-1, m) + t(n, m-1),  (其他情况)
{
if(!(mat
[m] == 0))//已经初始化的情形
return mat
[m];
if(n == 1)/*以下都是未初始化的情形*/
return mat
[m] = m;
if(n == m)
return mat
[m] = t(n - 1, m);
return mat
[m] = t(n - 1, m) + t(n, m - 1);
}
int main()
{
int n;
while(cin >> n)
cout << t(n, n) << '/n';
return 0;
}