很长时间都没写过代码了，试着写了这个常见的题目。整体思路：采用整形链表记录大整数的每一位，然后分别遍历乘数和被乘数的每一位，将每两个数字的乘积累加到结果的相应位上面。针对大整数类型，重载输入和输出流，重载乘法。
输出部分实现不是特别好。用STL的容器实现链表也许会简单的多。重载乘法的实现不合常理，本应该返回一个新的对象，为了简单起见，直接返回了新对象的指针；在计算乘法结果时使用了一个递归，理论上来说有可能深度过大，导致栈溢出；另外对于指针和应用的使用不规范。
具体代码：
/* main.cpp */ #include "utils.h" #include <iostream> #include <string> using std::cin; using std::cout; using std::endl; using std::string; int main(int argc, char** argv) { try { BigInt* pright = new BigInt; pright->data = 9; pright->next = new BigInt; pright->next->data = 9; pright->next->next = new BigInt; pright->next->next->data = 9; cout << "right is: " << *pright << endl; BigInt* pleft = new BigInt; pleft->data = 9; pleft->next = new BigInt; pleft->next->data = 9; pleft->next->next = new BigInt; pleft->next->next->data = 9; cout << "left is: " << *pleft << endl; BigInt* res = (*pleft) * (*pright); cout << "the result is: " << *res << endl; BigInt abc; cout << "pleas input the left num: "; cin >> abc; BigInt def; cout << "pleas input the right num: "; cin >> def; res = abc * def; cout << "the result is: " << *res << endl; } catch (string e) { cout << e; return -1; } return 0; } /* utils.h */ #ifndef __UTILS_MY__ #define __UTILS_MY__ #include <iostream> #include "bigint.h" using std::istream; using std::ostream; BigInt* operator* (BigInt& plhs, BigInt& prhs); istream& operator>> (istream& in, BigInt& n); ostream& operator<< (ostream& out, BigInt& n); #endif /* utils.cpp */ #include <string> #include "utils.h" using std::string; using std::cout; /* * d 的取值范围 [0, 9] */ BigInt* Addonenum (BigInt* pnum, unsigned char d) { if (d > 9 || d == 0) { return pnum; } if (!pnum) { pnum = new BigInt; pnum->data = d; return pnum; } d += pnum->data; if (d > 9) { pnum->next = Addonenum(pnum->next, d / 10); } pnum->data = d % 10; return pnum; } BigInt* operator* (BigInt& plhs, BigInt& prhs) { BigInt* preshead = new BigInt; if (plhs.iszero() || prhs.iszero()) { return preshead; } if ((plhs.isneg() && !prhs.isneg()) || (!plhs.isneg() && prhs.isneg())) { preshead->bneg = true; } BigInt* presnumpos1 = preshead; BigInt* presnumpos2 = preshead; BigInt* plhsnum = &plhs; unsigned char d = 0; while (plhsnum) { presnumpos1 = presnumpos2; BigInt* prhsnum = &prhs; while (prhsnum) { d = prhsnum->data * plhsnum->data; presnumpos1 = Addonenum(presnumpos1, d % 10); presnumpos1->next = Addonenum(presnumpos1->next, d / 10); presnumpos1 = presnumpos1->next; prhsnum = prhsnum->next; } plhsnum = plhsnum->next; presnumpos2 = presnumpos2->next; } return preshead; } istream& operator>> (istream& in, BigInt& n) { BigInt* pn = &n; BigInt* prenode = pn; string src; in >> src; string::reverse_iterator itrbegin = src.rbegin(); string::reverse_iterator itrend = src.rend(); if (src.at(0) == '-') { n.bneg = true; itrend--; } else if (src.at(0) == '+') { itrend--; } for (string::reverse_iterator it = itrbegin; it != itrend; it++) { if (*it > '9' || *it < '0') { throw "input num error: " + src + "!"; } if (!pn) { // new node; pn = prenode->next = new BigInt; prenode = pn; } pn->data = *it - '0'; pn = pn->next; } return in; } ostream& operator<< (ostream& out, BigInt& n) { BigInt* pn = &n; string outstr; BigInt* prev = NULL; BigInt* pout = NULL; while (pn) { pout = new BigInt; pout->data = pn->data; if (prev) { pout->next = prev; } prev = pout; pn = pn->next; } if (n.isneg()) { outstr += '-'; } while (pout) { outstr += (char)(pout->data + '0'); pout = pout->next; } return out << outstr; } /* bitint.h */ #ifndef __BIGINT__ #define __BIGINT__ struct BigInt { public: unsigned char data; BigInt* next; bool bneg; BigInt() { next = NULL; bneg = false; data = 0; } ~BigInt() { if (next) { delete next; next = NULL; } } bool iszero() { BigInt* pn = this; while (pn && pn->data != 0) { return false; pn = pn->next; } return true; } bool isneg() { if (iszero()) { return false; } return bneg; } }; #endif