Uva10106 - Product

求两个大数的乘积

#include <iostream>#include <cstdlib>#include <cstring>#include <string>#include <cstdio>const int maxn = 505;using namespace std;struct bign{  int len, s[maxn];  bign() {    memset(s, 0, sizeof(s));    len = 1;  }  bign(int num) {    *this = num;  }  bign(const char* num) {    *this = num;  }  bign operator = (int num) {    char s[maxn];    sprintf(s, "%d", num);    *this = s;    return *this;  }  bign operator = (const char* num) {    len = strlen(num);    for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';    return *this;  }  string str() const {    string res = "";    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;    if(res == "") res = "0";    return res;  }  bign operator + (const bign& b) const{    bign c;    c.len = 0;    for(int i = 0, g = 0; g || i < max(len, b.len); i++) {      int x = g;      if(i < len) x += s[i];      if(i < b.len) x += b.s[i];      c.s[c.len++] = x % 10;      g = x / 10;    }    return c;  }  void clean() {    while(len > 1 && !s[len-1]) len--;  }  bign operator * (const bign& b) {    bign c; c.len = len + b.len;    for(int i = 0; i < len; i++)      for(int j = 0; j < b.len; j++)        c.s[i+j] += s[i] * b.s[j];    for(int i = 0; i < c.len-1; i++){      c.s[i+1] += c.s[i] / 10;      c.s[i] %= 10;    }    c.clean();    return c;  }  bign operator - (const bign& b) {    bign c; c.len = 0;    for(int i = 0, g = 0; i < len; i++) {      int x = s[i] - g;      if(i < b.len) x -= b.s[i];      if(x >= 0) g = 0;      else {        g = 1;        x += 10;      }      c.s[c.len++] = x;    }    c.clean();    return c;  }  bool operator < (const bign& b) const{    if(len != b.len) return len < b.len;    for(int i = len-1; i >= 0; i--)      if(s[i] != b.s[i]) return s[i] < b.s[i];    return false;  }  bool operator > (const bign& b) const{    return b < *this;  }  bool operator <= (const bign& b) {    return !(b > *this);  }  bool operator == (const bign& b) {    return !(b < *this) && !(*this < b);  }  bign operator += (const bign& b) {    *this = *this + b;    return *this;  }};istream& operator >> (istream &in, bign& x) {  string s;  in >> s;  x = s.c_str();  return in;}ostream& operator << (ostream &out, const bign& x) {  out << x.str();  return out;}int main() {    bign a, b, ans;    while(cin >> a >> b) {        cout << a * b << endl;    }    return 0;}