UVa UVa 10494 - If We Were a Child Again

/* coder: ACboy date: 2010-3-3 result: 1AC description: UVa 10494 - If We Were a Child Again */ #include <iostream> using namespace std; int const MAXN = 800; class BigNumber { public: int s[MAXN]; int len; public: // 去除前导零 void cleanLeadZero(); // 乘以10的n次方 void multiplyTen(int n); // 除以10的n次方 void divisionTen(int n); // 将结果转换成字符串 string str() const; // 构造函数 BigNumber(); BigNumber(int); BigNumber(const char *); // 截取整数的前n位数（例如1234434 调用getSub(3)的话得到的结果是123） BigNumber getSub(int n) const; // 重载赋值运算符 BigNumber operator = (const char *); BigNumber operator = (int num); // 重载加减乘除 BigNumber operator + (const BigNumber &) const; BigNumber operator - (const BigNumber &) const; BigNumber operator * (const BigNumber &) const; BigNumber operator / (const BigNumber &) const; BigNumber operator % (const BigNumber &) const; BigNumber operator -= (const BigNumber &); BigNumber operator += (const BigNumber &); BigNumber operator *= (const BigNumber &); BigNumber operator /= (const BigNumber &); // 重载比较运算符 bool operator < (const BigNumber &) const; bool operator > (const BigNumber &) const; bool operator <= (const BigNumber &) const; bool operator >= (const BigNumber &) const; bool operator == (const BigNumber &) const; // 重载输入输出流 friend istream & operator >> (istream &, BigNumber &); friend ostream & operator << (ostream &, BigNumber &); }; void BigNumber::cleanLeadZero() { while (len > 1 && !s[len - 1]) len--; } void BigNumber::divisionTen(int n) { int i; if (n > len) { while (len >= 1) s[len--] = 0; } else { for (i = 0; i < len - n; i++) { s[i] = s[i + n]; } len -= n; } } void BigNumber::multiplyTen(int n) { if (n > 0) { int i; for (i = len - 1; i >= 0; i--) { s[i + n] = s[i]; } for (i = 0; i < n; i++) { s[i] = 0; } len += n; } } string BigNumber::str() const { string res = ""; // 每个位的数逆序添加到str末尾。 for (int i = 0; i < len; i++) { res = (char)(s[i] + '0') + res; } if (res == "") res = "0"; return res; } BigNumber::BigNumber() { memset(s, 0, sizeof(s)); len = 1; } BigNumber::BigNumber(int num) { *this = num; } BigNumber::BigNumber(const char * num) { *this = num; } BigNumber BigNumber::getSub(int n) const { BigNumber c; c.len = 0; for (int i = 0; i < n; i++) { c.s[c.len++] = s[len - n + i]; } return c; } BigNumber BigNumber::operator = (const char * num) { len = strlen(num); // 整数在s数组中是逆序存放的（如："456" 在s数组中是s[0] = 6, s[1] = 5, s[2] = 4） for (int i = 0; i < len; i++) { s[i] = num[len - i - 1] - '0'; } return *this; } BigNumber BigNumber::operator = (int num) { char s[MAXN]; sprintf(s, "%d", num); *this = s; return *this; } BigNumber BigNumber::operator + (const BigNumber & x) const { BigNumber r; r.len = 0; // up 是用来保持进位的 int i, up; int maxLen = max(len, x.len); for (i = 0, up = 0; up || i < maxLen; i++) { int temp = up; if (i < len) temp += s[i]; if (i < x.len) temp += x.s[i]; up = temp / 10; r.s[r.len++] = temp % 10; } // 去除前导零 r.cleanLeadZero(); return r; } // 减法在使用时要注意在计算a - b时要确保a >= b; // 如果a < b 则计算 先输出一个'-' 再输出b - a 的结果 BigNumber BigNumber::operator - (const BigNumber & b) const { BigNumber c; c.len = 0; int i; // 用来保存退位 int down; for (i = 0, down = 0; i < len; i++) { int temp = s[i] - down; if (i < b.len) temp -= b.s[i]; if (temp >= 0) down = 0; else { down = 1; temp += 10; } c.s[c.len++] = temp; } c.cleanLeadZero(); return c; } BigNumber BigNumber::operator * (const BigNumber & b) const { int i, j; BigNumber c; c.len = len + b.len; for (i = 0; i < len; i++) { for (j = 0; j < b.len; j++) { c.s[i + j] += s[i] * b.s[j]; } } for (i = 0; i < c.len - 1; i++) { c.s[i + 1] += c.s[i] / 10; c.s[i] %= 10; } c.cleanLeadZero(); return c; } BigNumber BigNumber::operator / (const BigNumber & b) const { int i, j; BigNumber r; r.len = 0; // 模拟除法的过程 // 先取blen - 1位 BigNumber temp = this->getSub(b.len - 1); // 一位一位的除从而取得完整的答案 for (i = len - b.len; i >= 0; i--) { // temp用来存储被除数的前blen位。 temp = temp * 10 + s[i]; // 如果temp < b则再在该位的结果为0 if (temp < b) { r.s[r.len++] = 0; } else { // 否则找到第一个j使得b * j的结果大于 temp for (j = 1; j <= 10; j++) { if (b * j > temp) break; } // 因为此时（j - 1） * b小于等于temp，所有j - 1就是在该位除的结果 r.s[r.len++] = j - 1; // temp 减去被减去部分为下一次迭代做准备 temp = temp - (b * (j - 1)); } } // 逆序（因为结果是要逆序存储的，而在求解过程中结果是顺序存储的） for (i = 0; i < r.len / 2; i++) { int temp = r.s[i]; r.s[i] = r.s[r.len - 1 - i]; r.s[r.len - 1 - i] = temp; } r.cleanLeadZero(); return r; } BigNumber BigNumber::operator % (const BigNumber & b) const { BigNumber r; r = *this / b; //cout << r << endl; r = *this - r * b; return r; } BigNumber BigNumber::operator += (const BigNumber & b) { *this = *this + b; return *this; } BigNumber BigNumber::operator -= (const BigNumber & b) { *this = *this - b; return *this; } BigNumber BigNumber::operator *= (const BigNumber & b) { *this = *this * b; return *this; } BigNumber BigNumber::operator /= (const BigNumber & b) { *this = *this / b; return *this; } bool BigNumber::operator < (const BigNumber & b) const { if (len != b.len) return len < b.len; else { for (int i = len - 1; i >= 0; i--) { if (s[i] != b.s[i]) return s[i] < b.s[i]; } } return false; } bool BigNumber::operator > (const BigNumber & b) const { return b < *this; } bool BigNumber::operator <= (const BigNumber & b) const { return !(b > *this); } bool BigNumber::operator >= (const BigNumber & b) const { return !(*this < b); } bool BigNumber::operator == (const BigNumber & b) const { return !(b < *this) && !(b > *this); } istream & operator >> (istream & in, BigNumber & x) { string s; in >> s; x = s.c_str(); return in; } ostream & operator << (ostream & out, BigNumber & x) { out << x.str(); return out; } char a[1000]; char b[100]; char op; int main() { #ifndef ONLINE_JUDGE freopen("10494.txt", "r", stdin); #endif while (scanf("%s %c %s", a, &op, b) != EOF) { BigNumber d(a); BigNumber k(b); d.cleanLeadZero(); k.cleanLeadZero(); if (op == '/') { cout << (d / k).str() << endl; } else { cout << (d % k).str() << endl; } } return 0; }