大数模板

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#include <bits/stdc++.h>
using namespace std;

struct BigInteger {
  typedef unsigned long long LL;

  static const int BASE = 100000000;
  static const int WIDTH = 8;
  vector<int> s;

  BigInteger& clean() { while(!s.back() && s.size() > 1) s.pop_back(); return *this; }
  BigInteger(LL num = 0) { *this = num; }
  BigInteger(string s) { *this = s; }
  BigInteger& operator = (long long num) {
    s.clear();
    do {
      s.push_back(num % BASE);
      num /= BASE;
    } while (num > 0);
    return *this;
  }
  BigInteger& operator = (const string& str) {
    s.clear();
    int x, len = (str.length() - 1) / WIDTH + 1;
    for (int i = 0; i < len; i++) {
      int end = str.length() - i * WIDTH;
      int start = max(0, end - WIDTH);
      sscanf(str.substr(start, end - start).c_str(), "%d", &x);
      s.push_back(x);
    }
    return (*this).clean();
  }

  BigInteger operator + (const BigInteger& b) const {
    BigInteger c; c.s.clear();
    for (int i = 0, g = 0; ; i++) {
      if (g == 0 && i >= s.size() && i >= b.s.size()) break;
      int x = g;
      if (i < s.size()) x += s[i];
      if (i < b.s.size()) x += b.s[i];
      c.s.push_back(x % BASE);
      g = x / BASE;
    }
    return c;
  }
  BigInteger operator - (const BigInteger& b) const {
    assert(b <= *this); // 减数不能大于被减数
    BigInteger c; c.s.clear();
    for (int i = 0, g = 0; ; i++) {
      if (g == 0 && i >= s.size() && i >= b.s.size()) break;
      int x = s[i] + g;
      if (i < b.s.size()) x -= b.s[i];
      if (x < 0) { g = -1; x += BASE; } else g = 0;
      c.s.push_back(x);
    }
    return c.clean();
  }
  BigInteger operator * (const BigInteger& b) const {
    vector<LL> v(s.size() + b.s.size(), 0);
    BigInteger c; c.s.clear();
    for (int i = 0; i < s.size(); i++)
      for (int j = 0; j < b.s.size(); j++)
        v[i + j] += LL(s[i]) * b.s[j];
    LL g = 0;
    for (int i = 0; ; i++) {
      if (g == 0 && i >= v.size()) break;
      LL x = v[i] + g;
      c.s.push_back(x % BASE);
      g = x / BASE;
    }
    return c.clean();
  }
  BigInteger operator / (const BigInteger& b) const {
    assert(b > 0);  // 除数必须大于0
    BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大
    BigInteger m;               // 余数:初始化为0
    for (int i = s.size() - 1; i >= 0; i--) {
      m = m * BASE + s[i];
      c.s[i] = bsearch(b, m);
      m -= b * c.s[i];
    }
    return c.clean();
  }
  BigInteger operator % (const BigInteger& b) const { //方法与除法相同
    BigInteger c = *this;
    BigInteger m;
    for (int i = s.size() - 1; i >= 0; i--) {
      m = m * BASE + s[i];
      c.s[i] = bsearch(b, m);
      m -= b * c.s[i];
    }
    return m;
  }
  // 二分法找出满足bx<=m的最大的x
  int bsearch(const BigInteger& b, const BigInteger& m) const {
    int L = 0, R = BASE - 1, x;
    while (1) {
      x = (L + R) >> 1;
      if (b * x <= m) { if (b * (x + 1) > m) return x; else L = x; }
      else R = x;
    }
  }
  BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
  BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
  BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
  BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
  BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}

  bool operator < (const BigInteger& b) const {
    if (s.size() != b.s.size()) return s.size() < b.s.size();
    for (int i = s.size() - 1; i >= 0; i--)
      if (s[i] != b.s[i]) return s[i] < b.s[i];
    return false;
  }
  bool operator >(const BigInteger& b) const {return b < *this;}
  bool operator<=(const BigInteger& b) const {return !(b < *this);}
  bool operator>=(const BigInteger& b) const {return !(*this < b);}
  bool operator!=(const BigInteger& b) const {return b < *this || *this < b;}
  bool operator==(const BigInteger& b) const {return !(b < *this) && !(b > *this);}
};

ostream& operator << (ostream& out, const BigInteger& x) {
  out << x.s.back();
  for (int i = x.s.size() - 2; i >= 0; i--) {
    char buf[20];
    sprintf(buf, "%08d", x.s[i]);
    for (int j = 0; j < strlen(buf); j++) out << buf[j];
  }
  return out;
}

istream& operator >> (istream& in, BigInteger& x) {
  string s;
  if (!(in >> s)) return in;
  x = s;
  return in;
}