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 {
int i, j; LL g;
vector<LL> v(s.size() + b.s.size(), 0);
BigInteger c; c.s.clear();
for (i = 0; i < s.size(); i++) for (j = 0; j < b.s.size(); j++) v[i + j] += LL(s[i]) * b.s[j];
for (i = 0, g = 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;
}
BigInteger gcd(BigInteger m, BigInteger n)
{
return n == 0 ? m : gcd(n, m % n);
}
BigInteger lcm(BigInteger m, BigInteger n)
{
return n * m / gcd(m, n);
}