在日常使用c++的過程中，經常會遇到數字太大越界的情況，對於這樣的大整數運算，我們可以用模擬比算的方法來實現，但是這樣每次運算都要實現這樣的演算法會帶來一定的不方便，我們希望能像int這樣的內建型別一樣使用大整數，所以我們實現一個大整數struct

感謝劉汝佳老師的演算法競賽入門經典一書

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

	BigInteger(long long num = 0) { *this = num; } // 建構函式
	BigInteger operator = (long long num) { // 賦值運算子
		s.clear();
		do {
			s.push_back(num % BASE);
			num /= BASE;
		} while (num > 0);
		return *this;
	}
	BigInteger operator = (const string& str) { // 賦值運算子
		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;
	}
	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;
	}
};

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;
}