ZOJ-1279

蛋疼的模拟题，大致就是模拟加法操作和左移右移操作，注意左移或右移后两个数位数可能不一样了，做加法时要小心，本题V,U,C,D分别可用0，1，2，3表示，然后加法是4进制的，观察那个表格就能得出结论。在一些小坑上错了半天，智商捉鸡啊

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

namespace
{
	string to_num(const string &s)
	{
		string t;
		for (size_t i = 0; i < s.size(); i++)
			if (s[i] == 'V')
				t += '0';
			else if (s[i] == 'U')
				t += '1';
			else if (s[i] == 'C')
				t += '2';
			else
				t += '3';
		return t;
	}

	string to_str(const string &s)
	{
		string t;
		for (size_t i = 0; i < s.size(); i++)
			if (s[i] == '0')
				t += 'V';
			else if (s[i] == '1')
				t += 'U';
			else if (s[i] == '2')
				t += 'C';
			else
				t += 'D';
		return t;
	}

	string add(string &s1, string &s2)
	{
		reverse(s1.begin(), s1.end());
		reverse(s2.begin(), s2.end());
		int carry = 0;
		string res;
		size_t i = 0;
		while (i < s1.size() || i < s2.size())
		{
			int a = i < s1.size() ? (s1[i] - '0') : 0;
			int b = i < s2.size() ? (s2[i] - '0') : 0;
			int sum = a + b + carry;
			carry = sum / 4;
			res += sum % 4 + '0';
			i++;
		}
		if (carry)
			res += '1';
		reverse(s1.begin(), s1.end());
		reverse(s2.begin(), s2.end());
		reverse(res.begin(), res.end());
		return res;
	}

	void shift_right(string &s)
	{
		s = '0' + s;
		s.erase(s.size() - 1, 1);
	}

	void shift_left(string &s)
	{
		s += '0';
	}

	void append_head(string &s)
	{
		while (s.size() < 8)
			s = '0' + s;
	}
}

int main()
{
	int N;
	cin >> N;
	cout << "COWCULATIONS OUTPUT" << endl;
	while (N--)
	{
		string s1, s2, t1, t2, op;
		cin >> s1 >> s2;
		t1 = to_num(s1);
		t2 = to_num(s2);
		for (int i = 0; i < 3; i++)
		{
			cin >> op;
			if (op == "A")
				t2 = add(t1, t2);
			else if (op == "R")
				shift_right(t2);
			else if (op == "L")
				shift_left(t2);
		}
		cin >> op;
		append_head(t2);
		cout << (to_str(t2) == op ? "YES" : "NO") << endl;
	}
	cout << "END OF OUTPUT" << endl;
	return 0;
}