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;
}