#69. 新年的QAQ

题目描述

http://uoj.ac/problem/69

题解

做之前瞄了先一眼题解，get到了核心部分

因为给的运算符很不靠谱，所以考虑实现一个靠谱的运算符

题解的"正常"做法是按照行号来编号，并不是很明白

正常做法（自己写的）：如果要实现c=a+b，那么搞一个d=c-a-b判断d是否为0来跳转即可，因为判断0的跳转是100%正确的，而且如果结果是错的但d=0的几率可以忽略不计，a%b同理

于是就没了

更加正常的做法：实现c=a+b，把运算做两次之后判断是否相等，给的逻辑运算符很辣鸡，所以直接用两次的差判断0即可

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define ll long long
//#define file
using namespace std;

int main()
{
    #ifdef file
    freopen("prog.txt","w",stdout);
    #endif
    
    cout<<"input n"<<endl;
	cout<<"input m"<<endl;
	cout<<"a = 0"<<endl;
	cout<<"b = 1"<<endl;
	cout<<"c = a + b"<<endl;
	cout<<"z = c - a"<<endl;
	cout<<"z = z - b"<<endl;
	cout<<"if z goto 5"<<endl;
	cout<<"x = c % m"<<endl;
	cout<<"y = c / m"<<endl;
	cout<<"z = y * m"<<endl;
	cout<<"z = z + x"<<endl;
	cout<<"z = z - c"<<endl;
	cout<<"if z goto 9"<<endl;
	cout<<"c = x"<<endl;
	cout<<"a = b"<<endl;
	cout<<"b = c"<<endl;
	cout<<"c = n - 1"<<endl;
	cout<<"z = c + 1"<<endl;
	cout<<"z = z - n"<<endl;
	cout<<"if z goto 18"<<endl;
	cout<<"n = c"<<endl;
	cout<<"if n goto 5"<<endl;
	cout<<"output a"<<endl;
}