这题非常难，其他的两篇题解使用了cout或者printf的高级算法，这里介绍一个基础的做法——手搓 64 位 0.5MB 内存的机器码虚拟机。

这是虚拟机框架：

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned int
#define N 65536
int memory[N];
ull code[N] = {

};
/*
0 Input
1 Output
2 Write
3 Copy
4 Calculate (& | ~ ^ << >>)
5 Calculate (+ - * / %)
6 Goto
7 If-goto
8 x++
9 x--
f Exit
*/
signed main()
{
	for(int i=0;;i=(i+1)%N)
	{
		int op=code[i]>>60;
		if(op==0)		//Input
		{
			int p=code[i]&65535;
			cin >> memory[ (code[i]>>56)&1 ? memory[p] : p ];
		}
		else if(op==1)	//Output
		{
			int p=code[i]&65535, val = memory[ (code[i]>>56)&1 ? memory[p] : p ];
			if((code[i]>>57)&1)
			{
				cout<<char(val&127);
			}
			else
			{
				cout<<val;
			}
		}
		else if(op==2)	//Write
		{
			int p = (code[i]>>32)&65535, val = code[i]&((1ll<<32)-1);
			memory[ (code[i]>>56)&1 ? memory[p] : p ] = val;
		}
		else if(op==3)	//Copy
		{
			int pf = (code[i]>>16)&65535, pt = code[i]&65535;
			int val = memory[ (code[i]>>55)&1 ? memory[pf] : pf ];
			memory[ (code[i]>>56)&1 ? memory[pt] : pt ] = val;
		}
		else if(op==4)	//Calculate (& | ~ ^ << >>)
		{
			int p1 = (code[i]>>32)&65535, p2 = (code[i]>>16)&65535, p3 = code[i]&65535;
			int val1 = memory[ (code[i]>>58)&1 ? memory[p1] : p1 ],
			val2 = memory[ (code[i]>>57)&1 ? memory[p2] : p2 ], val3;
			int op2 = (code[i]>>52)&15;
			if(op2==0)
			{
				val3=val1&val2;
			}
			else if(op2==1)
			{
				val3=val1|val2;
			}
			else if(op2==2)
			{
				val3=~val2;
			}
			else if(op2==3)
			{
				val3=val1^val2;
			}
			else if(op2==4)
			{
				val3=((val2>>6)?0:val1<<val2);
			}
			else
			{
				val3=((val2>>6)?0:val1>>val2);
			}
			memory[ (code[i]>>56)&1 ? memory[p3] : p3 ] = val3;
		}
		else if(op==5)	//Calculate (+ - * / %)
		{
			//想要乘方的建议自己写一个快速幂, 应该是能写的 
			int p1 = (code[i]>>32)&65535, p2 = (code[i]>>16)&65535, p3 = code[i]&65535;
			int val1 = memory[ (code[i]>>58)&1 ? memory[p1] : p1 ],
			val2 = memory[ (code[i]>>57)&1 ? memory[p2] : p2 ], val3;
			int op2 = (code[i]>>52)&15;
			if(op2==0)
			{
				val3=val1+val2;
			}
			else if(op2==1)
			{
				val3=val1-val2;
			}
			else if(op2==2)
			{
				val3=val1*val2;
			}
			else if(op2==3)
			{
				val3=(val2?val1/val2:0);
			}
			else
			{
				val3=(val2?val1%val2:0);
			}
			memory[ (code[i]>>56)&1 ? memory[p3] : p3 ] = val3;
		}
		else if(op==6)	//Goto
		{
			int p=code[i]&65535;
			i = (code[i]>>56)&1 ? memory[p] : p;
			i = (i+N-1)%N;
		}
		else if(op==7)	//If-goto (> < == >= <= !=)
		{
			int p1 = (code[i]>>32)&65535, p2 = (code[i]>>16)&65535, p3 = code[i]&65535;
			int val1 = memory[ (code[i]>>58)&1 ? memory[p1] : p1 ],
			val2 = memory[ (code[i]>>57)&1 ? memory[p2] : p2 ];
			int op2 = (code[i]>>52)&15;
			bool flag;
			if(op2==0)
			{
				flag=(val1>val2);
			}
			else if(op2==1)
			{
				flag=(val1<val2);
			}
			else if(op2==2)
			{
				flag=(val1==val2);
			}
			else if(op2==3)
			{
				flag=(val1>=val2);
			}
			else if(op2==4)
			{
				flag=(val1<=val2);
			}
			else
			{
				flag=(val1!=val2);
			}
			if(flag)
			{
				i = (code[i]>>56)&1 ? memory[p3] : p3;
				i = (i+N-1)%N;
			}
		}
		else if(op==8)	//x++
		{
			int p=code[i]&65535;
			memory[ (code[i]>>56)&1 ? memory[p] : p ]++;
		}
		else if(op==9)	//x--
		{
			int p=code[i]&65535;
			memory[ (code[i]>>56)&1 ? memory[p] : p ]--;
		}
		else if(op==15)	//Exit
		{
			break;
		}
		else
		{
			cout<<"\n\nError: Invalid code\n\n";
		}
	}
	return 0;
}

这是这题的实现：

// 将其复制进代码的第 8 行即可
0x2000000000000048, 0x1200000000000000,	//H
0x2000000000000065, 0x1200000000000000,	//e
0x200000000000006c, 0x1200000000000000,	//l
0x200000000000006c, 0x1200000000000000,	//l
0x200000000000006f, 0x1200000000000000,	//o
0x200000000000002c, 0x1200000000000000,	//,
0x2000000000000057, 0x1200000000000000,	//W
0x200000000000006f, 0x1200000000000000,	//o
0x2000000000000072, 0x1200000000000000,	//r
0x200000000000006c, 0x1200000000000000,	//l
0x2000000000000064, 0x1200000000000000,	//d
0x2000000000000021, 0x1200000000000000,	//!
0x200000000000000a, 0x1200000000000000,	//[\n]
0xf000000000000000	//exit

拓展阅读：

将 code 数组设为 0x0000000000000000, 0x0000000000000001, 0x5000000000010002, 0x1000000000000002, 0xf000000000000000 即可达到 A+B Problem 的要求。逐行解析：

    0x0000000000000000：输入一个整数，将其存放于内存的 0 号位置；
    0x0000000000000001：输入一个整数，将其存放于内存的 1 号位置；
    0x5000000000010002：将内存的 0,1 号位置的数相加，并将结果存放在内存的 2 号位置；
    0x1000000000000002：将 2 号位置的数以数字形式输出；
    0xf000000000000000：退出程序。

AC记录