题目描述

输入

输出

样例输入

样例输出

源代码

方法1

#include<iostream>
#include <bitset>
using namespace std;
int D2G(int x)
{
    
 	return x ^ (x >> 1);
}



int main()

{
    
 	int x;
 	cin >> x;
    int y=D2G(x);
    int  i, j = 0;
    int a[1000];
     i = y;
      while (i)
      {
    
            a[j] = i % 2;
            i /= 2;
            j++;

        }
        for (i = j - 1; i >= 0; i--)
            cout << a[i];
        cout << endl;
    
}

方法2

#include <stdio.h>
#include <string.h>
#include<iostream>
#include<stack>
using namespace std;
int main() 
{
    
	int n;
	stack<int>q;
	while (cin >> n)
	{
    
		int a[10000];
		int k = 0;
		int t = n;
		while (t > 0)
		{
    
			q.push(t % 2);
			t /= 2;
		}
		while (!q.empty())
		{
    
			a[k++] = q.top();
			q.pop();
		}
		cout << a[0];
		for (int i = 1; i < k; i++)
		{
    
			if (a[i - 1] == a[i])
			{
    
				cout << '0';
			}
			else
			{
    
				cout << '1';
			}
		}
		cout << endl;
	}
	return 0;
}

关于这题

第一种解题方法 较为简单 The first method is explained here
Implementation of Gray Code 有很多办法 But in fact there are only two operations,就能产生：one is right shift 一个是异或 这里都用D2G 函数来实现了
This time we gety 已经是 Gray code for decimal digits We just need to convert it to binary

异或 ^
In fact, it can be regarded as non-carrying2进制加法 (Or summed up as the same0 不同为1）
1^0 1
0^1 1
0^0 0
1^1 0