1263:GrayCode

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最新推荐文章于 2022-04-01 10:45:23 发布

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本文链接：https://blog.youkuaiyun.com/happylaoxu/article/details/50909750

本文介绍如何将十进制数转换为格雷码，详细解释了格雷码的规则，并通过示例演示了转换过程。

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1263: Gray Code

时间限制: 1 Sec 内存限制: 128 MB

提交: 204 解决: 115

[提交][状态][讨论版]

题目描述

Gray code is wellknown for its stable. In the machine , gray code can ensure every neighbor number only change one binary bit. Such as the number 3 code as 010 and 4 code as 110 (but in normal binary code 3 is 011 and 4 is 100 ,change two bit) .

Gray code rule is as follows : First turn a number to a normal binary code such as 5 is 101 , and every bit XOR with its left bit , change 101 to 111 )

输入

There are several test cases,every test case has a decimal number n ( n<2^30)

输出

Please output the given number’s binary gray code . Extra leading zero is unexpected.

样例输入

样例输出

111101

提示

来源

2013东软杯

//
//  main.cpp
//  十进制转换为二进制
//
//  Created by xiaohongqi on 16/3/16.
//  Copyright © 2016年 xiaohongqi. All rights reserved.
//  yaowen1996@hotmail.com

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <memory.h>
using namespace std;

int a[50] = {0}; // 注意自己的数组的开的区间的大小，再来进行判断题目之中的解；

int print(int a[], int n){
    int i  =0;
    for(i = n-1 ; i >=0; i--){
        cout << a[i];
    }
    cout << endl;
    return 0;
}

int main(){
    int T;
    while (cin >> T) {
        memset(a, 0, 50 * sizeof(int));
        int count = 0;
        while (T) {
            a[count++] =  T%2;
            T /= 2;
        }
        //print(a, 8);
        // 看了讨论版，发现它的前导零竟然不要，这个不看还真的是坑啊；
        int i = 0;
        for(i = 0; i <count; i++){
            a[i] =  a[i+1]^a[i];
        }
        print(a, count);
        
    }
    return 0;
}