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偶数大翻转

[编程题]偶数大翻转

热度指数：3462 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

今天的计算机课上，老师给同学们出了一道题：
输入n个数，请将其中的偶数的二进制反转。
eg：输入1 6 5
其中6是偶数，二进制表示为110，反转后为011，代表3，所以最终输出1 3 5.
小贱君最近脑子不怎么好使，想了半天也没想出来如何做，最后他向你寻求帮助了，帮帮可怜的小贱君吧！

输入描述:


输入包含多组测试数据。

对于每组测试数据：

N --- 输入的数字个数

N个数：a0,a1,...,an-1

保证：

1<=N<=3000,0<=ai<=INT_MAX.

输出描述:

对于每组数据，输出N个整数。

示例1

输入

5
1 3 10 6 7
6
26 52 31 45 82 34

输出

1 3 5 3 7
11 11 31 45 37 17

算法知识视频讲解

Allen()

#include <iostream>
using namespace std;
int flipNum(int x){
    if (x & 1) return x;
    int n = 0;
    while (x){
        n = (n << 1) | (x & 1);
        x = x >> 1;
    }
    return n;
}
int main(){
    int N;
    int x;
    while (cin >> N){
        for (int i = 0; i < N; ++i){
            cin >> x;
            cout << flipNum(x);
            if (i != N - 1) cout << " ";
        }
        cout << endl;
    }
}

编辑于 2016-06-22 12:56:52 回复(11)

iniesta2016

import java.util.Scanner;
//注意输出的格式。。。。思路：位操作，count数组保存二进制形式的每一位。欢迎交流~
public class Main {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		while(scanner.hasNext()) {
			int n = scanner.nextInt();
			int[] num = new int[n];
            int[] res = new int[n];
			for(int i = 0; i < n; i ++) {
				num[i] = scanner.nextInt();
				if((num[i]& 1) == 0) {
					res[i] = reverse(num[i]);
				} else {
					res[i] = num[i];
				}
			}
            for(int i = 0; i < n - 1; i ++) {
               System.out.print(res[i] + " ");
            }
             System.out.print(res[n - 1]);
             System.out.println();
		}
	}

	private static int reverse(int a) {
		int res = 0;
		//count数组存储a的二进制表示，从低位到高位，如10，count[1] = 1,count[3] = 1,其余为0
		int[] count = new int[32];
		for(int i = 0; i < 32; i ++) {
			 count[i] = (a >> i) & 1;
		}
		int k = 0;
		//最高位起，除符号位外（第31位），寻找a的二进制表示中，第一个为1的位
		for(k = 30; k >= 0; k --) {
			if(count[k] == 1)
				break;
		}
		//反转
		for(int i = k; i >= 0; i --) {
			res += (count[i] << k - i);
		}
		//符号位
		res += (count[31] << 31);
        return res;
		
	}
	
}

编辑于 2016-06-24 10:55:10 回复(5)

aidim78

为何没人用JavaScript？
"use strict";
const readline = require('readline');
const rl = readline.createInterface(process.stdin,process.stdout);
rl.on("line",function(ans){
        var arr = ans.split(/\s+/);
        arr = arr.slice(0,n);
        let newarr = arr.map(function(e){
            return e-0;
        }).map(function(e){
            return e.toString('2');
        }).map(function(e){
            let res ;
            if(e%2==0){
                e = e.split("").reverse().join("");

            }
            res = parseInt(e,2)
            return res;
        })
        console.log(newarr.join(" "));

        });
       rl.close();
    }else{
    throw new Error();
    }
    
})

编辑于 2016-09-08 21:58:04 回复(3)

weeknet

#include <iostream>
#include <vector>
 
using namespace std;
 
int main()
{
    int num;
    while(cin>>num)
    {
        int arr;
        vector<int> vec;
        for(int i=0; i<num; ++i)
        {
            cin >> arr;
            vec.push_back(arr);
        }
 
        for(int i=0; i<num; ++i)
        {
            if(vec[i]%2== 0)
            {
                int temp = vec[i];
                int str[32] = {0};
                int j=1;
                int count = 0;
                while(temp)
                {
                    ++count;
                    str[32-j] = temp%2;
                    temp/=2;
                    j++;
                }
                int* p = new int[count];
                for(int m=31,k=0;k<count;m--,++k)
                {
                          p[k] = str[m];
 
                }
                int tmp = 0;
                for(int jj=count-1; jj>=0;--jj)
                {
                    int aaa;
                    int sum = 0;
                    if(p[jj] == 1)
                    {
                        aaa = 1;
                        for(int bb=0; bb<count-jj-1; ++bb)
                            aaa *= 2;
                        sum += aaa;
                    }
                    tmp += sum;
                }
                vec[i] = tmp;
 
 
                delete [] p;
                p = NULL;
            }
        }
        for(int i=0; i<num-1; ++i){
            cout << vec[i] << " ";
			if(i==num-2)
				cout << vec[num-1] << endl; 
        }
            
        
    }
    
 
    return 0;
}

编辑于 2016-07-29 16:45:30 回复(9)

wcz2023

#include <iostream>
#include <string>
#include <math.h>
using namespace std;
int main() {
	unsigned int N, a;
	while (cin >> N) {
		for (int i = 1; i <= N; i++) {
			cin >> a;
			if (!(a % 2)) {
				string str;
				while (a) {
					str.insert(str.length(), 1, a % 2);
					a /= 2;
				}
				for (int i = 0; i < str.length(); i++)
					a += str[i] * pow(2, str.length() - 1 - i);
			}
			cout << a << " ";
		}
		cout << endl;
	}
	return 0;
}

//思考五分钟，写代码十分钟，压缩代码一小时。。。。PS:直接模拟算了

发表于 2016-07-14 18:16:49 回复(1)

发愤图强的小菜

#include <iostream>
using namespace std;
int flipNum(int x){
    if (x & 1) return x;//x&1实际上就是取x二进制最左边的数，即判断奇偶
    int n = 0;//n先定义位0，作为x的翻转
    while (x){
        n = (n << 1) | (x & 1);//将x的左边的数放到n上并且左移n
        x = x >> 1;//将x右移使得下次循环在获得左边的数
    }
    return n;
}
int main(){
    int N;
    int x;
    while (cin >> N){
        for (int i = 0; i < N; ++i){
            cin >> x;
            cout << flipNum(x);
            if (i != N - 1) cout << " ";
        }
        cout << endl;
    }
}

发表于 2016-11-04 15:39:07 回复(0)

Java

甫艾蒽廷

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N;
		int[] a;
		while(sc.hasNext()){
			N = sc.nextInt();
			a = new int[N];
			for(int i = 0; i< N; i++){
				a[i] = sc.nextInt();
				if(a[i] % 2==0){
				  a[i] = toBinary(a[i]);
				}
			}
			for(int i =0; i < N-1;i++){
				System.out.print(a[i] + " ");
			}
			System.out.print(a[N-1]);
			System.out.println();
		}
		
	}
	
	public static int toBinary(int x){
		int m =0;
		StringBuilder yu = new StringBuilder();
		for(int i = 0; x > 0; i++){
			yu.append( x % 2);
			x = x / 2;
		}
		int index = 0;
		for(int i = yu.length()-1; i >= 0;i--){
			m += (yu.charAt(i)-'0')* (1<<(yu.length()-i-1));
		}
		return m;
	}
}

发表于 2016-08-27 13:42:57 回复(0)

aoho求索


	importjava.util.Scanner;



	 



	publicclassMain {



	    publicstaticvoidmain(String[] args) {



	        Scanner in = newScanner(System.in);



	        while(in.hasNext()) {



	            intnum = in.nextInt();



	            inta[] = newint[num];



	            for(inti = 0; i<num;i++) {



	                a[i] = in.nextInt();



	            }



	            print(num,a);



	        }



	    }



	 



	    publicstaticvoidprint(intnum,int[]a) {



	        for(inti= 0;i<num;i++) {



	            if(a[i]%2== 0) {



	                String binary = Integer.toBinaryString(a[i]);



	                chartmp[] = newchar[binary.length()];



	                for(intj=0;j<binary.length();j++) {



	                    tmp[j] = binary.charAt(j);



	                }



	                String res = reverse(tmp);



	                res = Integer.valueOf(res,2).toString();



	                a[i] = Integer.valueOf(res);



	            }



	            System.out.print(a[i]);



	            if(i<num-1){



	                System.out.print(" ");



	            }



	        }



	        System.out.println();



	    }



	    publicstaticString reverse(char[] str) {



	        for(inti = 0;i<str.length/2;i++) {



	            chartmp = str[i];



	            str[i] = str[str.length - i - 1];



	            str[str.length - i - 1] = tmp;



	        }



	        String res = "";



	        for(inti = 0;i<str.length;i++ ) {



	            res += str[i];



	        }



	        returnres;



	    }



	}

发表于 2016-08-06 14:09:53 回复(0)

某同学

思路就是先转String，记下当前二进制位数，然后反序右移32-len。
之前直接又while循环做左右移动操作，O（n^2）超出了时间。 另外就是输出格式，很坑，，不能多最后那个空格。。

importjava.util.*;

publicclassMain {

publicstaticvoidmain(String[] args) {

// TODO Auto-generated method stub

Scanner scanner=newScanner(System.in);

while(scanner.hasNext())

{

intn=scanner.nextInt();

intarr[]=newint[n];

for(inti = 0; i < arr.length; i++)

{

arr[i] = scanner.nextInt();

if(arr[i] % 2==0)

{

intlen=String.valueOf(Integer.toBinaryString(arr[i])).length();

arr[i] =(Integer.reverse(arr[i]))>>(32-len);

}

for(inti = 0; i < arr.length; i++)

{

System.out.print(arr[i]);

if(i!=n-1)

{

System.out.print(' ');

}

System.out.println();

}

scanner.close();

}

发表于 2016-08-02 11:04:32 回复(0)

C/C++

不负时光

#include <iostream>
 
using namespace std;

unsigned int reverseBin(unsigned int num){
    if(num%2 != 0){
        return num;
    }
	unsigned int ret = 0;
	for(int i = 0; i < 32; i++){
		ret <<= 1;
		ret = ret | (num & 1);
		num >>= 1;
	}
	while(!(ret & 1)){
		ret >>= 1;
	}
	return ret;
}

int main(){
    int n = 0;
    unsigned int num;
    while(cin >> n){
        for(int i = 0; i < n; i++){
            cin >> num;
            cout << reverseBin(num);
            if(i != n-1) cout << " ";
        }
        cout << endl;
    }
    return 0;
}

发表于 2016-07-12 15:05:45 回复(0)

SnowMonkey

我想问一下，这个二进制是怎么反转的，比如6 为110，反转以后为011，，，，这个是把0变为1,1变为0？？？还是直接倒过来？？比如8 二进制位1000，第一种反转结果是0111，第二种反转结果还是0001。

发表于 2016-06-27 11:32:56 回复(1)

schuman

VS2013调的没有问题，提交测试就是通不过，也是醉了。。

#include<iostream>

#include<math.h>

using namespace std;

int main(){

int i, j, N, len, temp;

while (cin >> N && N > 1 && N < 3000){

int *input = new int[N];

int *output = new int[N];

for (i = 0; i < N; i++){

cin >> input[i];

}

for (i = 0; i < N; i++){

if (input[i] % 2 == 0){

len = 1;

while (pow(2, len + 1) <= input[i])

len++;

int *revert = new int[len];

temp = input[i] / 2;

j = 0;

while (temp != 1){

if (temp % 2 == 0)

revert[j] = 0;

else

revert[j] = 1;

temp = temp / 2;

j++;

}

revert[j] = 1;

for (output[i] = 0, j = len - 1, temp = 1; j >= 0; j--){

output[i] += temp*revert[j];

temp *= 2;

}

delete [] revert;

}

else

output[i] = input[i];

}

for (i = 0; i < N; i++)

cout << output[i] << " ";

delete [] input;

delete [] output;

}

return 0;

}

编辑于 2016-06-25 11:34:22 回复(3)

bboyzqh

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

Scanner scan = new Scanner(System.in);

while(scan.hasNext()){

int N = scan.nextInt();

int[] a = new int[N];

for(int i = 0;i<N;i++){

a[i] = scan.nextInt();

if(a[i]%2==0){

char bArr[] = Integer.toBinaryString(a[i]).toCharArray();

int sum = 0;

int flag = 0;

for(int j = 0;j<bArr.length;j++){

int tempJ = Integer.valueOf(String.valueOf(bArr[j]));

sum += Math.pow(2, flag)*tempJ;

flag++;

}

a[i] = sum;

} //if

}

for(int k = 0;k<N-1;k++){

System.out.print(a[k]+" ");

}

System.out.println(a[N-1]);

} // while

} //main

}

发表于 2016-06-21 23:26:55 回复(0)

cpeng

var readline = require('readline'); 
rl = readline.createInterface({ 
      input: process.stdin, 
      output: process.stdout 
  }); 
  var n = 0; 
  rl.on('line', function(line) { 
      var data = line.trim().split(' '); 
      var result=[];
      if(data.length == 1) { 
          n = parseInt(data[0]); 
      }else if(data.length == n) { 
          result = data.map(function(val, index) { 
              if(val % 2== 0) { 
                  return parseInt(parseInt(val).toString(2).split('').reverse().join(''), 2).toString(10);
              }else{ 
                  return val; 
              } 
          }); 
          console.log(result.join(' ')); 
          n = 0; 
      } 
});

发表于 2017-08-07 20:40:32 回复(0)

Rabbit***ingQ

我用java代码实现如下：

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

Scanner scan = new Scanner(System.in);

while(scan.hasNextInt()){

int a = Integer.parseInt(scan.nextLine());

String[] s = scan.nextLine().split(" ");//利用空格进行拆分

for(int i = 0;i<s.length;i++){

int b = Integer.parseInt(s[i]);//取每一个数判断奇偶

if(b%2==0){//此时为偶数 ,并计算他的二进制存入字符串ss中

String ss = "";

while(b!=0){

ss = b%2+ss;

b=b/2;

}//算完时b为0

for(int j=0;j<ss.length();j++){

if(ss.charAt(j)=='1'){

b = (int) (b + Math.pow(2, j));//并计算它反向二进制的十进制数值

}

s[i] = ""+b;//把整型转换为String

}

System.out.print(s[0]);//每两个数字之间有一个空格

for(int k=1;k<s.length;k++){

System.out.print(" "+s[k]);

}

System.out.println();

}

发表于 2017-02-23 20:03:24 回复(0)

sherryyes

public static void main(String[] args) {
  
  Scanner sc = new Scanner(System.in);
  while (sc.hasNext()) {
   String str = sc.nextLine();
   int n = Integer.parseInt(str);
   String[] str2 = sc.nextLine().split(" ");
   for (int i = 0; i < n; i++) {
    int c = Integer.parseInt(str2[i]);
    if (c % 2 == 0) {
     String binary = Integer.toBinaryString(c);
     char[] r = binary.toCharArray();
     char[] ne = new char[binary.length()];
     int q = 0;
     for (int t = r.length - 1; t >= 0; t--) {
      ne[q] = r[t];
      q++;
     }
     int s = Integer.parseInt(String.valueOf(ne), 2);
     str2[i] = String.valueOf(s);
    }
   }
   for (String string : str2) {
    System.out.print(string+" ");
   }
   System.out.println();
  }
 }

发表于 2016-10-20 15:48:24 回复(0)

chasel86


import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            int N = scanner.nextInt();
            int[] firstArr = new int[N];
            for (int i = 0; i < N; i++) {
                firstArr[i] = scanner.nextInt();
            }
//只转变偶数，将其二进制各位倒序存在字符串temp中，然后调用Integer.valueOf(temp, 2);方法转为整数 for (int j = 0; j < N; j++) {
                if (firstArr[j] % 2 == 0) {
                    String temp = "";
                    while (firstArr[j] != 0) {//一直到整数部分为0
                        temp = temp + (firstArr[j] % 2);//余数，并且拼接起来,注意已经反过来了
                        firstArr[j] = firstArr[j] / 2;//除2取整
                    }
                    firstArr[j] = Integer.valueOf(temp, 2);
                } else continue;
            }

//输出要严格按照题目意思才不会出错
            for (int i = 0; i < (N-1); i++) {
                System.out.print(firstArr[i] + " ");
            }
            System.out.print(firstArr[N - 1]);
            System.out.println();
            
        }
        scanner.close();
    }

}

发表于 2016-09-28 22:25:17 回复(0)

牛客178061号

//模拟10金子转二进制的过程，输出格式太坑

import java.util.Scanner;

public class Main {

public static void main(String[] args){

Scanner scanner=new Scanner(System.in);

while(scanner.hasNextInt()){

int N=scanner.nextInt();

for(int i=0;i<N;i++){

int temp=scanner.nextInt();

if (temp%2!=0) {

if(i==(N-1)){

System.out.print(temp);

}else{

System.out.print(temp+" ");

}

}else{

if(i==(N-1)){

System.out.print(getFZ(temp));

}else{

System.out.print(getFZ(temp)+" ");

}

System.out.println();

}

public static int getFZ(int i){

int out=0;

int yushu=0;

while(i>0){

yushu=i%2;

i=i/2;

out=out*2+yushu;

}

return out;

}

发表于 2016-09-14 18:09:25 回复(0)

笑遇微尘

#include<stdio.h>

#include<math.h>

int main()

{

int i,j,k,n,m=0,t,count;

int a[10][20],b[100],c[10];

while(scanf("%d\n",&n)!=EOF)

{

c[m]=n;

for(i=0;i<n;i++)

{

scanf("%d",&a[m][i]);

}

m=m+1;

}

for(t=0;t<m;t++)

{

for(i=0;i<c[t];i++)

{

count=0;

if(a[t][i]%2==0)

{

k=a[t][i];

for(j=0;k/2!=0;j++)

{

b[j]=k%2;

if(k%2==0)

count=count+pow(2,j);

k=k/2;

}

a[t][i]=count;

}

for(i=0;i<m;i++)

{

for(j=0;j<c[i];j++)

printf("%5d",a[i][j]);

printf("\n");

}

求问，为什么会提示发生段错误，数组越界？有大神能给解答一下么？

发表于 2016-09-02 18:58:22 回复(0)

牛客583092号

    <script type="text/javascript">
    
    function reverse(num) {
    var i, len, numArray, result = 0;
    numArray = [];
    //分解位
    while (num != 0) {
        numArray.push(num % 2);
        num = parseInt(num / 2);
      
    }
    
     for (i = 0, len = numArray.length; i <len; i++) {
        result += numArray[i] * Math.pow(2, len - i - 1);
    }
    return result; 
}
window.onload = function(){
 

   var input =["1","6","5"];
   for (var i = 0; i < input.length; i++) {
       if(input[i]%2==0){
           input[i] = reverse(input[i]);
        }
   };
   alert(input);
}
    </script>

发表于 2016-08-31 10:41:51 回复(0)