Daily question - xiaolele changes the number

The topic comes from niuke.com

describe

Little Lele likes numbers , Especially like 0 and 1. He now has a number , I want to change the number of each person into 0 or 1. If a bit is odd , Just turn it into 1, If it's even , Then turn it into 0. Please answer what he finally got .

Input description ：

Input contains an integer n (0 ≤ n ≤ 109)

Output description ：

Output an integer , That is, the number obtained by xiaolele after modification .

Example 1

Input ：

222222

Output ：

0

Example 2

Input ：

123

Output ：

101

Thinking analysis ：

First we enter an integer , Then we need to convert the even bits of the integer into 0, Odd digits become 1; Then it naturally involves retrieval conversion , Next, bloggers will share two ideas for solving problems .

The first is to use arrays , The advantage is data retrieval 、 It is easy to modify data , The disadvantage is that the program is cumbersome , Poor versatility , According to the instance , Make adjustments accordingly .

The second is making numbers , Through data retrieval , And then according to the question , Generate a new number , Until each item of the input integer is fetched , The new number is manufactured . Very easy to use .

Numeration method ：

#include<stdio.h>
int main()
{
    int a=0;// Integers a
    int k=0;// Store the number of bits of each bit 
    int sum=0;// Make numbers 
    scanf("%d",&a);
    // Define a i Used to control the data size of each bit , for example ：123=1*100+2*10+3*1;
    // The loop end condition is an integer a Everyone of has taken ;
    //i*=10, amount to i=i*10; Because it's making numbers , So we should control the size of each bit 
    for(int i=1;a!=0;i*=10)
    {
        // And 10 You can get the last one by finding the module 
        k=a%10;
        // Judge the parity of this bit 
        if(k%2==0)
        {
            k=0;// If it's even , Then this one is 0
            sum+=k;
        }
        else
        {
            k=1;// If it's odd , Then this one is 1
            sum+=k*i;// By riding i To adjust the size of digits , adopt sum Storage 
        }
        // Divide 10, Remove the last digit from the approximate number , So we can judge the next 
        a=a/10;
    }
      // Print the number of manufacturing sum
      printf("%d",sum);
    return 0;
}

Array method ：

#include<stdio.h>
int main()
{
    int i = 0, j = 0;
    int t[16] = { 0 };
    int n;
    float sum = 0;
    scanf("%d",&n);                 
    while (n)
    {
        // Judge parity , And the corresponding treatment 
        if (n % 2 == 0||n==0)
        { 
            t[i] = 0;
            sum += t[i];
        }
        else
        {
            t[i] = 1;
            sum += t[i];
        }
        ++i;
        n = n / 10;
    }
 
    // Every bit of the data is the same ( identical ） The situation of 
    if (sum /( i +1)== t[0])
        printf("%d", t[0]);
    // Data starts with 0, We output non 0 Future data 
    else if (t[i] == 0)
    {
        for (j = i; j >=0; j--)
        {
            if (t[j] % 2 != 0)
                while (j >=0)
                {
                    printf("%d", t[j]);
                    j--;
                }
        }
    }
    // Print the converted data 
    else
        while (i>=0)
        {
            printf("%d",t[i]);
            --i;
        }
    return 0;
}

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