Bc116 xiaolele changed to digital

（ come from Cattle from Beginner programming training ）

1. problem ：

（ Link up ： Change the number ）

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 ≤ 10^9)

Output description ：

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

Example 1

Input ：

222222

Output ：

0

2. Ideas ：

  First Enter an integer , According to the meaning , need Use / and % as well as while loop The whole number is divided into numbers for odd and even judgment （（）%2 whether ==0）, if Odd numbers make the new variable 1, Even numbers are 0 

secondly The main point is The presentation of the final number is an integer of the final variables of each digit , So you need an integer output at this time （ Instead of outputting as an array ！）, This Integer source On ： The resulting number on each bit *pow And to present , and pow The power number is required （ Number of digits -1）, The number of digits is count Record

3. Code ：

#include<stdio.h>
#include<math.h>
int main()
{
	long int n = 0;
	//printf(" Please enter an integer n：");
	scanf("%ld", &n);
	//long int ret=Change(n);
	//printf(" The result is ：");
	long int m = 0;
	long int sum = 0;
	long int ret = 0;
	float count = 0;
	while(n)// Use it carefully while  use for You need to know how many numbers an integer consists of   We need to ask for it again 
	{
		m = n % 10;// Start with the last one 
		n /= 10;
		if (0 == (m % 2))// even numbers - It can be 2 to be divisible by   Remainder is 0！
		{
			count++;
			 ret = 0;
			 sum += ret*(long int )(pow(10, (count - 1)));// Pay attention to the combination ： No arrays   It is calculated in exponential form ！！
			 //pow(float x,float y)
		}
		else
		{
			count++;
			ret = 1;
			sum += ret * (long int)(pow(10, (count - 1)));
		}
	}
	printf("%ld\n", sum);
	return 0;
}

4. Be careful ：

pow（float x, float y）  float form , So this problem involves cast ！