We are given an integer variable as N storing the positive integer type value. The task is to recursively print all the numbers less than given value N having digit 1, 3 or the combination of both.
Let us see various input output scenarios for this −
Input − int num = 40
Output − Recursive program to print all numbers less than N which consist of digits 1 or 3 only are: 33 31 13 11 3 1
Explanation − we are given a positive integer value as 40 stored in a variable num. Now, we will recursively find out all the numbers containing digits 1, 3 or both and those numbers less than 40 are 1, 3, 11, 13, 31, 33
Input − int num = 5
Output − Recursive program to print all numbers less than N which consist of digits 1 or 3 only are: 3 1
Explanation − we are given a positive integer value as 40 stored in a variable num. Now, we will recursively find out all the numbers containing digits 1, 3 or both and those numbers less than 5
are 1 and 3.
Input − int num = 1
Output − Wrong Input
Explanation − 我们给定了一个存储在变量num中的正整数值为1。现在，我们将递归地找出所有包含数字1、3或两者的数字，并且这些数字小于1
are 0 because the only positive integer less than 1 is 0 therefore, output is wrong input.
Approach used in the below program is as follows
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Input an integer variable as num. Pass it to the function Recursive_Numbers(num) by passing num as the parameter to the function.
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Inside the function Recursive_Numbers(num)
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Declare a variable as check of type bool and set it with 1.
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Check IF num greater than 0 then start WHILE temp greater than 0 AND check to 1. Set digit to temp % 10.
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Check IF digit not equals 1 AND digit not equals to 3 then set check to 0. Set temp = temp / 10.
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Check IF check is 1 then print num.
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Make a recursive call to the function Recursive_Numbers(num - 1)
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Example
#include
using namespace std;
void Recursive_Numbers(int num){
bool check = 1;
int temp = num;
if(num > 0){
while(temp > 0 && check == 1){
int digit = temp % 10;
if (digit != 1 && digit != 3){
check = 0;
}
temp = temp / 10;
}
if(check == 1){
cout
