Mirsella - 1 year ago 68

C++ Question

I've just recently started to dabble in coding, and I ran into a problem that I haven't been able to solve for days, and the closest thing I've been able to find online is a program checking whether a number contains a specific digit, but that doesn't really apply in my case, I don't think. The problem is to let the user enter two positive numbers and check whether the reverse of the second number is contained within the first one. For example if you enter 654321 and 345, it would say say that it contains it because the reverse of 345 is 543 and 654321 contains that. Here's what I've been trying, but it has been a disaster.

P.S: The variables should stay integer through the program.

`#include <iostream>`

using namespace std;

bool check(int longer, int shorter)

{

int i = 1;

int rev=0;

int digit;

while (shorter > 0)

{

digit = shorter%10;

rev = rev*10 + digit;

shorter = shorter/10;

}

cout << rev << endl;

bool win=0;

int left = longer / 10; //54321

int right = longer % 10; // 65432

int middle = (longer /10)%10; // 5432

int middle1;

int middle2;

int trueorfalse = 0;

while (left > 0 && right > 0 && middle1 > 0 && middle2 >0)

{

left = longer / 10; //4321 //321

right = longer % 10; //6543 //654

middle1 = middle%10; //543

middle2= middle/10; //432

if (rev == left || rev == right || rev == middle1 || rev == middle2 || rev == middle)

{

win = true;

}

else

{

win = false;

}

}

return win;

}

int main ()

{

int longer;

int shorter;

int winorno;

cout << "Please enter two numbers, first of which is longer: ";

cin >> longer;

cin >> shorter;

winorno = check(longer,shorter);

if (winorno==true)

{

cout << "It works.";

}

else

{

cout << "It doesn't work.";

}

return 0;

}

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Answer Source

The more you overthink the plumbing, the easier it is to stop up the drain. -- Scotty, Star Trek III.

This becomes much easier if you divide this task in two parts:

Reverse the digits in an integer.

Search the second integer for the reversed integer calculated by the first part.

For the first part, assume that `n`

contains the number to reverse.

```
int modulo=1;
int reversed_n=0;
do
{
reversed_n = reversed_n * 10 + (n % 10);
modulo *= 10;
} while ( (n /= 10) != 0);
```

The end result is if `n`

contained 345, `reversed_n`

will end up with 543, and `modulo`

will be 1000. We'll need `modulo`

for the second part.

The reason the loop is structured this way is intentional. If the original number is 0, we want to wind up with `reversed_n`

also 0, and `modulo`

as 10.

And now, we can take a similar approach to search the second number, called `search`

, whether it contains `reversed_n`

:

```
for (;;)
{
if ((search % modulo) == reversed_n)
{
std::cout << "Yes" << std::endl;
return 0;
}
if (search < modulo)
break;
search /= 10;
}
std::cout << "No" << std::endl;
```

Complete program:

```
#include <iostream>
int main()
{
int search=654321;
int n=345;
int modulo=1;
int reversed_n=0;
do
{
reversed_n = reversed_n * 10 + (n % 10);
modulo *= 10;
} while ( (n /= 10) != 0);
for (;;)
{
if ((search % modulo) == reversed_n)
{
std::cout << "Yes" << std::endl;
return 0;
}
if (search < modulo)
break;
search /= 10;
}
std::cout << "No" << std::endl;
return 0;
}
```

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