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# C++ Looping | Find output programs | Set 5

This section contains the C++ find output programs with their explanations on C++ Looping (set 5).
Submitted by Nidhi, on June 07, 2020

Program 1:

```#include <iostream>
using namespace std;

int main()
{
int num = 15673;
int R1 = 0, R2 = 0;

do {
R1 = num % 10;
R2 = R2 * 10 + R1;

num = num / 10;
} while (num > 0);

cout << R2 << " ";

return 0;
}
```

Output:

```37651
```

Explanation:

Here, we declared three local variables num, R1, and R2, and we are calculating the reverse of variable num, and R1 is used to extract the last digit in each iteration and R2 to store the result.

```Iteration 1:
num=15673, R1=0, R2=0.
After executing loop statements R1=3, R2=3, and num=1567.

Iteration 2:
num=1567, R1=3, R2=3.
After executing loop statements R1=7, R2=37, and num=156.

Iteration 3:
num=156, R1=6, R2=37.
After executing loop statements R1=6, R2=376, and num=15.

Iteration 4:
num=15, R1=5, R2=376.
After executing loop statements R1=5, R2=3765, and num=1.

Iteration 5:
num=1, R1=1, R2=3765.
After executing loop statements R1=1, R2=37651 and num=0.

Then the condition will false and print "37651"
```

Program 2:

```#include <iostream>
using namespace std;

int main()
{
int I = 1;
int D = 0;
int R = 0;

do {
R = I++ * D++;

cout << R << " ";
} while (I <= 5);

return 0;
}
```

Output:

```0 2 6 12 20
```

Explanation:

In the above program, we declared three local variables I, D, and R.

```Iteration 1:
I=1, D=0, R=0
R = 1*0
R = 0
Then I=2 and D=1 and loop condition is true.

Iteration 2:
I=2, D=1, R=0
R = 2*1
R = 2
Then I=3 and D=2 and loop condition is true.

Iteration 3:
I=3, D=2, R=2
R = 3*2
R = 6
Then I=4 and D=3 and loop condition is true.

Iteration 4:
I=4, D=3, R=6
R = 4*3
R = 12
Then I=5 and D=4 and loop condition is true.

Iteration 5:
I=5, D=4, R=12
R = 5*4
R = 20
Then I=6 and D=5 and loop condition is false.
And program terminates.
```

Program 3:

```#include <iostream>
using namespace std;

int main()
{
int I = 1;
int D = 1;
int R = 0;

do {
R = I++ * D++;
if (I == 3)
continue;
cout << R << " ";
} while (I <= 5);

return 0;
}
```

Output:

```1 9 16 25
```

Explanation:

In the above program, we declared three local variables I, D, and R.

```Iteration 1:
I=1, D=1, R=0
R = 1 * 1
R = 1
Print the value of R that is 1.
Then I=2 and D=2 and loop condition is true.

Iteration 2:
I=2, D=2, R=1
R = 2 * 2
R = 4
But it will skip "cout" statement because
of the continue statement.
Then I=3 and D=3 and loop condition is true.

Iteration 3:
I=3, D=3, R=4
R = 3 * 3
R = 9
Print the value of R that is 9.
Then I=4 and D=4 and loop condition is true.

Iteration 4:
I=4, D=4, R=9
R = 4 * 4
R = 16
Print the value of R that is 16.
Then I=5 and D=5 and loop condition is true.

Iteration 5:
I=5, D=5, R=16
R = 5 * 5
R = 25

Print the value of R that is 25.
Then I=6 and D=6 and loop condition is false.
Then the loop will terminate.
```
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