Practice Programming Assignment (PPA 03)
Happy Numbers: A number is called a happy number, if you repeat the process, of squaring the sum of the digits, till the value 1 is obtained. E.g. You need to do the following to perform this check: (a) compute the sum of the squares of its digits (b) if the resultant value is 1, then the number is a happy number, else execute point (a). If a number is not a happy number, there will be an endless loop/cycle to this execution.
Task: In this programming assignment, you are required to write code that checks whether the number is a happy number or not, for 10 cycles (iterations) only. 2 examples of happy numbers (limited to 10 cycles ) are given below:
You are required to do the following:
Find the sum of square of the digits of the number.
Check the result obtained in point 1. If it is 1, assign value 1 to the variable 'finalNumber', else again execute point 1, till the number obtained is 1 or till the number of cycle increases to 10.
Assign the iteration value to the variable 'cycle_no'.
Write the required code in C++. My code so far:
int number, finalnumber, a, cycle_no;
cin>>number;
for (cycle_no=0,finalnumber=0;cycle_no<=10;cycle_no+=1)
{
for (a=0;number>0;number/=10)
a=number%10;
finalnumber+=(a*a);
if (finalnumber==1)
break;
else
number=finalnumber;
continue;
}
cout<<finalnumber;
using namespace std;
int a, number ;
int cycle_no=1;
int sumdigits( int number)
{
int sum=0;
while(number>0)
{a=number%10;
number/=10;
sum+=(a*a);}
return sum;
}
int main(){
cin>>number;
while(cycle_no<=10)
{cycle_no+=1;
if(sumdigits(number)==1)
break;
else
number=sumdigits(number);
}if( sumdigits(number)==1)
cout<<sumdigits (number );
else cout<<number;
}
Related
This is the question Im trying to solve: Link
Im running this code in an online editor and it gives a memory limit exceeded, even though I have used str+=c instead of str=str+c. And I cant seem to figure out why. Could anyone help me wth this?
#include <bits/stdc++.h>
using namespace std;
void solve(){
int a,b,x;
cin>>a>>b>>x;
string res="";
res+='0';
a--;
while(x--){
cout<<res;
res+=res.back()=='0'?'1':'0';
if(res.back()=='0')
a--;
else
b--;
}
string ans="";
for(char ch: res){
ans+=ch;
if(ch=='0'){
while (a--){
ans+='0';
}
}
else{
while(b--){
ans+='1';
}
}
}
cout<<ans;
}
int main() {
int t;
t=1;
while(t--){
solve();
}
return 0;
}
The input I give is 3 3 3
and output I expect is 101100
Basically your solution idea is nearly correct.
The most important requirement here is the number of tansitions. So, when we go from a 1 to a 0 or from a 0 to a 1. These transitions must exist. And the number of transisitions also determines the minimum numbers of 0es or 1s needed.
If more 0es or 1s should be present, then you can simply repeat any 0 or 1 with the same value. This will have no impact on the transistion.
Let's have a closer look. Below is an example for the minimum number of 0es or 1s for a given number of transitions
Transitions Sequence Min 0es Min 1s
1 01 1 1
2 010 2 1
3 0101 2 2
4 01010 3 2
5 010101 3 3
6 0101010 4 3
You immediately see that there is a simple mathematical relation between the number of transitions and the minimum number of needed 1s or 0es. It is:
(Number of Transitions + 1)/2 rounded up
(Number of Transitions + 1)/2 rounded down
For odd number of transisitions, the minimum numbers of 1s or 0es are always the same. For even numbers of transitions however, it depends on the starting value.
The reverse conclusion is that it does not matter for odd transitions, if you start with a 0 or a 1. For an even number of transitions it is important.
Example:
Input 1 2 2, meaning one 0, two 1s and 2 transitions.
With the above formula, we calculate that we need two digits of the one and 2 digits for the other, so theoretically 010 or 101, But since we shall use only one 0, it can only be 101
Resulting in: If we have an even number of transitions, then the start value may depend from other input parameters. And more precicely: If the minimum number needed for a digit is equal to the given number for that digit, then we must start with the other digit.
Example:
1 2 2 must be 101
2 2 2 can be 0110 or 1001
Knowing that we can now draft an algorithm. We will work only one one of the many solutions.
Check, if the number of transitions is odd or even
If even, then determine the start digit with above condition
create a sequence of 010101... or 10101... depending on the start digit and the given number of sequences
Add the not yet consumed 0es or 1s to the sequence by simply duplicating or repeating existing 0es or ones.
This can then be implemented in a similar way like your approach:
#include <iostream>
#include <string>
int main() {
// Here we will store the input parameters
int numberOfZeroes{}, numberOfOnes{}, numberOfTransitions{};
// The input will always be correct, so need to check it
std::cin >> numberOfZeroes >> numberOfOnes >> numberOfTransitions;
// Start digit
char digit{ '0' };
// Check, if the number of transitions is even, then we need a special additional check
if (numberOfTransitions % 2 == 0) {
// Calculate the minimum number of needed 0es or 1s
const int minimum = (numberOfTransitions + 1) / 2;
// Check, if we need to start with digit 1
if (minimum == numberOfZeroes)
digit = '1';
}
// Now we want to create a string starting made of alternating 0es and 1s, so transitions
std::string sequenceWithTransitions{};
do {
// Build string
sequenceWithTransitions += digit;
// Update counters and digits
if (digit == '1') {
digit = '0'; // Make transition
--numberOfOnes; // Update counter
}
else {
digit = '1'; // Make transition
--numberOfZeroes; // Update counter
}
} while (numberOfTransitions--);
// Fill in the remaining 0es and 1s
std::string result{};
for (char c : sequenceWithTransitions) {
result += c; // Copy value
if (c == '1') // Potential replications of 1
while (numberOfOnes-- > 0)
result += '1';
if (c == '0') // Potential replications of 0
while (numberOfZeroes-- > 0)
result += '0';
}
std::cout << result << '\n';
}
Of course this code can be optimzied in many ways
i'm going to learn C++ at the very beginning and struggling with some challenges from university.
The task was to calculate the cross sum and to use modulo and divided operators only.
I have the solution below, but do not understand the mechanism..
Maybe anyone could provide some advice, or help to understand, whats going on.
I tried to figure out how the modulo operator works, and go through the code step by step, but still dont understand why theres need of the while statement.
#include <iostream>
using namespace std;
int main()
{
int input;
int crossSum = 0;
cout << "Number please: " << endl;
cin >> input;
while (input != 0)
{
crossSum = crossSum + input % 10;
input = input / 10;
}
cout << crossSum << endl;
system ("pause");
return 0;
}
Lets say my input number is 27. cross sum is 9
frist step: crossSum = crossSum + (input'27' % 10 ) // 0 + (modulo10 of 27 = 7) = 7
next step: input = input '27' / 10 // (27 / 10) = 2.7; Integer=2 ?
how to bring them together, and what does the while loop do? Thanks for help.
Just in case you're not sure:
The modulo operator, or %, divides the number to its left by the number to its right (its operands), and gives the remainder. As an example, 49 % 5 = 4.
Anyway,
The while loop takes a conditional statement, and will do the code in the following brackets over and over until that statement becomes false. In your code, while the input is not equal to zero, do some stuff.
To bring all of this together, every loop, you modulo your input by 10 - this will always return the last digit of a given Base-10 number. You add this onto a running sum (crossSum), and then divide the number by 10, basically moving the digits over by one space. The while loop makes sure that you do this until the number is done - for example, if the input is 104323959134, it has to loop 12 times until it's got all of the digits.
It seems that you are adding the digits present in the input number. Let's go through it with the help of an example, let input = 154.
Iteration1
crossSum= 0 + 154%10 = 4
Input = 154/10= 15
Iteration2
crossSum = 4 + 15%10 = 9
Input = 15/10 = 1
Iteration3
crossSum = 9 + 1%10 = 10
Input = 1/10 = 0
Now the while loop will not be executed since input = 0. Keep a habit of dry running through your code.
#include <iostream>
using namespace std;
int main()
{
int input;
int crossSum = 0;
cout << "Number please: " << endl;
cin >> input;
while (input != 0) // while your input is not 0
{
// means that when you have 123 and want to have the crosssum
// you first add 3 then 2 then 1
// mod 10 just gives you the most right digit
// example: 123 % 10 => 3
// 541 % 10 => 1 etc.
// crosssum means: crosssum(123) = 1 + 2 + 3
// so you need a mechanism to extract each digit
crossSum = crossSum + input % 10; // you add the LAST digit to your crosssum
// to make the number smaller (or move all digits one to the right)
// you divide it by 10 at some point the number will be 0 and the iteration
// will stop then.
input = input / 10;
}
cout << crossSum << endl;
system ("pause");
return 0;
}
but still dont understand why theres need of the while statement
Actually, there isn't need (in literal sense) for, number of digits being representable is limited.
Lets consider signed char instead of int: maximum number gets 127 then (8-bit char provided). So you could do:
crossSum = number % 10 + number / 10 % 10 + number / 100;
Same for int, but as that number is larger, you'd need 10 summands (32-bit int provided)... And: You'd always calculate the 10 summands, even for number 1, where actually all nine upper summands are equal to 0 anyway.
The while loop simplifies the matter: As long as there are yet digits left, the number is unequal to 0, so you continue, and as soon as no digits are left (number == 0), you stop iteration:
123 -> 12 -> 1 -> 0 // iteration stops, even if data type is able
^ ^ ^ // to store more digits
Marked digits form the summands for the cross sum.
Be aware that integer division always drops the decimal places, wheras modulo operation delivers the remainder, just as in your very first math lessons in school:
7 / 3 = 2, remainder 1
So % 10 will give you exactly the last (base 10) digit (the least significant one), and / 10 will drop this digit afterwards, to go on with next digit in next iteration.
You even could calculate the cross sum according to different bases (e. g. 16; base 2 would give you the number of 1-bits in binary representation).
Loop is used when we want to repeat some statements until a condition is true.
In your program, the following statements are repeated till the input becomes 0.
Retrieve the last digit of the input. (int digit = input % 10;)
Add the above retrieved digit to crosssum. (crosssum = crosssum + digit;)
Remove the last digit from the input. (input = input / 10;)
The above statements are repeated till the input becomes zero by repeatedly dividing it by 10. And all the digits in input are added to crosssum.
Hence, the variable crosssum is the sum of the digits of the variable input.
I am trying to find the sum of each digit in an integer squared, and for any integer that is input to sqdnumber, it outputs 0 to sqdNumber_result, and I can't figure out why.
Also, this is through edX, but I have been stuck for a week or so on this problem, and I have looked at a lot of different topics, but haven't found anything of use to me.
I used codeblocks to write this, but the system testing it uses codeboard
void squaredSum(int sqdnumber,int &sqdNumber_result) {
for (int i=1; i>1; i++){
if (sqdnumber >= ((10^(i-1))-1)){
int rem = (sqdnumber % (10^i));
int rem1 = (sqdnumber % (10^(i-1)));
int temp = (rem - rem1);
sqdNumber_result = sqdNumber_result + (temp^2);
}
else{
break;
}
}
}
I am new to coding, and just learning to do loops in C++.
This is the first iteration of the loop I have gotten their system to actually give me an output for it(I've written and rewritten it 20 or so times), but it isn't giving me an output that makes sense.
I wouldn't ask but I am at my wit's end.
In C++, ^ is the xor operator, not the nth power. for that, you should use pow.
The for statement does not loop. The condition is false the first iteration
There are two issues:
for (int i=1; i>1; i++){
This loop will not loop at all, since the condition i>1 is never met.
The second issue is the usage of ^ to do a power operation. The ^ in C++ is not a power operator, it is the exclusive-or operator.
So the answer at first glance would be to use the std::pow function to compute powers. However there can be drawbacks using it if the exponent is an integer. The reason is that pow is not guaranteed to work perfectly for integer powers.
See this as to dangers of using pow() for integral exponents
It is advised to just use a simple array of values with the powers of 10 and doing a lookup.
you said you were new to C++ so I tried to get a solution without using the for loop and tried to make it as simple as I could.
Let me know if this was any help.
//Code to calculate the sum of each digit squared//
#include<iostream>
using namespace std;
int main ()
{
int integer1,integer2,sum, square;
cout<<"Please enter two integers"<<endl;
cin>>integer1>>integer2 ;
cout<<"The sum of your integers is"<<" "<<endl;
sum = (integer1+integer2);
cout<<sum<<endl;
cout<<"The square of your sum is"<<" "<<endl;
square = (sum*sum);
cout<<square<<endl;
return 0;
}
Hi Im trying to translate this code to TI-BASIC. Im having problems with how to change for loop into while loop and also with incrementing a number in TI-BASIC.
#include <stdio.h>
int main()
{
int n, i, flag=0;
printf("Enter a positive integer: ");
scanf("%d",&n);
for(i=2;i<=n/2;++i)
{
if(n%i==0)
{
flag=1;
break;
}
}
if (flag==0)
printf("%d is a prime number.",n);
else
printf("%d is not a prime number.",n);
return 0;
}
You can efficiently use a While loop in this situation:
Input "NUMBER: ",A
1->B
3->I
√(A->D
If not(fPart(A/2
DelVar BWhile I<=D and B
fPart(A/I->B
I+2->I
End
If not(B
Disp "NOT
Disp "PRIME
In TI-Basic a While loop works as you would expect and you can have conditions for it.
Incrementing a number is as simple as saying
X+i->X
Where 'i' is the incrementer.
To change a For loop into a While loop, you'll have to set up the While loop to constantly check to see if the number and increment have passed the upper bound while increasing the increment each run through.
If you wanted to mimic i++ or ++i in TI-Basic (Using a While loop), all you would have to change would be the arrangement of the code. Please note that TI-Basic For statements always operates under ++i.
Example (i++):
0->X
While X<10
Disp X
X+1->X
End
This will display (With each number on a new line)
0 1 2 3 4 5 6 7 8 9
Example (++i):
0->X
While X<10
X+1->X
Disp X
End
This will display (With each number on a new line)
1 2 3 4 5 6 7 8 9 10
Let it be noted that TI-Basic For statements are much much faster than While loops when it comes to incrementing and should almost always be considered superior for the task.
Integrating Timtech's idea to skip even numbers effectively halves the required time to check the primality of the number with the addition of only a few extra lines.
I expanded the idea to skip multiples of two and multiples of three.
Input "Number:",X:abs(X->X
0
If not(fPart(X/2)) or not(fPart(X/3:Return
For(B,5,sqrt(X),6)
If not(fPart(X/B)) or not(fPart((X+2)/B:Return
End
1
Test Number: 1003001
Time Required: ~4 Seconds (So much better than 15 :D)
Size: 65 Bytes
I dont see why you would want to use a while loop as ti-basic has for loops:
0->F
Input "ENTER NUMBER:",N
For(I,2,int(N/2
If N/I=int(N/I
Then
int(N/2->I
1->F
End
End
If F
Then
Disp "NUMBER IS PRIME
Else
Disp "NUMBER IS NOT PRIME
End
N/I=int(N/I is a way to check for a number's remainder (another way of saying N%I==0 but ti basic does not have modulus). Another trick here is setting I to its maximum bound (int(N/2) as a sort of "break" like other languages would have
The keypad is broken so the input numbers 1, 4, and 7 aren't working. In turn the computer outputs the next lowest and next highest number where 1, 4, and 7 are none of the digits.
My goal is to check out the digits and output true using a boolean function and then output the next highest number and next lowest number. I'm pretty sure I did most of what I need to do, but it isn't working out.
I have inputted the number 444, and the results that came out were 443, and 445.
Thank you for your help.
#include <iostream>
#include <conio.h>
#include <cmath>
using namespace std;
bool containDigit(int number, int digit);
int main()
{
int number, digit, lowNum, highNum;
cout<<"Enter a number between 1 and 999 for the oven temperature: ";
cin>>number;
//1st digit
digit = number / 100;
containDigit(number, digit);
if (containDigit(number, digit) == true)
{
number = number - 100;
}
//2nd digit
digit = (number / 10) % 10;
containDigit(number, digit);
if (containDigit(number, digit) == true)
{
number = number - 10;
}
//3rd Digit
digit = number % 10;
containDigit(number, digit);
if (containDigit(number, digit) == true)
{
number = number - 1;
}
cout<<number<<endl;
getche();
return 0;
}
bool containDigit(int number, int digit)
{
if ((digit == 1) || (digit == 4) || (digit == 7))
{
return true;
}
else
{
return 0;
}
}
The bug is in containDigit function. Try this:
bool containDigit(int number, int digit) {
if(digit == 1 || digit == 4 || digit == 7) return true;
return false;
}
You must use == instead of =.
Also you actually don't need number argument there.
Also there can be done several optimizations. Please look at it yourself (it's your homework) and think about repeated code.
Since this looks like homework, I will refrain from doing it for you and give you these hints:
It looks like you're not really clear on what "1st digit" is. Is it the first one from the left (hundreds) or the right (ones)? Look at your code and tell yourself how each portion of it would answer my question.
Is it ever possible for lowNum or highNum to have more than one digit different than number with your code as it is? How? Where are lowNum and highNum changed, and how?
Also, to expand on what #Al Kepp has said: When you have a function like that, try to test it with some very simple inputs rather than straight out assuming it works. This is called (or is similar to) "unit testing", which dictates that you divide your program into simple, independent units and test them separately. A simple call like containDigit(999, 4) returning true would've rang warning bells.
And speaking of warnings, always, always compile with as many of them as you can stand. (e.g. -Wall for gcc) Doing such might've warned you of the fact that you're not using the parameter number inside containDigit at all.
Your function containDigit has two problems:
It shouldn't need to receive the variable "number" as it doesn't use it
You want to compare with == not =