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;
}
Related
In C++, I should write a program where the app detects which numbers are divisible by 3 from 1 till 10 and then multiply all of them and print the result. That means that I should multiply 3,6,9 and print only the result, which is 162, but I should do it by using a "While" loop, not just multiplying the 3 numbers with each other. How should I write the code of this? I attached my attempt to code the problem below. Thanks
#include <iostream>
using namespace std;
int main() {
int x, r;
int l;
x = 1;
r = 0;
while (x < 10 && x%3==0) {
r = (3 * x) + 3;
cout << r;
}
cin >> l;
}
Firstly your checking the condition x%3 == 0 brings you out of your while - loop right in the first iteration where x is 1. You need to check the condition inside the loop.
Since you wish to store your answer in variable r you must initialize it to 1 since the product of anything with 0 would give you 0.
Another important thing is you need to increment the value of x at each iteration i.e. to check if each number in the range of 1 to 10 is divisible by 3 or not .
int main()
{
int x, r;
int l;
x = 1;
r = 1;
while (x < 10)
{
if(x%3 == 0)
r = r*x ;
x = x + 1; //incrementing the value of x
}
cout<<r;
}
Lastly I have no idea why you have written the last cin>>l statement . Omit it if not required.
Ok so here are a few hints that hopefully help you solving this:
Your approach with two variables (x and r) outside the loop is a good starting point for this.
Like I wrote in the comments you should use *= instead of your formula (I still don't understand how it is related to the problem)
Don't check if x is dividable by 3 inside the while-check because it would lead to an too early breaking of the loop
You can delete your l variable because it has no affect at the moment ;)
Your output should also happen outside the loop, else it is done everytime the loop runs (in your case this would be 10 times)
I hope I can help ;)
EDIT: Forget about No.4. I didn't saw your comment about the non-closing console.
int main()
{
int result = 1; // "result" is better than "r"
for (int x=1; x < 10; ++x)
{
if (x%3 == 0)
result = result * x;
}
cout << result;
}
or the loop in short with some additional knowledge:
for (int x=3; x < 10; x += 3) // i know that 3 is dividable
result *= x;
or, as it is c++, and for learning purposes, you could do:
vector<int> values; // a container holding integers that will get the multiples of 3
for (int x=1; x < 10; ++x) // as usual
if ( ! x%3 ) // same as x%3 == 0
values.push_back(x); // put the newly found number in the container
// now use a function that multiplies all numbers of the container (1 is start value)
result = std::accumulate(values.begin(), values.end(), 1, multiplies<int>());
// so much fun, also get the sum (0 is the start value, no function needed as add is standard)
int sum = std::accumulate(values.begin(), values.end(), 0);
It's important to remember the difference between = and ==. = sets something to a value while == compares something to a value. You're on the right track with incrementing x and using x as a condition to check your range of numbers. When writing code I usually try and write a "pseudocode" in English to organize my steps and get my logic down. It's also wise to consider using variables that tell you what they are as opposed to just random letters. Imagine if you were coding a game and you just had letters as variables; it would be impossible to remember what is what. When you are first learning to code this really helps a lot. So with that in mind:
/*
- While x is less than 10
- check value to see if it's mod 3
- if it's mod 3 add it to a sum
- if not's mod 3 bump a counter
- After my condition is met
- print to screen pause screen
*/
Now if we flesh out that pseudocode a little more we'll get a skeletal structure.
int main()
{
int x=1//value we'll use as a counter
int sum=0//value we'll use as a sum to print out at the end
while(x<10)//condition we'll check against
{
if (x mod 3 is zero)
{
sum=x*1;
increment x
}
else
{
increment x
}
}
//screen output the sum the sum
//system pause or cin.get() use whatever your teacher gave you.
I've given you a lot to work with here you should be able to figure out what you need from this. Computer Science and programming is hard and will require a lot of work. It's important to develop good coding habits and form now as it will help you in the future. Coding is a skill like welding; the more you do it the better you'll get. I often refer to it as the "Blue Collar Science" because it's really a skillset and not just raw knowledge. It's not like studying history or Biology (minus Biology labs) because those require you to learn things and loosely apply them whereas programming requires you to actually build something. It's like welding or plumbing in my opinion.
Additionally when you come to sites like these try and read up how things should be posted and try and seek the "logic" behind the answer and come up with it on your own as opposed to asking for the answer. People will be more inclined to help you if they think you're working for something instead of asking for a handout (not saying you are, just some advice). Additionally take the attitude these guys give you with a grain of salt, Computer Scientists aren't known to be the worlds most personable people. =) Good luck.
Let suppose a user enters a number below:
54353325421435
i want the variables below to store parts of the number above
Eg.
int part[3]
part[0]=54353
part[1]=32
part[2]=5421435
and then add up the digits and store it the variable like below:
Eg.
int sum[3]
sum[0]=5+4+3+5+3 //sum up part[0]
sum[1]=3+2 //sum up part[1]
sum[2]=5+4+2+1+4+3+5 //sum up part[2]
Sorry guys! I don't know, how to explain this better! I hope you understand my question.
Thanks for reading
Problem
You want to divide your integer to 3 different parts. Basically, you have a number 54353325421435, and you want to divide it up into:
part[0]=54353
part[1]=32
part[2]=5421435
Then add them up.
Solution
A for loop will do best. If you don't know what a for loop is, basically it's a means of iteration with a defined starting and ending point. For example, here is a simple iteration that prints out "hello world" 2 times:
for(int i=0; i<2; i++)
cout << "Hello World" << endl;
You can learn more about for loops here. In your case, what you want to do is iterate through this. So basically, first you store the variable in an integer. (I'm sure you can do that.)
const unsigned long long NUM = 54353325421435; //Make it a constant to not change it
And then you have an array of parts as you mentioned above:
int part[3]
What you can do now is, loop through the NUM. So let me show you how to do the first one:
int access_digits(const unsigned long long int n, int index)
{
int digit_array[64] = {0};
unsigned long long digits = 0, digit, number = n;
while (number) {
digit = number % 10;
number /= 10;
digit_array[digits] = digit;
digits++;
}
return digit_array[digits - index - 1];
}
std::string digits;
for(int i=0; i<=4; i++)
{
digits.append(std::to_string(access_digits(NUM,i)));
}
int digit_int = std::stoi( digits );
You can see above, first that there is an access_digits function. You can use that function to access digits by index. (Credit goes toward Slayther.) Anyway, after that, you can see I am looping from 0 to 4 to get the first 5 digits for part[0]. The first 5 digits bring 54353.
Now finally you want to add them up. Well, again that's pretty easy. Just loop through the digits, and have an accumulator add them up like so:
int accum=0;
for(int i=0; i<4; i++)
{
accum += access_digits(digit_int,i);
}
Exercise
Now edit this to include part[1] and part[2] below on the exercise section.
References
Iterating through digits in integer in C
Teenage Territory chat
for loop
std::string
string::append
Glossary
For Loops:
Executes init-statement once, then executes statement and iteration_expression repeatedly until the value of condition becomes false. The test takes place before each iteration.
Syntax
formal syntax:
attr(optional) for ( init-statement condition(optional) ; iteration_expression(optional) ) statement
informal syntax:
attr(optional) for ( declaration-or-expression(optional) ; declaration-or-expression(optional) ; expression(optional) ) statement
if you want the sum of the digits of no. as you mentioned above then
int number,digit,sum=0;
while(number!=0)
{
digit=number%10;
sum=s+digit;
number=number/10;
}
this code segment will calculate the sum of the digits of the no.
if number=123;
it will calculate sum=3+2+1=6;
This question already has answers here:
Making a square() function without x*x in C++
(7 answers)
Closed 4 years ago.
I'm a beginner in programming and trying to learn C++ by the book Programming principles and practice using C++. In some parts of the book there are little exercises that you can try to do, one of this exercises is about calculating the square of a number, here is what my book says :
Implement square() without using the multiply operator, that is, do the x * x by repetead addition (start a variable result to 0 and add x to it x times).
I've already found a solution for this program but my first tentative was something like this :
#include <iostream>
int main()
{
int a = 0;
std::cout << "Enter an integer value : ";
std::cin >> a;
while (a < a * a)
{
a += a;
std::cout << a << "\n";
}
}
I know this code is wrong but I can't understand the output of the progam, if I enter 5 the program prints 10 20 30 40 50 until 8000, why the for loop doesn't stop when a is greater than its square ? I'm just curious to undersant why
Using multiplication when trying to avoid multiplication seems broken. What about this:
int r = 0;
for (int n = 0; n < a; ++n) {
r += a;
}
why the for loop doesn't stop when a is greater than its square ?
Because it never is. If you compare the graph of y=x^2 against the graph of y=x, you will see that the only time y=x is above, is when 0 < x < 1. That's never the case for integers1. Now, since we're talking about computers with limited storage here, there is a thing called overflow, which will cause a very large number to become a very small number. However, signed integer overflow is undefined behavior in C++. So once your loop gets to the point where overflow would happen, you cannot rely on the results.
1. Note that your loop is not set to stop just when a is greater than its square, but when it is greater than or equal to its square. So, your loop will actually stop if a is 0 or 1.
Input
The input begins with two positive integers n k (n, k<=10^7). The next n lines of input contain one positive integer ti, not greater than 10^9, each.
Output
Write a single integer to output, denoting how many integers ti are divisible by k.
Example
Input:
7 3
1
51
966369
7
9
999996
11
Output:
4
My Code:
#include <iostream>
using namespace std;
int main()
{
long long n,k, i;
cin>>n;
cin>>k;
int count=0;
for(i=0;i<n;i++)
{
int z;
cin>>z;
if(z%k == 0) count++;
}
cout<<count;
return 0;
}
Now this code produces the correct output. However, its not being accepted by CodeChef(http://www.codechef.com/problems/INTEST) for the following reason: Time Limit Exceeded. How can this be further optimized?
As said by caleb the problem is labeled "Enormous Input Test" so it requires you to use some better/faster I/O methods
just replacing cout with printf and cin with scanf will give you an AC but to improve your execution time you need to use some faster IO method for example reading character by character using getchar_unlocked() will give you a better execution time
so you can read the values by using a function like this , for a better execution time.
inline int read(){
char c=getchar_unlocked();
int n=0;
while(!(c>='0' && c<='9'))
c=getchar_unlocked();
while(c>='0' && c<='9'){
n=n*10 + (c-'0');
c=getchar_unlocked();
}
return n;
}
The linked problem contains the following description:
The purpose of this problem is to verify whether the method you are
using to read input data is sufficiently fast to handle problems
branded with the enormous Input/Output warning. You are expected to be
able to process at least 2.5MB of input data per second at runtime.
Considering that, reading values from input a few bytes at a time using iostreams isn't going to cut it. I googled around a bit and found a drop-in replacement for cin and cout described on CodeChef. Some other approaches you could try include using a memory-mapped file and using stdio.
It might also help to look for ways to optimize the calculation. For example, if ti < k, then you know that k is not a factor of ti. Depending on the magnitude of k and the distribution of ti values, that observation alone could save a lot of time.
Remember: the fact that your code is short doesn't mean that it's fast.
I am in the midst of solving a simple combination problem whose solution is 2^(n-1).
The only problem is 1 <= n <= 2^31 -1 (max value for signed 32 bit integer)
I tried using Java's BigInteger class but It times out for numbers 2^31/10^4 and greater, so that clearly doesn't work out.
Furthermore, I am limited to using only built-in classes for Java or C++.
Knowing I require speed, I chose to build a class in C++ which does arithmetic on strings.
Now, when I do multiplication, my program multiplies similarly to how we multiply on paper for efficiency (as opposed to repeatedly adding the strings).
But even with that in place, I can't multiply 2 by itself 2^31 - 1 times, it is just not efficient enough.
So I started reading texts on the problem and I came to the solution of...
2^n = 2^(n/2) * 2^(n/2) * 2^(n%2) (where / denotes integer division and % denotes modulus)
This means I can solve exponentiation in a logarithmic number of multiplications. But to me, I can't get around how to apply this method to my code? How do I choose a lower bound and what is the most efficient way to keep track of the various numbers that I need for my final multiplication?
If anyone has any knowledge on how to solve this problem, please elaborate (example code is appreciated).
UPDATE
Thanks to everyone for all your help! Clearly this problem is meant to be solved in a realistic way, but I did manage to outperform java.math.BigInteger with a power function that only performs ceil(log2(n)) iterations.
If anyone is interested in the code I've produced, here it is...
using namespace std;
bool m_greater_or_equal (string & a, string & b){ //is a greater than or equal to b?
if (a.length()!=b.length()){
return a.length()>b.length();
}
for (int i = 0;i<a.length();i++){
if (a[i]!=b[i]){
return a[i]>b[i];
}
}
return true;
}
string add (string& a, string& b){
if (!m_greater_or_equal(a,b)) return add(b,a);
string x = string(a.rbegin(),a.rend());
string y = string(b.rbegin(),b.rend());
string result = "";
for (int i = 0;i<x.length()-y.length()+1;i++){
y.push_back('0');
}
int carry = 0;
for (int i =0;i<x.length();i++){
char c = x[i]+y[i]+carry-'0'-'0';
carry = c/10;
c%=10;
result.push_back(c+'0');
}
if (carry==1) result.push_back('1');
return string(result.rbegin(),result.rend());
}
string multiply (string&a, string&b){
string row = b, tmp;
string result = "0";
for (int i = a.length()-1;i>=0;i--){
for (int j= 0;j<(a[i]-'0');j++){
tmp = add(result,row);
result = tmp;
}
row.push_back('0');
}
return result;
}
int counter = 0;
string m_pow (string&a, int exp){
counter++;
if(exp==1){
return a;
}
if (exp==0){
return "1";
}
string p = m_pow(a,exp/2);
string res;
if (exp%2==0){
res = "1"; //a^exp%2 is a^0 = 1
} else {
res = a; //a^exp%2 is a^1 = a
}
string x = multiply(p,p);
return multiply(x,res);
//return multiply(multiply(p,p),res); Doesn't work because multiply(p,p) is not const
}
int main(){
string x ="2";
cout<<m_pow(x,5000)<<endl<<endl;
cout<<counter<<endl;
return 0;
}
As mentioned by #Oli's answer, this is not a question of computing 2^n as that's trivially just a 1 followed by 0s in binary.
But since you want to print them out in decimal, this becomes a question of how to convert from binary to decimal for very large numbers.
My answer to that is that it's not realistic. (I hope this question just stems from curiosity.)
You mention trying to compute 2^(2^31 - 1) and printing that out in decimal. That number is 646,456,993 digits long.
Java BigInteger can't do it. It's meant for small numbers and uses O(n^2) algorithms.
As mentioned in the comments, there are no built-in BigNum libraries in C++.
Even Mathematica can't handle it: General::ovfl : Overflow occurred in computation.
Your best bet is to use the GMP library.
If you're just interested in seeing part of the answer:
2^(2^31 - 1) = 2^2147483647 =
880806525841981676603746574895920 ... 7925005662562914027527972323328
(total: 646,456,993 digits)
This was done using a close-sourced library and took roughly 37 seconds and 3.2 GB of memory on a Core i7 2600K # 4.4 GHz including the time needed to write all 646 million digits to a massive text file.
(It took notepad longer to open the file than needed to compute it.)
Now to answer your question of how to actually compute such a power in the general case, #dasblinkenlight has the answer to that which is a variant of Exponentiation by Squaring.
Converting from binary to decimal for large numbers is a much harder task. The standard algorithm here is Divide-and-Conquer conversion.
I do not recommend you try to implement the latter - as it's far beyond the scope of starting programmers. (and is also somewhat math-intensive)
You don't need to do any multiplication at all. 2^(n-1) is just 1 << (n-1), i.e. 1 followed by (n-1) zeros (in binary).
The easiest way to apply this method in your code is to apply it the most direct way - recursively. It works for any number a, not only for 2, so I wrote code that takes a as a parameter to make it more interesting:
MyBigInt pow(MyBigInt a, int p) {
if (!p) return MyBigInt.One;
MyBigInt halfPower = pow(a, p/2);
MyBigInt res = (p%2 == 0) ? MyBigInt.One : a;
return res * halfPower * halfPower;
}