I am trying to teach myself c++.
On Sololearn I have a task, which is
You are making a program for a bus service.
A bus can transport 50 passengers at once.
Given the number of passengers waiting in the bus station as input, you need to calculate and output how many empty seats the last bus will have.
Sample Input: 126
Sample Output: 24
It also says I should use the "%" operator.
This is the code I created:
int bus = 50;
int stop;
cin >> stop;
cout<< stop % bus;
return 0;
I get 12.
What is the correct way? I'm finding it difficult to understand what the modulo operator does. My understanding is that it divides as many times as it can and leaves the remainder (i.e. 16 % 3 = 1).
First you have to get the number of passengers
int passengers; cin >> passengers;
then you have to find how many passengers are left
int remainPass = passengers % 50;
then you have to find how many seats are left
int remainSeats = 50 - remainPass;
Modulos operator basically represents the leftover from the division
so what we need to do is take the number of people that will remain in the last bus travel which is stop % bus and compute bus - (stop % bus)
that way we know the number of empty seats on the last travel
This is like each bus was filled to the fullest (50 people per bus) what will remain is 26 and so on the last bus the number of empty seats will be 50 - 26 = 24
PS: 12 doesn't seem to be the right output of 126 % 50 it should be 26
Related
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'm looking at figuring out this program for a lab, I have been working on it for a while now, but cannot for the life of me figure out how this works.
The objective is to display an active equation as it decrements from from the initial value.
For example:
20000.00 + 83.33 - 150.00 = 19933.33
19933.33 + 83.06 - 150.00 = 19866.39
19866.39 + 82.78 - 150.00 = 19799.17
19799.17 + 82.50 - 150.00 = 19731.66
19731.66 + 82.22 - 150.00 = 19663.88
19663.88 + 81.93 - 150.00 = 19595.81
19595.81 + 81.65 - 150.00 = 19527.46
And so forth. I have to display that as an output on the screen. But am not sure how to keep a decrementing total like that and how to display it as an active equation like that in cout form.
The number on the far left is an initial loan that a user inputs, the number in the middle is the interest rate which is calculate using p*r/p (initial loan * interestrate(user will input this as well)/ initial loan). And the number on the right just before the equal sign is a payment which the usual will enter.
The goal is to get it to perform 10 iterations or once the initial loan is fully paid off whichever comes first.
Here is a little guidance
Finance
First you got your basics wrong. If this is finance and it looks this way :-), p*r/p is crap.
The second column in your plot is not a rate, neither is it an interest rate, it is an interest.
P is the loan ammount
r is an annual interest rate
The interest is calculated using P times r/12 since the payments you show are monthly in case r is entered mathematically (e.g. 0.05 for 5 %) or P*r/1200 in case r is counterconventional entered as percentage.
C++
The input of the parameters could be done e.g.
double P, r, q;
std::cout << "Enter P, q and r:\t";
std::cin >> P >> r >> q;
you will need to have the numbers printed fixed precision, this can by done with
std::cout << std::fixed << std::setprecision(2)
one last hint: The needed include files will be
#include <iostream>
#include <iomanip>
last you will need a loop have a look for for-loops or do-while loops.
This should help you to get your homework a good start.
I'm currently working on a homework assignment where we have to read from a text file which houses an ID number, and that ID's blood pressure. With that, part of the assignment asks us to create a count controlled loop to read the blood pressures listed and sum them up. This is what I've come up with so far.
text file:
1234 4 100 101 102 103
3625 6 105 120 154 132 188 153
part of the code pertaining to these items:
while (count > 0)
{
inData >> BP;
total = total + BP;
count--;
cout << total << endl;
}
and what this comes up for me is it just adds up the last blood pressure (BP) together 6 times. Any help is appreciated as I just started c++ this semester I am fairly nooby with it. Thank you.
If you only want to calculate a sum you need to place the output of the total outside the loop
while (count > 0)
{
inData >> BP;
total = total + BP;
count--;
}
cout << total << endl;
Then you will output the result only once.
So this is the actual Problem
Can anyone tell me that how I read the repective Data from the file, and how would I able to store it in variables (without using array) also the code should be generic, That if the number of series will incresed or decresed.. Code will not be affected... I Just can't understand that how would I store sata in variables and how.. Please Help.. :(
Problem
A file contains information of a batsman. Information is no of series
played by the batsman. No of matches played in each series & score in
each match by the batsman. You have to read the data (without using
any array) and find average score and maximum score in all matches of
a series. In the end find overall average score and max score in all
matches.
Input:
Read data from file "cricket.txt". First line contains no of seasons/
series played by the player. Next pair of lines contains matches
played by the batsman followed in next line scores by batsman in
different matches of a season. See sample "cricket.txt"
5
6
93 75 41 40 90 19
5
45 86 30 60 29
3
47 90 33
4
22 2 92 5
5
88 67 96 91 90
First 5 shows player has played 5 seasons/ series
Next 6 show in first series player has played 6 matches
Next line has scores of player in 6 matches
Next 5 show in second series player has played 5 matches
Next line has scores of player in 5 matches
So on in second last line 5 shows player has played 5 matches in 5th
series
Last line has scores of player in 5 matches of last series
You're looking for an array.
int a[10];
// Loop that assigns all elements in array a to 0
for (int i = 0; i < 10; i++)
{
a[i] = 0;
}
// Array b will have all of it's members initialized to 0
int b[10]{};
// You can also assign different values to different elements of the array
b[0] = 6;
b[8] = 2;
// You can then use the array elements in operations
int c = b[0] * b[8];
If you want array like structure without compile time defined size, then use std::vector.
// An empty vector of ints
std::vector<int> d;
// A simple int
int e = 5;
// Push 2 values to the end of the vector
d.push_back(2);
d.push_back(e);
// Use the members for operations
int f = d.at(0) * d.at(1);
Since you've now described the problem you're trying to solve instead of just the problem with the solution you came up with:
You don't need to invent variable names or use arrays to compute averages and maximums.
Here's an example of how you can compute an average of the numbers a user inputs:
float sum = 0;
int elements = 0;
float input = 0;
while (cin >> input)
{
sum += input;
elements += 1;
}
std::cout << "Average: " << sum / elements << std::endl;
It's easy to expand this to also keep track of the maximum value so far.
To expand to the average and maximum of a number of series, add another loop "around" it.
Hope you all are having a great day!
I love programming, but these past days I am having sleepless nights, with CodeChef always returning SIGSEGV errors on my Dynamic Programming solutions.
I am solving this question right now. Here's the question -
In Byteland they have a very strange monetary system. Each Bytelandian
gold coin has an integer number written on it. A coin n can be
exchanged in a bank into three coins: n/2, n/3 and n/4. But these
numbers are all rounded down (the banks have to make a profit). You
can also sell Bytelandian coins for American dollars. The exchange
rate is 1:1. But you can not buy Bytelandian coins. You have one gold
coin. What is the maximum amount of American dollars you can get for
it?
Input
The input will contain several test cases (not more than 10). Each
testcase is a single line with a number n, 0 <= n <= 1 000 000 000. It
is the number written on your coin. Output
For each test case output a single line, containing the maximum amount
of American dollars you can make. Example
Input: 12 2
Output: 13 2 You can change 12 into 6, 4 and 3, and then change these
into $6+$4+$3 = $13. If you try changing the coin 2 into 3 smaller
coins, you will get 1, 0 and 0, and later you can get no more than $1
out of them. It is better just to change the 2 coin directly into $2.
Now I know that it's very easy. And I did get stuck initially when I was declaring a big 10^9 integer long array (Over 1GB of memory..whoo!), but coming back to my senses - I decided to do memoization till 10001, and after that simple recursion. But still - I am making a mistake, and it's giving SIGSEGV error.
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
long long n[100001];
long long calc(long long x) {
if (x < 10001) {
if(n[x] != 0) return n[x];
n[x] = max(x, calc(x/2) + calc(x/3) + calc(x/4));
return n[x];
}
else return max(x, calc(x/2) + calc(x/3) + calc(x/4));
}
int main() {
memset(n, 0, sizeof(n));
n[1] = 1;
n[2] = 2;
n[3] = 3;
n[4] = 4;
n[5] = 5;
n[6] = 6;
for (int i = 7; i < 10001; i++)
n[i] = calc(i);
int t = 10;
while (t--) {
long long c;
scanf("%lld", &c);
printf("%lld\n", calc(c));
}
return 0;
}
I have solved some previous questions too - and all of them gave me this error once or twice. I know this error means that I am trying to access memory that hasn't been allocated, but what is wrong in my approach that I always get this error?
The problem is with the corner case n=0.
calc(0) recurses indefinitely because 0<10001 and n[0]=0. You need to add the terminating condition that calc(0)=0.
Takeaways:
Always check your programming competition solutions on corner cases.
Always ensure that your recursion does not result in an infinite loop.