Is this program OK, or can it be improved (but simply)? How do I make sure no repeat numbers are entered?
int n;
int array[9];
cout<<"Enter Number Between 9-0 Only"<<endl;
for(int i = 0; i<=10; i++){
cout<<"Enter Number" <<(i+1)<<endl;
cin >> n;
if((n >= 0) && (n <=9)){
array[i]=n;
}
else{
cout<<"Numbers from 0-9 only\n"<<endl;
break;
}
}
(edit) complete, compiling code
To check if the numbers are used with higher performance, try something like this (using the working code from Jack Radcliffe):
#include <iostream>
using namespace std;
int main()
{
int n = 0;
int array[9] = {0};
bool isUsed[10] = {0};
for(int i = 0; i < 9; i++)
{
cout << "Enter Number " << (i + 1) << endl;
cin >> n;
if((n >= 0) && (n <= 9))
{
if (isUsed[n] == false)
{
array[i] = n;
isUsed[n] = true;
}
else
{
cout << "Number has already been used." << endl;
i--;
}
}
else
{
cout << "Numbers from 0-9 only." << endl;
i--;
}
}
return 0;
}
Optimization isn't exactly necessary with this simple of code, but it's this seems to be an exercise of practice, so why not practice optimized code, too?
Most of it is fine, though there are two problems standing out.
First, you have an array of size 9, but you are taking in 11 numbers since you're starting the for loop at 0 and going through to 10.
Second, since you have it so if the entered number is not between 0 and 9, inclusive, the for loop breaks. This entails that fewer than 9 numbers will be put into the array if an invalid number is entered. Change the entire loop to read this and you should be good:
for(int i = 0; i < 9; i++) {
cout << "Enter Number " << (i + 1) << endl;
cin >> n;
if((n >= 0) && (n <= 9))
array[i] = n;
else {
cout << "Numbers from 0-9 only\n" << endl;
i--;
}
}
The whole fire part was right, but I removed the break in the else-statement and added in the i--. I added that in so when the user is prompted to re-enter the number, the entry number will be at the correct index.
I hope this was helpful.
Related
Let me preface this by saying I am fairly new to functions and arrays.
I have to make 3 functions: Function1 will be user input, Function2 will determine even/odd numbers, and Function3 will display the contents. I have Function1 and Function3 complete, and will post below, but I'm having a difficult time with Function2. What I have now will give the user an error message if they enter an even number, but it's messed up, and I just can't seem to figure out.
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
void getUsrInput(int num[], int size) //function for user input (function 1 of 3)
{
int n;
for (int i = 0; i < size; i++)
{
cout << "Enter five odd numbers: ";
cin >> n;
num[i] = n;
if (num[i] % 2 != 0) //if the number is odd then store it, if it is even: //function for even or odd (function 2 of 3 *doesn't work)
{
i++;
}
else
{
cout << "Invalid input. Please only enter odd numbers!" << endl; //let the user know to enter only odd numbers.
}
}
}
int main()
{
const int size = 5; //array size is 5 numbers
int num[size];
getUsrInput(num, size);
cout << "D I S P L A Y - PART C/B" << endl;
cout << "========================" << endl;
for (int i = 0; i < size; i++)
{
cout << num[i] << endl; //function for display (function 3 of 3)
}
}
for (int i = 0; i < size; i++)
increments i each time through the loop.
if (num[i] % 2 != 0) {
i++;
}
increments i each time the number is odd. So each time the user inputs an odd number, i gets incremented twice. Change the loop control to
for (int i = 0; i < size; }
so that i only gets incremented on valid input.
It' perfect for a while loop and you only count up if the insert is ok:
void getUsrInput(int num[], int size) //function for user input (function 1 of 3)
{
int n, i=0;
while( i < size )
{
cout << "Enter five odd numbers: ";
cin >> n;
if (n % 2 == 0)
cout << "Invalid input. Please only enter odd numbers!" << endl;
else
num[i++] = n; // ok then store and count up
}
}
I've been working on a lottery checking program. The goal is to create a function that finds duplicates and allows the user to choose another number in the case of userTicket, and generate another random number in case of winningNums. So this function must be reusable between the two, or with any array for the matter. I am not familiar with sorting and scanning through the array just yet. I have created a nested for loop to go through each index and compare the two between [i] and [j]. My function only works on the first number for some reason. Any ideas are greatly appreciated.
void getLottoPicks(int userArray[])
{
cout << "Please enter your 7 lotto number picks between 1 and 40.\n";
for (int i = 0; i < NUMS; i++)
{
cout << "selection #" << i + 1 << ":";
cin >> userArray[i];
cin.ignore();
if (noDuplicates(userArray) == true)
{
cout << "You already picked this number. Please enter a different number: " << endl;
cin >> userArray[i];
}
}
}
void genWinNums(int winArray[])
{
srand((int)time(NULL));
for (int i = 0; i < NUMS; i++)
{
winArray[i] = rand() % 40 + 1;
if (noDuplicates(winArray) == true)
{
winArray[i] = rand() % 40 + 1;
}
}
}
bool noDuplicates(int dupArray[])
{
int temp = 0;
for (int i = 0; i < NUMS; i++)
{
//temp += dupArray[i];
for (int j = 0; j < i; j++)
{
if (dupArray[i] == dupArray[j])
{
return true;
}
else
{
return false;
}
}
}
}
You can use std::set, it's faster and much less code
void getLottoPicks(std::set<int> userArray)
{
cout << "Please enter your 7 lotto number picks between 1 and 40.\n";
for (int i = 0; i < NUMS; i++)
{
cout << "selection #" << i + 1 << ":";
int num;
cin >> num;
cin.ignore();
// while the num is already in the set
while (userArray.find(num) != userArray.end())
{
cout << "You already picked this number. Please enter a different number: " << endl;
cin >> num;
}
}
}
In both cases your calls to noDuplicates() are nested inside the loops which are constructing the array that you pass to noDuplicates(). That won't work. First you need to construct the array, and once you are done constructing it, then you pass it to noDuplicates().
I'm trying to build a program which will accept numbers from user and create Floyd triangle.
I tried using the logic of Floyd triangle, but its printing as a line.
Example:
Enter total numbers: 5
Enter the numbers: 3,8,2,4,9
O/p:
3
82
249
Here's my code:
#include <iostream>
using namespace std;
int main()
{
int totalnos, j, i;
cout << "Enter total numbers: ";
cin >> totalnos;
int numbers[totalnos];
cout << "Enter the numbers: ";
for (i = 1; i <= totalnos; i++)
{
cin >> numbers[i];
}
for (i = 1; i <= totalnos; i++)
{
for (j = 1; j <= 1; j++)
{
cout << numbers[i];
}
}
}
You have a problem with the kind of loops shown below. I don't know wether this kind of solution is due to you coming from the Pascal world, or because you've seen it elsewhere. Anyway, you should not make loops start in 1 and go to i, or at least, you should take into account that in the C-like world (C, C++, Java, C#, and many others), arrays start at index 0, and end at index n - 1, being n the size of the array.
int numbers[totalnos];
cout << "Enter the numbers: ";
for (i = 1; i <= totalnos; i++)
{
cin >> numbers[i];
}
The problem is actually not what indexes you use for loops, but that you must always use 0..n-1 when accessing arrays. So you can change your loop to just access the array correctly:
int numbers[totalnos];
cout << "Enter the numbers: ";
for (i = 1; i <= totalnos; i++)
{
cin >> numbers[ i - 1 ];
}
Or you can do as all programmers in the C-like world, and directly start your indexes at 0:
int numbers[totalnos];
cout << "Enter the numbers: ";
for (i = 0; i < totalnos; i++)
{
cin >> numbers[i];
}
Instead of going from 1 to totalnos, now you go from 0 to totalnos - 1 (notice the i < totalnos instead of the i <= totalnos, that's a sutil change).
You were accessing memory past the limit of the array, which means that your program will show undefined behaviour (this means that it will probably crash, though under some conditions, nothing seems to happen, which is even more dangerous).
Now the algorithm itself. I haven't heard about the Floyd triangle. It seems that it is built with the natural numbers starting from 1. However, you are asking for totalnos numbers. You will need more than totalnos numbers in order to build a Floyd triangle with totalnos rows. That's why you need to adjust the position of the number being shown taking into account the number of columns for each row (numPos starts with 0).
cout << endl;
for (i = 0; i < totalnos; i++)
{
if ( ( totalnos - i ) < numPos ) {
numPos = totalnos - i;
}
for (j = 0; j < i; j++)
{
cout << numbers[numPos] << ' ';
++numPos;
}
cout << endl;
}
You can find the whole code here: http://ideone.com/HhjFpz
Hope this helps.
Internal loop can be modified as below :
for (i=0; i < 3; i++)
{
for (j=0; j<=i; j++)
{
cout << numbers[i+j];
}
cout<<" ";
}
Hard coded value "3" can be replaced with the "number of rows of Floyd triangle .
I think this will do the trick .
In inner loop you made mistake with j <= 1; should be j <= i;
And you missed '\n' char for new line.
Here is fix:
#include <iostream>
using namespace std;
int main()
{
int totalnos, j, i, k = 0;
cout << "Enter total numbers: ";
cin >> totalnos;
//int numbers[totalnos];
//cout << "Enter the numbers: ";
// for (i = 1; i <= totalnos; i++)
// {
// cin >> numbers[i];
// }
for (i = 1; i <= totalnos; i++)
{
// your code for (j = 1; j <= 1; j++)
for(j=1; j<=i; ++j) // fixed
cout << k+j << ' ';
++k;
cout << endl; // fix
}
}
I am suppose to make a program that asks a user to input an integer 'n' between 1 and 100, and have an input validation loop. The program will then calculate the first 'n' prime numbers and print them. So I figured out how to calculate the prime numbers and display them, 10 numbers per line, all in the main function. What I have to do is to have a function called isPrime() that takes an integer and returns true if it is prime and false otherwise. I'm not sure how to go about this. This is the code I have for function main().
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int number;
int count = 0;
cout << "Enter an integer between 1 and 100: ";
cin >> number;
while (number < 0 || number > 100)
{
cout << "Invalid number." << endl;
cout << "Enter an integer between 1 and 100: ";
cin >> number;
}
cout << "The first " << number << " primes: \n" << endl;
for (int i = 2; number > 0; ++i)
{
bool isPrime = true;
for (int j = 2; j < i; ++j)
{
if (i % j == 0)
{
isPrime = false;
break;
}
}
if (isPrime)
{
count++;
--number;
cout << setw(5) << i;
if (count % 10 == 0)
cout << endl;
}
}
cout << endl;
system("pause");
return 0;
}
Any help is good, thanks in advance.
You seem to be testing i % j for every j lower then i. But if you tested for j = 3 then you can exclude all multiples of 3. When a number can't be divided by 3 it can't be divided by 6. This will improve performance.
You can also implement the AKS primalty test https://en.wikipedia.org/wiki/AKS_primality_test
This might be a little more work. I don't know the details of this test but it is deterministic and works on all numbers.
If you want a function bool isPrime(int number) you can just extract this code from your main method, it looks something like this:
bool isPrime(int number){
for (int j = 2; j < number/2; ++j)
{
if (number % j == 0)
{
return false;
}
}
return true;
}
The for loop in the main function will be something like:
for (int i = 2; number > 0; ++i)
{
if (isPrime(i))
{
count++;
--number;
cout << setw(5) << i;
if (count % 10 == 0)
cout << endl;
}
}
I am trying to write a program to count each number the program has encountered. by putting M as an input for the number of the array elements and Max is for the maximum amount of number like you shouldn't exceed this number when writing an input in the M[i]. for some reason the program works just fine when I enter a small input like
Data input:
10 3
1 2 3 2 3 1 1 1 1 3
Answer:
5 2 3
But when I put a big input like 364 for array elements and 15 for example for max. the output doesn't work as expected and I can't find a reason for that!
#include "stdafx.h"
#include <iostream>
#include<fstream>
#include<string>
#include <stdio.h>
#include<conio.h>
using namespace std;
int ArrayValue;
int Max;
int M[1000];
int checker[1000];
int element_cntr = 0;
int cntr = 0;
int n = 0;
void main()
{
cout << "Enter the lenght of the Elements, followed by the maximum number: " << endl;
cin >> ArrayValue>> Max;
for (int i = 0; i < ArrayValue; i++)
{
cin >> M[i];
checker[i]= M[i] ;
element_cntr++;
if (M[i] > Max)
{
cout << "the element number " << element_cntr << " is bigger than " << Max << endl;
}
}
for (int i = 0; i < Max; i++)
{
cntr = 0;
for (int j = 0; j < ArrayValue; j++)
{
if (M[n] == checker[j])
{
cntr+=1;
}
}
if (cntr != 0)
{
cout << cntr << " ";
}
n++;
}
}
You have general algorithm problem and several code issues which make code hardly maintainable, non-readable and confusing. That's why you don't understand why it is not working.
Let's review it step by step.
The actual reason of incorrect output is that you only iterate through the first Max items of array when you need to iterate through the first Max integers. For example, let we have the input:
7 3
1 1 1 1 1 2 3
While the correct answer is: 5 1 1, your program will output 5 5 5, because in output loop it will iterate through the first three items and make output for them:
for (int i = 0; i < Max; i++)
for (int j = 0; j < ArrayValue; j++)
if (M[n] == checker[j]) // M[0] is 1, M[1] is 1 and M[2] is 1
It will output answers for first three items of initial array. In your example, it worked fine because the first three items were 1 2 3.
In order to make it work, you need to change your condition to
if (n == checker[j]) // oh, why do you need variable "n"? you have an "i" loop!
{
cntr += 1;
}
It will work, but both your code and algorithm are absolutely incorrect...
Not that proper solution
You have an unnecessary variable element_cntr - loop variable i will provide the same values. You are duplicating it's value.
Also, in your output loop you create a variable n while you have a loop variable i which does the same. You can safely remove variable n and replace if (M[n] == checker[j]) to if (M[i] == checker[j]).
Moreover, your checker array is a full copy if variable M. Why do you like to duplicate all the values? :)
Your code should look, at least, like this:
using namespace std;
int ArrayValue;
int Max;
int M[1000];
int cntr = 0;
int main()
{
cout << "Enter the lenght of the Elements, followed by the maximum number: " << endl;
cin >> ArrayValue >> Max;
for (int i = 0; i < ArrayValue; i++)
{
cin >> M[i];
if (M[i] > Max)
{
cout << "the element number " << i << " is bigger than " << Max << endl;
}
}
for (int i = 0; i < Max; i++)
{
cntr = 0;
for (int j = 0; j < ArrayValue; j++)
{
if (i == M[j])
{
cntr ++;
}
}
if (cntr != 0)
{
cout << cntr << " ";
}
}
return 0;
}
Proper solution
Why do you need a nested loop at all? You take O(n*m) operations to count the occurences of items. It can be easily counted with O(n) operations.
Just count them while reading:
using namespace std;
int arraySize;
int maxValue;
int counts[1000];
int main()
{
cout << "Enter the lenght of the Elements, followed by the maximum number: " << endl;
cin >> arraySize >> maxValue;
int lastReadValue;
for (int i = 0; i < arraySize; i++)
{
cin >> lastReadValue;
if (lastReadValue > maxValue)
cout << "Number " << i << " is bigger than maxValue! Skipping it..." << endl;
else
counts[lastReadValue]++; // read and increase the occurence count
}
for (int i = 0; i <= maxValue; i++)
{
if (counts[i] > 0)
cout << i << " occurences: " << counts[i] << endl; // output existent numbers
}
return 0;
}