I just finished going through this pascal triangle project and the the output is not a proper triangle , anyone have any idea why this is happening , the spacing looks about right its just the alignment that's way off
#include <string>
#include <iostream>
using namespace std;
int factorial(int numPar); //the factorial function will recieve a number, and return that number's factorial
int pascalNum(int rowPar, int elementPar);
int main(){
//declarations
string strOutput = "";
int userNum;
int rowCounter = 0;
int elementCounter = 0;
int columnsPerRow = 1;
int spaces;
int initialSpaces;
int counter = 0;
//get user input
cout << "Enter an integer from 1 to 10: ";
cin >> userNum;
while (userNum > 10 || userNum < 1){
cout << "Invalid entry: " << endl;
cin >> userNum;
}
initialSpaces = userNum + 4; //needed to make the triangle an isoscoles, and keep it away from left
//calculations
while ((rowCounter + 1) <= userNum){
for (counter = initialSpaces; counter > 0; counter--){
strOutput = strOutput + " ";
}
while (elementCounter < columnsPerRow){
strOutput = strOutput + to_string(pascalNum(rowCounter, elementCounter));
if (pascalNum(rowCounter, elementCounter) < 10){
spaces = 3;
}
else if (pascalNum(rowCounter, elementCounter) < 100){
spaces = 2;
}
else if (pascalNum(rowCounter, elementCounter) < 1000){
spaces = 1;
}
while (spaces > 0){
strOutput = strOutput + " ";
spaces = spaces - 1;
}
elementCounter = elementCounter + 1;
}
cout << strOutput << endl;
columnsPerRow = columnsPerRow + 1;
elementCounter = 0;
rowCounter = rowCounter + 1;
//initialSpaces--; //this makes there be less and less space until the triangle starts
strOutput = "";
}
system("pause>nul");
return 0;
}
int factorial(int numPar) {
//declarations
int counter = 1;
int numResult = 1;
int initial = numPar;
if (numPar > 1){
while (counter <= numPar){
numResult = numResult * counter;
counter++;
}
return numResult;
}
else
return 1;
}
int pascalNum(int rowPar, int elementPar){
int answer;
answer = factorial(rowPar) / (factorial(elementPar) * factorial(rowPar - elementPar));
return answer;
}
Well, you're going about the layout wrong. First you've got to work out your number cell layout. You've got the start of code that does this, but you put all the extra spaces on the right.
What you probably want is something that's centered. Which means that, for a maximum of three digits, you'll want the one padding space on the left, and for one digit, you want a space on either side. And overall you want one space between cells. So your padding code changes to:
for(int elementCounter = 0; elementCounter < columnsPerRow;
elementCounter = elementCounter + 1){
// Don't Repeat Yourself
int num = pascalNum(rowCounter, elementCounter);
string numStr = to_string(num);
if (num < 10){
numStr = numStr + " ";
}
if (num < 100){
numStr = string(" ") + numStr;
}
strOutput += numStr;
}
Now that you know your code has cells with three possible digits and one padding space, draw out what it should look like for a small test case:
###
### ###
### ### ###
Now look at the pattern, and lo, there are two indent spaces on the left per row, or 2 * (9 - r) total, where r goes from 0 to 9. Fix your outer (row) loop accordingly and get rid of the initialSpaces stuff:
while ((rowCounter + 1) <= userNum){
for (counter = 2 * (9 - rowCounter); counter > 0; counter--){
strOutput = strOutput + " ";
}
// ... continue as above ...
And that should fix things. Moral of the story:
Don't forget to draw up your expected output format.
Use graph paper if you have to.
Related
So my problem is bigger but I just do not know what to do with my code. I can do what I want if I use an array works just fine but we are not using arrays yet so I have no idea how to do it. So I have to take user input as a string validate that the string is 16 characters long, all of them are digits, and most importantly I have to multiply every other or even character by 2. Then if it is a double digit add the two digit (ex. 10 1+0). Oh by the way I do not know why but every time I do i%2 == 0 I get the odd numbers. Is it because i is unsigned?
for(unsigned i = 1; i < card.length(); i++){
if (i % 2 == 1){
}
else {
}
}
return sum;
}
You could use an array of strings where each string contains a number.
Go through them checking for 2 conditions:
Double the number if it is even (i.e., i % 2 == 0)
Add the digits if the number has 2 digits (i.e., string's length is 2)
Code:
#include <iostream>
#include <iterator> using namespace std;
int TOTAL_CARDS = 16;
void printCards(string msg, string *array) {
cout<<msg<<endl;
for(int i = 0; i < TOTAL_CARDS; i++) {
cout<<"array["<<i<<"]="<<array[i]<<endl;
}
cout<<"\n"<<endl; }
int main() {
string cards[TOTAL_CARDS];
// hardcoded numbers 0 up to TOTAL_CARDS for demo purposes
for(int i = 0; i < TOTAL_CARDS; i++) {
cards[i] = to_string(i);
}
printCards("Before:", cards);
for (unsigned i = 1; i < TOTAL_CARDS; i++){
// double if even
if (i % 2 == 0){
cards[i] = to_string(stoi(cards[i]) * 2);
}
// add digits if double digit number
if (cards[i].length() == 2) {
// get each digit
string currentNum = cards[i];
int firstDigit = currentNum[0] - '0'; // char - '0' gives int
int secondDigit = currentNum[1] - '0';
// do sum and put in array
int sum = firstDigit + secondDigit;
cards[i] = to_string(sum);
}
}
printCards("After:", cards); }
Output:
Before:
array[0]=0
array[1]=1
array[2]=2
array[3]=3
array[4]=4
array[5]=5
array[6]=6
array[7]=7
array[8]=8
array[9]=9
array[10]=10
array[11]=11
array[12]=12
array[13]=13
array[14]=14
array[15]=15
After:
array[0]=0
array[1]=1
array[2]=4
array[3]=3
array[4]=8
array[5]=5
array[6]=3
array[7]=7
array[8]=7
array[9]=9
array[10]=2
array[11]=2
array[12]=6
array[13]=4
array[14]=10
array[15]=6
If you wanted to get user input for the numbers:
// get user to enter numbers
cout<<"Please enter "<<TOTAL_CARDS<<" numbers: "<<endl;
for(int i = 0; i < TOTAL_CARDS; i++) {
cin>>cards[i];
}
I found the answer to it. I first needed to create char variable named num. Convert the char to an int using chnum and then multiply.
for(unsigned i = 0; i < card.length(); i++){
if (i % 2 == 1){
num = card.at(i);
chnum = (num -'0');
add = chnum * 2;
if(add >= 10){
char ho = (add + '0');
string str(1,ho);
for (unsigned j = 0; j < str.length();j++){
char digi = str.at(j);
int chub = (digi - '0');
cout << digi;
//add = (chub) + (chub);
}
}
sum += add;
}
Beginner here, and I'm stuck. The main program is provided to us, and we're supposed to write 3 functions. readBig(), addBig(), and printBig(). I'm stuck on the addBig() function. It's supposed to sum two arrays, and perform the carry operation. I cannot figure out where I'm going wrong. The carry operation is working out for me.
Any help/direction is appreciated.
#include <iostream>
using namespace std;
//This program will test three functions capable of reading, adding,
//and printing 100-digit numbers.
// Do not change these function prototypes:
void readBig(int[]);
void printBig(int[]);
void addBig(int[], int[], int[]);
// This constant should be 100 when the program is finished.
const int MAX_DIGITS = 100;
//There should be no changes made to the main program when you turn it
in.
int main(){
// Declare the three numbers, the first, second and the sum:
int num1[MAX_DIGITS], num2[MAX_DIGITS], sum[MAX_DIGITS];
bool done = false;
char response;
while (not done)
{
cout << "Please enter a number up to "<<MAX_DIGITS<< " digits: ";
readBig(num1);
cout << "Please enter a number up to "<<MAX_DIGITS<< " digits: ";
readBig(num2);
addBig(num1, num2, sum);
printBig(num1);
cout << "\n+\n";
printBig(num2);
cout << "\n=\n";
printBig(sum);
cout << "\n";
cout <<"test again?";
cin>>response;
cin.ignore(900,'\n');
done = toupper(response)!='Y';
}
return 0;
}
//ReadBig will read a number as a string,
//It then converts each element of the string to an integer and stores
it in an integer array.
//Finally, it reverses the elements of the array so that the ones digit
is in element zero,
//the tens digit is in element 1, the hundreds digit is in element 2,
etc.
void readBig(int num[])
{
for(int i = 0; i < MAX_DIGITS; i++){
num[i] = 0;
}
string numStr;
getline(cin,numStr);
string temp;
//store into array
for (int i = 0; i < numStr.length(); i++){
temp = numStr.at(i);
num[i] = stoi(temp);
}
int arrayLength = MAX_DIGITS;
int temp2;
for (int i = 0; i < (arrayLength/2); i++){
temp2 = num[i];
num[i] = num[(arrayLength - 1) - i];
num[(arrayLength - 1) - i] = temp2;
}
}
//AddBig adds the corresponding digits of the first two arrays and
stores the answer in the third.
//In a second loop, it performs the carry operation.
void addBig(int num1[], int num2[], int sum[])
{
for (int i = 0; i < MAX_DIGITS; i++){
sum[i] = num1[i] + num2[i];
if (sum[i] > 9){
sum[i] = sum[i] - 10;
sum[i+1] = sum[i+1] + 10;
}
}
}
//PrintBig uses a while loop to skip leading zeros and then uses a for
loop to print the number.
void printBig(int array[])
{
int i = 0;
while (array[i] == 0){
i++;
}
for (int j = i; j < MAX_DIGITS;j++){
cout << array[j] << endl;
}
}
Looks like readBig function isn't correct, it stores least significiant digit into num[numStr.length()-1], after reversing it became num[MAX_DIGITS -1 - ( numStr.length()-1], but addNum assumes last digit is num[0].
Correct variant:
void readBig(int num[])
{
//clear num, read numStr...
//store into array
int count = 0;
for (int i = numStr.length()-1; i >= 0; --i){
temp = numStr.at(i);
num[count++] = stoi(temp);
}
|
So this
sum[i] = sum[i] - 10;
sum[i+1] = sum[i+1] + 10;
should most likely be this
sum[i] = sum[i] - 10;
sum[i+1] = sum[i+1] + 1;
Since its the next decimal place it shouldnt be incremented by 10
Also when you get to the last cell in your array
sum[i+1] = sum[i+1] + 1;
this will be out of bounds, so depending on the requirments you will want to change this
I am trying to write a code that takes a binary number input as a string and will only accept 1's or 0's if not there should be an error message displayed. Then it should go through a loop digit by digit to convert the binary number as a string to decimal. I cant seem to get it right I have the fact that it will only accept 1's or 0's correct. But then when it gets into the calculations something messes up and I cant seem to get it correct. Currently this is the closest I believe I have to getting it working. could anyone give me a hint or help me with what i am doing wrong?
#include <iostream>
#include <string>
using namespace std;
string a;
int input();
int main()
{
input();
int decimal, x= 0, length, total = 0;
length = a.length();
// atempting to make it put the digits through a formula backwords.
for (int i = length; i >= 0; i--)
{
// Trying to make it only add the 2^x if the number is 1
if (a[i] = '1')
{
//should make total equal to the old total plus 2^x if a[i] = 1
total = total + pow(x,2);
}
//trying to let the power start at 0 and go up each run of the loop
x++;
}
cout << endl << total;
int stop;
cin >> stop;
return 0;
}
int input()
{
int x, x2, count, repeat = 0;
while (repeat == 0)
{
cout << "Enter a string representing a binary number => ";
cin >> a;
count = a.length();
for (x = 0; x < count; x++)
{
if (a[x] != '0' && a[x] != '1')
{
cout << a << " is not a string representing a binary number>" << endl;
repeat = 0;
break;
}
else
repeat = 1;
}
}
return 0;
}
I don't think that pow suits for integer calculation. In this case, you can use shift operator.
a[i] = '1' sets the value of a[i] to '1' and return '1', which is always true.
You shouldn't access a[length], which should be meaningless.
fixed code:
int main()
{
input();
int decimal, x= 0, length, total = 0;
length = a.length();
// atempting to make it put the digits through a formula backwords.
for (int i = length - 1; i >= 0; i--)
{
// Trying to make it only add the 2^x if the number is 1
if (a[i] == '1')
{
//should make total equal to the old total plus 2^x if a[i] = 1
total = total + (1 << x);
}
//trying to let the power start at 0 and go up each run of the loop
x++;
}
cout << endl << total;
int stop;
cin >> stop;
return 0;
}
I would use this approach...
#include <iostream>
using namespace std;
int main()
{
string str{ "10110011" }; // max length can be sizeof(int) X 8
int dec = 0, mask = 1;
for (int i = str.length() - 1; i >= 0; i--) {
if (str[i] == '1') {
dec |= mask;
}
mask <<= 1;
}
cout << "Decimal number is: " << dec;
// system("pause");
return 0;
}
Works for binary strings up to 32 bits. Swap out integer for long to get 64 bits.
#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
string getBinaryString(int value, unsigned int length, bool reverse) {
string output = string(length, '0');
if (!reverse) {
for (unsigned int i = 0; i < length; i++) {
if ((value & (1 << i)) != 0) {
output[i] = '1';
}
}
}
else {
for (unsigned int i = 0; i < length; i++) {
if ((value & (1 << (length - i - 1))) != 0) {
output[i] = '1';
}
}
}
return output;
}
unsigned long getInteger(const string& input, size_t lsbindex, size_t msbindex) {
unsigned long val = 0;
unsigned int offset = 0;
if (lsbindex > msbindex) {
size_t length = lsbindex - msbindex;
for (size_t i = msbindex; i <= lsbindex; i++, offset++) {
if (input[i] == '1') {
val |= (1 << (length - offset));
}
}
}
else { //lsbindex < msbindex
for (size_t i = lsbindex; i <= msbindex; i++, offset++) {
if (input[i] == '1') {
val |= (1 << offset);
}
}
}
return val;
}
int main() {
int value = 23;
cout << value << ": " << getBinaryString(value, 5, false) << endl;
string str = "01011";
cout << str << ": " << getInteger(str, 1, 3) << endl;
}
I see multiple misstages in your code.
Your for-loop should start at i = length - 1 instead of i = length.
a[i] = '1' sets a[i] to '1' and does not compare it.
pow(x,2) means and not . pow is also not designed for integer operations. Use 2*2*... or 1<<e instead.
Also there are shorter ways to achieve it. Here is a example how I would do it:
std::size_t fromBinaryString(const std::string &str)
{
std::size_t result = 0;
for (std::size_t i = 0; i < str.size(); ++i)
{
// '0' - '0' == 0 and '1' - '0' == 1.
// If you don't want to assume that, you can use if or switch
result = (result << 1) + str[i] - '0';
}
return result;
}
The program builds and runs, however after entering the first integer and pressing enter then the error pop up box appears, then after pressing ignore and entering the second integer and pressing enter the pop up box appears and after pressing ignore it returns the correct answer. I am at my wits end with this can somebody help me fix the pop up box thing.
#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <string>
using namespace std;
#define numbers 100
class largeintegers {
public:
largeintegers();
void
Input();
void
Output();
largeintegers
operator+(largeintegers);
largeintegers
operator-(largeintegers);
largeintegers
operator*(largeintegers);
int
operator==(largeintegers);
private:
int integer[numbers];
int len;
};
void largeintegers::Output() {
int i;
for (i = len - 1; i >= 0; i--)
cout << integer[i];
}
void largeintegers::Input() {
string in;
int i, j, k;
cout << "Enter any number:";
cin >> in;
for (i = 0; in[i] != '\0'; i++)
;
len = i;
k = 0;
for (j = i - 1; j >= 0; j--)
integer[j] = in[k++] - 48;
}
largeintegers::largeintegers() {
for (int i = 0; i < numbers; i++)
integer[i] = 0;
len = numbers - 1;
}
int largeintegers::operator==(largeintegers op2) {
int i;
if (len < op2.len) return -1;
if (op2.len < len) return 1;
for (i = len - 1; i >= 0; i--)
if (integer[i] < op2.integer[i])
return -1;
else if (op2.integer[i] < integer[i]) return 1;
return 0;
}
largeintegers largeintegers::operator+(largeintegers op2) {
largeintegers temp;
int carry = 0;
int c, i;
if (len > op2.len)
c = len;
else
c = op2.len;
for (i = 0; i < c; i++) {
temp.integer[i] = integer[i] + op2.integer[i] + carry;
if (temp.integer[i] > 9) {
temp.integer[i] %= 10;
carry = 1;
} else
carry = 0;
}
if (carry == 1) {
temp.len = c + 1;
if (temp.len >= numbers)
cout << "***OVERFLOW*****\n";
else
temp.integer[i] = carry;
} else
temp.len = c;
return temp;
}
largeintegers largeintegers::operator-(largeintegers op2) {
largeintegers temp;
int c;
if (len > op2.len)
c = len;
else
c = op2.len;
int borrow = 0;
for (int i = c; i >= 0; i--)
if (borrow == 0) {
if (integer[i] >= op2.integer[i])
temp.integer[i] = integer[i] - op2.integer[i];
else {
borrow = 1;
temp.integer[i] = integer[i] + 10 - op2.integer[i];
}
} else {
borrow = 0;
if (integer[i] - 1 >= op2.integer[i])
temp.integer[i] = integer[i] - 1 - op2.integer[i];
else {
borrow = 1;
temp.integer[i] = integer[i] - 1 + 10 - op2.integer[i];
}
}
temp.len = c;
return temp;
}
largeintegers largeintegers::operator*(largeintegers op2) {
largeintegers temp;
int i, j, k, tmp, m = 0;
for (i = 0; i < op2.len; i++) {
k = i;
for (j = 0; j < len; j++) {
tmp = integer[j] * op2.integer[i];
temp.integer[k] = temp.integer[k] + tmp;
temp.integer[k + 1] = temp.integer[k + 1] + temp.integer[k] / 10;
temp.integer[k] %= 10;
k++;
if (k > m) m = k;
}
}
temp.len = m;
if (temp.len > numbers) cout << "***OVERFLOW*****\n";
return temp;
}
using namespace std;
int main() {
int c;
largeintegers num1, num2, result;
num1.Input();
num2.Input();
num1.Output();
cout << " + ";
num2.Output();
result = num1 + num2;
cout << " = ";
result.Output();
cout << "\n\n";
num1.Output();
cout << " - ";
num2.Output();
result = num1 - num2;
cout << " = ";
result.Output();
cout << "\n\n";
num1.Output();
cout << " * ";
num2.Output();
result = num1 * num2;
cout << " = ";
result.Output();
cout << "\n\n";
c = num1 == num2;
num1.Output();
switch (c) {
case -1:
cout << " is less than ";
break;
case 0:
cout << " is equal to ";
break;
case 1:
cout << " is greater than ";
break;
}
num2.Output();
cout << "\n\n";
system("pause");
}
It seems you are falling victim to the difference between C-style strings and C++ strings. C-style strings are a series of chars followed by a zero (or null) byte. C++ strings are objects that contain a series of characters (usually char, but eventually this will be an assumption you should break) and that know their own length. C++ strings can contain null bytes in the middle of themselves without problem.
To loop through all of the characters of a C++-style string, you can do one of a number of things:
You can use the .size() or .length() members of a string variable to find the number of characters in it, as in for (int i=0; i<str.size(); i++) { char c = str[i];
You can use .begin() and .end() to get iterators to the beginning and end of the string, respectively. A for loop in the form for (std::string::iterator it=str.begin(); it!=str.end(); ++it) will loop you through the members of the string by accessing *it.
If you're using C++11, you can use the for loop construct as follows: for (auto c: str) where c will be of the type of a character of the string str.
In the future, to solve problems like these, you can try using the debugger to see what happens when your program crashes or hits an exception. You likely would find that inside of largeintegers::Input() you running into either a memory access violation or some other problem.
Finally, as a future-looking criticism, you should not use C-style arrays (where you say int integer[ numbers ];) in favor of using C++-style containers, such as vector. A vector is a series of objects (such as ints) that can expand as needed.
I need to rewrite the program to use a function isPalindrome. It needs to input a 5 digit integer and return a boolean (true if it is a palindrome, false if it is not), and it cannot contain any cout statements. I am not sure how I would do this without a cout function. Here is my code:
#include <iostream>
using namespace std;
int main()
{
int number, digit1, digit2, digit3, digit4, digit5;
cout << "\nEnter a 5-digit integer: ";
cin >> number;
//Break down input number into individual digits:
digit1 = number / 10000;
digit2 = number % 10000 / 1000;
digit3 = number % 10000 / 100;
digit4 = number % 10000 / 10;
digit5 = number % 10;
if ( digit1 == digit5 && digit2 == digit4 )
cout << number <<" is a palindrome.\n";
else
cout << number << " is not a palindrome.\n";
return 0;
}
int isPalindrome ()
{
}
This should help get you started (without ruining too much of the fun)
int main(){
//accept integer input using cin
if(isPalindrome(input))
cout << "yaay!";
else
cout << "nooouuu!";
}
bool isPalindrome (int input)
{
//test things here
return ( digit1 == digit5 && digit2 == digit4 )
// ^ returns true if condition satisfied
}
Additionally, your way of separating out the digits is incorrect. It should be:
digit1 = number/10000 % 10;
digit2 = number/1000 % 10;
digit3 = number/100 % 10;
digit4 = number/10 % 10;
digit5 = number % 10;
Ofcourse, the above should actually be in a loop.
It doesn't have to be specified how many digits does the number contain. You can try something like this:
bool isPalindrome(int number) {
int reverse = 0, copy = number;
while(copy != 0) {
reverse = reverse*10 + copy%10;
copy /= 10;
}
return number == reverse;
}
string s;
cout<<"\nEnter a string : " ;
cin>>s;
int length = s.length();
char* arr = new char();
int k = length;
for(int i = 0 ; i <= length ; i++)
{
arr[i] = s[k];
k -= 1;
}
if(!palindrome(s, arr, length))
{
cout<<"\nNot Palindrome\n";
}
else
{
cout<<"\nPalindrome\n";
}
}
bool palindrome(string& s, char* arr, int length)
{
int j = 0;
for(int i = 1 ; i <= length; i++)
{
if(arr[i]!= s[j])
{
return false;
}
j++;
}
return true;
}