My tutor gave me an assignment where I have to write a code that only contains while loop and prints:
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
I tried it 100 times and failed 100 times. Due to my limited knowledge, I began to think that my tutor is just messing with my brain. If it's possible, please introduce me a code that prints the numbers in that order.
Thanks...
int i = 1;
int LIMIT = 5;
while (i <= LIMIT)
{
int j = 1;
while (j <= LIMIT -i) //Loop to print the desired space.
{
cout << " ";
j++;
}
int k = i;
while(k)
{
cout<<k; //Printing the digits
k--;
}
cout << endl; //Adding new line character at the end.
i++;
}
Say hello to your tutor :)
int main(void)
{
char str[11] = " ";// Filled with 10 blank spaces
int i=0;
while(i < 5)
{
str[9 - 2*i] = (i+1) + 48;// +48 will give equivalent ASCII code
printf("%s\n",str);
i++;
}
return 0;
}
Try some code like below. i've also create a sample working code here http://goo.gl/gJqias
#include <iostream>
using namespace std;
int main()
{
int start_Index=1, maximum_Index=5;
while(start_Index<=maximum_Index)
{
//below loop prints leading whitespaces
//note that there are two whitespaces per number
int temp_var=start_Index;
while (maximum_Index-temp_var>0)
{
cout <<" ";
temp_var++;//note the post increment operator.
}
//below whiel loop prints lagging numbers with single whitespace before them
temp_var=start_Index;
while(temp_var>0)
{
cout<<" "<<temp_var--;//note the post decrement operator.
}
//Now we start over to next line
cout<<endl;
//Increment the start_index by 1
start_Index++;
}
return 0;
}
Related
I'm trying to solve Codewars task and facing issue that looks strange to me.
Codewars task is to write function digital_root(n) that sums digits of n until the end result has only 1 digit in it.
Example: 942 --> 9 + 4 + 2 = 15 --> 1 + 5 = 6 (the function returns 6).
I wrote some bulky code with supporting functions, please see code with notes below.
The problem - digital_root function works only if I put cout line in while loop. The function returns nonsense without this cout line (please see notes in the code of the function).
My questions are:
Why isn't digital_root working without cout line?
How cout line can effect the result of the function?
Why does cout line fix the code?
Thanks a lot in advance! I'm a beginner, spent several days trying to solve the issue.
#include <iostream>
#include <cmath>
#include <string>
using namespace std;
int getDigit (int, int);
int sumDigits (int);
int digital_root (int);
int main()
{
cout << digital_root (942); // expected output result is 6 because 9 + 4 + 2 = 15 -> 1 + 5 = 6
}
int getDigit (int inputNum, int position) // returns digit of inputNum that sits on a particular position (works)
{
int empoweredTen = pow(10, position-1);
return inputNum / empoweredTen % 10;
}
int sumDigits (int inputNum) // returns sum of digits of inputNum (works)
{
int sum;
int inLen = to_string(inputNum).length();
int i = inLen;
while (inLen --)
{
sum += getDigit(inputNum, i);
i --;
}
return sum;
}
int digital_root (int inputNum) // supposed to calculate sum of digits until number has 1 digit in it (abnormal behavior)
{
int n = inputNum;
while (n > 9)
{
n = sumDigits(n);
cout << "The current n is: " << n << endl; // !!! function doesn't work without this line !!!
}
return n;
}
I've tried to rewrite the code from scratch several times with Google to find a mistake but I can't see it. I expect digital_root() to work without any cout lines in it. Currently, if I delete cout line from while loop in digital_root(), the function returns -2147483647 after 13 seconds of calculations. Sad.
Here is an implementation using integer operators instead of calling std::to_string() and std::pow() functions - this actually works with floating-point numbers. It uses two integer variables, nSum and nRem, holding the running sum and remainder of the input number.
// calculates sum of digits until number has 1 digit in it
int digital_root(int inputNum)
{
while (inputNum > 9)
{
int nRem = inputNum, nSum = 0;
do // checking nRem after the loop avoids one comparison operation (1st check would always evaluate to true)
{
nSum += nRem % 10;
nRem /= 10;
} while (nRem > 9);
inputNum = nSum + nRem;
std::cout << "The current Sum is: " << inputNum << endl; // DEBUG - Please remove this
}
return inputNum;
}
As for the original code, the problem was the uninitialized sum variable, as already pointed out by other members - it even generates a compiler error.
int sumDigits (int inputNum) // returns sum of digits of inputNum (works)
{
int sum = 0; // MAKE SURE YOU INITIALIZE THIS TO 0 BEFORE ADDING VALUES TO IT!
int inLen = to_string(inputNum).length();
int i = inLen;
while (inLen --)
{
sum += getDigit(inputNum, i);
i --;
}
return sum;
}
Initialize your variables before adding values to them, otherwise you could run into undefined behaviour. Also for the record, adding the cout line printed out something, but it wasn't the correct answer.
N = Input How much attempt (First Line).
s = Input How much value can be added (Second, fourth and sixth lines).
P = Input of numbers separated with space.
Example :
3 ( Input N )
2 ( s 1 )
2 3
3 ( s 2 )
1 2 3
1 ( s 3 )
12
Example :
Read #1: 5 (Output s1 = 2 + 3)
Read #2: 6 (Output s2 = 1+2+3)
Read #3: 12 (Output s3 = 12)
I've been searching and trying for very long but couldn't figure out such basic as how to cin based on given numbers, with spaces and add all values into a variable. For example:
#include <iostream>
using namespace std;
int main() {
int l, o[l], r, p[r], i;
cin >> l;
for(i = 0; i < l; i++) {
cin>>o[l];
r = o[l]; // for every o[0] to o[l]
}
while (cin>>o[l]) {
for (i = 0; i < l; i++){
cin>>p[o]; // for every o[0] to o[l]
// i.e o[l] = 1 then 2 values can be added (because it starts from zero)
// input 1 2
// o[1] = {1, 2}
int example += o[1];
cout<< "Read#2: " << example;
}
}
}
And it doesn't work. Then i found getline(), ignoring the s and just input anything that will finally be added to a number, turned out it is only usable for char string. I tried scanf, but I'm not sure how it works. So im wondering if it's all about s(values) × 1(column) matrix from a looping but sill not sure how to make it. Any easy solutions to this without additional libraries or something like that? Thanks in advance.
#include <iostream>
using namespace std;
int main() {
int t; //number of attempts
cin >> t;
while(t--) { // for t attempts
int n, s = 0; //number of values and initial sum
cin >> n;
while (n--) { //for n values
int k; //value to be added
cin >> k;
s += k; //add k to sum
}
cout << s << "\n"; //print the sum and a newline
}
return 0;
}
If you want to add more details, (i.e. print Read#n on the nth attempt), you can always use
for (int i = 1; i <= n; i++)
to replace while(t--) and at the end of the attempt just print
cout << "Read#" << i << ": " << s << "\n";
I am trying to solve a problem where every letter has a respective number such as a-1,b-2....z-26.
Now given a number, in how many ways can the number be decoded is the question. consider an example where 25114 can be decoded as 'BEAN',‘BEAAD’, ‘YAAD’, ‘YAN’, ‘YKD’ and ‘BEKD’. this could be decoded in 6 ways.
I have written code in c++ but I am getting the wrong answer. Please correct my code.
#include<bits/stdc++.h>
using namespace std;
int total = 0;
int arr[100001];
void func(int start,int end,int factor){
if(start==end)
return;
int j =start;
if(factor==2&&j==end-1)//if j is the last element and factor is 2,accessing j+1 element is illegual
return;
if(factor==2){
if((arr[j]*10+arr[j+1])>26)
return;
else{
total++;
func(start+2,end,1);
func(start+2,end,2);
}
}
else{//factor is 1
total++;
func(start+1,end,1);
func(start+1,end,2);
}
}
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
int p;
cin>>p;
arr[i]=p;
}
func(0,n,1);
func(0,n,2);
cout<<total<<endl;
return 0;
}
essentially what my code is doing is that it fixes one number from the given array(using one digit or two digits from the the given array) and recurses until all the combinations are covered. for example considering the above case, I first choose '2' as my first digit and decode it as 'B'(factor = 1) and then choose '25' and decode it as 'E'(factor = 2).
**following are the input and output from the following code
input : 25114
expected output : 6
my output : 15
input : 3333333333(10 digits)
expected output : 1
my output : 10
Based on the original program from the question I suggest to count the encodings when you reach the end only (if(start==end)).
As func will always be called twice with factor=1 and factor=2, I can freely choose either condition for counting.
Here is the modified code:
#include<bits/stdc++.h>
using namespace std;
int total = 0;
int arr[100001];
void func(int start,int end,int factor){
if(start==end) {
if(factor == 1) total++; // count once when reaching the end
return;
}
int j =start;
if((factor==2) && (j==end-1))//if j is the last element and factor is 2,accessing j+1 element is illegal
return;
if(factor==2){
if((arr[j]*10+arr[j+1])>26)
return;
else{
//total++;
func(start+2,end,1);
func(start+2,end,2);
}
}
else{//factor is 1
//total++;
func(start+1,end,1);
func(start+1,end,2);
}
return;
}
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
int p;
cin>>p;
arr[i]=p;
}
func(0,n,1);
func(0,n,2);
cout<<total<<endl;
return 0;
}
This calculates the expected results from the example input in the question.
$ echo 5 2 5 1 1 4|./program
6
$ echo 10 3 3 3 3 3 3 3 3 3 3|./program
1
There is room for improvement.
Instead of modifying a global variable I would return the number of combinations from func and add the values in the higher level.
I would also handle the distinction between 2-digit and 1-digit numbers in the called func instead of in the caller.
Something like this pseudo code:
int func(int start, int end)
{
if(remaining length is <2) {
// we reached the end, so this is one combination
return 1;
}
if(two-digit number is >26) {
// only a 1-digit number is possible, count remaining combinations
return func(start+1, end);
}
// both a 1-digit or 2-digit number is possible, add the remaining combinations for both cases
return func(start+1) + func(start+2);
}
Your question is tagged as "dynamic-programming", but it is anything but.
Instead, think about the state space and its boundary conditions:
The empty string has zero encodings;
A single digit has a single encoding;
An n-digit string has as many encodings as an (n-1)-digit substring plus as many encodings as an (n-2)-digit substring if the first two digits are <= 26.
Thus, we can walk the string from back to front and store the intermediate results for reuse:
uint64_t solve(std::vector<int>& digits) {
const int n = digits.size();
std::vector<int> encodings(n+1);
encodings[n] = 1;
for (int i = n-1; i >= 0; i--) {
bool two_digits_fit = (i < n - 1) && (digits[i] * 10 + digits[i+1]) <= 26; // What if digits[i] == 0?
encodings[i] = encodings[i+1] + (two_digits_fit ? encodings[i+2] : 0);
}
return encodings[0];
}
I'm trying to solve the following problem:
What is the smallest number of factoriais summed that are needed to be equal an given number a? (1 ≤ a ≤ 10^5)
Example:
Input: 10, Output: 3. (10 = 3! + 2! + 2!)
Input: 25, Output: 2. (25 = 4! + 1!)
My code:
#include<bits/stdc++.h>
using namespace std;
int a;
int rec(int vet){
int count = 0;
a = a - vet;
if(a >= vet){
count++;
rec(vet);
}
count++;
return count;
}
int main(){
int vet[8] = {1}, count = 0;
cin >> a;
for(int i = 2; i <= 8; i++){
vet[i-1] = vet[i-2]*i;
}
for(int i = 7; i >= 0; i--){
if(a < vet[i]){
continue;
}
count += rec(vet[i]);
}
cout << count << endl;
}
My logic:
1°: a max is equal to 100000, so the maximum fatorial we have to
compare is 8!;
2°: I take a factioral that is equal or nearest small to a,
subtract the factorial from it and count++; If after the subtraction,
a still bigger then my factorial, I do the same step recursively.
This code pass on the base cases, but I got a wrong answer. I wasn't capable to find what case it didn't pass, so I'm here.
Can you find where am I wrong? Or if my solution is not good and I should try another approach.
Thanks for the help!
The problem is easily solved by a recursive approach.
Here is checked code:
#include <iostream>
using namespace std;
int factorial(int n) {
return n<=1 ? 1 : n * factorial(n-1);
}
int MinFact(int number)
{
static int num_of_facts;
int a = 1;
if (number)
{
while(factorial(a+1)<=number)a++;
cout << a << "!" << endl;
num_of_facts++;
MinFact((number-factorial(a)));
}
return num_of_facts;
}
int main()
{
int num;
cout << "Enter number" << endl;
cin >> num;
num = MinFact(num);
cout << "Number of factorials: " << num;
return 0;
}
As I mentioned in the comment, the issue is with the rec function. Due to rec being local, the count is not being incremented correctly.
A simple solution would be to replace the rec function as follows
int rec(int vec) {
int count = a / vec;
a = a % vec;
return count;
}
Edit : for a failing case try 18. The solution will be 3 but you will get 2.
I guess you can figure out how this logic works. If not you could do it with a loop.
I have an assignment in which I am supposed to
1.) Read in 2 char arrays of numbers
2.) add together the arrays, performing carry addition, carrying the tens place
3.) print out the newly added array.
I am also supposed to print an error if I need to carry on the last digit of the array (overflow)
so something like 99999999999999999999 +
1 =
________________________
ERROR
That's the part I'm having trouble with.
The above outputs something like "99999999999999999:0" so I have no idea what's going wrong.
I'll post my code, but please be nice :( I know it certainly isn't the most efficient, but I'm just trying to lay things out in a way that is easy for my brain to understand right now.
And yes, I HAVE to use char arrays. I guess it's to help us understand the ascii table.
#include <iostream>
using namespace std;
void InitNumber(char[]);
int AddArrays(char first[], char second[], char combined[]);
void OutputNumber (char[]);
const int LENGTH = 20; // global variable
int main()
{
char set1[LENGTH];
char set2[LENGTH];
char sum[LENGTH];
InitNumber (set1);
InitNumber (set2);
if(AddArrays (set1, set2, sum)) {
cout << "overflow!" << endl;
}
OutputNumber(sum);
}
void InitNumber (char list[])
{
int numberOfDigits;
cout << "Please enter the number of digits in the number: ";
cin >> numberOfDigits;
cout << "Please enter the digits in the number with the LEAST significant first: ";
for (int i = 0; i < numberOfDigits; i++) {
cin >> list [i];
}
for (int l=(numberOfDigits); l < LENGTH; l++) {
list [l] = '0';
}
}
int AddArrays(char first[], char second[], char combined[])
{
for (int h = 0; h < LENGTH; h++)
combined[h]= '0';
int overflow = 0;
for (int i = 0; i < LENGTH; i++) {
int currentSum = (first[i]-'0' + second[i]-'0');
cout << "currentSum = " << currentSum << endl;
if(currentSum / 10 == 0 )
combined[i] += currentSum;
else if (currentSum/10 !=0) {
if (i == LENGTH-1 && currentSum/10 !=0){
overflow = 1;
}
else{
combined [i] += currentSum%10;
cout << "current i: " << combined[i] << endl;
combined [i+1] += currentSum/10;
cout << "current i+1: " << combined[i+1] << endl;
}
}
}
return overflow;
}
void OutputNumber(char arrayOut[])
{
for (int l=LENGTH - 1; l >= 0; l--)
cout << arrayOut[l];
}
working input
input
6
1 2 3 4 5 6
7
1 2 3 4 5 6 7
output
00000000000008308642
Not-working output
input
20
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
1
1
output
999999999999999999:0
I'm going to reproduce small parts of your inner loop where you are carrying out the addition, in order to explain why your overflow detection is broken.
for (int i = 0; i < LENGTH; i++) {
int currentSum = (first[i]-'0' + second[i]-'0');
if(currentSum / 10 == 0 )
combined[i] += currentSum;
Here you're adding the corresponding pair of digits from the two numbers you're adding, and checking (in your particular style) if they overflowed. Try to remember this part, it's important. In order to check for overflow, you're only checking the result of adding the pair of digits from the two large numbers you're adding.
else if (currentSum/10 !=0) {
This is a completely useless check. You can only get here if the division is known to produce a non-zero result. This if() can be removed completely. Now, the relevant part of your code is
combined [i] += currentSum%10;
combined [i+1] += currentSum/10;
Do you see the problem yet?
Your approach is, once overflow is detected, is to increment the next higher order digit in the result.
Unfortunately, on the next loop iteration, in order to check for carry over, you're just checking the sum of the next corresponding digit pair, from the two large numbers you're adding. The carry-over you're saving here is going to get completely ignored.
Say your numbers are two digits long max, rather than 20, and you entered the numbers 99 and 1.
On the first iteration, you'll add 9 and 1, save 0 as the first digit, and add 1 to the second digit in the sum.
On the second iteration, you'll add 9 and 0, and, given your logic above, conclude that nothing overflowed.