1
232
34543
I have divided the pattern in two parts. first part is prints increment numbers and second part prints decrement numbers.
1st loop is for printing rows.
2nd loop is for printing spaces. 3rd
loop is for printing increment numbers.
4th loop is for decrement
numbers.
Code works fine up-to the 3rd loop. But the 4th loop is not giving the required output.
#include<stdio.h>
int main()
{
int i,j,n;
scanf("%d",&n);
/*Print row*/
for(i=1;i<=n;i++)
{
/*Printing Spaces*/
for (j = 1; j <= n-i; j++)
{
printf(" ");
}
/*Printing increasing number values*/
for ( int num = i; num <= 2*i-1; num++)
{
printf("%d",num);
}
/*Printing decreasing numbers value*/
for ( int num = 2*i-2; num <= i-1; num++)
{
printf("%d",num);
}
/*New line*/
printf("\n");
}
return 0;
}
Output is :
10
23
345
Welcome to SO.
Your first error was on the your counter on your 4th loop. Your loop should display decreasing numbers but you keep incrementing your counter (that you use to print too).
First error, decrease the counter in the 4th loop.
Second error is your condition on this first loop. You decrease a counter so your condition "While my counter is under a value" can never be false if the counter, at start of the loop, is already before the value. So you have to replace <= by >=.
And after test you will see that i-1 always print one number too far of remove your -1.
Here is your code corrected :
#include<stdio.h>
int main()
{
int i,j,n;
scanf("%d",&n);
/*Print row*/
for(i=1;i<=n;i++)
{
/*Printing Spaces*/
for (j = 1; j <= n-i; j++)
{
printf(" ");
}
/*Printing increasing number values*/
for ( int num = i; num <= 2*i-1; num++)
{
printf("%d",num);
}
/*Printing decreasing numbers value*/
for ( int num = 2*i-2; num >= i; num--)
{
printf("%d",num);
}
/*New line*/
printf("\n");
}
return 0;
}
I am writing a code to give numbers in a line and the inputs finish with zero then wirtes the highest power of 2 smaller or equal the inputs in a line.
it doesn't work.
#include<iostream>
#include<stdio.h>
using namespace std;
int highestPowerof2( int n)
{
static int result = 0;
for (static int i=n; i>=1; i--)
{
if ((i & (i-1)) == 0)
{
result = i;
break;
}
}
return result;
}
int main() {
static int num ;
do{
cin>>num ;
}
while(num=!0);
cout<<highestPowerof2(num)<<"\n";
return 0;
}
The most surprising thing in your code is this:
do{
cin>>num ;
}
while(num=!0);
You keep reading num from user input until num == 0. I have to admit that I dont really understand the rest of your code, but for num == 0 calling the function highestPowerof2(num) will always result in 0.
Perhaps you wanted to repeat the program until the user decides to quit, that could be
do{
cin>>num ;
cout<<highestPowerof2(num)<<"\n";
} while(num=!0);
PS: the other "surprising" thing is that you use static in places where it does not really make sense. Better simply remove it.
Here is another approach that is a little bit faster for large n. For example if n = 2^31 - 1, then the original loop would need to iterate 2^30 - 1 = 1,073,741,823 times, whereas this loop only needs a single iteration (provided sizeof(int) == 4):
#include <iostream>
#include <stdio.h>
using namespace std;
int highestPowerof2( int n)
{
if (n < 0) return 0;
int result = 0;
int num_bits = sizeof(int) * 8;
unsigned int i = 1 << (num_bits - 1);
while(i > 0) {
if (n >= i) return i;
i >>= 1;
}
return 0;
}
int main() {
int num ;
while (1) {
cin >> num;
cout << highestPowerof2(num) << "\n";
if (num == 0) break;
}
return 0;
}
I have to create a program which calculates the factorial of any number, the problem is if I input any number above 20 it just returns that number. What in my else if statement could be causing this and is there a better way to solve this? ( this function is called in main and works if num <= 20)
void factorial() {
//User input for number
long long num;
std::cout << "Input any positive integer to find its factorial: ";
std::cin >> num;
unsigned long long numFact = 1;
if (num <= 20) {
while (num > 0) {
numFact = numFact * num;
num = num - 1;
}
std::cout << numFact;
}
else if (num > 20) {
std::vector<int> multFactorial;
//stores num as seperate elements in vector multFactorial
while (num > 0) {
int remain = num % 10;
num = num / 10;
multFactorial.insert(multFactorial.begin(), remain);
}
std::vector<int> answer;
std::vector<int> answerFinal;
//Manually multiplies elements in multFactorial
//Then adds new vectors created by multiplying to get final answer
//Repeats until factorial is solved
//Ex: 21 * 20; 0 * 1 and 0 * 2 stored as {0 , 0}
//2*1 and 2*2 stored as {4, 2, 0}
//Vectors will be addes to get {4, 2, 0} and then that will be multiplied
by 19 until num = 1
while (num > 1) {
for (int i = multFactorial.size() - 1; i >= 0; i--) {
int remain1 = ((num - 1) % 10) * multFactorial[i];
answer.insert(answer.begin(), remain1);
int remain2 = (((num - 1) / 10) * multFactorial[i]);
answerFinal.insert(answerFinal.begin(), remain2);
}
answerFinal.insert(answerFinal.begin(), 0);
//Adds vectors to get final value seperate as digits
for (int i = multFactorial.size() - 1; i >= 0; i--) {
multFactorial[i] = answer[i] + answerFinal[i];
}
num = num - 1;
}
//Prints what should be the factorial of the number input
for (size_t i = 0; i < multFactorial.size(); i++) {
std::cout << multFactorial[i];
}
}
}
Factorials of large numbers results in huge numbers. This can be accommodated in languages like C, C++ etc by putting the results into arbitrary length strings.
Here is an algorithm for that - similar to yours.
https://www.geeksforgeeks.org/factorial-large-number/
Best advice is to check your code against this.
Use a debugger if you have one and step through the code line by line.
If not print out intermediate results and compare with expected.
EDIT: As per review comment, the code at above ref is similar to below- just in case link is broken in future.
// C++ program to compute factorial of big numbers
#include<iostream>
using namespace std;
// Maximum number of digits in output
#define MAX 100 // change to whatever value you need
int multiply(int x, int res[], int res_size);
// Calculate factorial of large number
void factorial(int n)
{
int res[MAX];
// Initialize result
res[0] = 1;
int res_size = 1;
// Apply factorial formula
for (int x=2; x<=n; x++)
res_size = multiply(x, res, res_size);
// print out the result
cout << "Factorial is \n";
for (int i=res_size-1; i>=0; i--)
cout << res[i];
}
// Multiplies x with the number represented by res[].
// res_size is size of res[] or number of digits in the
// number represented by res[].
int multiply(int x, int res[], int res_size)
{
int carry = 0; // Initialize carry
// One by one multiply n with individual digits of res[]
for (int i=0; i<res_size; i++)
{
int prod = res[i] * x + carry;
// Store last digit of 'prod' in res[]
res[i] = prod % 10;
// Put rest in carry
carry = prod/10;
}
// Put carry in res and increase result size
while (carry)
{
res[res_size] = carry%10;
carry = carry/10;
res_size++;
}
return res_size;
}
// Main program
int main()
{
//put code here to read a number
factorial(50); // take 50 for example
return 0;
}
#include <iostream>
using namespace std;
int main(){
int ctr = 0;
int count = 1; //Counts the nth prime number
int num = 3;
int div = 2; //Potential factors of the number
while(count <= 1000){
while(div < num){
if(num%div == 0){
ctr += 1; //If ctr is equal to 0, then num is prime
}
div += 1;
}
if(ctr == 0){ //If num is prime, count increases by 1
count += 1;
}
num += 1;
}
cout << num;
}
This is the code that I made to output the 1000th prime number. However, there must be something wrong with my program since it does not output 7919, which is the 1000th prime number.
It usually helps to refactor code like this into functions that have a clearly defined and testable behavior. For instance, the inner part of your code is a 'isPrime' function, and if you define it like this:
bool isPrime(int n) {
int div = 2; //Potential factors of the number
while (div < n) {
if (n % div == 0) {
return false;
}
++div;
}
return div == n;
}
It is easy to test, either through unit testing, or just manually checking if isPrime() works ok.
That makes the rest of the code more easy to write (and more importantly, read):
int primeCount = 0;
int n = 1;
while (primeCount < 1000) {
if (isPrime(n++)) {
++primeCount;
}
}
--n;
std::cout << n << std::endl;
As for why your code doesn't work. You should debug it. Go through line by line and see where it deviates from your expectations. Start out with finding the 3rd prime number, and not the 1000th.
Your isPrime part does not do what it is supposed to. Finding out why isn't hard, and you should definitely do that as a debugging-exercise, and not go with an easy answer from stackoverflow.
#include <stdio.h>
int main(){
int ctr = 0;
int count = 1; //Counts the nth prime number
int num = 3;
int div = 2; //Potential factors of the number
while(count <= 1000){
while(div < num){
if(num%div == 0){
ctr += 1; //If ctr is equal to 0, then num is prime
}
div += 1;
}
if(ctr == 0){ //If num is prime, count increases by 1
count += 1;
}
num += 1;
ctr=0;
div=2;
}
printf("%d",num);
}
A number is called digit-increasing if it is equal n + nn + nnn + ... for some digit n between 1 and 9. For example 24 is digit-increasing because it equals 2 + 22 (here n = 2).
Actually, a friend of mine asked me this question and i am stuck thinking about it but couldn't find the exact solution so far. Can anyone help ? I needed the function that returns true if it is digit-increasing else false.
There are only relatively few numbers with this property: Within the range of unsigned long long (64 bits), there are only 172 digit-increasing numbers.
Therefore, in terms of a practical solution, it makes sense to pre-compute them all and put them in a hash. Here is Python code for that:
# Auxiliary function that generates
# one of the 'nnnn' elements
def digits(digit,times):
result = 0
for i in range(times):
result += digit*(10**i)
return result
# Pre-computing a hash of digit-increasing
# numbers:
IncDig = {}
for i in range(1,30):
for j in range(1,10):
number = reduce(lambda x,y:x+y,[digits(j,k) for k in range(1,i+1)])
IncDig[number] = None
Then the actual checking function is just a look-up in the hash:
def IncDigCheck(number):
return (number in IncDig)
This is virtually O(1), and the time and space taken for the pre-calculation is minimal, because there are only 9 distinct digits (zero doesn't count), hence only K*9 combinations of type n + nn + ... for a sum of length K.
General representation is:
n + (n*10 + n) + (n*100+n)...
If number look like sum of same digits then any digit can be represented as
(1+111+...) * base_digit
. Assuming this we can use simple algorithm:
bool isDigitIncreasing(const int num)
{
int n = 1;
int sum = 1; //value to increase n
while (n <= num) {
//if num is (111...) * base_digit and base_digit is < 10
if (num % n == 0 && n * 10 > num) return true;
sum = sum * 10 + 1; //N*10+N where n is 1 as was assumed
n += sum; //next step
}
return false;
}
Simple exhaustive search will work.
def is_digit_increasing_number(x):
# n = 1, 1+11, 1+11+111, ...
n = 1
i = 1
while n <= x:
if x % n == 0 and n * 10 > x:
return True
i += 1
n = n * 10 + i
return False
Simplest possible way is do the addition (bottom-up), I'll use simple for loop:
List<int> numbersSum = new List<int>{1,2,3,4,5,6,7,8,9};
List<int> lastNumber = new List<int>{1,2,3,4,5,6,7,8,9};
for(int i=0;i<= lg n + 1;i++)
{
for(int j=0;j<9;j++)
{
if(list[j] < n)
{
var lastNumberJ = lastNumber[j]*10+j+1;
list[j] += lastNumberJ; // add numbers to see will be same as n.
if (list[j] == n)
return j+1;
lastNumber[j] = lastNumberJ;
}
}
}
return -1;
The important part is you just need at most log n iteration and also you can return sooner if all numbers are bigger than given number, this is O(log n) algorithm.
Here is a python code.The basic logic here is that a digit increasing number if divided by a specific number between 1-9 gives a digit increasing number made of only ones.All the digit increasing numbers of 1 follow a specific pattern ie 12345678...
import sys
for n in range(1,10):
a=1
if k%n!=0:
a=0
else:
g=str(k/n)
j=int(g[0])
for i in range(1,len(g)):
if int(g[i])==j+1:
j=int(g[i])
else:
a=0
break
if a==1:
print "Yes,it is a digit increasing number"
sys.exit(0)
print "No,it is not a digit increasing number"
I have done in this way. Check out once.
int sum = 0, count =0;
bool flag = false;
public bool isDigitIncreasing(int input_number)
{
int n= get_number_of_digit(input_number); // Gets number of digits
int sum = 0;
for(int i=0;i<n;i++)
{
sum = sum*10+1;
count = count + sum;
}
for(int i=1; i<=9;i++)
{
if((input_number)==count*i)
{
flag = true;
break;
}
else
flag = false;
}
return flag;
}
public int get_number_of_digit(int num)
{
int size = 0;
do
{
num = num/10;
size++;
}while(num>0);
return size;
}
Here is the shortest solution
public static int isDigitIncreasing (int n)
{
if(n<10)
{
return 1;
}
for(int i=1;i<=9;i++)
{
int tempsum=i;
int previous=i;
while(tempsum<=n)
{
previous=previous*10 + i;
tempsum=tempsum + previous;
if(tempsum==n)
{
return 1;
}
}
}
return 0;
}
Ambiguitiy: Are the values 1-9 repeating for themselves? (too lazy to google this myself)
If 1-9 are repeating then following should work. If not, and you want the code to work only on values > 10 then you can initialize mult with 10.
int i, mult = 1, result, flag;
for( i=1; i<9; i++ )
{
flag = 0;
while( result < TARGET )
{
result = result+(i*mult);
mult = mult*10;
if( result == TARGET )
{
flag = 1;
break;
}
}
if( flag == 1 )
break;
}
After execution, i must contain the values for which RESULT is a repeating number IF the flag is 1. If flag is zero after execution then the TARGET isn't a repeating number.
I wonder if its possible that a number could be repeating for multiple values, just curious.
Here num is the number and n is the digit
#include<stdio.h>
int f(int num,int n)
{
int d=n;
while(num>0)
{
num-=n;
n=d+n*10;
}
if(num==0)
return 1;
else
return 0;
}
int main()
{
int num;
int n;
int flag;
printf("Enter the number :");
scanf("%d",&num);
printf("Enter the digit :");
scanf("%d",&n);
flag = f(num,n);
if(flag == 1)
printf("It's in n+nn+nnn+...\n");
if(flag ==0)
printf("It's not\n");
return 0;
}
Let d(k) be 1+11+111+...+(11...11) where the last number has k digits. Then d(1)=1, and d(k+1)=10d(k)+k+1.
We want to test if d(k)*i = n, for some k, and for some i=1..9.
If we've computed d(k), then i (if it exists) must be n/d(k). We can check if n/d(k) is correct, by comparing n with ((n/d(k))%10)*d(k). The %10 makes the test fail if i is larger than 9.
This gives us a relatively terse solution: compute subsequent d(k) until they are bigger than n, and at each point check to see if n is a digit-multiple of d(k).
Here's a very lightly code-golfed implementation of that idea:
#include <stdio.h>
int is_digit_increasing(int n) {
for(int d=1,k=1;d<=n;d=d*10+ ++k)if(n==(n/d)%10*d)return 1;
return 0;
}
int main(int argc, char**argv) {
for (int i=0; i<10000; i++) {
if (is_digit_increasing(i)) {
printf("%d\n", i);
}
}
return 0;
}
// Example program
#include <iostream>
#include <string>
int isDigitIncreasingNo(int n) {
if(n<=0)
return 0;
int len = std::to_string(n).length();
int vector1 = 0;
int vector2 = 0;
for(int i=1;i<=len;i++)
vector2 = (vector2*10)+i;
vector1 = vector2/10;
if(n % vector2 == 0 && (n / vector2)<=9 )
return 1;
if(n % vector1 == 0 && (n / vector1)<=9 )
return 1;
return 0;
}
int main()
{
for (int i=0; i<10000000; i++) {
if (isDigitIncreasingNo(i)) {
printf("%d\n", i);
}
}
return 0;
}
public boolean isDigitIncreasing(int number)
{
int sum;
int size=calculateNumberOfDigits(number);
for(int i=1;i<=9;i++)
{
sum=0;
int temp=size;
while(temp>=1)
{
for(int j=temp;j<=1;j--)
{
sum=sum+i*(int)Math.pow(10,j-1);
}
temp--;
}
if(sum==number)
{
return true;//Its a digit increasing
}
}
return false;
}
public int calculateNumberOfDigits(int number)
{
int size=0;
do
{
number=number/10;
size++;
}while(size>0);
return size;
}