I am unable to find the logic for recursion in the problem mentioned below.
Given a non-negative int n, return the count of the occurrences of 7 as a digit
Eg. Count7(777)=3
Eg. Count7(123)=0
Eg. Count7(171)=1
Here is the logic that i applied
count number of 7 in ones place + count no of 7 in all the other place.
eg 13767
count number of 7 in (1)+ count number off 7 in (3767),just like factorial program where 5!=5*4!
count7(n)
{
if (n==0) {
return 0;
}
if (n==7) {
return 1;
}
if (n!=7) {
return 0;
}
return count7(n%10)+count7(n/10)
}
Any suggestion or help is appreciated .
You have
if (n==7) return 1;
followed by
if (n!=7) return 0;
As n is either 7 or not 7, everything that comes after will never be called. Instead you probably want
if (n<10) return 0;
as you want to call the recursion for numbers with more than 1 digit, and break here if you reached the last digit.
BTW, you could also remove the if (n==0) part, as this is also covered by the n<10.
The base case checks if there is just a single digit and if it is a 7 or not. Then, we recursively add the number of 7s in the remaining digits to whatever was found before (rem == 7 ? 1 : 0). This is just another bare-bones program. I am sure you can pick up from here.
#include <stdio.h>
int count7(int i)
{
int rem = i % 10; // remainder
int quo = i / 10; // quotient
if (rem == i) { // base case - check for a single digit.
if (rem == 7)
return 1;
else
return 0;
} else {
return (rem == 7 ? 1 : 0) + count7(quo);
}
}
int main(void) {
int i = 12;
printf("%d\n", count7(i));
}
The following code is completely functional.
#include<stdio.h>
void main()
{
printf("%d\n",count7(717));
}
int count7(int n)
{
if(n==7)
return 1;
else if(n<10)
return 0;
else
return count7(n%10)+count7(n/10);
}
There are some problems that I found in your code,
The critical problem is that you check if n!=7 and befor you check if n==7
that combination will equal to Ask the condition if() {} else {} that mean that in this case
you will not get to the next line - return count7(n%10)+count7(n/10) Never!!!!
Another thing that I saw is that you call at the return to count7 with the remainder of the number, it will work, but if you need to do a simple operation (check only if the remainder ==7 ) so my advice is to use to ? : condition and not to jump to function if you don't need.
My Code to your problem:
int count7(int n)
{
return n==0 ? 0 : (n==7 ? 1 : count7(n/10) + (n%10 == 7 ? 1 : 0) );
}
And another code more readably (do the same work) :
int count7(int curNum)
{
int curModNum=n%10;
int curDivNum=n/10;
if (currentNum==EMPTY_NUM) {
return EMPTY_NUM;
}
if (curNum==LUCKY_NUM) {
return ADD_ONE;
}
return count7(curDivNum) + (curModNum == LUCKY_NUM ? ADD_ONE : EMPTY_NUM);
}
Related
this is my code, I want to make a function that when it is called will generate a number between 1111 to 9999, I don't know how to continue or if I've written this right. Could someone please help me figure this function out. It suppose to be simple.
I had to edit the question in order to clarify some things. This function is needed to get 4 random digits that is understandable from the code. And the other part is that i have to make another function which is a bool. The bool needs to first of get the numbers from the function get_random_4digits and check if there contains a 0 in the number. If that is the case then the other function, lets call it unique_4digit, should disregard of that number that contained a 0 in it and check for a new one to use. I need not help with the function get_random_4digitsbecause it is correct. I need helt constructing a bool that takes get_random_4digits as an argument to check if it contains a 0. My brain can't comprehend how I first do the get_random_4digit then pass the answer to unique_4digits in order to check if the random 4 digits contains a 0 and only make it print the results that doesn't contain a 0.
So I need help with understanding how to check the random 4 digits for the integer 0 and not let it print if it has a 0, and only let the 4 random numbers print when it does not contain a 0.
the code is not suppose to get more complicated than this.
int get_random_4digit(){
int lower = 1000, upper = 9999,answer;
answer = (rand()%(upper-lower)1)+lower;
return answer;
}
bool unique_4digits(answer){
if(answer == 0)
return true;
if(answer < 0)
answer = -answer;
while(answer > 0) {
if(answer % 10 == 0)
return true;
answer /= 10;
}
return false;
}
printf("Random answer %d\n", get_random_4digit());
printf("Random answer %d\n", get_random_4digit());
printf("Random answer %d\n", get_random_4digit());
Instead of testing each generated code for a disqualifying zero just generate a code without zero in it:
int generate_zero_free_code()
{
int n;
int result = 0;
for (n = 0; n < 4; n ++)
result = 10 * result + rand() % 9; // add a digit 0..8
result += 1111; // shift each digit from range 0..8 to 1..9
return result;
}
You can run the number, dividing it by 10 and checking the rest of it by 10:
int a = n // save the original value
while(a%10 != 0){
a = a / 10;
}
And then check the result:
if (a%10 != 0) printf("%d\n", n);
Edit: making it a stand alone function:
bool unique_4digits(int n)
{
while(n%10 != 0){
n = n / 10;
}
return n != 0;
}
Usage: if (unique_4digits(n)) printf("%d\n", n);
To test if the number doesn't contain any zero you can use a function that returns zero if it fails and the number if it passes the test :
bool FourDigitsWithoutZero() {
int n = get_random_4digit();
if (n % 1000 < 100 || n % 100 < 10 || n % 10 == 0) return 0;
else return n;
}
"I need not help with the function get_random_4digits because it is correct."
Actually the following does not compile,
int get_random_4digit(){
int lower = 1000, upper = 9999,answer;
answer = (rand()%(upper-lower)1)+lower;
return answer;
}
The following includes modifications that do compile, but still does not match your stated objectives::
int get_random_4digit(){
srand(clock());
int lower = 1000, upper = 9999,answer;
int range = upper-lower;
answer = lower + rand()%range;
return answer;
}
" I want to make a function that when it is called will generate a number between 1111 to 9999,"
This will do it using a helper function to test for zero:
int main(void)
{
printf( "Random answer %d\n", random_range(1111, 9999));
printf( "Random answer %d\n", random_range(1111, 9999));
printf( "Random answer %d\n", random_range(1111, 9999));
printf( "Random answer %d\n", random_range(1111, 9999));
return 0;
}
Function that does work follows:
int random_range(int min, int max)
{
bool zero = true;
char buf[10] = {0};
int res = 0;
srand(clock());
while(zero)
{
res = min + rand() % (max+1 - min);
sprintf(buf, "%d", res);
zero = if_zero(buf);
}
return res;
}
bool if_zero(const char *num)
{
while(*num)
{
if(*num == '0') return true;
num++;
}
return false;
}
I have below code which works fine.
#include<stdio.h>
int calculateSum(int);
int main() {
int num;
int result;
printf("Input number = ");
scanf("%d", &num);
result = calculateSum(num);
printf("\nResult from 1 to %d = %d", num, result);
return (0);
}
int calculateSum(int num) {
int res;
if (num == 1) {
return (1);
}
else {
res = num + calculateSum(num - 1);
}
return (res);
}
Input number = 5
Result from 1 to 5 = 15
Now I am trying to give the program 2 inputs, from and to numbers.
Example: first input = 5, second = 8 and result should be = 26 (5 + 6 + 7 + 8)
Any ideas of how to go about this? failing thus far.
int calculateSum(int fromNum, int toNum) {
int res;
if (fromNum == toNum) {
return (fromNum);
}
else {
res = fromNum + calculateSum((fromNum + 1), toNum);
}
return (res);
}
At the moment, you are hard-coding 1 as the terminating point of the recursion.
What you need is to be able to use a different value for that, and the following pseudo-code shows how to do it:
def calculateSum(number, limit):
if number <= limit:
return limit
return number + calculateSum(number - 1, limit)
For efficiency, if you break the rules and provide a limit higher than the starting number, you just get back the number. You could catch that and return zero but I'll leave that as an exercise if you're interested.
It should be relatively easy for you to turn that into real code, using your own calculateSum as a baseline.
I should mention that this is a spectacularly bad use case for recursion. In general, recursion should be used when the solution search space reduces quickly (such as a binary search halving it with each recursive level). Unless your environment does tail call optimisation, you're likely to run out of stack space fairly quickly.
Instead of stopping when you reach 1, stop when you reach from.
int calculateSum(from, to) {
if (to == from) {
return from;
} else {
return to + calculateSum(from, to-1);
}
}
change 1 to from:
int calculateSum(int from,int to) {
int res;
if (to== from) {
return (from);
}
else {
res = to+ calculateSum(from,to - 1);
}
return (res);
}
You can use ternary operator.
int calculateSum(int from, int to) {
return from == to ? from : from + calculateSum(from + 1, to);
}
I want to know how to find the number of times that 4 occurs within a certain range; lets say 50 for example.
Therefore if the input is the range 50, then output should look like
4,14,24,34,40,41,...,49.
Now its pretty easy to find the first half of the output that is 4,14,..so on,i.e digits ending with 4,
for(i=0;i<=50;i++)
if(i%10==4)
{
printf("%d",i);
}
but what about starting with 4? Can the problem be solved using regular expressions?
You could try to use a function that repeatedly tests each digit in the number to see if it is equal to 4, like so:
int check4(int x)
{
//Units place
if (x==0)
{
return 0;
}
else if (x%10 == 4)
{
return 1;
}
//Divide by 10 so that tens place is now unit place; repeat till no more tens place
else
{
return check4(x/10);
}
}
This solution completely eschews the need for string processing.
Demo
No need for regex at all. Store the integer as a string in temporary variable, and then use strstr.
for( i = 0; i <= input; i++ ) {
char str[20];
sprintf( str, "%d", i );
if( strstr(str, "4") )
printf( "%d ", i );
}
Here's a demo on codepad.
The idea is simple: check the ending digit to see if it is 4, if not, keep dividing the number by 10 until the last digit is 4, or the number is 0.
It is equally easy to do in iterative or recursive approach:
Pseudo code (Iterative):
boolean contains4(int input) {
while (input > 0) {
if (i % 10 == 4) {
return true;
}
input /= 10;
}
return false;
}
Pseudo-code (Recursive):
boolean contains4(int input) {
if (input == 0) {
return false;
} else if (input % 10 == 4) {
return true;
} else {
return contains4(input / 10);
}
}
This programming I wrote below is used to display prime number from a list (20 numbers) which keyed in by user. But it only can detect 2 and 3 as prime number. I don't know why it doesn't work. Please tell me where is the errors and help me improve it. TQ.
#include <iostream>
#include <conio.h>
using namespace std;
void main ()
{
int i,number,list[20];
int t,p,prime=0;
cout<<"please key 20 numbers from 0 to 99"<<endl;
for(i=1;i<21;i++)
{
cin>>number;
if((number<0)||(number>99))
{
cout<<"Please key in an integer from 0 to 99"<<endl;
}
list[i]=number;
}
for(p=1;p<21;p++)
{
for(t=2;t<list[p];t++)
{
if ( list[p]%t==0)
{
prime=prime+1;
}
}
if (prime==0&&list[p]!=1)
{
cout<<"Prime numbers:"<<list[p]<<endl;
}
}
getch();
}
So there are a few issues with your code, but the one that will solve your issue is simply algorithmic.
When you start at the next iteration of p, you don't reset the value of prime and therefore it's always > 0 after we detect the second prime number and you'll never print out any again.
Change this:
for(p=1;p<21;p++)
{
for(t=2;t<list[p];t++)
{
if ( list[p]%t==0)
{
prime=prime+1;
}
}
if (prime==0&&list[p]!=1)
{
cout<<"Prime numbers:"<<list[p]<<endl;
}
}
To this (I've added some brackets for clairty and so we're certain the condition evaluates as we expect it to):
for(p=0;p<20;p++)
{
for(t=2;t<list[p];t++)
{
if ( list[p]%t==0)
{
prime=prime+1;
}
}
if ( (prime==0) && (list[p]!=1) )
{
cout<<"Prime numbers:"<<list[p]<<endl;
}
prime = 0;
}
And your issue will be solved.
HOWEVER: I would like to reiterate this does not solve all of your code issues. Make sure you think very carefully about the input part and what you are looping over (why is p 1 to 21? Why not 0 to 20 ;) arrays are zero indexed in C meaning that your list of 20 numbers goes from list[0] to list[19], you're currently looping from list[1] to list[20] which is actually out of range and I'm surprised you didn't get a segfault!)
What happens in your code is someone types in "123" or "-15"? Check and see if you can fix the error.
When you have fixed that, we can look at your prime checking code. Hint: there are a lot of prime testing code examples on the web.
Check this
#include<stdio.h>
int main()
{
int n, i = 3, count, c;
printf("Enter the number of prime numbers required\n");
scanf("%d",&n);
if ( n >= 1 )
{
printf("First %d prime numbers are :\n",n);
printf("2\n");
}
for ( count = 2 ; count <= n ; )
{
for ( c = 2 ; c <= i - 1 ; c++ )
{
if ( i%c == 0 )
break;
}
if ( c == i )
{
printf("%d\n",i);
count++;
}
i++;
}
return 0;
}
More Efficient Way
def print_hi(n):
if(n == 1 ):
return False;
if( n == 2 or n == 3):
return True;
if(n % 2 == 0 or n % 3 == 0):
return False;
for i in range (5,n,6):
if( i * i <= n):
if(n % i == 0 or n % (i+2) == 0):
return False
return True
if __name__ == '__main__':
x = print_hi(1032)
print(x)
UVA problem 100 - The 3n + 1 problem
I have tried all the test cases and no problems are found.
The test cases I checked:
1 10 20
100 200 125
201 210 89
900 1000 174
1000 900 174
999999 999990 259
But why I get wrong answer all the time?
here is my code:
#include "stdio.h"
unsigned long int cycle = 0, final = 0;
unsigned long int calculate(unsigned long int n)
{
if (n == 1)
{
return cycle + 1;
}
else
{
if (n % 2 == 0)
{
n = n / 2;
cycle = cycle + 1;
calculate(n);
}
else
{
n = 3 * n;
n = n + 1;
cycle = cycle+1;
calculate(n);
}
}
}
int main()
{
unsigned long int i = 0, j = 0, loop = 0;
while(scanf("%ld %ld", &i, &j) != EOF)
{
if (i > j)
{
unsigned long int t = i;
i = j;
j = t;
}
for (loop = i; loop <= j; loop++)
{
cycle = 0;
cycle = calculate(loop);
if(cycle > final)
{
final = cycle;
}
}
printf("%ld %ld %ld\n", i, j, final);
final = 0;
}
return 0;
}
The clue is that you receive i, j but it does not say that i < j for all the cases, check for that condition in your code and remember to always print in order:
<i>[space]<j>[space]<count>
If the input is "out of order" you swap the numbers even in the output, when it is clearly stated you should keep the input order.
Don't see how you're test cases actually ever worked; your recursive cases never return anything.
Here's a one liner just for reference
int three_n_plus_1(int n)
{
return n == 1 ? 1 : three_n_plus_1((n % 2 == 0) ? (n/2) : (3*n+1))+1;
}
Not quite sure how your code would work as you toast "cycle" right after calculating it because 'calculate' doesn't have explicit return values for many of its cases ( you should of had compiler warnings to that effect). if you didn't do cycle= of the cycle=calculate( then it might work?
and tying it all together :-
int three_n_plus_1(int n)
{
return n == 1 ? 1 : three_n_plus_1((n % 2 == 0) ? (n/2) : (3*n+1))+1;
}
int max_int(int a, int b) { return (a > b) ? a : b; }
int min_int(int a, int b) { return (a < b) ? a : b; }
int main(int argc, char* argv[])
{
int i,j;
while(scanf("%d %d",&i, &j) == 2)
{
int value, largest_cycle = 0, last = max_int(i,j);
for(value = min_int(i,j); value <= last; value++) largest_cycle = max_int(largest_cycle, three_n_plus_1(value));
printf("%d %d %d\r\n",i, j, largest_cycle);
}
}
Part 1
This is the hailstone sequence, right? You're trying to determine the length of the hailstone sequence starting from a given N. You know, you really should take out that ugly global variable. It's trivial to calculate it recursively:
long int hailstone_sequence_length(long int n)
{
if (n == 1) {
return 1;
} else if (n % 2 == 0) {
return hailstone_sequence_length(n / 2) + 1;
} else {
return hailstone_sequence_length(3*n + 1) + 1;
}
}
Notice how the cycle variable is gone. It is unnecessary, because each call just has to add 1 to the value computed by the recursive call. The recursion bottoms out at 1, and so we count that as 1. All other recursive steps add 1 to that, and so at the end we are left with the sequence length.
Careful: this approach requires a stack depth proportional to the input n.
I dropped the use of unsigned because it's an inappropriate type for doing most math. When you subtract 1 from (unsigned long) 0, you get a large positive number that is one less than a power of two. This is not a sane behavior in most situations (but exactly the right one in a few).
Now let's discuss where you went wrong. Your original code attempts to measure the hailstone sequence length by modifying a global counter called cycle. However, the main function expects calculate to return a value: you have cycle = calculate(...).
The problem is that two of your cases do not return anything! It is undefined behavior to extract a return value from a function that didn't return anything.
The (n == 1) case does return something but it also has a bug: it fails to increment cycle; it just returns cycle + 1, leaving cycle with the original value.
Part 2
Looking at the main. Let's reformat it a little bit.
int main()
{
unsigned long int i=0,j=0,loop=0;
Change these to long. By the way %ld in scanf expects long anyway, not unsigned long.
while (scanf("%ld %ld",&i,&j) != EOF)
Be careful with scanf: it has more return values than just EOF. Scanf will return EOF if it is not able to make a conversion. If it is able to scan one number, but not the second one, it will return 1. Basically a better test here is != 2. If scanf does not return two, something went wrong with the input.
{
if(i > j)
{
unsigned long int t=i;i=j;j=t;
}
for(loop=i;loop<=j;loop++)
{
cycle=0;
cycle=calculate(loop );
if(cycle>final)
{
final=cycle;
}
}
calculate is called hailstone_sequence_length now, and so this block can just have a local variable: { long len = hailstone_sequence_length(loop); if (len > final) final = len; }
Maybe final should be called max_length?
printf("%ld %ld %ld\n",i,j,final);
final=0;
final should be a local variable in this loop since it is separately used for each test case. Then you don't have to remember to set it to 0.
}
return 0;
}