When I try to execute this:
#include<stdio.h>
int byteland(int a)
{
int e,f,g;
if ((a/2 + a/3 + a/4) < a)
return a;
else
{
e = byteland(a/2);
f = byteland(a/3);
g = byteland(a/4);
return e + f + g;
}
}
int main()
{
int a, b;
scanf("%d", &a);
b = byteland(a);
return 0;
}
I get segmentation fault (core dumped). Any solution?
If you use 0 for a, you get infinite recursion -- stack overflow. You also get infinite recursion for many negative numbers.
Not sure what this function is supposed to do but there is nothing to break the recursion when a is equal to 0.
I would add a clause to break the recursion when a is equal to 0.
int byteland(int a)
{
int e,f,g;
if ( a == 0 )
{
return 0;
}
else if((a/2+a/3+a/4)<a)
{
return a;
}
else
{
e=byteland(a/2);
f=byteland(a/3);
g=byteland(a/4);
return e+f+g;
}
}
Related
I'm writing an recursion method to calculate collatz conjecture for a sequence of positive integers. However, instead of stopping the calculation when the value reaches 1, I need it to stop when the value become smaller than or equal to the original value. I can't figure out what condition I should put in the if statement.
int collatz (int n) {
printf("%d%s", n, " ");
if(n > collatz(n)) { // here I would get an error saying all path leads to the method itself
return n;
}
else {
if(n % 2 == 0) {
return collatz(n / 2);
}
else {
return collatz((3 * n) + 1);
}
}
}
I used two more parameters:
startValue, to pass through the recursive calls the initial value and
notFirstTime, to check if it is the first call (and not a recursive call). In this case a value n <= startValue is allowed.
Here the code:
int collatz (int startValue, int n, int notFirstTime){
printf("%d%s ", n, " ");
if(n <= startValue && !notFirstTime)
{ // here I would get an error saying all path
//leads to the method itself
return n;
}
else
{
if ( n%2==0 )
{
collatz(startValue, n/2, 0);
}
else
{
collatz(startValue, (3*n)+1, 0);
}
}
}
int main() {
int x = 27;
int firstTime = 1;
int test = collatz(x,x, firstTime);
printf("\nLast value: %d\n", test);
return 0;
}
Please note that I removed two return statements from the recursive calls.
This question already has answers here:
Check if one integer is an integer power of another
(13 answers)
Closed 8 years ago.
How to find out whether a number b is can be expressed as a power of another number c and to find the corresponding exponent? Without using math.h or recursion.
The numbers are of the type int!
This is the code I have written:
#include <stdbool.h>
bool is_apowb(int a, int b, int *e) {
int x =0;
if (a>b)
{
while (a%b==0 && b>0)
{
a=a/b;
x++;
}
*e=x;
return true;
}
else if (a==b)
{
*e=1;
return true;
}
else if (b<0)
{
while (a%b==0)
{
a=a/b;
x++;
}
*e=x;
if (x%2==0)
{
return true;
}
else return false;
}
return false;
}
The code is failing the following tests:
assert(is_apowb(9,-3,&e));
assert(e == 2);
assert(!is_apowb(8,-2,&e));
assert(e == 2);
Thanks in Advance!
If you have a number n and want to know if it is a power of b, you can compute e = log(n)/log(b) and see if it is close to an integer. If so, round e to that integer and compute b^e. If this equals n then you're done and the exponent is e. If not then n isn't a power of b.
You could also divide by b repeatedly in a loop, like so:
e = 0;
while (n%b == 0) {
n = n/b;
e++;
}
if (n == 1) return e; // n = b^e
return -1; // n is not a power of b
#include <stdio.h>
int main(int argc, char* argv[] ) {
int b = atoi(argv[1]);
int c = atoi(argv[2]);
int e;
for (e = 0; b > 1; ++e) {
if (b%c != 0) break;
b /= c;
}
if (b == 1)
printf("%d\n", e );
else
printf("nope\n");
return 0;
} // end main()
I tried a prime no generating question in SPOJ(link : http://www.spoj.com/problems/PRIME1/) .
I'm using seive algorithm . I get the SIGSEGV error when i use spoj gcc . But when i compiled using my ubuntu gcc it works for all the test cases.
Here is my source code . Plz Help
float sqroot( float x)
{
float a , b;
a = x; // copy given value to 'a'
do
{
b = a; // copy value of 'a' to 'b' before 'a' is modify
a = (a + x/a) / 2; // modify 'a' value until we reach sqroot result
}
while( a!= b); // execute loop until a == b
return( a); // 'a ' or 'b' is sqroot of 'x'
}
int main()
{
int prime[4000];
int prime_index=0;
bool find_prime[100001];
int i,j;
int m,n;
int iremainder;
int T,t_index;
int PRIME_FLAG=1;
float square;
int limit;
prime_index++;
prime[prime_index]=2;
for(i=3;i<=32000;i=i+2)
{
PRIME_FLAG=1;
square = sqroot((float)i);
limit = ((int)(square))+1;
for(j=1;j<=prime_index,prime[j]<=limit;j++)
{
if(prime[j]!=0)
{
if((i%prime[j]) == 0)
{
PRIME_FLAG = 0;
break;
}
}
}
if(PRIME_FLAG)
{
prime_index++;
prime[prime_index]=i;
printf("%d\n",i);
}
}
printf("Enter the no of test cases:");
scanf("%d",&T);
if(T<=10)
{
for(t_index=1;t_index<=T;t_index++)
{
printf("Enter the values of m and n :");
scanf("%d%d",&m,&n);
if((m>=1) && (n<=1000000000) && ((n-m)<=100000))
{
if(m == 1)
m=2;
//Set all numbers from m to n as prime
for(i=m;i<=n;i++)
find_prime[i]=true;
//Find the prime numbers between m to n
square = sqroot((float)n);
limit = ((int)(square))+1;
for(i=1;i<=prime_index,prime[i]<=limit;i++)
{
if(m>=prime[i])
{
if(prime[i]!=0)
iremainder=m%prime[i];
j=prime[i]*iremainder;
}
else
{
iremainder=prime[i]-m;
if(m+iremainder == prime[i])
j=2*(m+iremainder);
else
j=m+iremainder;
}
for(;j<=n;j=j+prime[i])
find_prime[j]=false;
}
//Print all prime no's
for(i=m;i<=n;i++)
{
if(find_prime[i])
printf("%d\n",i);
}
}
}
}
return 0;
}
your for loops are wrong.
for(j=1;j<=prime_index,prime[j]<=limit;j++)
should be
for(j=1;(j<=prime_index)&&(prime[j]<=limit);j++)
the first condition will be executed, but the result will be ignored.
So it is luck, that you find a number in the uninitialized array that is larger than limit before you go out of range of the array, which will lead to a SIGSEGV.
for(i=1;i<=prime_index,prime[i]<=limit;i++)
has the same problem.
Click here for more details
A code in C to find maximum of an array using divide and conquer but it keeps throwing
"stack overflow exception" . Help would be appreciated!
int a[10];
int find(int l,int h)
{
int x;
if(h==0)
{
x=0;
return x;
}
else
{
if(h==1)
{
if(a[0]>a[1])
{
x=0;
return x;
}
else
{
x=1;
return x;
}
}
else
{
int mid,z,y;
mid=(l+h)/2;
y=find(0,mid);
z=find(mid+1,h);
if(a[y]<a[z])
{
x=z;
}
else
{
x=y;
}
return x;
}
}
}
There are only limited variables and I don't see where the function can go into an infinite recursion.
int main()
{
int i,n,max,min,ans;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
ans=find(0,n-1);
printf("the maximum element is- %d\n",ans);
getch();
return 0;
}
Consider the case where you call find(0, 2). Since h > 1, you enter the second else clause, and mid is 1. Then on the second recursive call, it is to find(2, 2). On this recursive call, you again enter the second else, since h is still 2. But the mid is also 2. Now, the first recursive call goes to find(0, 2), which enters an infinite loop.
find(0, 2)
h not 0
h not 1
mid = 1
find(0, 1)
find(2, 2)
h not 0
h not 1
mid = 2
find (0, 2) <-- loop
It seems the intention of the if checks on h is to prevent the mid calculation from being the same as l. If so, then you can calculate the mid variable at the top of your function, and use that as the stopping condition.
It looks like this is an attempt to use divide and conquer to locate the position of the maximum element in the array a. If so, then your first recursive call should restrict itself to the range of [l..mid] instead of going back to 0.
Putting it all together:
int find(int l,int h)
{
int mid = (l+h)/2;
if (mid == l)
{
return (a[l] > a[h]) ? l : h;
}
else
{
int y = find(l, mid);
int z = find(mid+1, h);
return (a[y] > a[z]) ? y : z;
}
}
Here is your code modified, which is running successfully..
The problem is that you weren't checking the difference between l and h but only the value of h...
#include <iostream>
using namespace std;
int a[10];
int find(int l,int h)
{
int x;
if(h-l==0)
{
return h;
}
else
{
if(h-l==1)
{
if(a[l]>a[l+1])
{
return l;
}
else
{
return l+1;
}
}
else
{
int mid,z,y;
mid=(l+h)/2;
y=find(0,mid);
z=find(mid+1,h);
if(a[y]<a[z])
{
x=z;
}
else
{
x=y;
}
return x;
}}}
int main()
{
a[0]=3;
a[1]=7;
a[2]=5;
cout<<find(0,2)<<endl;
return 0;
}
you're using wrong conditions, try:
first if: if(h==l)
second if: if(h-l==1)
third if:
if(a[h]>a[l]) {
return h;
} else {
return l;
}
This is a function i made to count number of zeroes at the end of the factorial of a number b recursively.
However i'm getting runtime error due to the used code.Pardon my naivety but any help in this would be appreciated.
int noz(int b)
{
int c=0;
int e = b;
if(e < 5)
return 0;
while(e > 0)
c = c + (e/5) + noz(e/5);
return c;
}
You are encountering "runtime error" because:
int c;
...
while(e > 0)
c = c + (e/5) + noz(e/5); // <-- HERE
you are using uninitialized local variable c, which produces undefined behavior.
You could zero-initialize this variable to prevent it happen:
int c = 0;
And also note that in case that argument of your function is greater or equal than 5, this function doesn't return anything (thanks to #Paul R for pointing this out) and another problem is that you have loop with the condition e > 0 but the loop doesn't change the value of e making it infinite loop.
Your function could look like this instead (I'm not sure what exactly is the desired logic here):
int noz(int b)
{
int c = 0;
if (b < 5)
return 0;
else
return c + (b/5) + noz(b/5);
}
//count n! tail zero
int noz(int n){
int count;
if(n < 5) return 0;
count = n / 5;
return count + noz(count);
}