Well, I've been trying to rework this many times. Though, at one point I thought the longestSequence function would help, since it displays the longest hailstone sequence. Though, I can't seem to figure out how to find, or store the value it used to find that.If someone could explain how, I would appreciate it.
int longestSequence(int n)
{
int u = n;
if(u == 1)
{
return 1;
}
else
{
return max(hailstoneLength(u), longestSequence(u-1));
}
}
The part I'm having trouble with is my longest start sequence:
int hailLongestSeq(int n)
{
int k;
int longest = 0;
for(int j = 1; j <= n; j++)
{
if(hailstoneLength(j) > longest)
{
longest = hailstoneLength(j);
k = j;
}
}
return k;
}
I'm not sure how to make this into a recursion, I noticed for some recursions I saw people using for loops still, but I was sure we weren't supposed to use loops.
It may be a dumb question, but is there a formula to translating for loops to recursions, if anyone knows one?
The expected out put is like this:
The longest hailstone sequence starting with a number up to 10 has length 20.
The longest hailstone sequence starting with a number up to 10 begins with 9.
as 9's sequence has a length of 20 numbers, and is the longest from 1 to 10.
Yes every for loop can be translated to recursive call, obviously like this:
for (i=0; i<N; i++) {
body;
}
translate it to:
func(int i) {
if (i<N) { body; func(i+1) }
else return;
}
func(0);
This can be easily extended to any for loop computation (add parameters if needed, return value, etc).
You can add the local variables who's value you want to retain in the parameter signature of the function. I have tried to rewrite the code as recursive. Check and verify if this solves your problem.
int hailLongestSeq(int n, int j, int k, int longest)
{
if(j <= n)
{
if(hailstoneLength(j) > longest)
{
longest = hailstoneLength(j);
k = j;
}
k = hailLongestSeq(n, ++j, k, longest);
}
return k;
}
This may need a fix, however the logic remains the same, pass the variables as parameter to retain the value.
Related
Given a number n
Find the count of the sub numbers of size x in a number num which divides num.
For example, if the number is 250
and x=2
the answer will be 2
as 250%25==0
and 250 % 50==0.
Can anyone help me out with the cpp code ?
class Solution {
public:
int divisorSubstrings(int num, int k) {
string s=to_string(num);
int count=0;
int i=0;
int j=k-1;
string temp="";
for(int k=i;k<=j;k++)
{
temp.push_back(s[k]);
}
while(j<s.length())
{
if(num%stoi(temp)==0)
count++;
temp.erase(temp.begin() + i-1);
j++;
i++;
temp.push_back(s[j]);
}
return count;
}
};
this is showing runtime error
You have a number of problems. First, using simple letters for your variables means we don't have a clue what those variables are for. Use meaningful names.
Second, this:
for(int k=i;k<=j;k++)
You have an argument to your method called k. You've now shadowed it. Technically, you can do that, but it's a really really bad habit.
But the real problem is here:
temp.erase(temp.begin() + i-1);
i is initialize to 0 and never changed until AFTER this line runs the first time. So you're actually erasing a character before the start of the string.
The problem is simple. I'm given N - the number of digits in a number and then N digits of a number. I need to do exactly one digit-switch and get the highest number possible. I did do the problem right (as in gives out the right number) but it will be hitting the 1 second time restriction afaik. How do I improve on the efficiency of my program so it would go under the 1 second time restriction with N <= 10^6. New on Stack overflow so tell me if I did something wrong
with asking the question so I can fix it. Thanks. Here's my solution:
main:
int n;
cin >> n;
int a[n+1];
for(int i=0;i<n;++i)
cin >> a[i];
int maxofarray1;
bool changeHappened=false;
bool thereAreTwoSame=false;
for(int i=0;i<n;++i) //changing the two digits to make the highest number if possible
{
maxofarray1=maxofarray(a,i+1,n);
if(a[i]<maxofarray1)
{
int temp=a[a[n]];
a[a[n]]=a[i];
a[i]=temp;
changeHappened = true;
break;
}
}
for(int i=0;i<n;++i) //need to check if there are two of the same digit so I can change
//those two making the number the same instead of making it lower
for(int j=i+1;j<n;++j)
if(a[i]==a[j])
{
thereAreTwoSame=true;
break;
}
if(!changeHappened) //if the change has not been yet made, either leaving the number as is
//(changing two same numbers) or changing the last two to do as little "damage" to the number
{
if(!thereAreTwoSame)
{
int temp=a[n-1];
a[n-1]=a[n-2];
a[n-2]=temp;
}
}
for(int i=0;i<n;++i)
cout << a[i] << " ";
return 0;
maxofarray:
int maxofarray(int a[], int i,int n) //finding the maximum of the array from i to n
{
int max1=0;
int maxind;
for(int j=i;j<n;++j)
{
if(max1<a[j])
{
max1=a[j];
maxind=j;
}
}
a[n]=maxind; //can't return both the index and maximum (without complicating with structs)
//so I add it as the last element
return max1;
}
The problem in your code is complexity. I didn't fully understand your algorithm, but having nested loops is a red flag. Instead of trying to improve bits and pieces of your code you should rather rethink your overall strategy.
Lets start by assuming the digit 9 does appear in the number. Consider the number is
9...9 c ...9...
where 9...9 are the leading digits that are all 9 (possibly there are none of them). We cannot make the number bigger by swapping one of those.
c is the first digits !=9, ie its the place where we can put a 9 to get a bigger number. 9 is the digit that will make the number maximum when put in this place.
Last, ...9... denotes the last appearance of the digit 9 and digits sourrinding that. After that 9 no other 9 appears. While we increase the number by replacing c, the number will get smaller be replacing that 9, hence we have to choose the very last one.
For the general case only a tiny step more is needed. Here is a rough sketch:
std::array<size_t,10> first_non_appearance;
std::array<size_t,10> last_appearance;
size_t n;
std::cin >> n;
std::vector<int> number(n);
for (size_t i=0;i <n;++i) {
std::cin >> a[i];
for (int d=0;d<10;++d) {
// keep track of first and last appearance of each digit
}
}
size_t first = 0;
size_t second = 0;
for (int d=0;d<10;++d) {
// determine biggest digit that appeared and use that
}
std:swap( a[first],a[last] );
It is not complete, perhaps requires handling of special cases (eg number with only one digit), but I hope it helps.
PS: You are using a variable length array (int a[n+1];), this is not standard C++. In C++ you should rather use a std::vector when you know the size only at runtime (and a std::array when the size is known).
VLA (variable length arrays) are not standard. So instead of using this nonstandard feature, you might want to use a STL data type.
Given N is rather big, you also avoid stack overflow, given that VLA are allocated on the stack. And STL containers with variable length allocate on the heap.
Then, as you pointed out yourself, it makes sense to remember the index of the last occurrence of each digit, avoiding to search over and over again for a swap candidate index.
Your implementation idea is basically, to replace the first digit from the left, which has a bigger replacement to the right of it.
This is how I did it:
static void BigSwap(std::string& digits)
{
int64_t fromRight[10];
size_t ndigitsFound = 0;
for (size_t i = 0; i < 10; i++)
fromRight[i] = -1;
size_t i = digits.size() - 1;
while (ndigitsFound < 10 && i > 0)
{
if (-1 == fromRight[digits[i] - '0'])
{
fromRight[digits[i] - '0'] = static_cast<int64_t>(i);
ndigitsFound++;
}
i--;
}
for (size_t j = 0; j < digits.size(); j++)
{
char d = digits[j] - '0';
for (char k = 9; k > d; k--)
{
if (fromRight[k] != -1 && static_cast<size_t>(fromRight[k]) > j)
{
auto temp = digits[j];
digits[j] = k + '0';
digits[fromRight[k]] = temp;
return;
}
}
}
}
my program seems correct but I don't know why it has a logic problem
int main()
{
int r,s=0;
for(int i=10000;i<=998001;i++)
{
while (i>0)
{
r=i%10;
s=s*10+r;
i=i/10;
}
cout<<s<<endl;
}
Your question and code don't seem to match Ali. You have told that you need the largest possible palindrome, but you are printing each and every reversed number. And your code has glitches too. I will list the mistakes and corresponding changes to be made:
using i as both the for loop counter variable and as well as the number you're reversing. This causes i to become 0 after every reversal and hence the for loop never ends. The fix for this... use another variable, say num, and equate it to i at the start of the loop. This ensures that i remains unchanged and for goes unfazed.
use long int instead of int. This avoids any anomalies and chances of junk numbers.
s(the sum variable) is initialized only at the start. Hence every time you calculate a new reversed number, it is adding s to its previous value. The fix: Initialize s to 0 at the start of for loop so that you get a fresh reversed for every i value.
You are not checking for any palindrome condition. You are just printing the reversed number. The fix: Hence check if the the number is a palindrome, i.e, if the number ,i.e., i is equal to the reversed number, i.e., s
I have attached the code below. I am currently printing the greatest palindrome in the range in which you were checking. In case you need all the palindromes, just uncomment the commented cout line.
CODE:
#include <iostream>
using namespace std;
int main()
{
int r, s = 0;
long int num, max = 0;
for(long int i = 10000; i <= 998001; i++)
{
s = 0;
num = i;
while (num > 0)
{
r = num % 10;
s = s * 10 + r;
num = num / 10;
}
if(s == i) {
//cout<<s<<endl; //uncomment this line if you intend to display all palindromes
if(i > max)
max = i;
}
}
cout<<max<<endl;
}
I am a beginner in c++ and I am having problems with making this code work the way I want it to. The task is to write a program that multiplies all the natural numbers up to the loaded number n.
To make it print the correct result, I divided x by n (see code below). How can I make it print x and not have to divide it by n to get the correct answer?
#include<iostream>
using namespace std;
int main(){
int n,x=1;
int i=0;
cout<<"Enter a number bigger than 0:"<<endl;
cin>>n;
while(i<n){
i++;
x=i*x;
};
cout<<"The result is: "<<x/n<<endl;
return 0;
}
At very first a principle you best get used to as quickly as possible: Always check user input for correctness!
cin >> n;
if(cin && n > 0)
{
// valid
}
else
{
// appropriate error handling
}
Not sure, why do you need a while loop? A for loop sure is nicer in this case:
int x = 1;
for(int i = 2; i < n; ++i)
x *= i;
If you still want the while loop: Start with i == 2 (1 is neutral anyway) and increment afterwards:
i = 2;
while(i < n)
{
x *= i;
++i;
}
In case of n == 1, the loop (either variant) simply won't be entered and you are fine...
You already have two very good options, but here is an other one you might want to take a look at when you are at ease enough in programming :
unsigned factorial(unsigned value)
{
if (value <= 1)
{
return 1;
}
else
{
return value * factorial(value - 1);
}
}
It's a recursive function, which is kind of neat when used in proper moments (which could not be the case here unfortunately because the execution stack might get so big you fill your memory before you're done. But you can check it out to learn more about recursive functions)
When your memory is full, you then crash your app with what is called actually a stack overflow.
How can I make it so that in the last cout I can only put x and not have to divide x by n to get the correct answer?
It will be better to use a for loop.
// This stops when i reaches n.
// That means, n is not multiplied to the result when the loop breaks.
for (int i = 1; i < n; ++i )
{
x *= i;
}
cout << "The result is: " << x <<endl;
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int main() {
int t,m,n;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&m,&n);
int rootn=sqrt(double(n));
bool p[10000]; //finding prime numbers from 1 to square_root(n)
for(int j=0;j<=rootn;j++)
p[j]=true;
p[0]=false;
p[1]=false;
int i=rootn;
while(i--)
{
if(p[i]==true)
{
int c=i;
do
{
c=c+i;
p[c]=false;
}while(c+p[i]<=rootn);
}
};
i=0;
bool rangep[10000]; //used for finding prime numbers between m and n by eliminating multiple of primes in between 1 and squareroot(n)
for(int j=0;j<=n-m+1;j++)
rangep[j]=true;
i=rootn;
do
{
if(p[i]==true)
{
for(int j=m;j<=n;j++)
{
if(j%i==0&&j!=i)
rangep[j-m]=false;
}
}
}while(i--);
i=n-m;
do
{
if(rangep[i]==true)
printf("%d\n",i+m);
}while(i--);
printf("\n");
}
return 0;
system("PAUSE");
}
Hello I'm trying to use the sieve of Eratosthenes to find prime numbers in a range between m to n where m>=1 and n<=100000000. When I give input of 1 to 10000, the result is correct. But for a wider range, the stack is overflowed even if I increase the array sizes.
A simple and more readable implementation
void Sieve(int n) {
int sqrtn = (int)sqrt((double)n);
std::vector<bool> sieve(n + 1, false);
for (int m = 2; m <= sqrtn; ++m) {
if (!sieve[m]) {
cout << m << " ";
for (int k = m * m; k <= n; k += m)
sieve[k] = true;
}
}
for (int m = sqrtn; m <= n; ++m)
if (!sieve[m])
cout << m << " ";
}
Reason of getting error
You are declaring an enormous array as a local variable. That's why when the stack frame of main is pushed it needs so much memory that stack overflow exception is generated. Visual studio is tricky enough to analyze the code for projected run-time stack usage and generate exception when needed.
Use this compact implementation. Moreover you can have bs declared in the function if you want. Don't make implementations complex.
Implementation
typedef long long ll;
typedef vector<int> vi;
vi primes;
bitset<100000000> bs;
void sieve(ll upperbound) {
_sieve_size = upperbound + 1;
bs.set();
bs[0] = bs[1] = 0;
for (ll i = 2; i <= _sieve_size; i++)
if (bs[i]) { //if not marked
for (ll j = i * i; j <= _sieve_size; j += i) //check all the multiples
bs[j] = 0; // they are surely not prime :-)
primes.push_back((int)i); // this is prime
} }
call from main() sieve(10000);. You have primes list in vector primes.
Note: As mentioned in comment--stackoverflow is quite unexpected error here. You are implementing sieve but it will be more efficient if you use bistet instead of bool.
Few things like if n=10^8 then sqrt(n)=10^4. And your bool array is p[10000]. So there is a chance of accessing array out of bound.
I agree with the other answers,
saying that you should basically just start over.
Do you even care why your code doesn’t work? (You didn’t actually ask.)
I’m not sure that the problem in your code
has been identified accurately yet.
First of all, I’ll add this comment to help set the context:
// For any int aardvark;
// p[aardvark] = false means that aardvark is composite (i.e., not prime).
// p[aardvark] = true means that aardvark might be prime, or maybe we just don’t know yet.
Now let me draw your attention to this code:
int i=rootn;
while(i--)
{
if(p[i]==true)
{
int c=i;
do
{
c=c+i;
p[c]=false;
}while(c+p[i]<=rootn);
}
};
You say that n≤100000000 (although your code doesn’t check that), so,
presumably, rootn≤10000, which is the dimensionality (size) of p[].
The above code is saying that, for every integer i
(no matter whether it’s prime or composite),
2×i, 3×i, 4×i, etc., are, by definition, composite.
So, for c equal to 2×i, 3×i, 4×i, …,
we set p[c]=false because we know that c is composite.
But look closely at the code.
It sets c=c+i and says p[c]=false
before checking whether c is still in range
to be a valid index into p[].
Now, if n≤25000000, then rootn≤5000.
If i≤ rootn, then i≤5000, and, as long as c≤5000, then c+i≤10000.
But, if n>25000000, then rootn>5000,†
and the sequence i=rootn;, c=i;, c=c+i;
can set c to a value greater than 10000.
And then you use that value to index into p[].
That’s probably where the stack overflow occurs.
Oh, BTW; you don’t need to say if(p[i]==true); if(p[i]) is good enough.
To add insult to injury, there’s a second error in the same block:
while(c+p[i]<=rootn).
c and i are ints,
and p is an array of bools, so p[i] is a bool —
and yet you are adding c + p[i].
We know from the if that p[i] is true,
which is numerically equal to 1 —
so your loop termination condition is while (c+1<=rootn);
i.e., while c≤rootn-1.
I think you meant to say while(c+i<=rootn).
Oh, also, why do you have executable code
immediately after an unconditional return statement?
The system("PAUSE"); statement cannot possibly be reached.
(I’m not saying that those are the only errors;
they are just what jumped out at me.)
______________
† OK, splitting hairs, n has to be ≥ 25010001
(i.e., 50012) before rootn>5000.