Why we don't use gets function

Write a program in c++ to find the sum of n terms of the following notation = 3/4 + 6/8 + 9/12 + 12/16😢😢

Write a program in c++ to find the sum of n terms of the following notation = 3/4 + 6/8 + 9/12 + 12/16😢😢

Are you really sure that is the problem? The sum is very simple for that. You dont even need a computer. Before writing the program try some maths to see what the sum would be.

I'm trying to write the code but I'm having a hard time💔

I said try solving it mathematically by hand. You will then figure out that writing a program is as good as writing 2 lines of code.

Sum=3

Sum of 4 terms is 3.

Yes

Now while adding the fractions did you notice any similarity between the terms?

Now while adding the fractions did you notice any similarity between the terms?

#include<iostream> #include<cmath> using namespace std; int main() { double sum=0,a; float i; int n; cout<<"nEnter value of n"; cin>>n; for(i=0.75;i<=n;i++) { a=0.75; sum+=a; } cout<<"Sum="<<sum; }

= 3/4 + 6/8 + 9/12 + 12/16 The pattern looks like (numerator*i)/(denominator*i) where i=1,2,3,4,... which will result in 0.75+0.75+0.75+.... for n times... in this case( 3/4 + 6/8 + 9/12 + 12/16).. So what you could do is, something like result = (numerator/denominator)* no.of times

#include<iostream> #include<cmath> using namespace std; int main() { double sum=0,a; float i; int n; cout<<"nEnter value of n"; cin>>n; for(i=0.75;i<=n;i++) { a=0.75; sum+=a; } cout<<"Sum="<<sum; }

Just do a n×0.75 instead. Why do you need a loop? And this is why I asked you to try mathematically first. This is just an exercise in reducing fractions to its simplest form. It is something that we learn in primary schools

Is that true?

// C++ Program to find kth element from two sorted arrays #include <iostream> using namespace std; int kthElement(int arr1[], int arr2[], int n, int m, int k) { // C++ Program to find kth element from two sorted arrays // In case we have reached end of array 1 int st1=0, st2=0; while (1) { if (st1 == n) return arr2[st2 + k - 1]; // In case we have reached end of array 2 if (st2 == m) return arr1[st1 + k - 1]; // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (n - st1) + (m - st2)) return -1; // Compare first elements of arrays and return if (k == 1) return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; int curr = k / 2; // Size of array 1 is less than k / 2 if(curr-1>=n-st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[n - 1] < arr2[st2 + curr - 1]) return arr2[st2 + (k - (n - st1) - 1)]; else { st2 += curr; k -= curr; } } // Size of array 2 is less than k / 2 if(curr - 1 >= m - st2) { if (arr2[m - 1] < arr1[st1 + curr - 1]) return arr1[st1 + (k - (m - st2) - 1)]; else { st1 += curr; k -= curr; } } else { // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) 🥲🥲🥲🥲🥲🥲🥲//This condition fails if the size of the step ..i.e curr exceeds the size of array...so this code is giving me error if there is a skew in the size of the array { st1 += curr; k -= curr; } else { st2 += curr; k -= curr; } } } } // Driver code int main() { int arr1[7] = { 5 ,33, 55, 65, 76, 80, 90 }; int arr2[39] = { 10 ,13 ,14, 15, 15, 22, 27, 32, 34, 36, 36, 37, 39, 40, 42, 45, 49, 50, 50, 53, 56, 56, 57, 61, 65, 66, 70, 70 ,71, 74, 78, 84, 87, 90, 91, 94, 94 ,96 ,99 }; int k = 39; int p= kthElement(arr1, arr2, 7, 39, k); return 0; }

What exactly is your question? Is the element that you are looking for the k'th element in the array merged from the 2 sorted arrays? If yes, shouldn't you then be also checking the sum of the lengths to see if k is within that? And your code for merging doesn't seem right (I didnt go through it all. Just skimmed through).Check the code for merging in merge sort algorithm. That is exactly what you are looking for.

I didn't merge...I used a log k algorithm

no...see the line with sad emojiis

Is that true?

which ever element is small..go to that corresponding array and increment the starting index of that array

which ever element is small..go to that corresponding array and increment the starting index of that array

What if one array has less than k/2 elements but the elements towards the end are all greater than the k'th element?

yes see the algorithm...if(curr-1<n-start1)

This is why I asked you to give the outline. It is difficult to just understand it based on your code

What if one array has less than k/2 elements but the elements towards the end are all greater than the k'th element?

Hi I've got template function that I directly call somewhere: template<class First, class ...Rest> void foo(First && first, Rest && ...other) const { bar(first); bar(other...); } It invokes these: template<class Arg> void bar(Arg && single) const { std::cout << single << ' '; } void bar(Specific && specific) const { std::cout << "SPECIFIC"; } template<class ...Args> void bar(Args && ...args) const { (bar(std::forward<Args>(args)), ...); } I thought that it gonna work because C++ chooses the most specific overloading, but, as I see, it doesn't if I call it from another templated function, so bar(Specific) never be called. I know, that if I directly call bar then bar(Specific) will be used if I pass Specific type, anyway, from in templated function it is not. Help please 😔

Hi I've got template function that I directly call somewhere: template<class First, class ...Rest> void foo(First && first, Rest && ...other) const { bar(first); bar(other...); } It invokes these: template<class Arg> void bar(Arg && single) const { std::cout << single << ' '; } void bar(Specific && specific) const { std::cout << "SPECIFIC"; } template<class ...Args> void bar(Args && ...args) const { (bar(std::forward<Args>(args)), ...); } I thought that it gonna work because C++ chooses the most specific overloading, but, as I see, it doesn't if I call it from another templated function, so bar(Specific) never be called. I know, that if I directly call bar then bar(Specific) will be used if I pass Specific type, anyway, from in templated function it is not. Help please 😔

What is specific a template specialization?

Hi I've got template function that I directly call somewhere: template<class First, class ...Rest> void foo(First && first, Rest && ...other) const { bar(first); bar(other...); } It invokes these: template<class Arg> void bar(Arg && single) const { std::cout << single << ' '; } void bar(Specific && specific) const { std::cout << "SPECIFIC"; } template<class ...Args> void bar(Args && ...args) const { (bar(std::forward<Args>(args)), ...); } I thought that it gonna work because C++ chooses the most specific overloading, but, as I see, it doesn't if I call it from another templated function, so bar(Specific) never be called. I know, that if I directly call bar then bar(Specific) will be used if I pass Specific type, anyway, from in templated function it is not. Help please 😔

can you rewrite this - void bar(Specific && specific) const { std::cout << "SPECIFIC"; } as - template <> void bar(Specific && specific) const { std::cout << "SPECIFIC"; } and try again?

I've already tried this, but this functions are methods, so I cannot use template<>. I'll try again with separate struct to use template<>

https://en.cppreference.com/w/cpp/language/member_template#Member_function_templates A non-template member function and a template member function with the same name may be declared. In case of conflict (when some template specialization matches the non-template function signature exactly), the use of that name and type refers to the non-template member unless an explicit template argument list is supplied. 🤷

// C++ Program to find kth element from two sorted arrays #include <iostream> using namespace std; int kthElement(int arr1[], int arr2[], int n, int m, int k) { // C++ Program to find kth element from two sorted arrays // In case we have reached end of array 1 int st1=0, st2=0; while (1) { if (st1 == n) return arr2[st2 + k - 1]; // In case we have reached end of array 2 if (st2 == m) return arr1[st1 + k - 1]; // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (n - st1) + (m - st2)) return -1; // Compare first elements of arrays and return if (k == 1) return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; int curr = k / 2; // Size of array 1 is less than k / 2 if(curr-1>=n-st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[n - 1] < arr2[st2 + curr - 1]) return arr2[st2 + (k - (n - st1) - 1)]; else { st2 += curr; k -= curr; } } // Size of array 2 is less than k / 2 if(curr - 1 >= m - st2) { if (arr2[m - 1] < arr1[st1 + curr - 1]) return arr1[st1 + (k - (m - st2) - 1)]; else { st1 += curr; k -= curr; } } else { // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) 🥲🥲🥲🥲🥲🥲🥲//This condition fails if the size of the step ..i.e curr exceeds the size of array...so this code is giving me error if there is a skew in the size of the array { st1 += curr; k -= curr; } else { st2 += curr; k -= curr; } } } } // Driver code int main() { int arr1[7] = { 5 ,33, 55, 65, 76, 80, 90 }; int arr2[39] = { 10 ,13 ,14, 15, 15, 22, 27, 32, 34, 36, 36, 37, 39, 40, 42, 45, 49, 50, 50, 53, 56, 56, 57, 61, 65, 66, 70, 70 ,71, 74, 78, 84, 87, 90, 91, 94, 94 ,96 ,99 }; int k = 39; int p= kthElement(arr1, arr2, 7, 39, k); return 0; }

hi

hah, but I had an error like explicit specialization in non-namespace scope

are you sure you are even allowed to solve this is linear time. this looked like a leetcode problem so i checked. https://leetcode.com/problems/median-of-two-sorted-arrays/ imagine dividing the sorted array into two sections, with indices [0, k) and [k, m + n) (don't actually create the new array). find out how many elements the larger array contributes to the [0, k) part using binary search. use that information to deduce the kth element.

Sure I will check

I want to make a code in matlab gui to be a text watermarking in image

the cppreference example works for me :/

Hey guys I have a doubt can I post code here ??

just enclose it into monospace backticks

Sorry but I think you need to review full code

? I'm talking about ``````

Guys can u please tell me where is the error here in this code? If ("!image.data") { Cout << "could not open or" ; Cout << "find the image" ; Cout << endl; Return 0; }

it's just easier to read. you also can just select your code and Right-Click -> Format -> Monospace

u use string as boolean

Guys can u please tell me where is the error here in this code? If ("!image.data") { Cout << "could not open or" ; Cout << "find the image" ; Cout << endl; Return 0; }

C++ is not Pascal. It is case-sensitive. There's no Cout, Return and If there're cout, return and if

So, if will be after the return??

Guys can u please tell me where is the error here in this code? If ("!image.data") { Cout << "could not open or" ; Cout << "find the image" ; Cout << endl; Return 0; }

https://gist.github.com/aninditkarmakar/ac012e0459c9bf4b76ea