Hi everyone. I have a problem. I can find -74 but i dont know how to swipe big matris to right. Can you help me please. I cant send pic. Lets think matris1 is 21 19 17 25 28 71 76 73 68 59 153 164 157 155 200 201 190 185 180 205 210 215 230 232 Matris 2 -1 -1 -1 -1 8 -1 -1 -1 -1 First we multiple matris1 * matris2 but size is matris2size then we have to do swipe right. Like we multiple 19*-1 +17*-1+ 25 *-1 + 76*-1+73*8+ 68*-1+ 164*-1+164*-1+157*-1 I hope you guys understand. I cant share pic. So sorry.

gotcha, I think it's clear enough now. In your code you don't change k or m. So you probably need to more for loops to iterate over your temp matrix, e.g. for (int k = 0; k < ResultDimension; ++k) { for (int m = 0; m < ResultDimension; ++m) { /* In here compute the "multiplication" for every i and j in range Sum += matris1[i + k][j + m] * matris2[i][j]; */ Result[k][m] = Sum; } }

Please name few good and reliable resources.

Hi guys, could someone give an idea how to implement such thing 1 1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 in 2D array? I tried to do that here, but it was unsuccessful #include <stdio.h> #define n 7 void print(int a[n][n]) { for(int i = 0; i < n; i++) { printf("\n"); for(int j = 0; j < n; j++) { printf("%d ", a[i][j]); } } } void main() { int a[n][n] = { 0 }; for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { a[7-j][j] = 1; } } print(a); }

1. take the whole line 2. split by , 3. validate each number

Complex getComplex(){ Complex c; char ch; bool error; do{ error=false; c.real=c.imag=0; cout<<"Enter complex number(a + bi): "; cin>>c.real; if(!cin){ cin.clear(); if(cin.peek()=='i'){ //if i or i + 4 and any error situation cin.get(); removeWhite(); ch=cin.peek(); if(ch=='+'){ cin.get(); removeWhite(); if(isdigit(cin.peek())){ c.imag=1; cin>>c.real; removeWhite(); if(cin.peek() != '\n'){ error=true; } }else{ error=true; } }else if(ch=='\n'){ c.imag=1; }else{ error=true; } }else error=true; }else{ //case 4, 4i, 4i + 3 considering also all error situation removeWhite(); ch=cin.peek(); switch(ch){ case '\n': break; case 'i': cin.get(); removeWhite(); ch=cin.peek(); if(ch=='+' || ch=='\n'){ c.imag=c.real; c.real=0; if(ch=='+'){ cin.get(); removeWhite(); if(isdigit(cin.peek())){ cin >> c.real; removeWhite(); if(cin.peek()!='\n'){ error=true; } } } }else{ error=true; } break; case '+': cin.get(); //remove the + sign removeWhite(); ch=cin.peek(); if(isdigit(ch)){ cin>>c.imag; removeWhite(); ch=cin.peek(); if(ch=='i'){ cin.get(); //remove i removeWhite(); if(cin.peek()!='\n'){ error=true; } }else{ error=true; } }else if(ch=='i'){ cin.get(); //remove i removeWhite(); if(cin.peek()!='\n'){ error=true; }else{ c.imag=1; } }else{ error=true; } break; default: error=true; } } cin.ignore(INT_MAX,'\n'); //empty buffer if(error){ cout << "Invalid complex number. Please reenter!!!"<<endl; } }while(error); return c; }

I need help

with the details you have provided, i don't think anyone can

Complex getComplex(){ Complex c; char ch; bool error; do{ error=false; c.real=c.imag=0; cout<<"Enter complex number(a + bi): "; cin>>c.real; if(!cin){ cin.clear(); if(cin.peek()=='i'){ //if i or i + 4 and any error situation cin.get(); removeWhite(); ch=cin.peek(); if(ch=='+'){ cin.get(); removeWhite(); if(isdigit(cin.peek())){ c.imag=1; cin>>c.real; removeWhite(); if(cin.peek() != '\n'){ error=true; } }else{ error=true; } }else if(ch=='\n'){ c.imag=1; }else{ error=true; } }else error=true; }else{ //case 4, 4i, 4i + 3 considering also all error situation removeWhite(); ch=cin.peek(); switch(ch){ case '\n': break; case 'i': cin.get(); removeWhite(); ch=cin.peek(); if(ch=='+' || ch=='\n'){ c.imag=c.real; c.real=0; if(ch=='+'){ cin.get(); removeWhite(); if(isdigit(cin.peek())){ cin >> c.real; removeWhite(); if(cin.peek()!='\n'){ error=true; } } } }else{ error=true; } break; case '+': cin.get(); //remove the + sign removeWhite(); ch=cin.peek(); if(isdigit(ch)){ cin>>c.imag; removeWhite(); ch=cin.peek(); if(ch=='i'){ cin.get(); //remove i removeWhite(); if(cin.peek()!='\n'){ error=true; } }else{ error=true; } }else if(ch=='i'){ cin.get(); //remove i removeWhite(); if(cin.peek()!='\n'){ error=true; }else{ c.imag=1; } }else{ error=true; } break; default: error=true; } } cin.ignore(INT_MAX,'\n'); //empty buffer if(error){ cout << "Invalid complex number. Please reenter!!!"<<endl; } }while(error); return c; }

This is too complicated. What you want to do instead is to define a user defined literal. This code accepts literals in all the formats you described: https://pastebin.com/6hU3wNU3 It is just meant to show you how to do it. Think carefully about the operator overloads. Also I made all the data members public (which you normally do in such classes). Moreover there is a class called complex in the standard library already.

This is too complicated. What you want to do instead is to define a user defined literal. This code accepts literals in all the formats you described: https://pastebin.com/6hU3wNU3 It is just meant to show you how to do it. Think carefully about the operator overloads. Also I made all the data members public (which you normally do in such classes). Moreover there is a class called complex in the standard library already.

well what if i want the user to enter his own no instead of initializing

then i took the input separately but what if the user enter this 4 types how can i validate them 3 + 2i , 2i + 3 , 3 ,2i

this were just an examples that the user might enter the no.s i want the user to enter

Look at the code link I sent above. It already accepts literals in all the formats

Look at the code link I sent above. It already accepts literals in all the formats

i run it but since there's no cin the code cant accept any value from the user

it only displays 6 and 8

Ok. The code I gave is for literal values. If you want to accept user input in a+bi form, then this is not the solution. To be able to accept user inputs of a specific form, you must overload operator >> for your class and then parse it yourself. I misunderstood what you wanted.

Complex getComplex(){ Complex c; char ch; bool error; do{ error=false; c.real=c.imag=0; cout<<"Enter complex number(a + bi): "; cin>>c.real; if(!cin){ cin.clear(); if(cin.peek()=='i'){ //if i or i + 4 and any error situation cin.get(); removeWhite(); ch=cin.peek(); if(ch=='+'){ cin.get(); removeWhite(); if(isdigit(cin.peek())){ c.imag=1; cin>>c.real; removeWhite(); if(cin.peek() != '\n'){ error=true; } }else{ error=true; } }else if(ch=='\n'){ c.imag=1; }else{ error=true; } }else error=true; }else{ //case 4, 4i, 4i + 3 considering also all error situation removeWhite(); ch=cin.peek(); switch(ch){ case '\n': break; case 'i': cin.get(); removeWhite(); ch=cin.peek(); if(ch=='+' || ch=='\n'){ c.imag=c.real; c.real=0; if(ch=='+'){ cin.get(); removeWhite(); if(isdigit(cin.peek())){ cin >> c.real; removeWhite(); if(cin.peek()!='\n'){ error=true; } } } }else{ error=true; } break; case '+': cin.get(); //remove the + sign removeWhite(); ch=cin.peek(); if(isdigit(ch)){ cin>>c.imag; removeWhite(); ch=cin.peek(); if(ch=='i'){ cin.get(); //remove i removeWhite(); if(cin.peek()!='\n'){ error=true; } }else{ error=true; } }else if(ch=='i'){ cin.get(); //remove i removeWhite(); if(cin.peek()!='\n'){ error=true; }else{ c.imag=1; } }else{ error=true; } break; default: error=true; } } cin.ignore(INT_MAX,'\n'); //empty buffer if(error){ cout << "Invalid complex number. Please reenter!!!"<<endl; } }while(error); return c; }

have you seen the first one 👀

Yes. Now I understood what you wanted clearly. You should post huge chunks of code using sites like pastebin ratherthan pasting it here. Makes it easier for people to understand your code then.

Does anyone knows divide and conquer method nicely?

Does anyone knows divide and conquer method nicely?

Does anyone knows divide and conquer method nicely?

divide and conquer means dividing a problem into small or subproblems, solve the subproblems and finally combines the solutions to the subproblems to solve the overall problem

I will be needing some help in it

well if you need a power point document about it i can share you

Anyone here is quite pro in competitive programming?

Just ask your question, dont as meta questions

Does anyone of you know a YouTube channel or a website that teaches iteration in C? Thank youu

iteration as in development process or loops?

Actually my professor introduced us to iteration first before arrays that's why I'm having trouble coding an iterative program for padovan sequence

Well then you can learn about arrays online. There are a lot of sources available online and in YouTube. Stay away from Indian instructors as far as possible.

Thank you so muchh. I'll note this😊

bucky he is kinda good

Is he in YouTube?

yea

it's channel name is thenewboston

btw guys can you make a telegram bot using C++?

btw guys can you make a telegram bot using C++?

Telegram Bot API is a HTTP based API. You can use any language through which you can make a web request and receive a response and process it. There are tonnes of open source libraries for C++ that can be used for this purpose.

btw guys can you make a telegram bot using C++?

btw guys can you make a telegram bot using C++?

Oh sry, https://pastebin.com/X4a8aYTN. how can I create a .txt with the data that was entered in "getDatos()", I am using <fstream> but there is an error that I don't understand