LDF R2, id3 MULF R2, R2, #60.0 LDF R1, id2 ADDF R1, R1, R2 STF id1, R1
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- C++ PROGRAMMINGTopic: HashTable - PolyHash Quadratic Explain the c++ code below.: SEE ATTACHED PHOTO FOR THE PROBLEM INSTRUCTIONS It doesn't have to be long, as long as you explain what the important parts of the code do. (The code is already implemented and correct, only the explanation needed). You can also explain line by line for an upvote, thanks. EXPLAIN THE CODE BELOW: #include <cstdlib> #include <math.h> #include <cstring> #include <iostream> using namespace std; class HashTable { string* table; int N; int count; // TODO: Polynomial Hash Code using a=7 int hash_code(string key) { int code; int hash = 0; for (int i = 0; i < key.size(); i++) { char ch = key[i]; code += ((ch - 96) * pow(7, key.size() - (i + 1))); } return code; } // TODO: This hash table uses a MAD compression function // where a = 11, b = 461, p = 919 int compress(int code) { return (((11*code)+461) % 919) % N; }…IN c++ please ; apply_Irvine_libs.asm - A simple test program calling various Irvine library procedures,; code base modified from demo program AddTwo.asm ; include the Irvine's libraryINCLUDE Irvine32.inc .386.model flat,stdcall.stack 4096ExitProcess proto,dwExitCode:dword .data; strings(texts) for promptsprompt_1st_int BYTE "Please enter the first integer: ",0prompt_2nd_int BYTE "Please enter the second integer: ",0prompt_text_color BYTE "Please pick the text color (1=blue, 2=green, 4=red, or any number from 0-15): ",0prompt_proverb BYTE "Please teach me a proverb: ",0response_str BYTE "The proverb I just learned: ", 0 ; strings(texts) for output labels/descriptionsdec_sum_str BYTE "The sum in DEC is: ",0hex_sum_str BYTE "The sum in HEX is: ",0memory_dump_str BYTE "Memory Dump with operand_1, _2, and sum: ",0 ; variables definitionsoperand_1 SDWORD ?operand_2 SDWORD ?sum SDWORD ?color_code DWORD ?aString BYTE 80 DUP(0), 0 .codemain proc ;prompt the 1st number input mov edx,…c++ computer language what is the output of the following code segment? int main() { int myarray[]={-200, 33, 89, 50, 44}; sub(myarray, 5); for(int i =0; i<5; i++) cout<<myarray[i]<<" "; cout<<endl; system("pause"); return 0; }; void sub(int arr[], int z) { for(int i =0; i<z; i++) arr[i] = arr[i] - i; }
- C CodeApproved Libraries:<stdio.h><math.h><stdlib.h><time.h> (for srand(time(0)) only)C CodeApproved Libraries: <stdio.h><math.h><stdlib.h><time.h> (for srand(time(0)) only)C programming pseudocode pointers help. { totalInAGivenHour+=*(rawReadings+ monthNo*30*14*4 + 14*4*nThDateOfMonth+ 4*nThHourOfDateFrom7AM + nThReadingTimeOfHour); // the pointer is to assign the values in the bracket and storing them in totalInAGivenHour }
- Plz paraphrase this code #include<iostream>using namespace std;struct studmodule{char modulename[30];int totalmarks;}; struct Student{char name[30];int age;int studid;struct studmodule module[4];float avgmarks;}stud[5]; float avg;float classavg;int highmark = 0;int lowmark = 0;float passrate = 0; void inputdata(){int i = 0, tot;cout<<"-------------------------------------------"<<endl;cout<<"Welcome to Technology College Muscut"<<endl;cout<<"-------------------------------------------"<<endl;cout<<"Students Personnel data Entry"<<endl;cout<<"-------------------------------------------"<<endl;for (i=0; i<5; i++){tot = 0;cout<<"Enter Student "<<i+1 << "Name :";cin>>stud[i].name;cout<<"Enter Student "<<i+1 << "Id Number :";cin>>stud[i].studid;cout<<"Enter Student "<<i+1 << "Age…Visual Basic.net) Write a Code that defines one dimensional arrays of ten elements that sums the even numbers from them and a counter that counts how many elements is an even number and comes out from the Loop when the sum of the even values becomes greater than 30 and the counter for these values is greater than 4. Write a program in Visual Basic.net .A binary operator is an operation that is performed on two operands. For example, addition is a binary operator that adds two operands (e.g., 2 + 4). In the provided UML diagram, all binary operators have two operands and an execute method that performs the appropriate operation on the operands. The AddOperation class should add the two operands. The SubtractOperation class should subtract the two operands. Here is an example of how a binary operator could be used: IBinaryOperator operator = new AddOperator(8, 12);System.out.println(operator.execute()); // should print 20operator.setLeftOperand(18);System.out.println(operation.execute()); // should print 30 crate a Main.java to test your code
- Paraphrase this code plz #include<iostream>using namespace std;struct studmodule{char modulename[30];int totalmarks;}; struct Student{char name[30];int age;int studid;struct studmodule module[4];float avgmarks;}stud[5];float avg;float classavg;int highmark=0;int lowmark=0;float passrate=0;void inputdata(){int i=0,tot;cout<<"-------------------------------------------"<<endl;cout<<"Welcome to Technology College Muscut"<<endl;cout<<"-------------------------------------------"<<endl;cout<<"Students Personnel data Entry"<<endl;cout<<"-------------------------------------------"<<endl;for (i=0;i<5;i++){tot=0;cout<<"Enter Student "<<i+1 << "Name :";cin>>stud[i].name;cout<<"Enter Student "<<i+1 << "Id Number :";cin>>stud[i].studid;cout<<"Enter Student "<<i+1 << "Age…Design your own java console application about any valid problem that the application must solve the application must use arrays loops inheritance constructors and information hiding must be in a form of a report the consoleIN c++ pretty please ; apply_Irvine_libs.asm - A simple test program calling various Irvine library procedures, ; code base modified from demo program AddTwo.asm ; include the Irvine's libraryINCLUDE Irvine32.inc .386.model flat,stdcall.stack 4096ExitProcess proto,dwExitCode:dword .data; strings(texts) for promptsprompt_1st_int BYTE "Please enter the first integer: ",0prompt_2nd_int BYTE "Please enter the second integer: ",0prompt_text_color BYTE "Please pick the text color (1=blue, 2=green, 4=red, or any number from 0-15): ",0prompt_proverb BYTE "Please teach me a proverb: ",0response_str BYTE "The proverb I just learned: ", 0 ; strings(texts) for output labels/descriptionsdec_sum_str BYTE "The sum in DEC is: ",0hex_sum_str BYTE "The sum in HEX is: ",0memory_dump_str BYTE "Memory Dump with operand_1, _2, and sum: ",0 ; variables definitionsoperand_1 SDWORD ?operand_2 SDWORD ?sum SDWORD ?color_code DWORD ?aString BYTE 80 DUP(0), 0 .codemain proc ;prompt the 1st number…