Choose the statements to generate 4x4 A array [10 10 1010]1010 10 1010111010l10101010# include main (){int i, j;int A[4][4];for (i=0; i<=3; i++)for (j=0; j<=3; j++)for (i=0; <=3; i++)for (j=0; j<=3;j++)cout<

Given: The statements to generate 4x4 A array

Answered: # include main () {int i, j; int…

# include main () {int i, j; int A[4][4]; for (i=0; i<=3; i++) for (j=0; j<=3; j++) {- for (i=0; i<=3; i++) for (j=0; j<=3;j++) cout<

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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convert c code to mips these 2 functions are linked with each other. #define MAX_BOARD_SIZE 12 // Players #define PLAYER_EMPTY 0 #define PLAYER_BLACK 1 #define PLAYER_WHITE 2 int board_size; int current_player = PLAYER_BLACK; char board[MAX_BOARD_SIZE][MAX_BOARD_SIZE]; int main(void); void announce_winner(void); unsigned int count_discs(int player); void announce_winner(void) { int black_count = count_discs(PLAYER_BLACK); int white_count = count_discs(PLAYER_WHITE); if (white_count > black_count) { printf("The game is a win for WHITE!\n"); white_count += count_discs(PLAYER_EMPTY); } else if (black_count > white_count) { printf("The game is a win for BLACK!\n"); black_count += count_discs(PLAYER_EMPTY); } else { printf("The game is a tie! Wow!\n"); } printf("Score for black: %d, for white: %d.\n", black_count, white_count); } unsigned int count_discs(int player) { int count = 0; for (int row = 0; row < board_size; ++row) { for (int col = 0; col < board_size; ++col) { if…
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1、Counting Primitive Operations void STRAITMAXMIN(A,n,max,min) //Set the maximum value in A to max and the minimum value to min { int i,n max=min=A[1] for i=2 to n { if A[i]> max max=A[i]; if A[i]< min min=A[i]; } }
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C LANGUAGE ONLY. ANSWER ASAP Topic: pointers and dynamic memory allocation Make another version of the code below without copying it or using the same structure Thanks!
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JAVA LANGUAGE CODE Postfix Calculator by CodeChum Admin One good implementation of computing infix expressions is to transform them to postfix and then evaluate via the postfix expression. Infix expressions is the common way of writing arithmetic expressions. The binary operator come between them as shown below: 2 * 5 + 9 - 10 / 20 In postfix expressions, the operands come first before the operator: 2 5 * 9 + 10 20 / - A stack can be used to evaluate postfix expressions. The operands are pushed onto the Stack and when an operator is found two operands are popped and the operation is performed and finally the result is pushed back onto the Stack. The final answer will be the lone element of the Stack. Input The first line contains a positive integer n representing the number of postfix expressions. What follows are n postfix expressions themselves. 5 10 20 + 30 15 2 * + 100 20 30 + / 90 20 10 + + 0 / 9 3 - 10 + 2 * Output A single line containing the result of…
#include void main(void) { int c =2 ^3; cout<< c; }
Code: #include <bits/stdc++.h> using namespace std; void BUBBLE(int A[],int N){ for(int k=0;k<N-1;++k){ for(int ptr=0;ptr<N-k-1;++ptr){ if(A[ptr]>A[ptr+1]){ int temp = A[ptr]; A[ptr] = A[ptr+1]; A[ptr+1]=temp; } } }} //function to print the arrayvoid printArray(int arr[],int n){ int i; for(i=0;i<n;i++) cout<<arr[i]<<" "; cout<<endl;} //driver function to test the modulesint main(){ int arr[] ={15,17,5,3,25,66,14,7,59,100}; int n=sizeof(arr)/sizeof(arr[0]); cout<<"\nOriginal array: "; printArray(arr,n); cout<<"\n\nOutput of Bubble sort are shown below:\n"; BUBBLE(arr,n); printArray(arr, n); return 0;} Q: Remove the Function from the above code