Answer
Executable code
#include <stdio.h>
#include <stdlib.h>
int main() {
int array_size=20, i, j;
int array[array_size][array_size];
for(i=0; i<array_size; i++){
for(j=0; j<array_size; j++){
array[i][j]=rand()%1000 ;
//printf("%d\t",array[i][j]);
}
//printf("\n");
}
for(i=0; i<array_size; i++){
for(j=0; j<array_size; j++){
int min =0, max = 0;
if(i>0 && (array[i][j]>array[i-1][j]))
min =1;
else
max = 1;
if(j>0 && (array[i][j]>array[i][j-1]))
min =1;
else
max = 1;
if(i>0 && j>0 &&
(array[i][j]>array[i-1][j-1]))
min =1;
else
max = 1;
if(i<array_size-1 &&
(array[i][j]>array[i+1][j]))
min =1;
else
max = 1;
if(j<array_size-1 &&
(array[i][j]>array[i][j+1]))
min =1;
else
max = 1;
if(j<array_size-1 && i<array_size-1 &&
(array[i][j]>array[i+1][j+1]))
min =1;
else
max = 1;
if(j>0 && i<array_size &&
(array[i][j]>array[i+1][j-1]))
min =1;
else
max = 1;
if(j<array_size && i>0 &&
(array[i][j]>array[i-1][j+1]))
min =1;
else
max = 1;
if(min==0)
printf("element array[%d][%d] is peak at minimum\n", i, j);
else if(max==0)
printf("element array[%d][%d] is peak at maximum\n", i, j);
}
}
return 0;
}
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/***************************************************
Name:
Date:
Homework #7
Program name: HexUtilitySOLUTION
Program description: Accepts hexadecimal numbers as input.
Valid input examples: F00D, 000a, 1010, FFFF, Goodbye, BYE
Enter BYE (case insensitive) to exit the program.
****************************************************/
import java.util.Scanner;
public class HexUtilitySOLUTION {
public static void main(String[] args) {
// Maximum length of input string
final byte INPUT_LENGTH = 4;
String userInput = ""; // Initialize to null string
Scanner input = new Scanner(System.in);
// Process the inputs until BYE is entered
do {...
java
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