Rewrite this if/else if code segment into a switch statement int x = 0; int i = 1, j= 2, k = 3, m = 4, n = 5; if (x > 15 && x <= 19) { i += 1; } else if (x == 21) { j+= 2; n = 8; } else if (x > 21 && x < 23) { k *= 3; } else { m +=k + 5; }
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- int j; switch (i) { case 0: j = 2*i; break; case 1: j = -i; break; default: j = -2*i; } What is j if i = 1? What is j if i = -1?#include<stdio.h>#include<stdlib.h> int cent50=0;int cent20=0;int cent10=0;int cent05=0; void calculatechange(int* change){if(*change>0){if(*change>=50){*change-=50;cent50++;}else if(*change>=20){*change-=20;cent20++;}else if(*change>=10){*change-=10;cent10++;}else if(*change>=05){*change-=05;cent05++;}calculatechange(change);}}void printchange(){if(cent50)printf("\n50cents:%d coins",cent50);if(cent20)printf("\n20cents:%d coins",cent20);if(cent10)printf("\n10cents:%d coins",cent10);if(cent05)printf("\n05cents:%d coins",cent05);cent50=0;cent20=0;cent10=0;cent05=0;}void takechange(int* change){scanf("%d",change);getchar();}int main(){int change=0;int firstinput=0;while(1){if(!firstinput){printf("\nEnter the amount:");firstinput++;}else{printf("\n\nEnter the amount to continue or Enter -1 to…#include<stdio.h>#include<stdlib.h> int cent50=0;int cent20=0;int cent10=0;int cent05=0; void calculatechange(int* change){if(*change>0){if(*change>=50){*change-=50;cent50++;}else if(*change>=20){*change-=20;cent20++;}else if(*change>=10){*change-=10;cent10++;}else if(*change>=05){*change-=05;cent05++;}calculatechange(change);}}void printchange(){if(cent50)printf("\n50cents:%d coins",cent50);if(cent20)printf("\n20cents:%d coins",cent20);if(cent10)printf("\n10cents:%d coins",cent10);if(cent05)printf("\n05cents:%d coins",cent05);cent50=0;cent20=0;cent10=0;cent05=0;}void takechange(int* change){scanf("%d",change);getchar();}int main(){int change=0;int firstinput=0;while(1){if(!firstinput){printf("\nEnter the amount:");firstinput++;}else{printf("\n\nEnter the amount to continue or Enter -1 to…
- char ch=a'; switch(ch+1 ) { case 'a' :cout<<'a'; break; case 'b' :cout<#include<stdio.h>#include<stdlib.h> int cent50=0;int cent20=0;int cent10=0;int cent05=0; void calculatechange(int* change){if(*change>0){if(*change>=50){*change-=50;cent50++;}else if(*change>=20){*change-=20;cent20++;}else if(*change>=10){*change-=10;cent10++;}else if(*change>=05){*change-=05;cent05++;}calculatechange(change);}}void printchange(){if(cent50)printf("\n50cents:%d coins",cent50);if(cent20)printf("\n20cents:%d coins",cent20);if(cent10)printf("\n10cents:%d coins",cent10);if(cent05)printf("\n05cents:%d coins",cent05);cent50=0;cent20=0;cent10=0;cent05=0;}void takechange(int* change){scanf("%d",change);getchar();}int main(){int change=0;int firstinput=0;while(1){if(!firstinput){printf("\nEnter the amount:");firstinput++;}else{printf("\n\nEnter the amount to continue or Enter -1 to…#include<stdio.h>#include<stdlib.h> int cent50=0;int cent20=0;int cent10=0;int cent05=0; void calculatechange(int* change){if(*change>0){if(*change>=50){*change-=50;cent50++;}else if(*change>=20){*change-=20;cent20++;}else if(*change>=10){*change-=10;cent10++;}else if(*change>=05){*change-=05;cent05++;}calculatechange(change);}}void printchange(){if(cent50)printf("\n50cents:%d coins",cent50);if(cent20)printf("\n20cents:%d coins",cent20);if(cent10)printf("\n10cents:%d coins",cent10);if(cent05)printf("\n05cents:%d coins",cent05);cent50=0;cent20=0;cent10=0;cent05=0;}void takechange(int* change){scanf("%d",change);getchar();}int main(){int change=0;int firstinput=0;while(1){if(!firstinput){printf("\nEnter the amount:");firstinput++;}else{printf("\n\nEnter the amount to continue or Enter -1 to…Void Do1 (int: &, a. int &b) { a = 5; a = a + b; b = a + 2; } Int main() { Int x = 10; Do1 (x,x); Cout << x << endl; } The output of this program isC language Q4. Write a program that asks the user to enter an integer n, then prints the first n multiple of n. The program should keep asking to enter a number until -1 is entered. (0.5 point) Q5: a. What does the following program print? char c='Z'; for(i=1;i<= 4; i++) { for(j= 1; j<= i; j++) { printf("%c", C--); } printf("\n"); } b. Re-write the above code using while loop. Output: Sample Output: Enter a number: 4 48 12 16 Enter a number: 6 6 12 18 24 30 36 Enter a number: 1 1 Enter a number: -1 End Using while loop: (2x0.5 =1 point)22. What is the value of x after the following statements are executed? int x = 5; switch(x) { case 5: x += 2; case 6: x++; break; default: x *= 2; break; }Prompt Problem: A company wants a program that will calculate the weekly paycheck for an employee based on how many hours they worked. For this company, an employee earns $20 an hour for the first 40 hours that they work. The employee earns overtime, $30 an hour, for each hour they work above 40 hours. Example: If an employee works 60 hours in a week, they would earn $20/hr for the first 40 hours. Then they would earn $30/hr for the 20 hours they worked overtime. Therefore, they earned: ($20/hr 40hrs)+($30/hr 20 hrs) $800+ $600 = $1400 total. For this assignment, you must create pseudocode and a flowchart to design a program that will calculate an employee's weekly paycheck. - Write pseudocode to design a programming solution by outlining a series of steps and using appropriate indentation and keywords. As you write your pseudocode, be sure to consider the following: What input does the computer need? What steps does the program need to follow to process the input? What output should…int y=0,i; for (int i=0;i<10;++i) y+=i; 36 66 45 55Write a statement that assigns finalValue with the multiplication of userNum1 and userNum2. Ex: If userNum1 is 6 and userNum2 is 2, finalValue is 12. let userNum1 = 6; // Code tested with values: 6 and 4let userNum2 = 2; // Code tested with values: 2 and -2 let finalValue = 0;SEE MORE QUESTIONS
