Answer:
In order to make the following code in C it is necessary to take in consideration the possible tree of choices, or decision make trees.
Code:
#include <stdio.h>
int main(){
printf("Please answer with (y/n) \\");
printf("How about comedy movies? ");
char optional[2];
scanf("%s",&optional);
if(optional[0] == 'y'){
printf("Recommended: CENTRAL INTELLIGENCE\\");
} else if(optional[0] == 'n'){
printf("Then, how about a current movie ? ");
scanf("%s",&optional);
if(optional[0] == 'y' ){
printf("Recommended: HOBBS & SHAW\\");
}else if(optional[0] == 'n') {
printf("how about electronic games ? ");
scanf("%s",&optional);
if(optional[0] == 'y' ){
printf("Certified Fresh ? ");
scanf("%s",&optional);
if(optional[0] == 'y'){
printf("recommended JUMANJI : WELCOME TO THE JUNGLE\\");
}else if(optional[0] == 'n') {
printf("RAMPAGE");
}
}else if(optional[0] == 'n') {
printf("about franchises ? ");
scanf("%s",&optional);
if(optional[0] == 'y'){
printf("Recommended: FAST FIVE\\");
}else if(optional[0] == 'n') {
printf("How about animation ? ");
scanf("%s",&optional);
if(optional[0] == 'y'){
printf("recommended: MOANA\\");
}else if(optional[0] == 'n') {
printf("recommended: SKYSCRAPPER\\");
}
}
}
}
}
return 1;
}