Informatics3-2024/Solutions1
A MathWikiből
A lap korábbi változatát látod, amilyen Kkovacs (vitalap | szerkesztései) 2024. február 23., 14:08-kor történt szerkesztése után volt.
Tartalomjegyzék |
Solutions
Only the solutions we discussed on the practical will be here. All the other tasks are perfect to practice for the written exam.
1. Conditional print
#include<stdio.h> int main(void) { /* declare and give values to variables x and y */ int x; // this could be int x = 5; int y; // the it would be a variable definition x = 5; y = 2; printf("value of X: %d\n", x); // we print the values printf("value of Y: %d\n", y); // of the variables printf("\n"); // this is just an empty line in the output /* using conditionals make it so only true statements are printed */ if (x > y) { // a simple condition printf("%d is greater than %d\n", x, y); } else if (x < y) { // there's no elif, we need to use else if printf("%d is greater than %d\n", y, x); } else { printf("%d is equal to %d\n", x, x); } return 0; }
Now with using scanf:
#include<stdio.h> int main(void) { /* declare and give values to variables x and y */ int x; // this could be int x = 5; int y; // the it would be a variable definition x = 5; y = 2; printf("Provide a value for x: "); scanf("%d", &x); printf("Provide a value for y: "); scanf("%d", &y); printf("value of X: %d\n", x); // we print the values printf("value of Y: %d\n", y); // of the variables printf("\n"); // this is just an empty line in the output /* using conditionals make it so only true statements are printed */ if (x > y) { // a simple condition printf("%d is greater than %d\n", x, y); } else if (x < y) { // there's no elif, we need to use else if printf("%d is greater than %d\n", y, x); } else { printf("%d is equal to %d\n", x, x); } return 0; }
2. For cycle
#include<stdio.h> int main(void) { /* variable definitions */ int i; // no need to give any initial value to this, we'll do it in the cycle for (i = 1 ; i <= 100; i++) { // it's an equally good solution if we start if (i % 2 == 0) { // from 2, and use a step size of two printf("%d\n", i); // by using i += 2 instead of i++ } } return 0; }
3. Do / while cycle
#include<stdio.h> int main(void) { int x; // we'll read the input numbers into this variable float s = 0; // the sum will be stored here int n = 0; // we'll count how many numbers we've seen here scanf("%d", &x); while(x != 0) { s += x; n++; scanf("%d", &x); // we read a new number at the end of every cycle } printf("%f\n", s / n); // %f can be used to print a float // the float / int division results in a float // but for example int / int would result in an int // try to modify the type of s to int return 0; }