public class Main {
public static void main(String[] args) {
int a = 1, b=71, c=3, d= 10, max;
max = (a >b) ? a : b;
max = Math.max(max, c); //same thing
max = (max >d) ? max : d;
System.out.println(max);
}
}
71
כמה ספרות וכמה אפסים
קלוט מספר n
הדפס מספר הספרות
הדפס מספר האפסים
public class Main {
public static void main(String[] args) {
int num = MyConsole.readInt("enter a number: ");
System.out.println("You entered a " + (num >= 0 ? "positive" : "negative") +" number");
int countZeros=0;
int countDigits=0;
int absNum = num>=0 ? num : -num;
if (num==0){
countZeros=1;
countDigits=1;
}
while (absNum > 0) {
if (absNum%10==0)
countZeros++;
absNum /= 10;
countDigits++;
}
System.out.println("number of digits: " + countDigits);
System.out.println("number of zeros: " + countZeros);
}
}
enter a number: -30012
You entered a negative number
number of digits: 5
number of zeros: 2
זה if מקוצר
int absNum = num>=0 ? num : -num;
version 2
let num be a random number from -1000 until 9999 (including)
int max = 10_000, min = -1_000;
int num = (int) ((Math.random() * (max - min)) + min); //random number from -1,000 to 9,999 (including)
System.out.println("Random num:" + num);
Random num:6318
You entered a positive number
number of digits: 4
number of zeros: 0
Note: it includes the min but not the max [min, max)
Rotate Square
יש לסובב את הקוביה
public class Main {
public static void main(String[] args) {
int a = 1, b =2, c=3, d=4, temp;
temp = d;
d = c;
c = b;
b = a;
a = temp;
System.out.println("a: " + a + "\nb: " + b + "\nc: " + c + "\nd: " + d);
}
}
a: 4
b: 1
c: 2
d: 3
טבלת מעקב
Version 2 : n right or left shifts
קלוט מספר n:
אם הוא חיובי אז סובב n פעמים ימינה
אם הוא שלילי אז סובב n פעמים שמאלה
public class Main {
public static void main(String[] args) {
int a = 1, b =2, c=3, d=4, temp;
boolean isRightShift=true;
int n_shifts = MyConsole.readInt("enter:\n* posivite num for right shift" +
"\n* negative for left shift\ninput :");
if (n_shifts<0) {
isRightShift = false;
n_shifts = n_shifts * -1; //since while works on positive number, otherwise we would need 2 different whiles
System.out.println("Left shifting " + n_shifts +" times");
} else if (n_shifts==0)
System.out.println("No shifts");
else
System.out.println("Right shifting " + n_shifts +" times");
while (n_shifts>0){
if (isRightShift) {
temp = d;
d = c;
c = b;
b = a;
a = temp;
n_shifts = n_shifts-1;
System.out.println("a: " + a + "\tb: " + b + "\tc: " + c + "\td: " + d);
}else {
temp = a;
a = b;
b = c;
c = d;
d = temp;
n_shifts = n_shifts-1;
System.out.println("a: " + a + "\tb: " + b + "\tc: " + c + "\td: " + d);
}
}
}
}
enter:
* posivite num for right shift
* negative for left shift
input : -2
Left shifting 2 times
a: 2 b: 3 c: 4 d: 1
a: 3 b: 4 c: 1 d: 2
note
since while (n_shifts>0) checks for positive numbers and we do n_shifts = n_shifts-1; inside the while loop without n_shifts = n_shifts * -1 if we have a left shift we would need anotherwhile loop
Find Aprox. log2
מצא עבור קלט x את:
קודם כל בא נבין מה קורה כאן
אם הוא לא חזקה שלמה הדפס את חזקה שלמה ועוד .x
public class Main {
public static void main(String[] args) {
float x = (float)MyConsole.readDouble("enter positive num: ");
while (x<0.)
x = (float)MyConsole.readDouble("Try again\nPlease enter a positive num: ");
// float x = 1<<10;
// float x = 1_024f;
float original = x;
int count = 0;
boolean isPowOf2 = false;
while(x > 2) {
x = x / 2;
count +=1;
if (x==2.){
System.out.println("log_2("+original+")="+(count+1));
isPowOf2 = true;
}
}
if (isPowOf2==false)
System.out.println("log_2("+original+")="+count + ".x");
}
}
enter num: 1024
log_2(1024.0)=10
2nd run
enter num: 1025
log_2(1025.0)=10.x
To find the real answer we should use a Talyor series.
