Chapter 4 Last part

1. Chapter 4 Last part

2. Quiz • Print the integers from -7 to +7, one per line.

3. -7 to +7 Solution #include <stdio.h> int main() { int j; for (j= -7; j<=7; j++ ) { printf("%3d\n", j); } system("PAUSE"); return 0; }

5. What will be printed? • #include <stdio.h> • main(){ • inti=1; • for(i=1;i;){ • printf("The value of I = %d:\n",i); • i++; • if((i%10)==0){ • i=(!i); • } • } • system("pause"); • return 0; • }

7. What? #include <stdio.h> int main() { int sum = 0, /* Current sum */ num; /* Number to be read */ printf ( "Please enter the integers:\n" ); for ( scanf ( "%d", &num ); num; scanf ( "%d", &num ) ) sum += num; printf ( "The sum of the numbers is %d\n", sum ); system("pause"); return ( 0 ); }

9. GCD Solution //Greatest common divisor /* To compute gcd(48,18), divide 48 by 18 to get a quotient of 2 and a remainder of 12. Then divide 18 by 12 to get a quotient of 1 and a remainder of 6. Then divide 12 by 6 to get a remainder of 0, which means that 6 is the gcd.*/ #include <stdio.h> int main(){ int a=12; int b=8; // test case : 48,18 =6 // test case : 12,8 = 4 while(b){ int temp =b; //b is smaller b=a%b; a=temp; } printf("%d \n",a); system("pause"); return 0; }

10. "unsigned integer" and "signed integer" • All integer data types are signed data types, i.e. they have values which can be positive or negative.

11. Bits - Signed number representations  Hence in a byte with only 7 bits (apart from the sign bit), the magnitude can range from 0000000 (0) to 1111111 (127). Thus you can represent numbers from −12710 to +12710 once you add the sign bit (the eighth bit). Decimal/Binary Conversion Tool: http://acc6.its.brooklyn.cuny.edu/~gurwitz/core5/nav2tool.html

12. Integer Overflow • Integer Overflows are arithmetic errors. Integers have finite ranges in computers. •  For 32-bit signed integers, the minimum value is 0x80000000 (-2147483648) and the maximum value is 0x7fffffff (2147483647)

13. Integer Overflow • #include <stdio.h> • int main ( void ) { • inti=2147483647; • printf ("% d\n" , i+1); • system("pause"); • return 0; • }

14. What will be printed ? #include <stdio.h> int main(){ int a=0, b=0; a = 3; printf("Value of A :%d\n",a); b = a++; printf("Value of B :%d\n",b); system("pause"); return 0; }

16. Increment and Decrement Operators • The unusual feature of `++' and `--' is that they can be used either before or after a variable. • The value of ++k is the value of k after it has been incremented. The value of k++ is kbefore it is incremented. • Suppose k is 5. Then • x = ++k; • Increments k to 6 and then sets x to the resulting value, i.e., to 6. But • x = k++;

17. Increment and Decrement Operators • x = k++; • First sets x to to 5, and then increments k to 6. • The incrementing effect of ++k and k++ is the same, but their values are respectively 5 and 6.

18. The First 1,000 Primes • A prime number is a positive integer, which is divisible on 1 and itself.  • No even number can be a Primeexcept 2 • We only need to check the divisibility of number until square root of number (we call this sqrt(number)) • We need to check if number is even. If it is, then is not prime. 4. We need to keep count of Primary Numbers Found

19. Assignment-divide-by-0 error • With the continuestatement, you can avoid a divide-by-0 error without ending the loop completely. • Write a program to divide 4000by -4 to number 10 , except 0;

20. Assignment-divide-by-0 error Output • 4000 divided by -4 is... -1000 • 4000 divided by -3 is... -1333.33333333 • 4000 divided by -2 is... -2000 • 4000 divided by -1 is... -4000 • 4000 divided by 1 is... 4000 • 4000 divided by 2 is... 2000 • 4000 divided by 3 is... 1333.33333333 • 4000 divided by 4 is... 1000 • 4000 divided by 5 is... 800 • 4000 divided by 6 is... 666.666666667 • 4000 divided by 7 is... 571.428571429 • 4000 divided by 8 is... 500 • 4000 divided by 9 is... 444.44444444444 • 4000 divided by 10 is... 400

21. Prime numbers and cryptography (RSA) • To decode the secret message, you must first factor the number 5192...4257 of the public key into a product of primes, just as 39 factors as 3 x 13.

22. Theprime factors of a positive integer • The prime factors of a positive integer are the prime numbers that divide that integer exactly. • As you can see, every factor is a prime number

23. The prime factors of a positive integer

24. Example 1: What are the prime factors of 12 ? • It is best to start working from the smallest prime number, which is 2, so let's check: • 12 ÷ 2 = 6 • Yes, it divided evenly by 2. We have taken the first step! • But 6 is not a prime number, so we need to go further. Let's try 2 again: • 6 ÷ 2 = 3 • Yes, that worked also. And 3 is a prime number, so we have the answer: • 12 = 2 × 2 × 3 • As you can see, every factor is a prime number, so the answer must be right.

25. What is the prime factorization of 17 ? • Hang on ... 17 is a Prime Number. • So that is as far as we can go. • 17 = 17

26. Prime Factorization Code #include <stdio.h> int main(){ inti; int n=288; for ( i = 2; i <= n; i++) { while (n % i == 0) { printf("%d ,",i); n /= i; } } system("pause"); return 0; }