1. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 1 If statement (preview version) Syntax form: if ( condition) Statements else Statements /* example: compare if two given integers are equal */ #include main() { int a, b; printf("Input two integers>"); scanf("%d%d", &a, &b); /* comparison of the two integers */ if (b == a) { printf("The two numbers are equal. \n"); } /* { } can be omitted if there is only one statement */ else printf("The two numbers are not equal. \n"); }

2. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 2 Flowchart for if Statements Two AlternativesOne Alternative a == b a and b equal a and b not equal exit a == b A and b are equal F T T F

3. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 3 Example 1: Find Roots of a Quadratic Equation (Refined) Algorithm 3 1.Get a, b, c 2.Calculate the discriminant: disc = pow(b,2) – 4*a*c; 3.Square root of discriminant: sqrt_disc = sqrt(labs(disc)); 4.e = -b / (2*a), f = sqrt_disc /(2*a) 5.If disc >= 0 then root_1 = e + f, root_2 = e – f 6.else root_1 = e + f i, root_2 = e – f i Algorithm 2 1.Get a, b, c 2.Calculate the discriminant: disc = pow(b,2) – 4*a*c; 3.Calculate the square root of discriminant: sqrt_disc = sqrt(disc); 4.Calculate root_1 = (-b +sqrt_disc)/(2*a) 5.Calculate root_2 = (-b – sqrt_disc)/(2*a) 6.Display root_1, root_2 Algorithm 4 1.Get a, b, c 2.If a == 0, print the first coefficient must not be 0 3.else 4.disc = pow(b,2) – 4*a*c; 5.sqrt_disc = sqrt(labs(disc)); 6.e = -b / (2*a) 7.f = sqrt_disc /(2*a) 8.If disc >= 0 then root_1 = e + f, root_2 = e – f 9.else root_1 = e + f i, root_2 = e – f I See implementation

4. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 4 Example 2: find if a year is a leap year Analysis input: a positive integer year output: Yes if it is a leap year, and No otherwise relation: a year is a leap year if and only if it is divided by 4 and not by 100 except it is divided by 400. 1900? 2000? 2004? Algorithm 1 1.get a positive integer, assign it to year 2.If year % 4 ==0 and year % 100 == 0 and year % 400 == 0, then leap = 1 3.If year % 4 ==0 and year % 100 == 0 and year % 400 != 0, then leap = 0 4.If year % 4 ==0 and year % 100 != 0, then leap = 1 5.If year % 4 !=0, then leap = 0 6.Output

5. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 5 Implementation /* leap year tester implementation for algorithm 1 */ #include main(void) { int year, leap; /* input a year */ printf("Input a year>\n"); scanf("%d", &year); /* test if the year is leap */ if (year%4 == 0 && year%100 == 0 && year%400 == 0) leap = 1; if (year%4 == 0 && year%100 == 0 && year%400 != 0) leap = 0; if (year%4 == 0 && year%100 != 0) leap = 1; if (year%4 != 0) leap = 0; if (leap) printf("%d is a leap year\n", year); else printf("%d is not a leap year", year); } In this algorithm, more time is spent on condition expression evaluation Time can be saved by using the simple condition and expression and nested if statement

6. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 6 Refined Algorithm Using Nested if Algorithm 2 1.Get a positive integer year 2.If year % 4 ==0 3. if year % 100 == 0 4. if year % 400 == 0 5. leap = 1 6. else leap = 0 7. else leap = 1 8.else leap = 0 9.Output Nested if statements if ( condition) { if (condition) Statements else Statements } else statements

7. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 7 Implementation Using Nested if #include main(void) { int year, leap; printf("Input a year>\n"); scanf("%d", &year); if (year%4 == 0) { if (year%100 == 0) { if (year%400 == 0) leap = 1; else leap = 0; } else leap = 1; } else leap = 0; if (leap) printf("%d is a leap year\n", year); else printf("%d is not a leap year", year); } year%4 == 0 leap = 0 Output leap year%100 == 0 year%400 == 0 leap = 1 leap = 0leap = 1 FT FT FT Input year

8. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 8 Case Study: Program for Water Bill Problem

9. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 9 Program for Water Bill Problem (cont’d)

10. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 10 Program for Water Bill Problem (cont’d)

11. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 11 Program for Water Bill Problem (cont’d)

12. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 12 Example 3: Percentage Grade to Letter Grade Convert Problem: covert a percentage grade to a letter grade according the table: Analysis input: an integer between 0 and 100 output: a letter grade Algorithm 1.get a number grade 2.If 100>=grade >=90 then assign letter_grade A 3.else if 89>=grade >=80 then assign letter_grade B 4.else if 79>=grade >=70 then assign letter_grade C 5.else if 69>=grade >=50 then assign letter_grade D 6.else 49>=grade >=0 then assign letter_grade F 7.Print the letter_grade 100~9089~8079~7069-5049~0 ABCDF

13. CP104 Introduction to Programming Selection structures_2 Lecture 10 __ 13 Implementation for Letter Grade Convert with Nested if Statements #include main(void) { int grade; char letterGrade; /* input a percentage grade */ scanf(“Input a percentage grade: %d", &grade); /* find the corresonding letter grade */ if (100 >=grade && grade >=90) letterGrade = ‘A’; else if (89 >=grade && grade >=80) letterGrade = ‘B’; else if (79 >=grade && grade >=70) letterGrade = ‘C’; else if (69 >=grade && grade >=50) letterGrade = ‘D’; else letterGrade = ‘F’; /* output the result */ printf(“The corresponding letter grade is %c.”, letterGrade); } Syntax form for nested if if ( condition) Compound Statements; else if Compound statements; else if Compound statements; … else Compound statements;