CS1010E Programming MethodologyTutorial 3

Control Structures and Data Files

Question 1 (a) (b)

1. What is output ?

2. Correct the code to print [0, input)!

1. #include <stdio.h> 2. int main(void) { 3. int i, input; 4. scanf(“%d”, &input)5. while(i < input) 6. printf("%d\n", i); 7. i++; 8. return 0; 9. }

1. #include <stdio.h> 2. int main(void) { 3. int i = 0, input; 4. scanf(“%d”, &input);5. while(i < input) {6. printf("%d\n", i); 7. i++; 8. }9. return 0; 10.}

Question 1 (c)

Change to Do-while Loop

Which one is correct ?

1. #include <stdio.h> 2. int main(void) { 3. int i = 0, input; 4. scanf(“%d”, &input);5. while(i < input) {6. printf("%d\n", i); 7. i++; 8. }9. return 0; 10. }

int i = 0, input;scanf(“%d”, &input);do { printf("%d\n", i); i++;} while(i < input);

int i = 0, input;scanf(“%d”, &input);do { printf("%d\n", i);} while(i++ < input);

int i = 0, input;scanf(“%d”, &input);if ( i < input) {

do { printf("%d\n", i);

i++;} while(i < input);

}

√

Question 1 (d)

Rewrite using for-loop

1. #include <stdio.h> 2. int main(void) { 3. int i = 0, input; 4. scanf(“%d”, &input);5. while(i < input) {6. printf("%d\n", i); 7. i++; 8. }9. return 0; 10. }

for(i = 0;i < input; i++) { printf("%d\n", i);

}

Question 1 (d)

• Three loops are equivalent !!

• Why we have three loops instead of one ?!!• For convenience • While - used when you know stopping condition• For - used when you know the number of iterations

while(i < input){ printf("%d\n", i); i++; }

if ( i < input){ do{ printf("%d\n", i); i++; } while(i < input); }

for(i = 0;i < input; i++){ printf("%d\n", i);

}

Question 1 (e)

Print Out all the odd number from [0, input)• How to test parity of a number ?• num % 2 != 0

for(i = 0;i < input; i++) { printf("%d\n", i);

}

for(i = 0;i < input; i++){ if(i%2 != 0) {

printf("%d\n", i);

}}

Can you use num%2 == 1 for odd number test?

What is wrong here?

Question 1 (f)

Print Out all the odd number divisible by 5 from [0, input)• How to test such condition ?• num is a not multiple of 2 AND num is multiple of 5

for(i = 0;i < input; i++) { printf("%d\n", i);

}

for(i = 0;i < input; i++){ if(i%2 != 0 && i%5 == 0){

printf("%d\n", i); }

}

Note the difference between“&&” and “&”

Question 2

Iterate through a table (matrix)• Standard trick is to use a Nested-For-Loop

“i” indicate row number“j” indicate column number

i =1

j =1 j =3

i =2

j =2

for(i=1; i<=input; i++){ for(j=1; j<=input; j++){

do sth… }}

(1,1,)->(1,2)->(1,3)-> … -> (2,1) -> (2,2) ->(2,3)-> ….->….

Column First Order

Question 2 (a)

Write a program to take in an integer from 1 to 10 and print a multiplication table of that number.

• Analyze:

1.For every cell, cell value = row number * column number

for(i=1; i<=input; i++){ for(j=1; j<=input; j++){ printf("%4d", i * j); } printf(“\n”); }

for(i=1; i<=input; i++){ for(j=1; j<=input; j++){

do sth… }}

“i” indicate row number“j” indicate column number

End of each row, we need a newline

Question 2 (b)

Only the bottom triangle is obtained. • Analyze:

1.For every cell, cell value = row number * column number

2.Bottom triangle ≡ row number >= column number

for(i=1; i<=input; i++){ for(j=1; j<=i; j++){ printf("%4d", i * j); } printf(“\n”); }

for(i=1; i<=input; i++){ for(j=1; j<=input; j++){

do sth… }}

“i” indicate row number“j” indicate column number

Indicate row number >= column number

How to print out Top triangle ?How to print out Pascal's triangle ?

Question 2 (c)

Only the bottom triangle is obtained. Square Number -> ‘-’

Prime Number -> ‘p’• Analyze:

1.For every cell, cell value = row number * column number

2.Bottom triangle ≡ row number >= column number

3.We need to test the cell value:1.Square number

2.Prime number

for(i=1; i<=input; i++){ for(j=1; j<=i; j++){

cell = i *j;if(isPrime) { printf(“p”);} else if (isSquare) { printf(“-”);} else {

printf("%4d", cell); }

} printf(“\n”); }

Question 2 (c)

Test cell for prime

1. If cell is 1, it is not prime

2. Else If cell is divided by any number in [2, itself-1], it is not prime

3. Else it is primeint isPrime = 1; if(cell == 1) isPrime = 0; else{ for(k=2; k<= cell - 1; k++){ if(cell % k == 0) isPrime = 0; } }

Assume cell is prime

Find all cases where cell is not prime

Question 2 (c)

Test cell for square

1. If exists a number k in [1, cell), s.t. k * k == cell then cell is square

isSquare = 0; for(k=1; k<=cell; k++) {

if(k * k == cell) isSquare = 1;

}

Assume cell is not Square

Try to find a k

Question 2 (c)

Put all these together.int i, j, k, cell, input, isSquare, isPrime; scanf(“%d”, &input) for(i=1; i<=input; i++){ for(j=1; j<=i; j++){ cell = i * j; /* Square number check */ isSquare = 0; for(k=1; k<=cell; k++){ if(k * k == cell)

isSquare = 1; }

/* Prime number check */ isPrime = 1; if(cell == 1)

isPrime = 0; for(k=2; k<cell; k++){

if(cell % k == 0) isPrimeNumber = 0;

} /* Choose what to print */ if(isPrime == 1) printf(“ p”); else if(isSquare == 1) printf(“ -”); else printf(“%4d”, i*j);

} printf(“\n”); }

Can you improve your code to run faster ?

Question 3

Solution to Diophantine Equation• In mathematics, we build a linear system and solve it

using Guassian’s Elimination.• In Computer Science, we can do brute force using power

of computers.•Given a set of equations of X,Y. Test all the possible combinations of <X,Y> to find a solution

• Analyze:

1.Read input from file

2.Test all possible values of <X,Y>

3.Output to file

Question 3Read Input From File:

#define NOSOLUTION -99999FILE *fin, *fout;fin = fopen(“input.txt”, “r”);

int x, y, solutionX, solutionY, a1, a2, b1, b2, c1, c2;

fscanf(fin, “%d %d %d”, &a1, &b1, &c1);fscanf(fin, “%d %d %d”, &a2, &b2, &c2);

Will a1 b1 c1 has the same value of a2 b2 c2?

Other opening mode :w writea appendr+ read & write…

“w” will erase all the content of original file

Be careful when read and write on the same file!!!

Question 3

Test all combinations of possible X, Y’s value

Since we have two variable, we use nested loop

solutionX = NOSOLUTION; solutionY = NOSOLUTION; for(x=-100; x<=100; x++) { for(y=-100; y<=100; y++) { if(a1*x + b1*y == c1 && a2*x + b2 * y == c2) {

solutionX = x; solutionY = y;

} } }

Can we insert a break here ? solutionX = x; solutionY = y; break;

Question 3

Output to File Accordingly:

fout = fopen(“ouptut.txt”, “w”);if(solutionX != NOSOLUTION) fprintf(fout,“X = %d, Y = %d”, solutionX, solutionY); else fprintf(fout,“Integer solution not found in range [-100,100]”); fclose(fout);fclose(in);return 0;

If not closed, the result is not guaranteed to be written into file !

Thank You !

See you next week!