Cse115 lecture08repetitionstructures part02

Lecture 08
Repetition Structures
CSE115: Computing Concepts

Example
• Calculate the sum of the following series (𝑥 and 𝑚 are
user inputs).
𝑥0 + 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥 𝑚

Example
• Calculate the sum of the following series (𝑥 and 𝑚 are
user inputs).
𝑥0 + 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥 𝑚
#include <stdio.h>
int main()
{
int x, m, term = 1, sum = 1;
scanf("%d%d", &x, &m);
for(i=1; i<=m; i++)
{
term = term * x;
sum = sum + term;
}
printf("Sum = %lf", sum);
return 0;
}

•Read a positive integer and determine if it is
a prime number.
•Pseudo-code:
• Read integer and store it in number
• Set flag to 1
• For i = 2 to (number-1)
• If number is divisible by i then
• Set flag to 0
• If flag equals 1, then the number is a prime
number, otherwise not
Selection Inside Loop

•Read a positive integer and determine if it is
a prime number.
Selection Inside Loop
#include <stdio.h>
int main()
{
int number, i, flag = 1;
scanf("%d", &number);
for(i=2; i<number; i++)
{
if(number % i == 0)
flag = 0;
}
if(flag == 1)
printf("%d is a prime number",number);
else printf("%d is not a prime number",number);
return 0;
}

Example
• Calculate the sum of the following series (𝑛 is user input).
1
1
−
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+
1
7
− ⋯ ±
1
𝑛

Example
• Calculate the sum of the following series (𝑛 is user input).
1
1
−
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+
1
7
− ⋯ ±
1
𝑛
#include <stdio.h>
int main()
{
int n, i;
double term, sum = 0;
scanf("%d", &n);
for(i=1; i<=n; i++)
{
term = 1.0 / i;
if(i%2 == 0)
sum = sum - term;
else
sum = sum + term;
}
printf("Sum = %lf", sum);
return 0;
}

Using break Inside Loop
• In the prime number example, we do not need to
continue the loop till the end once the value of flag is set
to zero.
#include <stdio.h>
int main()
{
int number, i, flag = 1;
for(i=2; i<number; i++)
{
if(number % i == 0)
{
flag = 0;
break;
}
}
if(flag == 1)
printf("%d is a prime number",number);
else printf("%d is not a prime number",number);
return 0;
}
The break statement makes
the loop terminate
prematurely.

• Read 10 integers from the user and calculate the sum of the
positive numbers.
Using continue Inside Loop
#include <stdio.h>
int main()
{
int number, i, sum = 0;
for(i=0; i<10; i++)
{
printf("Enter a number: ");
if(number < 0)
continue;
sum = sum + number;
printf("%d is addedn", number);
}
printf("Total = %d",sum);
return 0;
}
The continue
statement forces next
iteration of the loop,
skipping any remaining
statements in the loop

• Read 10 integers from the user and calculate the sum of the
positive numbers.
Using continue Inside Loop
#include <stdio.h>
int main()
{
int number, i, sum = 0;
for(i=0; i<10; i++)
{
printf("Enter a number: ");
if(number < 0)
continue;
sum = sum + number;
printf("%d is addedn", number);
}
printf("Total = %d",sum);
return 0;
}
Output:
Enter a number: 1
1 is added
Enter a number: 2
2 is added
Enter a number: 3
3 is added
Enter a number: -4
Enter a number: -5
Enter a number: 6
6 is added
Enter a number: 7
7 is added
Enter a number: 8
8 is added
Enter a number: -9
Enter a number: 10
10 is added
Total = 37

• Suppose a man (say, A) stands at (0, 0) and waits for user to give
him the direction and distance to go.
• User may enter N E W S for north, east, west, south, and any value
for distance.
• When user enters 0 as direction, stop and print out the location
where the man stopped
N
E
S
W
Example: A Travelling Man

float x=0, y=0;
char dir;
float mile;
while (1) {
printf("Please input the direction as N,S,E,W (0 to exit): ");
scanf("%c", &dir); fflush(stdin);
if (dir=='0'){ /*stop input, get out of the loop */
break;
}
if (dir!='N' && dir!='S' && dir!='E' && dir!='W') {
printf("Invalid direction, re-enter n");
continue;
}
printf("Please input the mile in %c direction: ", dir);
scanf ("%f",&mile); fflush(stdin);
if (dir == 'N'){ /*in north, compute the y*/
y+=mile;
} else if (dir == 'E'){ /*in east, compute the x*/
x+=mile;
} else if (dir == 'W'){ /*in west, compute the x*/
x-=mile;
} else if (dir == 'S'){ /*in south, compute the y*/
y-=mile;
}
}
printf("nCurrent position of A: (%4.2f,%4.2f)n",x,y); /* output

Nested Loop
• What is the output of the following program?
for (i=1; i<=5; i++)
{
for (j=1; j<=4; j++)
{
printf("*");
}
printf("n");
}

Nested Loop
for (i=1; i<=5; i++)
{
for (j=1; j<=4; j++)
{
printf("*");
}
printf("n");
}
Output
****
****
****
****
****

Nested Loop
for (i=1; i<=5; i++)
{
for (j=1; j<=i; j++)
{
printf("*");
}
printf("n");
}

Nested Loop
for (i=1; i<=5; i++)
{
for (j=1; j<=i; j++)
{
printf("*");
}
printf("n");
}
Output
*
**
***
****
*****

Nested Loop
• Write a program that generates the following pattern.
Output
*
++
***
++++
*****

Nested Loop
int i, j;
for(i=1; i<=5; i++)
{
for(j=1; j<=i; j++)
{
if (i % 2 == 0)
printf("+");
else
printf("*");
}
printf("n");
}
Output
*
++
***
++++
*****

Nested Loop
Output
*
**
***
****
*****

Nested Loop
for(i=1; i<=5; i++)
{
for(j=5; j>i; j--)
printf(" ");
for(j=1; j<=i; j++)
printf("*");
printf("n");
}
Output
*
**
***
****
*****

Home-works
1. Calculate the sum of the following series, where 𝑛 is
provided as user input.
1 + 2 + 3 + 4 + ⋯ + 𝑛
2. Write a program that calculates the factorial of a positive
integer n provided as user input.
3. Write a program that calculates 𝑎 𝑥, where 𝑎 and 𝑥 are
provided as user inputs.
4. Calculate the sum of the following series, where 𝑥 and 𝑛
is provided as user input.
1 +
𝑥
1!
+
𝑥2
2!
+
𝑥3
3!
+
𝑥4
4!
+ ⋯ +
𝑥 𝑛
𝑛!

Home-works
5. Write programs that generate the following patterns. In
each case, the number of lines is the input.
**********
**** ****
*** ***
** **
* *
** **
*** ***
**** ****
**********
*
* *
* *
* *
* *
* *
* *
* *
*
*
**
***
****
*****
****
***
**
*
*****
****
***
**
*
*
***
*****
*******
*********
*********
*******
*****
***
*

Cse115 lecture08repetitionstructures part02

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