Tutorial_10
ICT/2022/139
R.S.R.Ranathunga
Print the following shapes using loop construct of C programming.
*
* *
* * *
* * * *
* * * * *
* * * * * *
#include <stdio.h>
int main(){
int i,x;
for(i=1;i<=6;i++){
for(x=1;x<=i;x++){
printf("*");
}
printf("\n");
}
return 0;
}
* * * * * * *
* * * * * * *
* * * * * * *
* * * * * * *
* * * * * * *
#include <stdio.h>
int main(){
int i,x;
for(i=1;i<=5;i++){
for(x=1;x<=7;x++){
printf("*");
}
printf("\n");
}
return 0;
}
* * * * * *
* * * * *
* * * *
* * *
* *
*
#include <stdio.h>
int main(){
int i,x;
for(i=6;i>=1;i--){
for(x=1;x<=i;x++){
printf("*");
}
printf("\n");
}
return 0;
}
*
* *
* * *
* * * *
* * * * *
* * * * * *
#include <stdio.h>
int main() {
int i, j, k;
for (i = 6; i >= 1; i--) {
for (j = 1; j <= i; j++) {
printf(" ");
}
for (k = 0; k <=6 - i; k++) {
printf("*");
}
printf("\n");
}
return 0;
}
* * * * * *
* * * * *
* * * *
* * *
* *
*
#include <stdio.h>
int main() {
int i, j;
for (i = 6; i >= 1; i--) {
for (j = 1; j <= 6 - i; j++) {
printf(" ");
}
for (j = 1; j <= i; j++) {
printf("* ");
}
printf("\n");
}
return 0;
}
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
#include <stdio.h>
int main(){
int i,x;
for(i=1;i<=5;i++){
for(x=1;x<=i;x++){
printf("%i", x);
}
printf("\n");
}
return 0;
}
3. Find the minimum and maximum of sequence of 10 numbers.
#include<stdio.h>
int main(){
int max, min, num1, num2;
int i;
printf("Enter first number from the sequence: ");
scanf("%i", &num1);
min=num1;
max=num1;
for(i=1; i<=9; i++){
printf("Enter next number from the sequence: ");
scanf("%i", &num2);
if(min > num2){
min=num2;
continue;}
else if(max < num2){
max=num2;
continue;}
}
printf("Minimum number is %i\n", min);
printf("Maximum number is %i", max);
return 0;
}
4. Find the total and average of a sequence of 10 numbers.
#include<stdio.h>
int main(){
int total=0, num1;
float avg=0;
int i;
for(i=1; i<=10; i++){
printf("Enter the number from the sequence: ");
scanf("%i", &num1);
total+=num1;
}
avg=(float)total/10;
printf("Total of the sequence= %i\n", total);
printf("Average of the sequence= %.2f", avg);
return 0;
}
5. Write a program to generate and display a table of n and n
2
, for integer values of n
ranging from 1 to 10. Be certain to print appropriate column headings.
#include <stdio.h>
int main(){
int n,c=0, x=10;
printf(" ________________________________\n");
printf(" |\tn\t|\tn^2\t|\n");
printf(" |______________|_______________|\n");
for(n=1;n<=x;n++){
c=n*n;
printf(" |\t%i\t|\t%i\t|\n", n,c);
}
printf(" |______________|_______________|\n");
return 0;
}
6. A triangular number can also be generated by the formula
triangularNumber = n (n + 1) / 2
for any integer value of n. For example, the 10th triangular number, 55, can be generated
by substituting 10 as the value for n in the preceding formula. Write a program that
generates a table of triangular numbers using the preceding formula.
#include <stdio.h>
int main(){
int n,Tn=0, i;
printf("Enter number count: ");
scanf("%i", &i);
printf("\n\tDisplay Tringular number list\n");
printf(" ________________________________________________\n");
printf(" |\tNumber\t|\tTriangular Number\t|\n");
printf(" |______________|_______________________________|\n");
for(n=1;n<=i;n++){
Tn=n*(n+1)/2;
printf(" |\t%i\t|\t\t%i\t\t|\n", n,Tn);
}
printf(" |______________|_______________________________|\n");
return 0;
}
7. The factorial of an integer n, written n!, is the product of the consecutive integers 1
through n. For example, 5 factorial is calculated as
5! = 5 x 4 x 3 x 2 x 1 = 120
#include <stdio.h>
int main(){
int i, n, fact=1;
printf("Enter number:");
scanf("%i", &i);
for(n=i; n>=1; n--){
fact=fact*n;
}
printf("Factorial of number %i is %i", i,fact);
return 0;
}
8. Write a program to generate and print a table of the first 10 factorials.
#include <stdio.h>
int main(){
int i,a=10, n, fact=1;
printf("\n\tDisplay first 10 factorial number list\n");
printf(" _______________________________________________\n");
printf(" |\tNumber\t|\tFactorial Number\t|\n");
printf(" |_____________|_______________________________|\n");
for(i=1; i<=a; i++){
for(n=i; n>=1; n--){
fact=fact*n;
}
printf(" |\t%i\t|\t\t%i\t\t|\n", i, fact);
printf(" |_____________|_______________________________|\n");
fact=1;
}
return 0;
}
9. Display the n terms of harmonic series and their sum.
1 + 1/2 + 1/3 + 1/4 + 1/5 ... 1/n terms
Test Data :
Input the number of terms : 5
Expected Output :
1/1 + 1/2 + 1/3 + 1/4 + 1/5 +
Sum of Series upto 5 terms : 2.283334
#include<stdio.h>
int main(){
int n,i=5;
float a=1, sum=0;
printf("Expected Output:\n");
for(n=1; n<=i; n++){
printf(" 1/%i +", n);
sum=sum+(a/n);
}
printf("\nSum of Series upto 5 terms : %.6f", sum);
return 0;
}
10. Write a program to generate students repot as shown below.
Index Number Name Mathes Physics Chemistry Total Average Grade
________________________________________________________________
In this program, for each student you have to input marks of three subjects. Students’
grade is determined according to following criteria:
a. Average >= 90% : Grade A
b. Average >= 80% : Grade B
c. Average >= 70% : Grade C
d. Average >= 60% : Grade D
e. Average >= 40% : Grade E
f. Average < 40% : Grade F
#include <stdio.h>
#include <string.h>
int main(){
int index, marks1, marks2, marks3;
int total = 0;
float avg = 0;
char name_1[]="Kumar";
char name_2[]="Mahela";
char name_3[]="Angelo";
char name_4[]="Dilshan";
char name_5[]="Invalid";
char temp_name[50];
int index_1=11, index_2=27, index_3=69, index_4=23;
printf("Enter Index number: ");
scanf("%i", &index);
if(index==11){
strcpy(temp_name, name_1);
}
else if(index==27){
strcpy(temp_name, name_2);
}
else if(index==69){
strcpy(temp_name, name_3);
}
else if(index==23){
strcpy(temp_name, name_4);
}
else{
strcpy(temp_name, name_5);
}
printf("Enter Maths subject marks: ");
scanf("%i", &marks1);
printf("Enter Physics subject marks: ");
scanf("%i", &marks2);
printf("Enter Chemistry subject marks: ");
scanf("%i", &marks3);
printf("\n");
printf(" Index Number \tName\t Mathes Physics Chemistry Total Average Grade \n");
printf(" ______________________________________________________________________________\n");
total = marks1 + marks2 + marks3 ;
avg = total/3;
avg = (avg/100)*100;
if(avg>=90)
printf("\n %i %s %i %i %i %i %.2f A\n", index, temp_name, marks1, marks2, marks3, total, avg);
else if(avg>=80)
printf("\n %i %s %i %i %i %i %.2f B\n", index, temp_name, marks1, marks2, marks3, total, avg);
else if(avg>=70)
printf("\n %i %s %i %i %i %i %.2f C\n", index, temp_name, marks1, marks2, marks3, total, avg);
else if(avg>=60)
printf("\n %i %s %i %i %i %i %.2f D\n", index, temp_name, marks1, marks2, marks3, total, avg);
else if(avg>=40)
printf("\n %i %s %i %i %i %i %.2f E\n", index, temp_name, marks1, marks2, marks3, total, avg);
else if(avg<40)
printf("\n %i %s %i %i %i %i %.2f F\n", index, temp_name, marks1, marks2, marks3, total, avg);
return 0;
}
11. Assume a village has 20 houses. Input electricity unit charges and calculate total
electricity bill according to the following criteria:
For first 50 units Rs. 0.50/unit
For next 100 units Rs. 0.75/unit
For next 100 units Rs. 1.20/unit
For unit above 250 Rs. 1.50/unit
An additional surcharge of 20% is added to the bill
3
The output should be as follows:
Serial Number: House Address Units Surcharge Amount to be paid
#include <stdio.h>
int main() {
int i, count;
float bill = 0, surcharge = 0, total_bill = 0;
char name_1[20];
int unit_charges[20];
int serial_count[20];
for (i = 0; i < 20; i++) {
printf("Enter serial number: ");
scanf("%i", &serial_count[i]);
printf("Enter unit charges %i: ", i + 1);
scanf("%i", &unit_charges[i]);
printf("Enter house address:: ");
scanf("%s", &name_1[i]);
}
printf("\n");
printf("\tSerial Number: House Address Units Surcharge Amount to be paid\n ");
printf("\t________________________________________________________________________\n");
for (count = 0; count < 20; count++) {
if (unit_charges[count] <= 50) {
bill = unit_charges[count] * 0.50;
} else if (unit_charges[count] <= 150) {
bill = (unit_charges[count] - 50) * 0.75 + 25;
} else if (unit_charges[count] <= 250) {
bill = (unit_charges[count] - 150) * 1.20 + 100;
} else if (unit_charges[count] > 250) {
bill = (unit_charges[count] - 250) * 1.50 + 220;
}
surcharge = bill * 0.20;
total_bill = bill + surcharge;
printf("\n\t %i %s %i %.2f %.2f \n", serial_count[count], name_1[count], unit_charges[count], surcharge, total_bill);
}
return 0;
}
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