Logic To Find Sum of All Even Numbers From 1 To N, using For loop
Since we are checking for even numbers from 1 to user entered number, we assign value of variable count to 1. For loop keeps iterating until value of count is less than or equal to value of user input number. For each iteration of for loop we increment the value of count by 1.
Inside for loop we check for the condition, count%2 == 0. If it’s true, then we add the value present inside variable count to previous value present in variable sum.
After the control exits for loop we display the value present in variable sum – which has the sum of all the even numbers from 1 to user entered number.
Source Code: C Program To Find Sum of All Even Numbers From 1 To N, using For loop
#include<stdio.h>
int main()
{
int count, num, sum = 0;
printf("Enter the limit\n");
scanf("%d", &num);
printf("Even numbers from 1 To %d are:\n", num);
for(count = 1; count <= num; count++)
{
if(count % 2 == 0)
{
printf("%d\n", count);
sum = sum + count;
}
}
printf("Sum of even numbers from 1 To %d is %d\n", num, sum);
return 0;
}
Output 1: Enter the limit 5 Even numbers from 1 To 5 are: 2 4 Sum of even numbers from 1 To 5 is 6
Output 2: Enter the limit 10 Even numbers from 1 To 10 are: 2 4 6 8 10 Sum of even numbers from 1 To 10 is 30
We ask the user to enter values for floating point variables p, n, r and q. We use the formula:
a = p (1 + r / q) nq
And display the result to the console window.
where, p – principal amount; n – number of years. r – rate of interest; q – number of times interest compounds per year.
Note: We need to convert the user entered interest into percentage by dividing the integer number by 100.
Source Code: C Program To Calculate Amount In Compound Interest
#include<stdio.h>
#include<math.h>
int main()
{
float p, n, r, q, a;
int count;
for(count = 1; count <= 10; count++)
{
printf("Enter principal amount\n");
scanf("%f", &p);
printf("Enter number of years\n");
scanf("%f", &n);
printf("Enter rate of interest\n");
scanf("%f", &r);
r = r / 100;
printf("Enter no of times you compound per year\n");
scanf("%f", &q);
a = p * pow( (1 + (r/q)), n * q );
printf("Compounded amount is %f\n\n", a);
}
return 0;
}
Output Enter principal amount 5000 Enter number of years 2 Enter rate of interest 8 Enter no of times you compound per year 4 Compounded amount is 5858.296875
Enter principal amount 1000 Enter number of years 7 Enter rate of interest 5 Enter no of times you compound per year 1 Compounded amount is 1407.100464
Enter principal amount 2000 Enter number of years 5 Enter rate of interest 12 Enter no of times you compound per year 12 Compounded amount is 3633.393311
Enter principal amount 5000 Enter number of years 5 Enter rate of interest 5 Enter no of times you compound per year 5 Compounded amount is 6412.160156
Enter principal amount
Note: Above program keeps on iterating 10 times and asks the input 10 times.
In above c program, we ask the user to input 2 integer value and store it in variables start and end. If value of start is greater than the value of end, then we swap the values.
For loop counter is initialized to start, and for loop executes until value of count is less than or equal to end. For each iteration of the for loop, count value increments by 1.
Inside for loop, for every value of count, we check if its not perfectly divisible by 2. If true, it’s a Odd number and we output that number to the console window.
Source Code: C Program To Find Odd Numbers Between Range using For Loop
#include<stdio.h>
int main()
{
int start, end, temp, count;
printf("Enter start and end value, to find odd numbers\n");
scanf("%d%d", &start, &end);
if(start > end)
{
temp = start;
start = end;
end = temp;
}
printf("Odd numbers between %d and %d are\n", start, end);
for(count = start; count <= end; count++)
{
if(count % 2 != 0)
printf("%d\n", count);
}
return 0;
}
Output 1 Enter start and end value, to find odd numbers 40 60 Odd numbers between 40 and 60 are 41 43 45 47 49 51 53 55 57 59
Output 2 Enter start and end value, to find odd numbers 60 40 Odd numbers between 40 and 60 are 41 43 45 47 49 51 53 55 57 59
In above c program, we ask the user to input 2 integer value and store it in variables start and end. For loop counter is initialized to start and for loop executes until loop counter is less than or equal to value of end. For each iteration of the for loop, loop counter value increments by 1.
Inside for loop, for every value of count, we check if its perfectly divisible by 2. If true, it’s a Even number and we output that number to the console window.
Source Code: C Program to Find Even Numbers Between Range using For Loop
#include<stdio.h>
int main()
{
int start, end, count, temp;
printf("Enter start value and end value to generate Even no's\n");
scanf("%d%d", &start, &end);
if(start > end)
{
temp = start;
start = end;
end = temp;
}
printf("Even numbers between %d and %d are:\n", start, end);
for(count = start; count <= end; count++)
{
if(count % 2 == 0)
{
printf("%d\n", count);
}
}
return 0;
}
Output 1 Enter start value and end value to generate Even no’s 10 20 Even numbers between 10 and 20 are: 10 12 14 16 18 20
Output 2 Enter start value and end value to generate Even no’s 50 40 Even numbers between 40 and 50 are: 40 42 44 46 48 50
According to a study, the approximate level of intelligence of a person can be calculated using the following formula:
i = 2 + ( y + 0.5x )
Write a C program that will produce a table of values of i, y and x, where y varies from 1 to 6, and, for each value of y, x varies from 5.5 to 12.5 in steps 0.5.
1. Problem statement snippet: “..y varies from 1 to 6”. That means, we can accomplish it using a looping statement. So we can take a for loop and initialize variable y to 1 and iterate the for loop until y is less than or equal to 6.
2. Problem statement snippet: “..for each value of y, x varies from”. That means we should write a nested for loop here.
3. Problem statement snippet: “..for each value of y, x varies from 5.5 to 12.5 in steps 0.5”. That means, we need to write a nested for loop and initialize x value to 5.5 and iterate through the for loop until x value is less than or equal to 12.5 And for each iteration of the inner for loop we need to increment the value of x by 0.5
Video Tutorial: C Program To Calculate Approximate Level of Intelligence of a Person
Source Code: C Program To Calculate Approximate Level of Intelligence of a Person
#include<stdio.h>
int main()
{
int y, count = 1;
float i, x;
for(y = 1; y <= 6; y++)
{
for(x = 5.5; x <= 12.5; x += 0.5)
{
i = 2 + (y + 0.5 * x);
printf("%d. i = %0.2f,\t x = %0.2f,\t y = %d\n", count++,i, x, y);
}
}
return 0;
}
Output:
1. i = 5.75, x = 5.50, y = 1
2. i = 6.00, x = 6.00, y = 1
3. i = 6.25, x = 6.50, y = 1
4. i = 6.50, x = 7.00, y = 1
5. i = 6.75, x = 7.50, y = 1
6. i = 7.00, x = 8.00, y = 1
7. i = 7.25, x = 8.50, y = 1
8. i = 7.50, x = 9.00, y = 1
9. i = 7.75, x = 9.50, y = 1
10. i = 8.00, x = 10.00, y = 1
11. i = 8.25, x = 10.50, y = 1
12. i = 8.50, x = 11.00, y = 1
13. i = 8.75, x = 11.50, y = 1
14. i = 9.00, x = 12.00, y = 1
15. i = 9.25, x = 12.50, y = 1
16. i = 6.75, x = 5.50, y = 2
17. i = 7.00, x = 6.00, y = 2
18. i = 7.25, x = 6.50, y = 2
19. i = 7.50, x = 7.00, y = 2
20. i = 7.75, x = 7.50, y = 2
21. i = 8.00, x = 8.00, y = 2
22. i = 8.25, x = 8.50, y = 2
23. i = 8.50, x = 9.00, y = 2
24. i = 8.75, x = 9.50, y = 2
25. i = 9.00, x = 10.00, y = 2
26. i = 9.25, x = 10.50, y = 2
27. i = 9.50, x = 11.00, y = 2
28. i = 9.75, x = 11.50, y = 2
29. i = 10.00, x = 12.00, y = 2
30. i = 10.25, x = 12.50, y = 2
31. i = 7.75, x = 5.50, y = 3
32. i = 8.00, x = 6.00, y = 3
33. i = 8.25, x = 6.50, y = 3
34. i = 8.50, x = 7.00, y = 3
35. i = 8.75, x = 7.50, y = 3
36. i = 9.00, x = 8.00, y = 3
37. i = 9.25, x = 8.50, y = 3
38. i = 9.50, x = 9.00, y = 3
39. i = 9.75, x = 9.50, y = 3
40. i = 10.00, x = 10.00, y = 3
41. i = 10.25, x = 10.50, y = 3
42. i = 10.50, x = 11.00, y = 3
43. i = 10.75, x = 11.50, y = 3
44. i = 11.00, x = 12.00, y = 3
45. i = 11.25, x = 12.50, y = 3
46. i = 8.75, x = 5.50, y = 4
47. i = 9.00, x = 6.00, y = 4
48. i = 9.25, x = 6.50, y = 4
49. i = 9.50, x = 7.00, y = 4
50. i = 9.75, x = 7.50, y = 4
51. i = 10.00, x = 8.00, y = 4
52. i = 10.25, x = 8.50, y = 4
53. i = 10.50, x = 9.00, y = 4
54. i = 10.75, x = 9.50, y = 4
55. i = 11.00, x = 10.00, y = 4
56. i = 11.25, x = 10.50, y = 4
57. i = 11.50, x = 11.00, y = 4
58. i = 11.75, x = 11.50, y = 4
59. i = 12.00, x = 12.00, y = 4
60. i = 12.25, x = 12.50, y = 4
61. i = 9.75, x = 5.50, y = 5
62. i = 10.00, x = 6.00, y = 5
63. i = 10.25, x = 6.50, y = 5
64. i = 10.50, x = 7.00, y = 5
65. i = 10.75, x = 7.50, y = 5
66. i = 11.00, x = 8.00, y = 5
67. i = 11.25, x = 8.50, y = 5
68. i = 11.50, x = 9.00, y = 5
69. i = 11.75, x = 9.50, y = 5
70. i = 12.00, x = 10.00, y = 5
71. i = 12.25, x = 10.50, y = 5
72. i = 12.50, x = 11.00, y = 5
73. i = 12.75, x = 11.50, y = 5
74. i = 13.00, x = 12.00, y = 5
75. i = 13.25, x = 12.50, y = 5
76. i = 10.75, x = 5.50, y = 6
77. i = 11.00, x = 6.00, y = 6
78. i = 11.25, x = 6.50, y = 6
79. i = 11.50, x = 7.00, y = 6
80. i = 11.75, x = 7.50, y = 6
81. i = 12.00, x = 8.00, y = 6
82. i = 12.25, x = 8.50, y = 6
83. i = 12.50, x = 9.00, y = 6
84. i = 12.75, x = 9.50, y = 6
85. i = 13.00, x = 10.00, y = 6
86. i = 13.25, x = 10.50, y = 6
87. i = 13.50, x = 11.00, y = 6
88. i = 13.75, x = 11.50, y = 6
89. i = 14.00, x = 12.00, y = 6
90. i = 14.25, x = 12.50, y = 6
Note: Inner for loop executes from 5.5 to 12.5 with a 0.5 step increment for each loop. So its 15 iterations. Outer for loop executes 6 times. So total iterations equals 15 x 6 = 90 times. So 90 values will be printed to the console window.