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The document contains C program code snippets for solving various math and logic problems. These include programs to find roots of a quadratic equation, generate the Fibonacci series, find Armstrong numbers between intervals, check if a number is prime, and convert between binary, decimal and other number bases. It also includes programs to check for palindromes, calculate power of a number using a while loop, and convert decimal numbers to hexadecimal representation.
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For example we want to convert decimal number 900 in the hexadecimal.
Step 1: 900 / 16 Remainder : 4 , Quotient : 56
Step 2: 56 / 16 Remainder : 8 , Quotient : 3
Step 3: 3 / 16 Remainder : 3 , Quotient : 0
So equivalent hexadecimal number is: 384
That is (900)10 = (384)16
C Program
1. #include<stdio.h>
2. int main() {
3. long int decimalNumber,remainder,quotient;
4. int i=1,j,temp;
5. char hexadecimalNumber[100];
6. printf("Enter any decimal number: ");
7. scanf("%ld",&decimalNumber);
8. quotient = decimalNumber;
9. while(quotient!=0) {
10. temp = quotient % 16;
11. //To convert integer into character
12. if( temp < 10)
13. temp =temp + 48; else
14. temp = temp + 55;
15. hexadecimalNumber[i++]= temp;
16. quotient = quotient / 16;
17. }
18. printf("Equivalent hexadecimal value of decimal number %d: ",decimalNumbe
r);
19. for (j = i -1 ;j> 0;j--)
20. printf("%c",hexadecimalNumber[j]);
21. return 0;
22.}
Output
1. Enter any decimal number: 45
2. Equivalent hexadecimal value of decimal number 45: 2D
Program to Check Palindrome
#include <stdio.h>
int main() {](https://image.slidesharecdn.com/program-flowchart-201123135930/85/Program-flowchart-12-320.jpg)
Program flowchart
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- 5. 5 Program to Find Roots of a Quadratic Equation #include <math.h> #include <stdio.h> int main() { double a, b, c, discriminant, root1, root2, realPart,imagPart; printf("Enter coefficients a, b and c: "); scanf("%lf %lf %lf", &a,&b, &c); discriminant = b * b - 4 * a * c; // condition for real and different roots if (discriminant > 0) { root1 = (-b + sqrt(discriminant)) / (2 * a); root2 = (-b - sqrt(discriminant)) / (2 * a); printf("root1 = %.2lf and root2 = %.2lf", root1, root2); } // condition for real and equal roots else if (discriminant == 0) { root1 = root2 = -b / (2 * a); printf("root1 = root2 = %.2lf;", root1); } // if roots are not real else { realPart = -b / (2 * a); imagPart = sqrt(-discriminant) / (2 * a); printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", realPart,imagPart, realPart,imagPart); }
- 6. 6 return 0; } Fibonacci Series up to n terms #include <stdio.h> int main() { int i, n, t1 = 0, t2 = 1, nextTerm; printf("Enter the number of terms: "); scanf("%d", &n); printf("Fibonacci Series: "); for (i = 1; i <= n; ++i) { printf("%d, ", t1); nextTerm = t1 + t2; t1 = t2; t2 = nextTerm; } return 0; } Armstrong Numbers Between Two Integers #include <math.h> #include <stdio.h> int main() { int low, high, i, temp1, temp2, rem, n = 0; float result = 0.0; printf("Enter two numbers(intervals): "); scanf("%d %d", &low, &high); printf("Armstrong numbers between %d and %d are: ", low, high); for (i = low + 1; i < high; ++i) { temp2 = i; temp1 = i; // number of digits calculation while (temp1 != 0) { temp1 /= 10; ++n; } // result contains sum of nth power of its digits while (temp2 != 0) { rem = temp2 % 10; result += pow(rem, n); temp2 /= 10; } // check if i is equal to the sum of nth power of its digits if ((int)result == i) {
- 7. 7 printf("%d ", i); } // resetting the values n = 0; result = 0; } return 0; } Program to Check Prime Number #include <stdio.h> int main() { int n, i, flag = 0; printf("Enter a positive integer: "); scanf("%d", &n); for (i = 2; i <= n / 2; ++i) { // condition for non-prime if (n % i == 0) { flag = 1; break; } } if (n == 1) { printf("1 is neither prime nor composite."); } else { if (flag == 0) printf("%d is a prime number.", n); else printf("%d is not a prime number.", n); } return 0; } Display Prime Numbers Between two Intervals #include <stdio.h> int main() { int low, high, i, flag; printf("Enter two numbers(intervals): "); scanf("%d %d", &low, &high); printf("Prime numbers between %d and %d are: ", low, high); while (low < high) { flag = 0;
- 8. 8 // if low is a non-prime number, flag will be 1 for (i = 2; i <= low / 2; ++i) { if (low % i == 0) { flag = 1; break; } } if (flag == 0) printf("%d ", low); ++low; } return 0; } Program to convert binary to decimal #include <math.h> #include <stdio.h> int convert(long long n); int main() { long long n; printf("Enter a binary number: "); scanf("%lld", &n); printf("%lld in binary = %d in decimal", n, convert(n)); return 0; } int convert(long long n) { int dec = 0, i = 0, rem; while (n != 0) { rem = n % 10; n /= 10; dec += rem * pow(2, i); ++i; } return dec; } #include <math.h> #include <stdio.h> int convert(long long n); int main() { long long n; printf("Enter a binary number: "); scanf("%lld", &n); printf("%lld in binary = %d in decimal", n, convert(n)); return 0; } int convert(long long n) { int dec = 0, i = 0, rem;
- 9. 9 while (n != 0) { rem = n % 10; n /= 10; dec += rem * pow(2, i); ++i; } return dec; } Output Enter a binary number: 110110111 110110111 in binary = 439 Program to convert decimal to binary #include <math.h> #include <stdio.h> long long convert(int n); int main() { int n; printf("Enter a decimal number: "); scanf("%d", &n); printf("%d in decimal = %lld in binary", n, convert(n)); return 0; } long long convert(int n) { long long bin = 0; int rem, i = 1, step = 1; while (n != 0) { rem = n % 2; printf("Step %d: %d/2, Remainder = %d, Quotient = %dn", step++, n, rem, n / 2); n /= 2; bin += rem * i; i *= 10; } return bin; } Output Enter a decimal number: 19 Step 1: 19/2, Remainder = 1, Quotient = 9 Step 2: 9/2, Remainder = 1, Quotient = 4 Step 3: 4/2, Remainder = 0, Quotient = 2 Step 4: 2/2, Remainder = 0, Quotient = 1 Step 5: 1/2, Remainder = 1, Quotient = 0 19 in decimal = 10011 in binary For example we want to convert decimal number 525 in the octal. Step 1: 525 / 8 Remainder : 5 , Quotient : 65 Step 2: 65 / 8 Remainder : 1 , Quotient : 8
- 10. 10 Step 3: 8 / 8 Remainder : 0 , Quotient : 1 Step 4: 1 / 8 Remainder : 1 , Quotient : 0 So equivalent octal number is: 1015 That is (525)10 = (1015)8 Example 1: Program to Convert Decimal to Octal #include <stdio.h> #include <math.h> int convertDecimalToOctal(int decimalNumber); int main() { int decimalNumber; printf("Enter a decimal number: "); scanf("%d", &decimalNumber); printf("%d in decimal = %d in octal", decimalNumber, convertDecimalToOctal(decimalNumber)); return 0; } int convertDecimalToOctal(int decimalNumber) { int octalNumber = 0, i = 1; while (decimalNumber != 0) { octalNumber += (decimalNumber % 8) * i; decimalNumber /= 8; i *= 10; } return octalNumber; } Output Enter a decimal number: 78 78 in decimal = 116 in octal Example 2: Program to Convert Octal to Decimal #include <stdio.h> #include <math.h> long long convertOctalToDecimal(int octalNumber); int main() { int octalNumber;
- 11. 11 printf("Enter an octal number: "); scanf("%d", &octalNumber); printf("%d in octal = %lld in decimal", octalNumber, convertOctalToDecimal(octalNumber)); return 0; } long long convertOctalToDecimal(int octalNumber) { int decimalNumber = 0, i = 0; while(octalNumber != 0) { decimalNumber += (octalNumber%10) * pow(8,i); ++i; octalNumber/=10; } i = 1; return decimalNumber; } Output Enter an octal number: 116 116 in octal = 78 in decimal Hexadecimal number system: It is base 16 number system which uses the digits from 0 to 9 and A, B, C, D, E, F. Decimal number system: It is base 10 number system which uses the digits from 0 to 9 Decimal to hexadecimal conversion method: Following steps describe how to convert decimal to hexadecimal Step 1: Divide the original decimal number by 16 Step 2: Divide the quotient by 16 Step 3: Repeat the step 2 until we get quotient equal to zero. Equivalent binary number would be remainders of each step in the reverse order. Decimal to hexadecimal conversion example:
- 12. 12 For example we want to convert decimal number 900 in the hexadecimal. Step 1: 900 / 16 Remainder : 4 , Quotient : 56 Step 2: 56 / 16 Remainder : 8 , Quotient : 3 Step 3: 3 / 16 Remainder : 3 , Quotient : 0 So equivalent hexadecimal number is: 384 That is (900)10 = (384)16 C Program 1. #include<stdio.h> 2. int main() { 3. long int decimalNumber,remainder,quotient; 4. int i=1,j,temp; 5. char hexadecimalNumber[100]; 6. printf("Enter any decimal number: "); 7. scanf("%ld",&decimalNumber); 8. quotient = decimalNumber; 9. while(quotient!=0) { 10. temp = quotient % 16; 11. //To convert integer into character 12. if( temp < 10) 13. temp =temp + 48; else 14. temp = temp + 55; 15. hexadecimalNumber[i++]= temp; 16. quotient = quotient / 16; 17. } 18. printf("Equivalent hexadecimal value of decimal number %d: ",decimalNumbe r); 19. for (j = i -1 ;j> 0;j--) 20. printf("%c",hexadecimalNumber[j]); 21. return 0; 22.} Output 1. Enter any decimal number: 45 2. Equivalent hexadecimal value of decimal number 45: 2D Program to Check Palindrome #include <stdio.h> int main() {
- 13. 13 int n, reversedN = 0, remainder, originalN; printf("Enter an integer: "); scanf("%d", &n); originalN = n; // reversed integer is stored in reversedN while (n != 0) { remainder = n % 10; reversedN = reversedN * 10 + remainder; n /= 10; } // palindrome if orignalN and reversedN are equal if (originalN == reversedN) printf("%d is a palindrome.", originalN); else printf("%d is not a palindrome.", originalN); return 0; } Power ofa Number Using the while Loop #include <stdio.h> int main() { int base,exp; long long result = 1; printf("Enter a base number: "); scanf("%d", &base); printf("Enter an exponent: "); scanf("%d", &exp); while (exp != 0) { result *= base; --exp; } printf("Answer = %lld", result); return 0; } Output Enter a base number: 3 Enter an exponent: 4 Answer = 81