10 basic algorithm C language
Algorithm is the soul of a program and software, as a good programmer, and only some of the basic algorithm has a comprehensive grasp, it would seem handy in the design process and the process of writing code. This article is nearly C-language algorithm is the second in the series, including the classic Fibonacci sequence, simple calculator, check palindrome, such as prime number check algorithm. Maybe they can design or interview come in handy in your graduation.
1, computing Fibonacci sequence
Also known as the Fibonacci sequence Fibonacci number, also known as golden columns, referring to such a series: 1,1,2,3,5,8,13,21.
C language code is as follows:
/* Displaying Fibonacci sequence up to nth term where n is entered by user. */
#include <stdio.h>
int main()
{
int count, n, t1=0, t2=1, display=0;
printf("Enter number of terms: ");
scanf("%d",&n);
printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */
count=2; /* count=2 because first two terms are already displayed. */
while (count<n)
{
display=t1+t2;
t1=t2;
t2=display;
++count;
printf("%d+",display);
}
return 0;
}
结果输出：
Number of Terms the Enter: 10
the Fibonacci Series: +. 1. 1 + 0 + 2 + +. 8. 3. 5 + + + 34 is 21 is 13 is + +
may use the following source code:
/* Displaying Fibonacci series up to certain number entered by user. */
#include <stdio.h>
int main()
{
int t1=0, t2=1, display=0, num;
printf("Enter an integer: ");
scanf("%d",&num);
printf("Fibonacci Series: %d+%d+", t1, t2); /* Displaying first two terms */
display=t1+t2;
while(display<num)
{
printf("%d+",display);
t1=t2;
t2=display;
display=t1+t2;
}
return 0;
}
结果输出：
Enter an integer: 200
Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+
2, check palindrome
Source:
/* C program to check whether a number is palindrome or not */
#include <stdio.h>
int main()
{
int n, reverse=0, rem,temp;
printf("Enter an integer: ");
scanf("%d", &n);
temp=n;
while(temp!=0)
{
rem=temp%10;
reverse=reverse*10+rem;
temp/=10;
}
/* Checking if number entered by user and it's reverse number is equal. */
if(reverse==n)
printf("%d is a palindrome.",n);
else
printf("%d is not a palindrome.",n);
return 0;
}
结果输出：
Enter an integer: 12321
12321 is a palindrome.
3, a prime number check
Note: 1 is neither bonded nor prime number.
Source:
/* C program to check whether a number is prime or not. */
#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)
{
if(n%i==0)
{
flag=1;
break;
}
}
if (flag==0)
printf("%d is a prime number.",n);
else
printf("%d is not a prime number.",n);
return 0;
}
结果输出：
Enter a positive integer: 29
29 is a prime number.
4, print pyramids and triangles
* The establishment of a triangle using
*
* *
* * *
* * * *
* * * * *
Source:
#include <stdio.h>
int main ()
{
int I, J, rows;
the printf ( "The the Enter Number of rows:");
Scanf ( "% D", & rows);
for (I =. 1; I <= rows; I ++)
{
for (J =. 1; J <= I; J ++)
{
the printf ( "*");
}
the printf ( "\ n-");
}
return 0;
}
below using the numbers shown in FIG. Print half pyramid.
. 1
. 1 2
. 1 2. 3
. 1 2. 4. 3
. 1 2. 4. 3. 5
Source:
#include <stdio.h>
int main()
{
int i,j,rows;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=1;i<=rows;++i)
{
for(j=1;j<=i;++j)
{
printf("%d ",j);
}
printf("\n");
}
return 0;
}
用 * 打印半金字塔
* * * * *
* * * *
* * *
* *
*
Source:
#include <stdio.h>
int main()
{
int i,j,rows;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=rows;i>=1;--i)
{
for(j=1;j<=i;++j)
{
printf("* ");
}
printf("\n");
}
return 0;
}
用 * 打印金字塔
*
* * *
* * * * *
* * * * * * *
* * * * * * * * *
Source:
#include <stdio.h>
int main()
{
int i,space,rows,k=0;
printf("Enter the number of rows: ");
scanf("%d",&rows);
for(i=1;i<=rows;++i)
{
for(space=1;space<=rows-i;++space)
{
printf(" ");
}
while(k!=2*i-1)
{
printf("* ");
++k;
}
k=0;
printf("\n");
}
return 0;
}
用 * 打印倒金字塔
* * * * * * * * *
* * * * * * *
* * * * *
* * *
*
Source:
#include<stdio.h>
int main()
{
int rows,i,j,space;
printf("Enter number of rows: ");
scanf("%d",&rows);
for(i=rows;i>=1;--i)
{
for(space=0;space<rows-i;++space)
printf(" ");
for(j=i;j<=2*i-1;++j)
printf("* ");
for(j=0;j<i-1;++j)
printf("* ");
printf("\n");
}
return 0;
}
5, simple addition and subtraction, multiplication and division calculator
Source:
/* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. */
# include <stdio.h>
int main()
{
char o;
float num1,num2;
printf("Enter operator either + or - or * or divide : ");
scanf("%c",&o);
printf("Enter two operands: ");
scanf("%f%f",&num1,&num2);
switch(o) {
case '+':
printf("%.1f + %.1f = %.1f",num1, num2, num1+num2);
break;
case '-':
printf("%.1f - %.1f = %.1f",num1, num2, num1-num2);
break;
case '*':
printf("%.1f * %.1f = %.1f",num1, num2, num1*num2);
break;
case '/':
printf("%.1f / %.1f = %.1f",num1, num2, num1/num2);
break;
default:
/* If operator is other than +, -, * or /, error message is shown */
printf("Error! operator is not correct");
break;
}
return 0;
}
结果输出：
Enter operator either + or - or * or divide : -
Enter two operands: 3.4
8.4
3.4 - 8.4 = -5.0
6, the number of checks can be expressed as a sum of two primes and
Source:
#include <stdio.h>
int prime(int n);
int main()
{
int n, i, flag=0;
printf("Enter a positive integer: ");
scanf("%d",&n);
for(i=2; i<=n/2; ++i)
{
if (prime(i)!=0)
{
if ( prime(n-i)!=0)
{
printf("%d = %d + %d\n", n, i, n-i);
flag=1;
}
}
}
if (flag==0)
printf("%d can't be expressed as sum of two prime numbers.",n);
return 0;
}
int prime(int n) /* Function to check prime number */
{
int i, flag=1;
for(i=2; i<=n/2; ++i)
if(n%i==0)
flag=0;
return flag;Result Output:
}
Enter a positive integer: 34
34 = 3 + 31
34 = 5 + 29
34 = 11 + 23
34 = 17 + 17
7, a reversed string recursively
Source:
/* Example to reverse a sentence entered by user without using strings. */
#include <stdio.h>
void Reverse();
int main()
{
printf("Enter a sentence: ");
Reverse();
return 0;
}
void Reverse()
{
char c;
scanf("%c",&c);
if( c != '\n')
{
Reverse();
printf("%c",c);
}
}
结果输出：
Enter a sentence: margorp emosewa
awesome program
8, to achieve conversion between binary and decimal
/* C programming source code to convert either binary to decimal or decimal to binary according to data entered by user. */
#include <stdio.h>
#include <math.h>
int binary_decimal(int n);
int decimal_binary(int n);
int main()
{
int n;
char c;
printf("Instructions:\n");
printf("1. Enter alphabet 'd' to convert binary to decimal.\n");
printf("2. Enter alphabet 'b' to convert decimal to binary.\n");
scanf("%c",&c);
if (c =='d' || c == 'D')
{
printf("Enter a binary number: ");
scanf("%d", &n);
printf("%d in binary = %d in decimal", n, binary_decimal(n));
}
if (c =='b' || c == 'B')
{
printf("Enter a decimal number: ");
scanf("%d", &n);
printf("%d in decimal = %d in binary", n, decimal_binary(n));
}
return 0;
}
int decimal_binary(int n) /* Function to convert decimal to binary.*/
{
int rem, i=1, binary=0;
while (n!=0)
{
rem=n%2;
n/=2;
binary+=rem*i;
i*=10;
}
return binary;
}
int binary_decimal(int n) /* Function to convert binary to decimal.*/
{
int decimal=0, i=0, rem;
while (n!=0)
{
rem = n%10;
n/=10;
decimal += rem*pow(2,i);
++i;
}
return decimal;
}
Output results:
9, multi-dimensional array implemented using the sum of two matrices
Source:
#include <stdio.h>
int main(){
int r,c,a[100][100],b[100][100],sum[100][100],i,j;
printf("Enter number of rows (between 1 and 100): ");
scanf("%d",&r);
printf("Enter number of columns (between 1 and 100): ");
scanf("%d",&c);
printf("\nEnter elements of 1st matrix:\n");
/* Storing elements of first matrix entered by user. */
for(i=0;i<r;++i)
for(j=0;j<c;++j)
{
printf("Enter element a%d%d: ",i+1,j+1);
scanf("%d",&a[i][j]);
}
/* Storing elements of second matrix entered by user. */
printf("Enter elements of 2nd matrix:\n");
for(i=0;i<r;++i)
for(j=0;j<c;++j)
{
printf("Enter element a%d%d: ",i+1,j+1);
scanf("%d",&b[i][j]);
}
/*Adding Two matrices */
for(i=0;i<r;++i)
for(j=0;j<c;++j)
sum[i][j]=a[i][j]+b[i][j];
/* Displaying the resultant sum matrix. */
printf("\nSum of two matrix is: \n\n");
for(i=0;i<r;++i)
for(j=0;j<c;++j)
{
printf("%d ",sum[i][j]);
if(j==c-1)
printf("\n\n");
}
return 0;
}
10, matrix transpose
Source:
#include <stdio.h>
int main()
{
int a[10][10], trans[10][10], r, c, i, j;
printf("Enter rows and column of matrix: ");
scanf("%d %d", &r, &c);
/* Storing element of matrix entered by user in array a[][]. */
printf("\nEnter elements of matrix:\n");
for(i=0; i<r; ++i)
for(j=0; j<c; ++j)
{
printf("Enter elements a%d%d: ",i+1,j+1);
scanf("%d",&a[i][j]);
}
/* Displaying the matrix a[][] */
printf("\nEntered Matrix: \n");
for(i=0; i<r; ++i)
for(j=0; j<c; ++j)
{
printf("%d ",a[i][j]);
if(j==c-1)
printf("\n\n");
}
/* Finding transpose of matrix a[][] and storing it in array trans[][]. */
for(i=0; i<r; ++i)
for(j=0; j<c; ++j)
{
trans[j][i]=a[i][j];
}
/* Displaying the transpose,i.e, Displaying array trans[][]. */
printf("\nTranspose of Matrix:\n");
for(i=0; i<c; ++i)
for(j=0; j<r; ++j)
{
printf("%d ",trans[i][j]);
if(j==r-1)
printf("\n\n");
}
return 0;
}