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10个经典的C语言面试基础算法及代码

2016-11-20 15:18:26 4046 C语言计算器源代码

0 算法是一个程序和软件的灵魂，作为一名优秀的程序员，只有对一些基础的算法有着全面的掌握，才会在设计程序和编写代码的过程中显得得心应手。本文包括了经典的Fibonacci数列、简易计算器、回文检查、质数检查等算法。也许他们能在你的毕业设计或者面试中派上用场。 1、计算Fibonacci数列 Fibonacci数列又称斐波那契数列，又称黄金分割数列，指的是这样一个数列：1、1、2、3、5、8、13、21。 C语言实现的代码如下： Enter number of terms: 10 Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+ 也可以使用下面的源代码： /* Displaying Fibonacci series up to certain number entered by user. / #include 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 { printf("%d+",display); t1=t2; t2=display; display=t1+t2; } return 0; } 复制代码* 结果输出： Enter an integer: 200 Fibonacci Series: 0+1+1+2+3+5+8+13+21+34+55+89+144+ 复制代码 2、回文检查源代码： /* C program to check whether a number is palindrome or not / #include int main() { int n, reverse=0, rem,temp; printf("Enter an integer: "); scanf("%d", &n); temp=n; while(temp!=0) { rem=temp%10; reverse=reverse10+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; } 复制代码* 结果输出： Enter an integer: 12321 12321 is a palindrome. 复制代码 3、质数检查注：1既不是质数也不是合数。源代码： /* C program to check whether a number is prime or not. / #include 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; } 复制代码* 结果输出： Enter a positive integer: 29 29 is a prime number. 复制代码 4、打印金字塔和三角形 *使用建立三角形** * * * * * * * * * * * * * * * 源代码： #include 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("* "); } printf("n"); } return 0; } 复制代码如下图所示使用数字打印半金字塔。 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 源代码： #include 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("* "); } printf("n"); } return 0; } *用打印半金字塔** * * * * * * * * * * * * * * * 源代码： #include 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** * * * * * * * * * * * * * * * * * * * * * * * * * 源代码： #include 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!=2i-1) { printf(" "); ++k; } k=0; printf("n"); } return 0; } *用打印倒金字塔** * * * * * * * * * * * * * * * * * * * * * * * * * 源代码： #include 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 printf(" "); for(j=i;j<=2i-1;++j) printf(" "); for(j=0;j printf("* "); printf("n"); } return 0; } 5、简单的加减乘除计算器源代码： /* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. / # include 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, num1num2); 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、检查一个数能不能表示成两个质数之和源代码： #include 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 + %dn", 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; } 结果输出：* Enter a positive integer: 34 34 = 3 + 31 34 = 5 + 29 34 = 11 + 23 34 = 17 + 17 7、用递归的方式颠倒字符串源代码： /* Example to reverse a sentence entered by user without using strings. / #include 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、实现二进制与十进制之间的相互转换 /* C programming source code to convert either binary to decimal or decimal to binary according to data entered by user. / #include #include 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+=remi; 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 += rempow(2,i); ++i; } return decimal; } 结果输出： 9、使用多维数组实现两个矩阵的相加源代码： #include 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* for(j=0;j { printf("Enter element a%d%d: ",i+1,j+1); scanf("%d",&a[j]); } /* Storing elements of second matrix entered by user. / printf("Enter elements of 2nd matrix:n"); for(i=0;i* for(j=0;j { printf("Enter element a%d%d: ",i+1,j+1); scanf("%d",&b[j]); } /Adding Two matrices / for(i=0;i for(j=0;j sum[j]=a[j]+b[j]; /* Displaying the resultant sum matrix. / printf("nSum of two matrix is: nn"); for(i=0;i for(j=0;j { printf("%d ",sum[j]); if(j==c-1)* printf("nn"); } return 0; } 结果输出： 10、矩阵转置源代码： #include 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 for(j=0; j { printf("Enter elements a%d%d: ",i+1,j+1); scanf("%d",&a[j]); } /* Displaying the matrix a[][] / printf("nEntered Matrix: n"); for(i=0; i for(j=0; j { printf("%d ",a[j]); if(j==c-1) printf("nn"); } /* Finding transpose of matrix a[][] and storing it in array trans[][]. / for(i=0; i for(j=0; j { trans[j]=a[j];* } /* Displaying the transpose,i.e, Displaying array trans[][]. / printf("nTranspose of Matrix:n"); for(i=0; i for(j=0; j { printf("%d ",trans[j]); if(j==r-1)* printf("nn"); } return 0; } 结果输出： 24
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