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, 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、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+=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;
}
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：
回帖（9）
2016-11-21 13:20:06
能问一个问题吗？
能问一个问题吗？
举报
2016-11-21 17:40:57
剩下的呢？
剩下的呢？
举报
2016-11-21 20:21:50
居然只有很小一部分
举报
2016-11-27 07:59:16
为何只看到一个？
为何只看到一个？
举报
2016-11-28 07:18:16
谢谢分享，谢谢分享，楼主辛苦了！
谢谢分享，谢谢分享，楼主辛苦了！
举报
2016-12-8 16:18:10
全是干货，棒棒滴
全是干货，棒棒滴
举报
2017-1-2 22:06:04
基础的算法，值得学习的
基础的算法，值得学习的
举报
2017-1-9 10:55:34
不错。。。。。。。。。。。。。。。。。。。
不错。。。。。。。。。。。。。。。。。。。
举报
2017-8-7 17:09:33
学习一下，谢谢分享！！！
学习一下，谢谢分享！！！
举报
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