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This article is a simple exercise in C language
1. Print a triangle
Input the number of lines, and print out the triangle code corresponding to the number of lines on the screen
:
#include<stdio.h>
void print(int n)
{
for (int i = 1; i <= n; i++)
{
for (int k = 1; k <= n - i; k++)
{
printf(" ");
}
for (int j = 1; j <= 2 * i - 1; j++)
{
printf("*");
}
printf("\n");
}
}
int main()
{
int a = 0;
printf("请输入一个整数：\n");
scanf_s("%d", &a);
print(a);
return 0;
}
operation result:
2. Print a rhombus
code:
#include<stdio.h>
void print(int n)
{
for (int i = 1; i <= n; i++)
{
for (int k = 1; k <= n - i; k++)
{
printf(" ");
}
for (int j = 1; j <= 2 * i - 1; j++)
{
printf("*");
}
printf("\n");
}
for (int i = n-1; i >= 1; i--)
{
for (int k = 0; k <= n-1 - i; k++)
{
printf(" ");
}
for (int j = 1; j <= 2 * i - 1; j++)
{
printf("*");
}
printf("\n");
}
}
int main()
{
int a = 0;
printf("请输入一个整数：\n");
scanf_s("%d", &a);
print(a);
return 0;
}
operation result:
3. String reverse order
the code
#include<stdio.h>
void nixu(char*a,int sz)
{
int left = 0;
int right = sz - 1;
while (left<right)
{
int tmp = *(a+right);
*(a + right) = *(a + left);
*(a + left) = tmp;
left++;
right--;
}
}
int main()
{
char arr[] = "ndscsnljrginv";
int sz = sizeof(arr) / sizeof(arr[0]);
nixu(arr,sz);
for (int i = 0; i < sz; ++i)
{
printf("%c", arr[i]);
}
return 0;
}
operation result
4. Print the number of daffodils
Calculate and output all "daffodil numbers" between 0 and 100,000.
the code
#include<stdio.h>
#include<math.h>
int main()
{
for (int i = 0; i <100000; ++i)
{
int tmp = i;
int count = 1;
int sum = 0;
while (tmp / 10)
{
count++;
tmp = tmp / 10;
}
tmp = i;
while (tmp)
{
sum += pow(tmp % 10, count);
tmp = tmp / 10;
}
if (sum == i)
{
printf("%d ", i);
}
}
return 0;
}
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