1. Find the sum of the first 5 items of Sn=a+aa+aaa+aaaa+aaaaa, where a is a number
For example: 2+22+222+2222+22222
int main()
{
int a = 0;
int n = 0;
scanf("%d%d", &a, &n);
int sum = 0;
int i = 0;
int ret = 0;
for (i = 0; i < n; i++)
{
ret = ret * 10 + a;
sum += ret;
}
printf("sum = %d", sum);
return 0;
}
2. Find and output all the "number of daffodils" between 0 and 100000. Also called the self-power number
"daffodil number" refers to an n-digit number, the sum of the n-th power of each digit is exactly equal to the number itself, such as: 153 = 1 ^ 3 + 5 ^ 3 + 3 ^ 3, then 153 is one Number of daffodils"
#include<stdio.h>
#include<math.h>
int main()
{
int i = 0;
for (i = 0; i <= 100000; i++)
{
//1.计算i的位数 n
int n = 1; //一个数至少是一位数，所以这里初始化为1
//但是这里你会发现如果使用i，那么就会在while里面改变循环变量i的值，需要重新定义一个临时变量，来暂时保管这个i的数值
int tmp = i;
int sum = 0;
while (tmp /= 10)
{
n++;
}
//2.计算每一位的n次方之和 sum
tmp = i;
while (tmp)
{
sum += pow(tmp % 10, n); //平方和的函数 pow得到的数是一个double类型，但是你赋给了一个整形的上面，所以为了避免警告可以强制转换类型
tmp /= 10;
}
//比较i和sum的值是否相等
if (i == sum)
{
printf("%d ", i);
}
}
return 0;
}
3. Print rhombus
int main()
{
int line = 0;
scanf("%d", &line); // 7
//打印上半部分
int i = 0;
for (i = 0; i < line; i++)
{
//打印空格
int j = 0;
for (j = 0; j <line-1-i; j++) // 6 5 4 3 2 1 0 每一行的空格数都是在减少的
{
printf(" ");
}
//打印'*'
for (j = 0; j < 2*i+1; j++) // '*' 是
{
printf("*");
}
printf("\n");
}
//打印下半部分
for (i = 0; i < line - 1; i++)
{
//打印空格
int j = 0;
for (j = 0; j <=i; j++)
{
printf(" ");
}
//打印'*'
for (j = 0; j <2*(line-1-i)-1 ; j++) // 下半部分的第一行是11个，第二行是9个
{
printf("*");
}
printf("\n");
}
return 0;
}
4. Rotate a string
Implement a function that can rotate k characters in a string to the left.
For example:
ABCD left-handed one character gets BCDA
ABCD left-handed two characters gets CDAB
First of all, your string should meet the condition that it can be modified. If it is a constant string, it cannot be modified, which is wrong.
#include<string.h>
#include<assert.h>
void left_move(char* arr, int k)
{
assert(arr != NULL);
int i = 0;
int len = strlen(arr);
for (i = 0; i < k; i++)
{
左旋转一个字符
1.先把第一个字符拿出来放在一个临时变量里面
char tmp = *arr;
2.把后面的元素都向前移动,所以这里需要知道字符串的长度
int j = 0;
for (j = 0; j <len-1 ; j++)
{
*(arr + j) = *(arr + j + 1);
}
3.
*(arr + len - 1) = tmp;
}
}
int main()
{
char arr[] = "abcdef";
left_move(arr, 2);
printf("%s\n", arr);
return 0;
}
The second realization method three-step flip method
//abcdef
//ba fedc 把这两个字符串分别逆序
//在整体逆序
void reverse(char* left, char* right)
{
assert(left != NULL);
assert(right != NULL);
while (left<right)
{
char tmp = *left;
*left = *right;
*right = tmp;
left++;
right--;
}
}
void left_move(char* arr, int k)
{
assert(arr != NULL);
int len = strlen(arr);
assert(k <= len);
reverse(arr,arr+k-1);//逆转左边
reverse(arr+k,arr+len-1);//逆转右边
reverse(arr,arr+len-1);//整体逆转
}
int main()
{
char arr[] = "abcdef";
left_move(arr, 2);
printf("%s\n", arr);
return 0;
}
5. String rotation result
Job content
Write a function to determine whether a character string is a character string rotated by another character string.
For example: given s1 = AABCD and s2 = BCDAA, return 1
Given s1 = abcd and s2 = ACBD, return 0.
AABCD rotates one character to the left to get ABCDA
AABCD left-handed two characters to get BCDAA
AABCD rotates one character to the right to get DAABC
void left_move(char* s1, int k)
{
assert(s1 != NULL);
int i = 0;
int len = strlen(s1);
for (i = 0; i < k; i++)
{
//左旋转一个字符
//1.先把第一个字符拿出来放在一个临时变量里面
char tmp = *s1;
//2.把后面的元素都向前移动,所以这里需要知道字符串的长度
int j = 0;
for (j = 0; j <len - 1; j++)
{
*(s1 + j) = *(s1 + j + 1);
}
//3.
*(s1 + len - 1) = tmp;
}
}
int is_left_move(char* s1, char* s2)
{
assert(s1 != NULL);
assert(s2 != NULL);
int len = strlen(s1);
int i = 0;
for (i = 0; i < len; i++)
{
left_move(s1, 1);
int ret = strcmp(s1, s2);
if (ret == 0)
return 1;
}
return 0;
}
int main()
{
char s1[] = "abcdef";
char s2[] = "cdefab";
int ret = is_left_move(s1, s2);
if (ret == 1)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
return 0;
}
The second method
//arr1："abcdefabcdef\0"
//arr2:"cdefab\0"
//看一下arr2是否是arr1的子串
//但是你会发现自己给自己追加的时候，会把后面的'\0'给覆盖掉，然后他就找不到挺值得标记了
#include<string.h>
#include<stdio.h>
int is_left_move(char* str1, char* str2)
{
assert(str1);
assert(str2);
int len1 = strlen(str1);
int len2 = strlen(str2);
if (len1 != len2) //当s2和s1的长度都不相等的时候，压根谈不是是不是旋转得来的结果。
return 0;
//1.在str1字符串中追加一个str1字符串
//但是strcat这个函数是不能给自己追加的
strncat(str1, str1, len1);
//2.判断str2指向的字符串是否是str1指向的字符串的子串
char* ret = strstr(str1,str2);
if (ret == NULL)
{
return 0;
}
else
{
return 1;
}
}
int main()
{
char s1[] = "abcdef";
char s2[] = "cdefab";
int ret = is_left_move(s1, s2);
if (ret == 1)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
return 0;
}
Young's matrix
There is a number matrix, each row of the matrix is increasing from left to right, and the matrix is increasing from top to bottom. Please write a program to find whether a certain number exists in such a matrix.
Requirements: time complexity is less than O(N);
//1 2 3
//4 5 6
//7 8 9
//这里面右上角的这个数具有特殊的意义，因为对于一行来说他是最大的值，但是对于一列来说他又是最小的值，所以那这个数去和我们需要查找的那个数
//去比较，就可以很快的去掉一行或者一列
int FindNum(int arr[3][3], int k, int row, int col)
{
int x = 0;
int y = col - 1;//定义右上角那个元素的坐标
while (x<=row-1 && y>=0)
{
if (arr[x][y] > k)
{
y--;
}
else if (arr[x][y]<k)
{
x++;
}
else
{
return 1;
}
}
return 0;
}
int main()
{
int arr[3][3] = {
{
1, 2, 3 }, {
4, 5, 6 }, {
7, 8, 9 } };
int k = 7;
int ret = FindNum(arr, k, 3, 3);
if (ret == 1)
{
printf("找到了\n");
}
else
{
printf("找不到\n");
}
}
A written test question from Baidu, the realization is a macro definition of OFFSETOF
struct S
{
char c1;
int a;
char c2;
};
#define OFFSETOF(struct_name,member_name) (int)&(((struct_name*)0)->member_name)
//(int)&((struct S*)0)->c1
//(struct S*)0相当于把0变成了一个结构体指针然后找到结构体内的成员变量，取其地址，然后和首地址相减就会得到一个指针类型，强制类型转换
int main()
{
printf("%d\n",OFFSETOF(struct S,c1));
printf("%d\n",OFFSETOF(struct S,a));
printf("%d\n",OFFSETOF(struct S,c2));
return 0;
}