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无货源网店怎么开,seo网站布局,网页设计各个部分的尺寸,找人做任务网站1.字符串左旋结果题目内容：写一个函数，判断一个字符串是否为另外一个字符串旋转之后的字符串。例：给定s1 AABCD和s2 BCDAA,返回1 给定s1 abcd和s2 ACBD,返回0 AABCD左旋一个字符得到ABCDA AABCD左旋两个字符得到BCDAA AABCD右旋一…

1.字符串左旋结果

题目内容：写一个函数，判断一个字符串是否为另外一个字符串旋转之后的字符串。

例：给定s1 = AABCD和s2 = BCDAA,返回1

给定s1 = abcd和s2 = ACBD,返回0

AABCD左旋一个字符得到ABCDA

AABCD左旋两个字符得到BCDAA

AABCD右旋一个字符得到DAABC

#include <stdio.h>
#include <string.h>int is_string_left_rotate(char* str1, char* str2)
{int i = 0;int len = strlen(str1);for (i = 0; i < len; i++){char tmp = *str1;int j = 0;for (j = 0; j < len - 1; j++){*(str1 + j) = *(str1 + j + 1);}*(str1 + len - 1) = tmp;if (strcmp(str1, str2) == 0){return 1;}}return 0;
}int main()
{char arr1[10] = "AABCD";char arr2[10] = "BCDAA";int ret = is_string_left_rotate(arr1,arr2);if (ret == 1){printf("YES\n");}elseprintf("NO\n");return 0;
}

2.offsetof宏的实现

写一个宏，计算结构体中变量相对于首地址的偏移

#include <stdio.h>#define OFFSETOF(struct_name,mem_name) (int)&(((struct_name*)0)->mem_name)struct A
{int a;short b;int c;char d;
};int main()
{printf("%d\n", OFFSETOF(struct A, a));printf("%d\n", OFFSETOF(struct A, b));printf("%d\n", OFFSETOF(struct A, c));printf("%d\n", OFFSETOF(struct A, d));return 0;
}

3.模拟实现atoi

#include <stdio.h>
#include <limits.h>
#include <ctype.h>enum State
{INVALID,VALID
};enum State state = INVALID;int my_atoi(const char* p)
{int flag = 1;if (p == NULL){return 0;}if (*p == '\0'){return 0;}while (isspace(*p)){p++;}if (*p == '+'){flag = 1;p++;}else if (*p == '-'){flag = -1;p++;}long long n = 0;while (isdigit(*p)){n = n * 10 + flag * (*p - '0');if (n < INT_MIN || n>INT_MAX){return 0;}p++;}if (*p == '\0'){state = VALID;return (int)n;}else{return (int)n;}
}int main()
{const char* p = "   -14a";int ret = my_atoi(p);if(state == VALID)printf("合法：%d\n", ret);elseprintf("非法：%d\n", ret);return 0;
}

4.n的k次方

编写一个函数实现n的k次方，使用递归实现。

#include <stdio.h>double Pow(int n, int k);int main()
{int n = 0;int k = 0;scanf("%d %d", &n, &k);double ret = Pow(n, k);printf("%lf\n", ret);return 0;
}double Pow(int n, int k)
{if (k == 0)return 1;else if (k > 0)return n * Pow(n, k - 1);elsereturn 1.0 / (Pow(n, -k));
}