Training4:字符串训练

1. 题目: 请编写一个C函数,该函数将一个字符串逆序                                                              

#include <stdio.h>#include <string.h>void reverse(char str[]){int len;len = strlen(str);while(len > 0){len--;printf("%c", str[len]);}printf("\n");}int main(){char str[20];printf("enter a string( within 20 bytes):\n");scanf("%s", str);reverse(str);    return 0;}

      

 

 

2. 题目: 请编写一个C函数,该函数可以实现将一个整数转为任意进制的字符串输出

 

#include <stdio.h>void int_char(int num){int tmp;int i;int n = 0;int a[1024];while(num){a[n++] = num % 10;num /= 10;}for(i = (n - 1); i >= 0; i--){printf("%d", a[i]);}printf("\n");}int main(){int num;printf("enter a number:\n");scanf("%d", &num);int_char(num);    return 0;}

#include <stdio.h>void int_float(char str[], int a[], float b[], int *m, int *n){*m = 0;*n = 0;int tmp1;float tmp2;int i = 0;while(str[i] != '\0'){while((str[i] <= '9' && str[i] >= '0') || str[i] == '.'){tmp1 = 0;tmp2 = 0;while(str[i] <= '9' && str[i] >= '0'){tmp1 = tmp1 * 10 + str[i] - '0';i++;}if(str[i] == '.'){tmp2 = tmp1;float k = 0.1;i++;while(str[i] <= '9' && str[i] >= '0'){tmp2 = tmp2 + (str[i] - '0') * k;k *= 0.1;i++;}b[(*n)++] = tmp2;}else{a[(*m)++] = tmp1;}i++;}i++;}}void display(int a[], float b[], int *m, int *n){int i;printf("the integers is:\n");for(i = 0; i < *m; i++){printf("%d", a[i]);}printf("\nthe float is:\n");for(i = 0; i < *n; i++){printf("%.3f", b[i]);}printf("\n");}int main(){int m, n;char str[1024];int a[1024];float b[1024];printf("enter a string:\n");scanf("%s", str);int_float(str, a, b, &m, &n);display(a, b, &m, &n);    return 0;}

 

 

3. 题目: 输入一个字符串,计算字符串中子串出现的次数

 

#include <stdio.h>#include <string.h>int func(char s[], char str[]){int i = 0;int j = 0;int num = 0;int count = 0;int len;len = strlen(str);while(s[i] != '\0'){if(str[j] == s[i]){num++;j++;i++;if((num % len) == 0){count++;j = 0;}}else if(num != 0){num = 0;j = 0;}else{i++;}}return count;}int main(){char s[1024];char str[1024];printf("enter a string :\n");scanf("%s", s);printf("enter a short string you want to find:\n");scanf("%s", str);printf("the string appears %d times\n", func(s, str));    return 0;}

 

 

4. 题目: 编写一个C函数,将”I am from shanghai ”倒置为”shanghai from am I”,及将句子中的单词位置倒置,而不改变单词内部结构.

 

#include <stdio.h>#include <string.h>void reverse(char a[], int len){int i, j;char tmp = 0;int n = len - 1;for(i = 0; i < n; i++){len--;for(j = 0; j < len; j++){tmp = a[j];a[j] = a[j + 1];a[j + 1] = tmp;}}}void re_reverse(char *a){int n = 0;while(1){n++;if((*a) == ' ' || (*a) == '\0'){reverse(a - n + 1, n - 1);n = 0;}if((*a) == '\0'){break;}a++;}}int main(){char a[] = "I am from shanghai";int len;len = strlen(a);reverse(a, len);re_reverse(a);printf("the string after change is:%s\n", a);    return 0;}

 

 

5. 题目: 输入一个字符串,同时输入帧头和帧尾(可以是多个字符),将该字符串中合法的帧识别出来.

提示:帧头和帧尾分别是head和tail  字符串”asdheadhauboisoktail”中headhauboisoktail是合法帧

 

#include <stdio.h>#include <string.h>int flag_h = 0;int flag_t = 0;char * head(char *s, char *h){int num_h = 0;int len_h = 0;char *tmp = h;len_h = strlen(h);if(s == NULL || h == NULL){return NULL;}if(strlen(s) < strlen(h)){return NULL;}while(*s != '\0' && flag_h == 0){while(*s == *tmp){num_h++;s++;tmp++;if(num_h == len_h){flag_h = 1;tmp = s - num_h;break;}}if(num_h != 0 && num_h != len_h){num_h = 0;tmp = h;s--;}s++;}if(flag_h == 1){return tmp;}else {printf("illegal string!\n");}}char * tail(char *s, char *t){int num_t = 0;int len_t;char *tmp = t;len_t = strlen(t);if(s ==NULL || t == NULL){return NULL;}if(strlen(s) < strlen(t)){return NULL;}if(flag_h == 1){while(*s != '\0' && flag_t == 0){while(*s == *tmp){s++;tmp++;num_t++;if(num_t == len_t){flag_t = 1;tmp = s - 1;break;}}if(num_t != 0 && num_t != len_t){num_t = 0;tmp = t;s--;}s++;}}if(flag_t == 1){return tmp;}else{printf("illegal string!\n");return NULL;}}void display(char *start, char *end){int n;int i;n = end - start + 1;if(n != 0){for(i = 0; i < n; i++){printf("%c", *start);start++;}printf("\n");}else{printf("no data!\n");}}int main(){char *start = NULL;char *end = NULL;char s[100];char h[] = "head";char t[] = "tail";printf("enter a string(<100):\n");scanf("%s", s);start = head(s, h);end = tail(s, t);display(start, end);    return 0;}