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C语言基础编程题（1月23日）

ps身份z • 2022-12-17 • 随笔 • 阅读 71

C语言基础编程题（1月23日）,第1张

C语言基础编程题（1月23日）

文章目录

1.从一个字符串中读出所有的整数（转为int）与实数（转为float）2.将整数转化为字符串3.字符串的冒泡排序（从小到大）4.查找介于1~32767之间所有满足下列条件的整数5.编写函数void change(char *a, char *b, char *c)6.对N行N列二维数组的每一行排序，偶数行（包括0）由小到大，奇数行由大到小7.找出1~10000之间的Troitsky数8.合并两个字符集合，每个字符串在新集合中仅出现一次，函数返回新集合中字符串9.找出7个默森尼数10.对一个二维数组按行最小值进行排序

1.从一个字符串中读出所有的整数（转为int）与实数（转为float） 
#include 
#include 

int is_num(char c);
int is_dot(char c);
void count_num(char *str);

int main() {
    char str[] = "Sakura.12.Uni.34.Pilot.56.468.Schneider.0.3";
    char *p = str;

    count_num(p);

    return 0;
}


int is_num(char c) {
    return (c >= 48 && c <= 57) ? 1 : 0;
}

int is_dot(char c) {
    return (c == 46) ? 1 : 0;
}


void count_num(char *str) {
    double x = 0;
    int tag_dot = 0;//identify whether '.' has occurred(0) or the offset to this '.'(>=1)
    int tag_num = 0;//identify whether the current position is a number

    while (*str != 0) {
        if (is_num(*str) == 1) {
            tag_num = 1;
            if (tag_dot == 0) {
                x = 10 * x + (*str - 48);//x = 10*x + i
            }
            else {
                x = x + pow(0.1, tag_dot) * (*str - 48);//x = x + 0.1^(offset)*i
                tag_dot ++;
            }
        }
        else if (is_dot(*str) == 1 && tag_num == 1 && is_num(*(str+1))) {
            tag_dot = 1;//type is double
        }
        else {
            if (tag_num == 1) {
                if (tag_dot == 0)
                    printf("%d ", (int)x);
                else
                    printf("%lf ", x);

                
                x = 0;
                tag_dot = 0;
                tag_num = 0;
            }
        }

        str++;
    }

    if (*str == 0 && is_num(*(str-1))) {
        if (tag_dot == 0)
            printf("%d ", (int)x);
        else
            printf("%lf ", x);
    }
}
2.将整数转化为字符串 
#include 
#include 

#define MAX_SIZE 100

void type_trans (int num, char *str);

int main() {
    char *str = (char*)malloc(MAX_SIZE * sizeof(char));//use this string to save result

    type_trans(12345, str);
    puts(str);

    return 0;
}

void type_trans (int num, char *str) {
    int digit = 1, tmp = 0;

    
    while(num / digit >= 10) {
        digit *= 10;
    }

    
    while (digit != 0) {
        *str = num / digit + 48;
        
        num = num % digit;
        digit /= 10;
        str ++;
    }

    *str = '';
}
3.字符串的冒泡排序（从小到大） 
#include 

int str_len(char *src) {
    int len;
    while (*src != '') {
        len++;
        src++;
    }
    return len;
}

void bubble_sort(char *str) {
    int len = str_len(str), tmp, flag;
    for (int i = 0; i < len-1; i++) {
        flag = 0;
        for (int j = 0; j < len-i-1; j++ ) {
            if (str[j] > str[j+1]) {
                tmp = str[j];
                str[j] = str[j+1];
                str[j+1] = tmp;
                flag = 1;
            }
        }
        if (flag == 0)
            break;
    }
}

int main() {
    char str[] = "2462435";

    bubble_sort(str);
    puts(str);
    
    return 0;
}
4.查找介于1~32767之间所有满足下列条件的整数

（1）该数的十进制表示中有且仅有两个相同的数字位；
（2）该数是素数。

#include 

int same_digit(int x);
int is_prime(int x);

int main() {
    int n1 = 1, n2 = 32767;
    int count = 0;

    for (int i = n1; i <= n2; ++i) {
        if (same_digit(i) && is_prime(i)) {
            printf("%d meets this two requirements.n", i);
            count++;
        }
    }

    printf("all %d numbers.", count);

    return 0;
}


int same_digit(int x) {
    int digit = 1;//represent the number of digits of x
    int count[10] = {0};//count the number of times of each number in 0~9

    while (x / digit >= 10)
        digit *= 10;

    for (int i = 0; digit != 0; ++i) {
        count[x/digit] ++;

        x = x % digit;
        digit /= 10;
    }

    for (int i = 0; i < 10; ++i) {
        if (count[i] == 2)
            return 1;
    }

    return 0;
}

int is_prime(int x) {
    if (x == 1)//1 isn't a prime number
        return 0;
    for (int i = 2; i < x; ++i) {
        if (x % i == 0)
            return 0;
    }
    return 1;
}
5.编写函数void change(char *a, char *b, char *c)

函数功能是：首先把b指向的字符串逆向存放，然后将a指向的字符串和b指向的字符串按排列顺序交叉合并到c指向的数组中，两个字符串中过长的剩余字符接在c指向数组的尾部。
例如，当a指向的字符串为"ABCDEFG"，b指向的字符串为"123456789"时，c指向的数组中字符串应为"A9B8C7D6E5F4G321"。

#include 
#include 

#define MAX_SIZE 100

void change(char *a, char *b, char *c);

int main() {
    char a[MAX_SIZE] = "ABCDEFG";
    char b[MAX_SIZE] = "123456789";
    char c[MAX_SIZE];

    change(a, b, c);
    puts(c);

    return 0;
}

void change(char *a, char *b, char *c) {
    char *invert_b = (char*)malloc(MAX_SIZE * sizeof(char));
    char *head_invert_b = invert_b;//the head of invert_b
    char *head_b = b;//the head of b
    int i = 0;

    
    while (*(b+1) != 0)
        b++;

    
    for (i = 0; b >= head_b; i++)
        *(invert_b+i) = *b--;
    *(invert_b+i) = 0;

    
    i = 0;
    while (*invert_b != 0 && *a != 0) {
        *(c + i++) = *a++;
        *(c + i++) = *invert_b++;
    }
    *(c+i) = 0;

    
    while (*invert_b != 0)
        *(c + i++) = *invert_b++;
    while (*a != 0)
        *(c + i++) = *a++;
    *(c+i) = 0;

    free(head_invert_b);
}
6.对N行N列二维数组的每一行排序，偶数行（包括0）由小到大，奇数行由大到小 
#include 
#include 
#include 

#define N 8

void arr_out(int arr[N][N]);//(*arr)[N] is also acceptable
void arr_creat(int arr[N][N]);
void sort_s_head(int *num);//sort one line with small number at head
void sort_b_head(int *num);//sort one line with big number at head

int main() {
    int arr[N][N];

    arr_creat(arr);
    arr_out(arr);

    for (int col = 0; col < N; col++)
        (col % 2 == 0) ? sort_s_head(arr[col]) : sort_b_head(arr[col]);

    arr_out(arr);

    return 0;
}


void arr_out(int arr[N][N]) {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            printf("arr[%d][%d]:%02d ", i, j, arr[i][j]);
        }
        printf("n");
    }
    printf("n");
}

void arr_creat(int arr[N][N]) {
    srand((unsigned)time(NULL));
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            arr[i][j] = rand() % 100;
        }
    }
}

void sort_s_head(int *num) {
    
    int min = 0, tmp = 0;
    for (int i = 0; i < N-1; i++) {
        min = i;
        for (int j = i+1; j < N; j++) {
            if (num[j] < num[min])
                min = j;
        }
        if (min != i) {
            tmp = num[i];
            num[i] = num[min];
            num[min] = tmp;
        }
    }
}

void sort_b_head(int *num) {
    
    int tmp = 0;
    for (int i = 0; i < N-1; i++) {
        for (int j = 0; j < N-1-i; j++) {
            if (num[j] < num[j+1]) {
                tmp = num[j+1];
                num[j+1] = num[j];
                num[j] = tmp;
            }
        }
    }
}
7.找出1~10000之间的Troitsky数

Troitsky数：该数首位数字移动到末位数字之后得到的数是原数的整数倍。
例如：将142857的首位数字1移动到末位之后得到的数是428571，而428571等于3倍142857，因此142857是Troitsky数。

#include 

#define MAX_SIZE 10000


int CountTroitsky(long a[]);
int IsTroitsky(long x);

int main() {
    long arr[MAX_SIZE] = {0};
    int i = 0;

    printf("All %d Troitsky numbers:n", CountTroitsky(arr));

    while (arr[i] != 0)
        printf("%ld is a Troitsky number,n", (int)arr[i++]);

    return 0;
}


int CountTroitsky(long a[]) {
    int count = 0;
    for (long i = 10; i <= 10000; i++) {
        if (IsTroitsky(i) == 1) {
            a[count] = i;
            count++;
        }
    }
    return count;
}

int IsTroitsky(long x) {
    int digit = 1;//represent the number of digits of x

    while (x / digit >= 10)
        digit *= 10;

    return (((x % digit) * 10 + (x / digit)) % x == 0) ? 1 : 0;
}
8.合并两个字符集合，每个字符串在新集合中仅出现一次，函数返回新集合中字符串

例如，
集合A：{"while", "for", "switch", "if", "continue"}，
集合B：{"for", "case", "do", "else", "char", "switch"}，
运行结果：{"while", "for", "switch", "if", "continue", "case", "do", "else", "char"}。

#include 

#define MAX_SIZE 100

void arr_out(char *src[MAX_SIZE]);
void arr_merge(char *tar[MAX_SIZE], char *src[MAX_SIZE]);
int str_cmp(char *str1, char *str2);

int main() {
    char *tar[MAX_SIZE] = {"while", "for", "switch", "if", "continue"};
    char *src[MAX_SIZE] = {"for", "case", "do", "else", "char", "switch"};

    arr_out(tar);
    arr_out(src);

    arr_merge(tar, src);

    arr_out(tar);

    return 0;
}


void arr_out(char *src[MAX_SIZE]) {
    for (int i = 0; src[i] != NULL; i++) {
        printf(""");
        for (int j = 0; src[i][j] != 0; j++) {
            printf("%c", src[i][j]);
        }
        printf("", ");
    }
    printf("bb n");
}

void arr_merge(char *tar[MAX_SIZE], char *src[MAX_SIZE]) {
    char **tar_tail = tar;//tail arr of tar
    int flag = 1;//'1' means insert allowed

    while (*tar_tail != NULL)
        tar_tail++;

    for (int i = 0; src[i] != NULL; i++) {
        flag = 1;
        for (int j = 0; tar[j] != NULL; j++) {
            if (str_cmp(src[i], tar[j]) == 1)
                flag = 0;//there is a repetition, cannot insert
        }
        if (flag == 1) {
            
            *tar_tail = src[i];
            tar_tail++;
        }
    }
}

int str_cmp(char *str1, char *str2) {
    while (*str1 != 0 || *str2 != 0) {
        if (*str1++ != *str2++)
            return 0;
    }

    return 1;
}
9.找出7个默森尼数

莫森尼数：对于 M p = 2 p − 1 M_p=2^p-1 Mp​=2p−1，当 p p p是素数并且 M p M_p Mp​也是素数时，称 M p M_p Mp​为默森尼数。

#include 

int is_prime(int x);
int pow_two(int x);//2^x

int main() {
    int count = 0;
    for (int i = 1; i < 1000; i++) {
        if (is_prime(i) && is_prime(pow_two(i)-1)) {
            printf("%d is a Monison number.n", pow_two(i)-1);
            count++;
        }
        if (count == 7)
            break;
    }

    return 0;
}


int is_prime(int x) {
    if (x == 1)
        return 0;
    for (int i = 2; i < x; i++) {
        if (x % i == 0)
            return 0;
    }
    return 1;
}

int pow_two(int x) {
    int ret = 1;
    while (x-- != 0)
        ret *= 2;
    return ret;
}
10.对一个二维数组按行最小值进行排序

排序规则：整个数组中值最小的元素所在行排在数组第一行，除第一行元素以外的最小元素所在行排在第2行，以此类推。

#include 
#include 
#include 

#define ROW_LENGTH 3
#define COL_LENGTH 3

void arr_out(int arr[ROW_LENGTH][COL_LENGTH]);
void arr_creat(int arr[ROW_LENGTH][COL_LENGTH]);
void arr_swap(int arr[ROW_LENGTH][COL_LENGTH], int min_row, int k);

int main() {
    int arr[ROW_LENGTH][COL_LENGTH];
    int min_row, min_num;

    arr_creat(arr);
    arr_out(arr);

    for (int k = 0; k < COL_LENGTH; k++) {
        min_num = arr[k][0];
        min_row = k;

        //find the minimum value
        for (int i = k; i < COL_LENGTH; i++) {
            for (int j = 0; j < ROW_LENGTH; j++) {
                if (arr[i][j] < min_num) {
                    min_row = i;
                    min_num = arr[i][j];
                }
            }
        }

        //swap row k with the row with the minimum value
        arr_swap(arr, min_row, k);
    }
    arr_out(arr);

    return 0;
}


void arr_out(int arr[ROW_LENGTH][COL_LENGTH]) {
    for (int i = 0; i < ROW_LENGTH; i++) {
        for (int j = 0; j < COL_LENGTH; j++) {
            printf("arr[%d][%d]:%d ", i, j, arr[i][j]);
        }
        printf("n");
    }
    printf("n");
}

void arr_creat(int arr[ROW_LENGTH][COL_LENGTH]) {
    srand((unsigned)time(NULL));
    for (int i = 0; i < ROW_LENGTH; i++) {
        for (int j = 0; j < COL_LENGTH; j++) {
            arr[i][j] = rand() % 100;
        }
    }
}

void arr_swap(int arr[ROW_LENGTH][COL_LENGTH], int min_row, int k) {
    int temp;
    for (int i = 0; i < ROW_LENGTH; i++) {
        temp = arr[min_row][i];
        arr[min_row][i] = arr[k][i];
        arr[k][i] = temp;
    }
}

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