Question

#include <stdio.h>

int main() {
    FILE *fb;
    char data[255];
    int c=0;
    int count[75] = {0};
    fb = fopen("Input.txt", "r");
    fgets(data, 255, fb);

    /* Start finding frequency*/
    while (data[c] != '\0')
    {
        if( data[c] >= '0' && data[c] <= 'z')
        count[data[c] - 'a']++;

        c++;
    }    
    for (c = 0; c < 75; c++)
    {
        /** Printing only those characters
            whose count is at least 1 */

        if (count[c] != 0)
            printf("%c occurs %d times in the entered string.\n",c+'a',count[c]);
    }

    return 0;
}

示例输入文件：＆＃34; Fred Fish 12345678＆＃34;

我能够处理输入文件中的空格，但程序不会读取大写字母的频率和数字字符。我可以在程序中更改什么有助于解决问题。阅读freq后，我的计划是保存文件，以便我可以使用HUffman进行压缩

Answer 1

我猜测代码

count[data[c] - 'a']++;

到

count[data[c] - '0']++;

如果数据[c]小于75，则计数[data [c] - 'a']不起作用

Answer 2

#include <stdio.h>

int main() {
    FILE *fb;
    char data[255];
    int c = 0;
    int count[75] = { 0 };
    fb = fopen("Input.txt", "r");
    fgets(data, 255, fb);

    /* Start finding frequency*/
    while (data[c] != '\0')
    {
        if (data[c] >= 'a' && data[c] <= 'z') // here you check normal letters
            count[data[c] - 'a']++;
        else if (data[c] >= 'A' && data[c] <= 'Z')
            count[data[c] - 'A' + 26]++; // Capital letters will be stored after lower cases
        else if (data[c] >= '0' && data[c] <= '9')
            count[data[c] - '0' + 51]++; // Numbers will be stored after capital letters
        c++;
    }

    // count[] is initialized as following : 
    // From count[0] to count[25] == Occurence of low case characters
    // From count[26] to count[51] == Occurence of capital characters
    // From count[52] to count[61] == Occurence of numbers from 0 to 9

    for (c = 0; c < 61; c++)
    {
        /** Printing only those characters
        whose count is at least 1 */
        if (count[c] != 0) {
            if (c < 26)
                printf("%c occurs %d times in the entered string.\n", c + 'a', count[c]); 

                // Starting from 'a', iterating until 'z'
            else if (c < 52)
                printf("%c occurs %d times in the entered string.\n", c + 'A' - 26, count[c]);

                // Same as for low case characters
                // Substracting 26 because we already have iterated through the first 26 low case characters
                // Starting from 'A' until 'Z'

            else if (c >= 52)
                printf("%c occurs %d times in the entered string.\n", c + '0' - 51, count[c]);

                // Same as for characters
                // Substracting 51 because we already have iterated through low cases and capital characters
                // Starting from '0' until '9'
        }
    }

    return 0;
}

问题在于，考虑到ascii table，您通过减去等于{{1}的count将数字和数字存储在数组'a'的负数中在ASCII中。这应该工作，但我无法测试它，所以要小心它。

对于打印，我们对小写字符，然后是大写字符和数字做同样的事情：我们从第一个开始：97，然后是'a'，然后是'A' ，迭代直到最后一个使用c并打印它们。我们使用C，'0'，'a' + 1 = 'b'等

这一事实

对于输入'A' + 1 = 'B'，此代码的输出为：

Fred Fish 12345678

Answer 3

您应该修改代码中的'a'到'0'，以便考虑所有大写字母和数字。

#include <stdio.h>

int main() {

       ...
    /* snippet */
       ...    

    while (data[c] != '\0')
    {   
        if( data[c] >= '0' && data[c] <= 'z')
        count[data[c] - '0']++;
        c++;                                                                                                                                                                                                 
    }   

    for (c = 0; c < 75; c++)
    {   
        /** Printing only those characters
         whose count is at least 1 */

        if (count[c] != 0)
            printf("%c occurs %d times in the entered string.\n",c+'0',count[c]);
    }   
    return 0;
}

无法读取数字char频率

3 个答案: