3

我正在尝试为每个空格添加一个空格，直到column = 0. 我不知道该怎么做。

问题如下。如果你看一份报纸，你会发现文章有理由适合列。编写一个程序，读取报纸的列宽，然后读取一行文本。对齐文本行以适合该宽度的列。当您的程序运行时，屏幕应如下所示：

Enter the width of the column: 40
Enter a line of text: Good morning how are you?
12345678901234567890123456789012345678901234567890...
Good     morning     how     are    you?

理由是通过计算文本中的空白数来完成的。在上面的示例中，有 4 个间隙。然后每个间隙都必须添加空格。额外空间的数量必须尽可能均匀地分配。在上面的示例中，前三个间隙各有 5 个空格，最后一个间隙有 4 个空格。

笔记：

  1. 如果文本比列长，那么您必须报告错误——不要尝试将其分成两行！
  2. 假设文本中包含多个单词。
  3. 请注意由 123456789012345678.... 组成的标题行，这对于检查您的结果很有用。您可以随意设置此标题行 - 70 个空格将是一个有用的长度。

谢谢

#include <stdio.h>

int clear_input_buffer(void);

int column;
int c;
int g;
int e;
int space;
int length;

char line[40];

int main(){

    g = 0;

    printf("enter width of column\n");
    scanf("%d", &column);

    printf("enter line of text\n");
    clear_input_buffer();
    gets(line);

    c = 0;

    while(c <= column){
        if(g <= 9)
        {
            printf("%d", g);

            g = g + 1;
            c = c + 1;
        }
        else
        {
            g = 0;
            printf("%d", g);
            g = g + 1;

            c = c + 1;
        }
    }

    printf("\n%s", line);

    space = 0;

    length = 0;

    for( e = 0; line[e] != '\0'; e++ )
    {
        length = length + 1;
        if( line[e] == ' ' )
        space = space + 1;
    }

    column = column - length;

    for( e = 0; line[e] != '\0'; e++ )
    {
        if((line[e] == ' ') && (column > 0))
        {
            add space to here
            column = column - 1;
        }
    }

    printf("%d\n", space);

    printf("%d", length);

    printf("%s", line);


}

int clear_input_buffer(void) {
    int ch;
    while (((ch = getchar()) != EOF) && (ch != '\n')) /* void */;
    return ch;
}
4

2 回答 2

2

这是我做的。这远非理想，但你明白了。您只需输入条件，例如输入的字符串大于或等于 40 个字符时，即可跳过该过程。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main(void)
{
int i = 0;  // first just initialiaze stuff
char ch[40]; // memset the arrays, get the string
memset(ch, '\0', 40);
gets(ch);

int diff = 40 - strlen(ch);
int spaces = 0;
while(i<strlen(ch))
{
    if(*(ch + i++) == ' ')  // count the number of words/spaces between words
    spaces++;
}
char finalt[40];
memset(finalt, '\0', 40);

i = 0;

diff /= spaces;  // diff is the number of spaces to be added between every word

i = 0;
int j = 0;  // j is for the finalt array
int k = 0;  // k  counts through the while, to put in spaces
printf("%d\n", diff);
while(i<40)    // just squeeze in the spaces
{
    if(ch[i] == ' ') {while(k<diff){ finalt[j++] = ' '; k++;} k = 0;}
    else {finalt[j] = ch[i]; j++;}
    i++;
}

printf("%s\n", finalt); // print the result
return 0;
}
于 2013-05-01T11:11:13.947 回答
0 
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define WIDTH 70
#define _(x) #x
#define str(x) _(x)

void ruler_print(int n){
    char ruler[] = "1234567890";

    while(n>9){
        printf(ruler);
        n -= 10;
    }
    ruler[n] = '\0';
    printf("%s\n", ruler);
}

int count_word(const char *text, size_t *count_char){
    int i;
    char *wk, *p;
    p=wk=strdup(text);
    *count_char=0;
    for(i=0;p=strtok(p, " ");++i,p=NULL){
        *count_char+=strlen(p);
    }
    free(wk);
    return i;
}

int main(void){
    int column, len, word_count;
    int i, spaces, between, remain;
    size_t count_char;
    char text[WIDTH + 1];
    char *p = text;

    printf("Enter the width of the column: ");scanf("%d%*c", &column);
    printf("Enter a line of text: ");scanf("%" str(WIDTH) "[^\n]", text);
    len=strlen(text);
    if(len > column || len > WIDTH){
        fprintf(stderr, "too long text!\n");
        return -1;
    }
    ruler_print(WIDTH);
    word_count = count_word(text, &count_char);
    spaces = column - count_char;
    between = spaces / (word_count -1);
    remain = spaces - (word_count -1)*between;
    strtok(text, " ");
    for(i=0;i<word_count-1;++i){
        printf("%s%*s", p, between + (remain ? 1 : 0), " ");
        if(remain) --remain;
        p=strtok(NULL, " ");
    }
    printf("%s\n", p);
    return 0;
}
于 2013-05-01T11:52:54.813 回答