有个c语言的编程问题..._C语言

有个c语言的编程问题...

to replace tab characters by a string of one or more ‘>’ characters so that the input line length is made up to the next multiple of 8, which means that characters following tabs will appear in columns 9, 17, 25, 33 and so on. For example, if the input looks like this (where the positions of tab characters are indicated by the symbol "这里是有个一个符号的键盘打不出来,我用%%代替")
Family%%Given%Phone
Simpson%%Homer%%555-1234
Gumble%%Barney%%555-9876
then the output should look like this (the ‘G’ of ‘Given’ is in column 9 and the ‘P’ of ‘Phone’ is in column 17)
Family>>Given>>>Phone
Simpson>Homer>>>555-1234
Gumble>>Barney>>555-9876

哪位高手可以帮忙解答下...

搜索更多相关的解决方案: c语言

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希望翻译一下

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长度不足8的用'>'来对齐嘛

什么符号打不出来？？打不出来那还怎么输入？

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程序代码：

#include <iostream.h>
#define MAX_LEN 256
#define MAX_NUM 100
#define MAX_WORD_LEN 8 //假定单词字母数字都是小于等于8的，多的没考虑！
int nCount = 0;

void SliptToWord(char *p,char **ret)
{
char *q = p;
int i = 0;
int j = 0;
while(q[i])
{
for(;!(((q[i]=='%')&&(q[i+1]=='%'))||q[i]=='\0');i++); //这边要看清楚
if(q[i] == '%')
{
q[i] = '\0';
ret[j++] = q;
q = &q[i+2];
nCount++;
i = 0;
}
else if(q[i] == '\0')
{
ret[j++] = q;
nCount++;
}
}
}

int main()
{
char *p = new char[MAX_LEN];
cin>>p;
char *ret[MAX_NUM];
for(int i = 0;i<MAX_NUM;i++)
{
ret[i] = new char[MAX_WORD_LEN];
}
SliptToWord(p,ret);
int nWordLen = 0;
for(int i = 0;i<nCount;i++)
{
cout<<ret[i];
nWordLen = strlen(ret[i]);
for(int j = 0;j<MAX_WORD_LEN-nWordLen;j++)
{
cout<<'>';
}
}

delete [] *ret;
return 0;
}

[此贴子已经被作者于2007-8-17 10:32:07编辑过]

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就是说用一个或者多个＂＞＂字符来代替tab字符．简单点说就是跟在tab字符后面的字母要出现在第９行．第１７行。第３５行．．．

楼上的。有这么复杂吗？？

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以下是引用c语言白痴在2007-8-17 12:58:54的发言：

就是说用一个或者多个＂＞＂字符来代替tab字符．简单点说就是跟在tab字符后面的字母要出现在第９行．第１７行。第３５行．．．

楼上的。有这么复杂吗？？

别光说不做，有简单的，就把代码贴出来。
还有，你是来请教的还是来消遣的？稍微谦虚一点。。。

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晕.你完全误解我的意思了．．我只是觉得应该不会很复杂．．因为我做了第一个．

copies each character from standard input directly to standard output, except that each space is replaced by a tilde (‘~’).if the input looks like this
The quick brown fox jumps
over the lazy dog.
then the output should look like this
The~quick~brown~fox~jumps
over~the~lazy~dog.
The program should continually process input until it reaches end-of-file. The core of the program will be a loop with the following structure:
while (get a character from stdin and test for end-of-file) {
put the character (or its replacement) to stdout
}

第二个应该只需要在这个基础上修改下便可，所以我觉得应该没那么复杂的．．你看看第一个．

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看了，那是完全不一样的题目！
如果用String来做的话，这题只要一条语句就可以完成了！

程序代码：

str.replace(\" \",\"~\"); //String str = \"quick brown fox jumps\";

而第二题要做完整还是比较复杂的，他要考察每个单词的长度！然后补齐！
“％”还要对只有一个％的情况的过滤！
完整的话，还要考虑，单词字长超过８的时候，就要考虑，后续单词的排列了！

一句话，自己动手，什么都清楚！

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回复：（c语言白痴）有个c语言的编程问题...

还有哪位高手可以解答下...
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