通过改变一个字符修复一个破碎的循环 [英] Fixing a broken loop by changing exactly one character

问题描述

我发现网站与一些复杂的难题Ç的。现在我正在处理这个：

I found a site with some complicated C puzzles. Right now I'm dealing with this:

下面是一个件C code，其意图是要打印减号20倍。但是你可以发现，这是行不通的。

The following is a piece of C code, whose intention was to print a minus sign 20 times. But you can notice that, it doesn't work.

#include <stdio.h>
int main()
{
    int i;
    int n = 20;
    for( i = 0; i < n; i-- )
        printf("-");
    return 0;
}

  
  

好固定高于code是直线前进。为了使问题有意思，你得解决上述code，通过改变一个字符。有三种已知的解决方案。看看你是否可以得到所有这三个。

Well fixing the above code is straight-forward. To make the problem interesting, you have to fix the above code, by changing exactly one character. There are three known solutions. See if you can get all those three.

我无法弄清楚如何解决。我知道，它可以通过改变是固定的 - 到 ++ ，但我不能找出单一性格改变，使其工作。

I cannot figure out how to solve. I know that it can be fixed by changing -- to ++, but I can't figure out what single character to change to make it work.

推荐答案

下面是一个解决办法：

for( i = 0; -i < n; i-- )
        printf("-");

这是第二个，这要归功于马克对我的帮助！

Here is a second one, thanks to Mark for helping me!

for( i = 0; i + n; i-- )
    printf("-");

和马克也有第三个是

for( i = 0; i < n; n-- )
    printf("-");

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