C语言编程. FizzBu​​zz程序 [英] C programming. The FizzBuzz program

问题描述

我进行了测验，并编写了以下代码:

I had a quiz and I wrote this code:

如果Fizz被3整除，则打印Fizz；如果是，则打印Buzz. 被5整除.如果是，则打印FizzBu​​ss 可被两者整除.否则，它将打印1到100之间的数字.

Print Fizz if it is divisible by 3 and it prints Buzz if it is divisible by 5. It prints FizzBuss if it is divisible by both. Otherwise, it will print the numbers between 1 and 100.

但是我回到家后，我想知道是否可以 用更少的代码写出来.可是我没出来 用较短的代码. 我可以用更短的代码吗?谢谢.

But after I arrived home, I wondered if could have writen it with less code. However, I could not come out with a shorter code. Can I do it with a shorter code? Thanks.

这是我写的，我认为它很好用.但是我能做到吗 用更少的代码.

This is what I wrote and I think it works well. But can I have done it with less code.

#include <stdio.h>

int main(void)
{
    int i;
    for(i=1; i<=100; i++)
    {
        if(((i%3)||(i%5))== 0)
            printf("number= %d FizzBuzz\n", i);
        else if((i%3)==0)
            printf("number= %d Fizz\n", i);
        else if((i%5)==0)
            printf("number= %d Buzz\n", i);
        else
            printf("number= %d\n",i);

    }

    return 0;
}

推荐答案

您也可以这样做:

#include <stdio.h>

int main(void)
{
    int i;
    for(i=1; i<=100; ++i)
    {
        if (i % 3 == 0)
            printf("Fizz");
        if (i % 5 == 0)
            printf("Buzz");
        if ((i % 3 != 0) && (i % 5 != 0))
            printf("number=%d", i);
        printf("\n");
    }

    return 0;
}

缩短几行，并且更容易阅读.

A few lines shorter, and a lot easier to read.

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