C语言-兔子繁殖规律

2021 年 04 月 30 日

991 次浏览

254字数

<div class="panel panel-default collapse-panel box-shadow-wrap-lg"><div class="panel-heading panel-collapse" data-toggle="collapse" data-target="#collapse-ca8db2b7341156abf7072c0d4c329fb572" aria-expanded="true"><div class="accordion-toggle"><span style="">兔子繁殖分析:</span>
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<div id="collapse-ca8db2b7341156abf7072c0d4c329fb572" class="collapse collapse-content"><p></p>在假设兔纸不死亡的前提下：

兔纸繁殖规律：第三代兔纸等于第一代与第二代兔纸之和`例如：2=1+1`

<div class="panel panel-default collapse-panel box-shadow-wrap-lg"><div class="panel-heading panel-collapse" data-toggle="collapse" data-target="#collapse-3fddc5207a9aa4fca70aa7543303c03e18" aria-expanded="true"><div class="accordion-toggle"><span style="">思路:</span>
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<div id="collapse-3fddc5207a9aa4fca70aa7543303c03e18" class="collapse collapse-content"><p></p>变量交换赋值<p></p></div></div></div>

```cpp
#include<stdio.h>
int main(){
	 int a , b , c , m , n;
	 b=1;
	 c=1;
	 for(a=1;a<=38;a++){
	 	m=b+c;
	 	printf("第%d天兔子数为%d\n",a,m);  //假设第一代共有2只兔纸
	 	b=c;
	 	c=m;
	 }
	 
}
```

最后修改：2021 年 04 月 30 日