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蓝桥杯 微生物增殖 C语言
2022-07-06 11:38:00 【大菜彩】
微生物增殖
假设有两种微生物 X 和 Y
X出生后每隔3分钟分裂一次(数目加倍),Y出生后每隔2分钟分裂一次(数目加倍)。
一个新出生的X,半分钟之后吃掉1个Y,并且,从此开始,每隔1分钟吃1个Y。
现在已知有新出生的 X=10, Y=89,求60分钟后Y的数目。
如果X=10,Y=90呢?
本题的要求就是写出这两种初始条件下,60分钟后Y的数目。
以下程序实现了这一功能,请你补全空白处内容:
提示:Y分别会在0.5,1.5,2.5······时被吃,所以,把60分钟分成120份,则在除以2余数为1时,Y的数目减少X个
当X=10, Y=89,代码:
#include<stdio.h>
int main()
{
int x=10,y=89;
for(int t=1;t<=120;t++) //半分钟一个单位 60分钟为120个单位
{
if(t%2==1)
y=y-x; //因为X出生半分钟后就要吃Y,之后每1分钟要吃Y,所以永远都是奇数个半分钟的时候吃Y
if(t%4==0) //每两分钟翻倍
y=y*2;
if(t%6==0) //每三分钟翻倍
x=x*2;
}
printf("%d",y);
return 0;
}
运行结果为负数,故y=0。
当X=10, Y=90,运行结果:
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