Title description
The 5 monkeys are good friends and fell asleep on the coconut tree by the sea. During this period, some merchant ships forgot a bunch of bananas and left on the beach.
The first monkey woke up and divided the bananas into 5 piles. Only one was left, and he ate it and hid his share and continued to sleep.
The second monkey woke up, divided the bananas into 5 piles again, and there were 2 remaining, so he ate it and hid his share and continued to sleep.
The third monkey woke up, divided the bananas into 5 piles again, and there were 3 left, so he ate it and hid his share and continued to sleep.
The fourth monkey woke up, divided the bananas into 5 piles again, and there were 4 remaining, so he ate it and hid his share and continued to sleep.
The fifth monkey wakes up and divides the bananas into 5 piles again, haha, just not left!
Please calculate how many bananas there are at least at the beginning.
Output
Output an integer to indicate the answer
code show as below:
#include <iostream>
using namespace std;
int main()
{
for(int i=1;;i++){
int sum = i; //用i表示香蕉的数量，因为后续会对i进行多次重新赋值，因此定义一个sum来接收i
if(sum%5==1){
sum = sum-1-sum/5;//猴子会吃掉多余的并藏起自己那份
if(sum%5==2){
sum = sum-2-sum/5;
if(sum%5==3){
sum = sum-3-sum/5;
if(sum%5==4){
sum = sum-4-sum/5;
if(sum%5==0&&sum!=0){
//最后香蕉数量不能为0，这一步很关键，没有这一步会得出错误答案
cout<<i<<endl;//最后输出的应当是i的值
return 0;
}
}
}
}}}}