每日一题3.25.1
每日一题3.25.1
汽水瓶问题
解题思路: 很简单的数学问题,空瓶数够3瓶换一瓶,直到最后剩下的空瓶够两瓶赊一瓶,不够就退出
代码实现:
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n;
vector<int>str;
while (cin>>n)
{
if (n == 0)
break;
int sum = 0;
while (n > 2)
{
sum += n / 3;
n = n / 3 + n % 3;
}
if (n == 2)
sum += 1;
str.push_back(sum);
}
for (int i = 0; i < str.size(); i++)
{
cout << str[i] << endl;
}
system("pause");
return 0;
}
参考答案: