题目难度：普及-
涉及知识点：换购
题意：
分析：一直换购直至瓶盖数<3为只
AC代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,sum=0,ans=0;
ios::sync_with_stdio(false);//加快cin，cout
cin>>a;
while(a!=0){//未兑换完时
a--;//减少一瓶可兑换的
ans++;//瓶数++
sum++;
a+=sum/3;//加上瓶盖可兑换的
sum%=3;}//加上未满3个瓶盖的数量
cout<<ans;
return 0;
}总结：一直换购直至瓶盖数<3为只
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