封装好的啤酒瓶问题代码

#include <bits/stdc++.h>using namespace std;
int bear(int x, int a, int b){ int emp = x; int res = x; int newb; while (emp >= a) { int r = emp % a; int q = emp / a; res += q * b; emp = r; emp += q * b; } return res;}
int main(){ int tot, a, b; // tot refers to the total number of the bottles which are consum scanf("%d%d%d", &tot, &a, &b); // a refers to the number of the bottles with which you can get some new bottles of bear int l = 1; // b emmm I believe you have get it. int r = 1000000; // this could be alternated accordingly while (l < r - 1) { int mid = (l + r) / 2; if (bear(mid, a, b) > tot) r = mid; else l = mid; } cout<< l <<" "<< bear(l, a, b)<<endl; cout<< r <<" "<< bear(r, a, b)<<endl;}

贴吧会吞格式
我放个pastebin

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做点补充，改进了二分法的写法，把注释翻译为小学生都看得懂的中文并且修正了某奇怪的拼写错误
#include <bits/stdc++.h>
using namespace std;
int beer(int x, int a, int b){ //本段代码计算购买x瓶啤酒实际可以喝到的酒数
int emp = x;
int res = x;
int newb;
while (emp >= a) {
int r = emp % a;
int q = emp / a;
res += q * b;
emp = r;
emp += q * b;
}
return res;
}
int main(){
int tot, a, b; // tot 表示总的消费的啤酒数
scanf("%d%d%d", &tot, &a, &b); // a和b表示可以用a个空瓶换b瓶啤酒
int l = 0; //为了以防**题目，你甚至应当考虑不用买酒的情况
int r = tot; // 哪怕不能额外换啤酒tot瓶也够了，所以这个上界没有问题
int ans;
while (l < r) { //二分法的可行性基于这样一个事实：
cout<<l<<' '<<r<<endl; //如果买n瓶酒可以满足喝酒的要求，那么买更多的酒肯定也是可以的
int mid = (l + r) / 2; //反之如果买n瓶酒不能满足要求，那么买更少的酒肯定也不行。
if (beer(mid, a, b) > tot) r = mid-1;
else l = mid+1,ans=mid;
}
cout<< ans <<" "<< beer(ans, a, b)<<endl;
cout<<ans+1<<" "<<beer(ans+1,a, b)<<endl;
}

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