Problem description

Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.

Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.

Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.

Input

The single line contains two integers, a and b (1 ≤ a ≤ 1000; 2 ≤ b ≤ 1000).

Output

Print a single integer — the number of hours Vasily can light up the room for.

Examples

Input

4 2

Output

7

Input

6 3

Output

8

Note

Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.

解题思路:题目的意思就是有a根蜡烛,每烧完一根蜡烛需要一个小时,而且每烧完b(b<=a)根蜡烛就能产生一根新的蜡烛,求这些蜡烛全部烧完需要多少个小时,水过!

AC代码:

 #include<bits/stdc++.h>
using namespace std;
int main(){
int a,b;cin>>a>>b;
for(int i=;i<=a;++i)
if(i%b==)a++;
cout<<a<<endl;
return ;
}

C - New Year Candles的更多相关文章

  1. codeforces A. New Year Candles 解题报告

    题目链接:http://codeforces.com/problemset/problem/379/A 题目意思:给定a支蜡烛(每支蜡烛可以燃烧1小时),可以在燃尽的a支蜡烛中看能组成多少组b支蜡烛, ...

  2. uvalive5810 uva12368 Candles

    题意:每组数据给出n个数,每个数在1-100,问组成这些数的蜡烛的权值的最小值.权值=把选的蜡烛从大到小排列组成的数 组成方式:比如有1 3两个蜡烛 可以组成13(1和3)或4(1+3) 只有一个加号 ...

  3. ARC 101 C - Candles

    题面在这里! 显然直接枚举左端点(右端点)就OK啦. #include<cstdio> #include<cstdlib> #include<algorithm> ...

  4. Gym - 101635K:Blowing Candles (简单旋转卡壳,求凸包宽度)

    题意:给定N个点,用矩形将所有点覆盖,要求矩形宽度最小. 思路:裸体,旋转卡壳去rotate即可. 最远距离是点到点:宽度是点到边. #include<bits/stdc++.h> #de ...

  5. Gym101635K Blowing Candles

    题目链接:http://codeforces.com/gym/101635 题目大意: 推荐一篇文章:https://blog.csdn.net/wang_heng199/article/detail ...

  6. Solution -「ABC 219H」Candles

    \(\mathcal{Description}\)   Link.   有 \(n\) 支蜡烛,第 \(i\) 支的坐标为 \(x_i\),初始长度为 \(a_i\),每单位时间燃烧变短 \(1\) ...

  7. Inverted sentences

    And ever has it been that love knows not its own depth until the hour of separation.  除非临到了别离的时候,爱永远 ...

  8. Good Bye 2013 A

    A. New Year Candles time limit per test 1 second memory limit per test 256 megabytes input standard ...

  9. HDU 5768:Lucky7(中国剩余定理 + 容斥原理)

    http://acm.hdu.edu.cn/showproblem.php?pid=5768 Lucky7 Problem Description   When ?? was born, seven ...

随机推荐

  1. 关于while((c=getchar()))的一些应用与思考

    最近做题发现一个特别牛逼又特别神奇的读取入字符串的方法 while((c=getchar())!=....) { //do something } 为什么说强大呢,首先这个表达式对空格回车都不怕,他不 ...

  2. wget扒网站

    wget神奇操作   整站复制 只限静态网页 wget -P 指定下载路径 -p 获取显示HTML页面所需的所有图像 -k  使链接指向本地文件 -H  递归时转到外部主机. wget --mirro ...

  3. nlogn求LIS(树状数组)

    之前一直是用二分 但是因为比较难理解,写的时候也容易忘记怎么写. 今天比赛讲评的时候讲了一种用树状数组求LIS的方法 (1)好理解,自然也好写(但代码量比二分的大) (2)扩展性强.这个解法顺带求出以 ...

  4. HDU 5343 MZL's Circle Zhou

    MZL's Circle Zhou Time Limit: 1000ms Memory Limit: 131072KB This problem will be judged on HDU. Orig ...

  5. PatentTips - Zero voltage processor sleep state

    BACKGROUND Embodiments of the invention relate to the field of electronic systems and power manageme ...

  6. 233 Matrix 矩阵快速幂

    In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233 ...

  7. [poj2417]Discrete Logging_BSGS

    Discrete Logging poj-2417 题目大意:求$a^x\equiv b(mod\qquad c)$ 注释:O(分块可过) 想法:介绍一种算法BSGS(Baby-Step Giant- ...

  8. Spring Data Jpa-动态查询条件

    /** * * 查看日志列表-按照时间倒序排列 * * @author: wyc * @createTime: 2017年4月20日 下午4:24:43 * @history: * @return L ...

  9. Oracle_Data_Gard Create a physical standby database

    创建之前要对DG的环境有一个总体的规划和了解.                                                   规划 IP 192.168.3.161 192.16 ...

  10. [MongoDB]mongo命令行工具

    1.use dbname 自动创建 2.db.user.find() 空 show collections 空 show dbs 3.db.user.save({name:'',age:20}) db ...