codeforces Gym 100187A A. Potion of Immortality
A. Potion of Immortality
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100187/problem/A
Description
The world famous scientist Innokentiy has just synthesized the potion of immortality. Unfortunately, he put the flask with this potion on the shelf where most dangerous poisons of all time were kept. Now there are n flasks on this shelf, and the scientist has no idea which of them contains the potion of immortality.
Fortunately, Innokentiy has an infinite amount of rabbits. But the scientist doesn't know how exactly these potions affect rabbits. There is the only thing he knows for sure. If rabbit tastes exactly k potions from the flasks on the shelf, it will survive if there was the immortality potion among them and die otherwise. If rabbit tastes the number of potions different from k, the result will be absolutely unpredictable, so the scientist won't make such experiments no matter what.
The scientist intends to minimize the amount of lost rabbits while searching for the potion of immortality. You should determine how many rabbits he has to sacrifice in the worst case.
Input
The only line contains two integers separated by space — n and k (1 ≤ n ≤ 2000, 1 ≤ k ≤ n) — the number of flasks on the Innokentiy's shelf and the number of potions Innokentiy will give to a single rabbit to taste.
Output
If the scientist cannot determine which flusk contains the potion of immortality, output «-1». Otherwise, output a single integer — the minimal number of rabbits that Innokentiy will sacrifice in the worst case while searching for the potion of immortality.
Sample Input
3 2
Sample Output
1
HINT
题意
有n瓶药,1瓶好的,其他都是毒药,一次性喝k瓶,问你在最坏情况下,最少死多少只兔子呢?
题解:
答案当然是n-1/k,很显然
讨论一下当n=k=1的情况就好了
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200101
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** ll n,k;
int main()
{
n=read(),k=read();
if(n==)
puts("");
else if(n==k)
puts("-1");
else
{
ll ans1=(n-k);
n--;
if(n%k!=)
n+=k;
printf("%lld\n",min(n/k,ans1));
}
}
codeforces Gym 100187A A. Potion of Immortality的更多相关文章
- codeforces Gym 100500H A. Potion of Immortality 简单DP
Problem H. ICPC QuestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/a ...
- Gym - 100187A A - Potion of Immortality —— 贪心
题目链接:http://codeforces.com/gym/100187/problem/A 题解: 光题意就想了很久:在最坏情况下的最小兔子数.其实就是至少用几只兔子就一定能找出仙药(答案存在的话 ...
- codeforces gym #101161H - Witcher Potion(状压DP)
题目链接: http://codeforces.com/gym/101161/attachments 题意: 总共有n瓶药可供选择 每瓶药可以增加$e_i$点体力,和$p_i$点毒性 每分钟消耗1点毒 ...
- Codeforces Gym 101252D&&floyd判圈算法学习笔记
一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈 ...
- Codeforces Gym 101190M Mole Tunnels - 费用流
题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...
- Codeforces Gym 101623A - 动态规划
题目传送门 传送门 题目大意 给定一个长度为$n$的序列,要求划分成最少的段数,然后将这些段排序使得新序列单调不减. 考虑将相邻的相等的数缩成一个数. 假设没有分成了$n$段,考虑最少能够减少多少划分 ...
- 【Codeforces Gym 100725K】Key Insertion
Codeforces Gym 100725K 题意:给定一个初始全0的序列,然后给\(n\)个查询,每一次调用\(Insert(L_i,i)\),其中\(Insert(L,K)\)表示在第L位插入K, ...
- Codeforces gym 101343 J.Husam and the Broken Present 2【状压dp】
2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Hu ...
- codeforces gym 100553I
codeforces gym 100553I solution 令a[i]表示位置i的船的编号 研究可以发现,应是从中间开始,往两边跳.... 于是就是一个点往两边的最长下降子序列之和减一 魔改树状数 ...
随机推荐
- ctype库试运行
from ctypes import * msvcrt=cdll.msvcrt message_string="Hello world!\n" msvcrt.wprintf(&qu ...
- linux笔记_20150825_linux有什么好处
那么多人在用,linux到底有毛好处? 其实我也不太清楚,有人说免费,可是大家用windows也不要钱的.我想在天朝,要钱的软件不多吧.一个子也不用花.真心感谢为人民服务的那些大牛. 现在,除了在ub ...
- 数往知来C#面向对象〈三〉
C# 基础方法篇 一.复习 1)方法的重载 方法的重载并不是一个方法,实际上方法的重载就是一些不同的 方法,目的是为了方便程序员编码,所以将功能相近的方法命名相同 根据参数,编译器自动的去匹配方法体, ...
- 判断浏览器是IE的几种方式
<script> if(!+[1,])alert("这是ie浏览器"); else alert("这不是ie浏览器"); </script&g ...
- [转] AOP面向切面编程
AOP面向切面编程 AOP(Aspect-Oriented Programming,面向切面的编程),它是可以通过预编译方式和运行期动态代理实现在不修改源代码的情况下给程序动态统一添加功能的一种技术. ...
- TopFreeTheme精选免费模板【20130827】
今天我们整理了一些关于WordPress, Joomla, Drupal, Magento, OpenCart的最新免费模板主题,绝大多数都是来自ThemeForest的响应式风格模板主题.题材多样化 ...
- GitHub--git push出错解决
当使用GitHub进行代码push是,出现了下面的错误,该如何是好? 错误描述: ! [rejected] master -> master (non-fast-forward)e ...
- Camel In Action 阅读笔记 第一部分概述 + 第一章概述 认识Camel
第一部分: 最开始的一小步 Apache Camel 是一个开源集成框架,其目的是让系统集成变得更加简便,在本书的第一章中,我们会为您介绍它并向您展示它是如何在大型企业应用中做好集成工作.您也会了解到 ...
- Mapreduce执行过程分析(基于Hadoop2.4)——(二)
4.3 Map类 创建Map类和map函数,map函数是org.apache.hadoop.mapreduce.Mapper类中的定义的,当处理每一个键值对的时候,都要调用一次map方法,用户需要覆写 ...
- Spark RDD概念学习系列之RDD的操作(七)
RDD的操作 RDD支持两种操作:转换和动作. 1)转换,即从现有的数据集创建一个新的数据集. 2)动作,即在数据集上进行计算后,返回一个值给Driver程序. 例如,map就是一种转换,它将数据集每 ...