POJ 1013:Counterfeit Dollar
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 42028 | Accepted: 13369 |
Description
from the other coins but Sally does not know if it is heavier or lighter than the real coins.
Happily, Sally has a friend who loans her a very accurate balance scale. The friend will permit Sally three weighings to find the counterfeit coin. For instance, if Sally weighs two coins against each other and the scales balance then she knows these two coins
are true. Now if Sally weighs
one of the true coins against a third coin and the scales do not balance then Sally knows the third coin is counterfeit and she can tell whether it is light or heavy depending on whether the balance on which it is placed goes up or down, respectively.
By choosing her weighings carefully, Sally is able to ensure that she will find the counterfeit coin with exactly three weighings.
Input
by two strings of letters and then one of the words ``up'', ``down'', or ``even''. The first string of letters will represent the coins on the left balance; the second string, the coins on the right balance. (Sally will always place the same number of coins
on the right balance as on the left balance.) The word in the third position will tell whether the right side of the balance goes up, down, or remains even.
Output
Sample Input
1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even
Sample Output
K is the counterfeit coin and it is light.
题意是从A到L的dollar,其中有一个跟他们其它的不同,可能是轻了也可能是重了。给了三个在秤上的关系,要你判断出哪一个是伪造的dollar,是轻是重。
做的时候还WA了几次,但AC之后觉得这个题真是水啊。。。不知道为什么只要自己好不容易A完之后就觉得这个题真的很水,为什么当时没有想到呢?就是这种感觉。
咳咳,因为这个题就A到L之中有问题嘛,数据量太小,导致枚举就OK了。所以从A开始就一个一个试吧,如果在even中的语句里出现了,那就排除它的嫌疑。在up语句中出现了,它的flag加3,代表可能是轻的标记。在down语句中出现了,它自己的flag加4,代表可能是重的标记。
然后对12个砝码标记完之后,有标记为1的排除,标记为7 10 11这样的排除,因为不可能出现一个砝码既轻又重的情况,然后找最独特的那一个就是伪造的砝码了。输出即可。
代码:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
using namespace std; int num[15];
int Test,i,j,flag[15],sum=0;
string left_s[5],right_s[5],bala[5];
char temp_c; void solve()
{
if(bala[1]=="up")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"heavy."<<endl;
return ;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"light."<<endl;
return;
}
}
}
if(bala[2]=="up")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"heavy."<<endl;
return;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"light."<<endl;
return;
}
}
}
if(bala[3]=="up")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"heavy."<<endl;
return ;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"light."<<endl;
return ;
}
}
} if(bala[1]=="down")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"light."<<endl;
return;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"heavy."<<endl;
return ;
}
}
}
if(bala[2]=="down")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"light."<<endl;
return;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"heavy."<<endl;
return ;
}
}
}
if(bala[3]=="down")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"light."<<endl;
return;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
sum=1;
cout<<temp_c<<" is the counterfeit coin and it is ";
cout<<"heavy."<<endl;
return ;
}
}
}
} bool pend()
{
int pend_flag=0;
if(bala[1]=="up")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
pend_flag=1;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
pend_flag=1;
}
}
if(pend_flag==0)
return false;
}
if(bala[2]=="up")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
pend_flag=1;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
pend_flag=1;
}
}
if(pend_flag==0)
return false;
}
if(bala[3]=="up")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
pend_flag=1;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
pend_flag=1;
}
}
if(pend_flag==0)
return false;
} if(bala[1]=="down")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
pend_flag=1;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
pend_flag=1;
}
}
if(pend_flag==0)
return false;
}
if(bala[2]=="down")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
pend_flag=1;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
pend_flag=1;
}
}
if(pend_flag==0)
return false;
}
if(bala[3]=="down")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
pend_flag=1;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
pend_flag=1;
}
}
if(pend_flag==0)
return false;
}
return true;
}
int main()
{
cin>>Test; while(Test--)
{
memset(flag,0,sizeof(flag));
memset(num,0,sizeof(num));
sum=0; for(i=1;i<=3;i++)
{
cin>>left_s[i]>>right_s[i]>>bala[i];
}
for(temp_c='A';temp_c<='L';temp_c++)
{
if(bala[1]=="even")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
flag[temp_c-'A']=1;
break;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
flag[temp_c-'A']=1;
break;
}
}
}
if(bala[2]=="even")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
flag[temp_c-'A']=1;
break;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
flag[temp_c-'A']=1;
break;
}
}
}
if(bala[3]=="even")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
flag[temp_c-'A']=1;
break;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
flag[temp_c-'A']=1;
break;
}
}
} if(bala[1]=="up")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] += 4;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] += 3;
}
}
}
if(bala[2]=="up")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] += 4;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] += 3;
}
}
}
if(bala[3]=="up")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=4;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=3;
}
}
} if(bala[1]=="down")
{
for(j=0;j<left_s[1].length();j++)
{
if(left_s[1][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=3;
}
}
for(j=0;j<right_s[1].length();j++)
{
if(right_s[1][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=4;
}
}
}
if(bala[2]=="down")
{
for(j=0;j<left_s[2].length();j++)
{
if(left_s[2][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=3;
}
}
for(j=0;j<right_s[2].length();j++)
{
if(right_s[2][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=4;
}
}
}
if(bala[3]=="down")
{
for(j=0;j<left_s[3].length();j++)
{
if(left_s[3][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=3;
}
}
for(j=0;j<right_s[3].length();j++)
{
if(right_s[3][j]==temp_c)
{
if(flag[temp_c-'A']==1)
continue;
else
flag[temp_c-'A'] +=4;
}
}
}
}
for(temp_c='A';temp_c<='L';temp_c++)
{
num[flag[temp_c-'A']]++;
}
for(i=0;i<=12;i++)
{
if(num[i]==1 && i!=7 && i!=10 && i!=11)
{
for(temp_c='A';temp_c<='L';temp_c++)
{
if(flag[temp_c-'A']==i)
{
if(pend())
{
solve();
}
}
}
}
}
} return 0;
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
POJ 1013:Counterfeit Dollar的更多相关文章
- 九度OJ 1150:Counterfeit Dollar(假美元) (分析、检验)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:485 解决:215 题目描述: Sally Jones has a dozen Voyageur silver dollars. Howev ...
- Poj 1013 Counterfeit Dollar / OpenJudge 1013(2692) 假币问题
1.链接地址: http://poj.org/problem?id=1013 http://bailian.openjudge.cn/practice/2692 http://bailian.open ...
- POJ 1013 Counterfeit Dollar
Counterfeit Dollar Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 36206 Accepted: 11 ...
- Counterfeit Dollar 分类: POJ 2015-06-12 15:28 19人阅读 评论(0) 收藏
Counterfeit Dollar Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 41559 Accepted: 13 ...
- POJ1013 Counterfeit Dollar
题目来源:http://poj.org/problem?id=1013 题目大意:有12枚硬币,其中有一枚假币.所有钱币的外表都一样,所有真币的重量都一样,假币的重量与真币不同,但我们不知道假币的重量 ...
- POJ 1013 小水题 暴力模拟
Counterfeit Dollar Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 35774 Accepted: 11 ...
- Counterfeit Dollar -----判断12枚钱币中的一个假币
Counterfeit Dollar Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u ...
- poj1013.Counterfeit Dollar(枚举)
Counterfeit Dollar Time Limit: 1 Sec Memory Limit: 64 MB Submit: 415 Solved: 237 Description Sally ...
- POJ 3321:Apple Tree + HDU 3887:Counting Offspring(DFS序+树状数组)
http://poj.org/problem?id=3321 http://acm.hdu.edu.cn/showproblem.php?pid=3887 POJ 3321: 题意:给出一棵根节点为1 ...
随机推荐
- impala invalidate metadata和impala-shell -r作用相同
impala的invalidate metadata内部命令,是否和外部命令impala-shell -r的作用相同的? 这个问题的回答: 在invalidate metadata 和 impala- ...
- 第3节 sqoop:3、sqoop的入门测试使用
3.5. Sqoop的数据导入 “导入工具”导入单个表从RDBMS到HDFS.表中的每一行被视为HDFS的记录.所有记录都存储为文本文件的文本数据(或者Avro.sequence文件等二进制数据) 列 ...
- 基于 若依 ,或者使用 LayUi ,用来展示数据表,同时要 转换字典数据时的转换的建议
原作者在做这个项目时,他把所有数据字典都放在数据库表中了,这种方法的确比较好,适用于中大型项目,方便统一管理字典:而且优点突出,字典值变化后不需要调整前端代码: 但是在实际开发项目中,一些小型的项目, ...
- 安卓Recycleview简单的网格布局-初学者的关键点
导包 def supportVersion = '28.0.0' 定义常量我的sdk版本为28 implementation "com.android.support:recyclervie ...
- Django(八)模型:Mysql8.0日志文件配置开启/关闭、查询
Mysql日志文件开启.配置.查看 mysql.log是mysql的日志文件,里面记录的对MySQL数据库的操作记录.默认情况下mysql的日志文件没有产生,需要修改mysql的配置文件,步骤如下: ...
- Spark学习入门(让人看了想吐的话题)
这是个老生常谈的话题,大家是不是看到这个文章标题就快吐了,本来想着手写一些有技术深度的东西,但是看到太多童鞋卡在入门的门槛上,所以还是打算总结一下入门经验.这种标题真的真的在哪里都可以看得到,度娘一搜 ...
- centos7上安装mysql8
话不多说仍然是更换虚拟机的系列安装. 一.首先下载最先版的mysql.到官网下载https://dev.mysql.com/downloads/file/?id=477146 下载后上传linux到相 ...
- 19 03 13 关于 scrapy 框架的 对环球网的整体爬取(存储于 mongodb 数据库里)
关于 spinder 在这个框架里面 和不用数据库 相同 # -*- coding: utf-8 -*- import scrapy from yang_guan.items import ...
- Oracle 序列(查询序列的值,修改序列的值)
1.序列的语法形式 create sequence 序列名 increment by n start with n maxvalue n | nomaxvalue minvalue n | nomin ...
- JAVA - 创建SpringBoot项目
JAVA - 创建SpringBoot项目 Spring Boot是由Pivotal团队提供的全新框架,设计目的是用来简化新Spring应用的初始搭建以及开发过程.它主要推崇的是'消灭配置’,实现零配 ...