Description

Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coin has a different weight 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

The first line of input is an integer n (n > 0) specifying the number of cases to follow. Each case consists of three lines of input, one for each weighing. Sally has identified each of the coins with the letters A--L. Information on a weighing will be given 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

For each case, the output will identify the counterfeit coin by its letter and tell whether it is heavy or light. The solution will always be uniquely determined.

Sample Input

1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even

Sample Output

K is the counterfeit coin and it is light. 

Source

 
 
#include <stdio.h>
#include <string.h> struct Balance
{
char left[], right[], status[];
}balance[]; bool view[]; int main()
{
int light[], heavy[], t;
scanf("%d", &t);
while(t--)
{
memset(view, , sizeof(view));
memset(light, , sizeof(light));
memset(heavy, , sizeof(heavy));
for(int i = ; i < ; i++)
{
scanf("%s %s %s", balance[i].left, balance[i].right, balance[i].status);
if(balance[i].status[] == 'e')
{
for(int j = ; balance[i].left[j]; j++)
view[balance[i].left[j]-'A'] = ;
for(int j = ; balance[i].right[j]; j++)
view[balance[i].right[j]-'A'] = ;
}
} for(int i = ; i < ; i++)
{
if(balance[i].status[] == 'u')
{
for(int j = ; balance[i].left[j] != '\0'; j++)
if(view[balance[i].left[j]-'A'] == )
heavy[i] |= ( << (balance[i].left[j]-'A')); for(int j = ; balance[i].right[j] != '\0'; j++)
if(view[balance[i].right[j]-'A'] == )
light[i] |= ( << (balance[i].right[j]-'A'));
}
else if(balance[i].status[] == 'd')
{
for(int j = ; balance[i].left[j] != '\0'; j++)
if(view[balance[i].left[j]-'A'] == )
light[i] |= ( << (balance[i].left[j]-'A')); for(int j = ; balance[i].right[j] != '\0'; j++)
if(view[balance[i].right[j]-'A'] == )
heavy[i] |= ( << (balance[i].right[j]-'A'));
}
} int heavy_final = -, light_final = -;
for(int i = ; i < ; i++)
{
if(light[i] != )
light_final &= light[i];
if(heavy[i] != )
heavy_final &= heavy[i];
} if(light_final > )
{
int x = ;
while((light_final & ( << x)) == )
x ++;
printf("%c is the counterfeit coin and it is light.\n", 'A'+x);
}
else
{
int x = ;
while((heavy_final & ( << x)) == )
x ++;
printf("%c is the counterfeit coin and it is heavy.\n", 'A'+x);
}
}
return ;
}

POJ 1013 Counterfeit Dollar 集合上的位运算的更多相关文章

  1. 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 ...

  2. POJ 1013 Counterfeit Dollar

    Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 36206   Accepted: 11 ...

  3. 思维+模拟--POJ 1013 Counterfeit Dollar

    Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver d ...

  4. 深入理解计算机系统(2.2)---布尔代数以及C语言上的位运算

    布尔代数上的位运算 布尔代数是一个数学知识体系,它在0和1的二进制值上演化而来的. 我们不需要去彻底的了解这个知识体系,但是里面定义了几种二进制的运算,却是我们在平时的编程过程当中也会遇到的.这四种运 ...

  5. POJ 2777 Count Color(线段树+位运算)

    题目链接:http://poj.org/problem?id=2777 Description Chosen Problem Solving and Program design as an opti ...

  6. POJ - 3074 Sudoku (搜索)剪枝+位运算优化

    In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For exa ...

  7. [poj 1185] 炮兵阵地 状压dp 位运算

    Description 司令部的将军们打算在N*M的网格地图上部署他们的炮兵部队.一个N*M的地图由N行M列组成,地图的每一格可能是山地(用"H" 表示),也可能是平原(用&quo ...

  8. leetcode上的位运算

    136-只出现过一次的数字 思路:可以考虑到数字以二进制形式存储,当两个不同的数字异或的时候会是true,所以把数组里的数字都一一处理一遍就可以了. class Solution { public: ...

  9. C语言中的位运算的技巧

    一.位运算实例 1.用一个表达式,判断一个数X是否是2的N次方(2,4,8,16.....),不可用循环语句. X:2,4,8,16转化成二进制是10,100,1000,10000.如果减1则变成01 ...

随机推荐

  1. Yii2 自定义Gii生成代码模板

    我们使用一个例子来介绍如何定制代码模板.假设我们想要定制由 model 生成器生成的代码. 我们首先创建一个名为 protected/gii/model/templates/customer 的目录. ...

  2. centos中使用python遇到的几个问题

    用python搞了一个从excel中读取cobbler节点信息并加入cobbler中的脚本,运行的过程中出了不少问题,这里记录下来,方便日后查找! 一.yum install python,我通过这个 ...

  3. ZBar只扫描二维码/条形码

    You can add these codes for ImageScanner scanner.setConfig(0, Config.ENABLE, 0); //Disable all the S ...

  4. requireJS(二)

    一.前言 requireJS(一) 本篇主要整理requirejs的一些用法,相对比较零散. 实例目录 二.优化 requirejs建议我们给每一个模块书写一个js文件.但是这样会增加网站的http请 ...

  5. fscanf函数

    函数定义: int fscanf( FILE *stream, const char *format [, argument ]... ); 以下是csdn的样例: /* FSCANF.C: This ...

  6. &&与&

    if((2>1)&&(4>3))System.out.printf("两边都是true"); else   System.out.println(&qu ...

  7. android开发之merge结合include优化布局

    merge结合include优化android布局,效果不知道,个人感觉使用上也有很大的局限,不过还是了解一下,记录下来. 布局文件都要有根节点,但android中的布局嵌套过多会造成性能问题,于是在 ...

  8. [转]cookie、session、sessionid 与jsessionid

    cookie.session.sessionid 与jsessionid,要想明白他们之间的关系,下面来看个有趣的场景来帮你理解. 我们都知道银行,银行的收柜台每天要接待客户存款/取款业务,可以有几种 ...

  9. Android 5.0 新特性

    Material Design Material Design简介 Material Design是谷歌新的设计语言,谷歌希望寄由此来统一各种平台上的用户体验,Material Design的特点是干 ...

  10. 25、Javascript 事件

    Javascript 事件 是指 Javascript 捕获到用户的操作,并做出正确的相应. Javascript 事件一般与DOM元素绑定. Javascript处理事件的基本机制 1.对DOM元素 ...