poj 1013

Counterfeit Dollar

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 36646 Accepted: 11717

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. 

题目大意：

有十二个硬币 ，其中有一个是假币，用大写字母A-L分别标号， 假币可能重也可能轻， 任务就是  通过三次称重找出假币，并判断是轻还是重了。 每次称重左右两边的硬币数相等，要注意的是  up,even ,down 针对的是右边的部分的状态

解题思路：

假币是重还是轻

1.重了  有两种可能 1. 假币在左边，且右边的状态为up   2. 假币在右边 右边状态为down

2.轻了  也有两种就不多说了  和上面相反

可以针对两种情况写两个函数 对每一个字母逐一判断

代码如下：

#include <iostream>#include <cstring>using namespace std;char Left[3][7] ={0}, Right[3][7] = {0};int result[3];bool isLight(char);bool isHeavy(char);int main(){char ch, resul[5];int n, i;cin >> n;while(n--){for(i = 0; i < 3; i++){cin >> Left[i] >> Right[i] >> resul;if(strcmp(resul, "up") == 0)result[i] = 1;if(strcmp(resul, "even") == 0)result[i] = 2;if(strcmp(resul, "down") == 0)result[i] = 3;}for(ch = 'A'; ch <= 'L'; ch++){if(isLight(ch))//如果该币是假币  且轻了{cout << ch << " is the counterfeit coin and it is light.\n";break;}else if(isHeavy(ch))//如果该币是假币 且重了{cout << ch << " is the counterfeit coin and it is heavy.\n";break;}}}return 0;}bool isLight(char ch){int i;for(i = 0; i < 3; i++){switch(result[i]){case 1:if(strchr(Right[i], ch) ==0)//轻了而且状态为up，那么如果该币为假币则必定在右边部分   return 0;break;case 2:if(strchr(Left[i], ch) != 0 || strchr(Right[i], ch) != 0)//轻了状态为even，则该币必定不在任何一边   return 0;break;case 3:if(strchr(Left[i], ch) == 0)//轻了且状态为down 那么如果该币为假币则必定在左边部分return 0;break;}}return 1;//当以上情况都不满足时   则该币必定为假币且轻了}bool isHeavy(char ch){int i;for(i = 0; i < 3; i++){switch(result[i]){case 1:if(strchr(Left[i], ch) == 0)   return 0;break;case 2:if(strchr(Right[i], ch) != 0 || strchr(Left[i], ch) != 0)   return 0;break;case 3:if(strchr(Right[i], ch) == 0)   return 0;break;}}return 1;}