POJ 1013 模拟题硬币称量-优快云博客

POJ 1013

Counterfeit Dollar

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27117		Accepted: 8460

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

East Central North America 1998

AC代码

//经典模拟题 //关键思路还是假想如果硬币重或者轻时放在左右两边24种情况应该的results，一一对比 //这里先假设硬币偏重，0表示不在，1表示在左边，-1表示在右边 //硬币偏轻的情况就恰好与预想情况相反 #include <iostream> #include <string> #include <fstream> #include <memory.h> using namespace std; int main(){ int exp[12][3]; int result[3]; int n,i,j; string s1,s2,s3; cin>>n; while(n--){ memset(exp,0,sizeof(exp)); memset(result,0,sizeof(result)); //下面开始三次试验，模拟出每次的结果 for (i=0;i<3;i++) { cin>>s1>>s2>>s3; for (j=0;j<s1.size();j++) { exp[s1[j]-'A'][i]=1;//表明此币在左边 exp[s2[j]-'A'][i]=-1;//表明此币在右边 } if (s3 == "even") { result[i] = 0;//表示此时重币应该不在天平上 } else if (s3 == "down") { result[i] = -1;//表示此时重币应该在右边出现才对 } else result[i] = 1; } for (j=0;j<12;j++) { if (exp[j][0] == result[0] && exp[j][1] == result[1] && exp[j][2] == result[2]) { cout<<char('A'+j)<<" is the counterfeit coin and it is heavy."<<endl; break; } else if(exp[j][0] == -result[0] && exp[j][1] == -result[1] && exp[j][2] == -result[2]) { cout<<char('A'+j)<<" is the counterfeit coin and it is light."<<endl; break; } } } return 0; }