1013 Counterfeit Dollar

Counterfeit Dollar

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27942		Accepted: 8742

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

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/* 一个重量数组int num[13] 先把全部硬币重量赋值为1 遍历全部硬币 { 每次把其中一个硬币重量赋值为2，检查3个表达式是否出现矛盾，如果没有矛盾就找到解了，break 再把同一个硬币重量赋值为0，检查3个表达式是否出现矛盾，如果没有矛盾就找到解了，break 再把同一个硬币重量恢复1，检查下一个 } 所谓矛盾，只要把表达式两边硬币重量加起来比较看看是不是和even、up、down一致 现在下标对应的就是假币 而这个下标对应的W[]值说明了它是heavy（为2时）或是light（为0时）*/ #include<iostream> #include<string> using namespace std; int num[13]; string a,b,c,d,e,f; string s[3]= {"even","up","down"}; string h1,h2,h3; int cal(string s) { int sum=0; for(int i=0; i<s.length(); i++) sum+=num[s[i]-'A'+1]; return sum; } int main() { int x,y; int cas; cin>>cas; while(cas--) { cin>>a>>b>>h1; cin>>c>>d>>h2; cin>>e>>f>>h3; bool flag=true; for(int i=1; i<=12; i++) num[i]=1; for(int i=1; i<=12; i++) { num[i]=2; x=cal(a); y=cal(b); if((x>y&&h1!=s[1])||(x<y&&h1!=s[2])||(x==y&&h1!=s[0])) { num[i]=1; continue; } x=cal(c); y=cal(d); if((x>y&&h2!=s[1])||(x<y&&h2!=s[2])||(x==y&&h2!=s[0])) { num[i]=1; continue; } x=cal(e); y=cal(f); if((x>y&&h3!=s[1])||(x<y&&h3!=s[2])||(x==y&&h3!=s[0])) { num[i]=1; continue; } printf("%c is the counterfeit coin and it is heavy./n",'A'+i-1); flag=false; break; } if(flag) { for(int i=1; i<=12; i++) { num[i]=0; x=cal(a); y=cal(b); if((x>y&&h1!=s[1])||(x<y&&h1!=s[2])||(x==y&&h1!=s[0])) { num[i]=1; continue; } x=cal(c); y=cal(d); if((x>y&&h2!=s[1])||(x<y&&h2!=s[2])||(x==y&&h2!=s[0])) { num[i]=1; continue; } x=cal(e); y=cal(f); if((x>y&&h3!=s[1])||(x<y&&h3!=s[2])||(x==y&&h3!=s[0])) { num[i]=1; continue; } printf("%c is the counterfeit coin and it is light./n",'A'+i-1); flag=false; break; } } } return 0; }