地址：http://acm.pku.edu.cn/JudgeOnline/problem?id=1013
Counterfeit Dollar
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 18115 |
|
Accepted: 5528 |
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.
看了discuss里的讨论：设置一个real数组标记even的硬币，hy数组放置怀疑的硬币，其中up的情况s2的硬币++，s1硬币--；down的情况相反，更新完后将同在real和hy里的硬币的hy值置零，最后再查找hy里绝对值最大的一个硬币就是了。
ps：如果一个硬币在一次测试中up，另一次测试中down，反向，则抵消怀疑值。
Source Code
Problem: 1013 User: Headacher
Memory: 412K Time: 0MS
Language: G++ Result: Accepted
Source Code
#include<iostream>
#include<cstring>
using namespace std;
int real[12]=
{0},hy[12]={0};
char s1[10],s2[10],judge[10];
int main()
{ int cas;
scanf("%d",&cas);
while(cas--)
{ memset(real,0,sizeof(real));
memset(hy,0,sizeof(hy));
int i,j,k;
int n=3;
while(n--)
{ scanf("%s %s %s",s1,s2,judge);
if(judge[0]=='e')
{
for(i=0;s1[i]!='\0';i++)
{ real[s1[i]-'A']++;
}
for(i=0;s2[i]!='\0';i++)
{ real[s2[i]-'A']++;
}
}
if(judge[0]=='u')
{ for(i=0;s1[i]!='\0';i++)
{ hy[s1[i]-'A']--;
}
for(i=0;s2[i]!='\0';i++)
{ hy[s2[i]-'A']++;
}
}
if(judge[0]=='d')
{ for(i=0;s1[i]!='\0';i++)
{ hy[s1[i]-'A']++;
}
for(i=0;s2[i]!='\0';i++)
{ hy[s2[i]-'A']--;
}
}
}
for(i=0;i<=11;i++)
{ if(real[i])
hy[i]=0;
}
int max=0;
int v=0;
for(i=0;i<=11;i++)
{ if(max<abs(hy[i]))
{ max=abs(hy[i]);
v=i;
}
}
if(hy[v]>0)
printf("%c is the counterfeit coin and it is light.\n",'A'+v);
else printf("%c is the counterfeit coin and it is heavy.\n",'A'+v);
}
system("pause");
return 0;
}