Counterfeit Dollar(POJ1013

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.
该题的题意大体就是:Sally Jones有12个硬币,而其中有一个假币,他的朋友借给他了一个天平,用天平去称量这12个硬币,称三次就能找出假币.
思路:先初始所有的硬币的重量值为0.若两边的硬币重量相等就说明这两边的硬币都是真币并进行标记且其值仍为0;若不相等,轻的一边的硬币的重量值就减1,重的一边的硬币的重量值就加1.最后遍历所有的硬币,谁的重量值的jue对值最大谁就是假币.

Sample Input

1 ABCD EFGH even ABCI EFJK up ABIJ EFGH even 

Sample Output

K is the counterfeit coin and it is light. 
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <math.h>using namespace std;int main(){    int T;    int flag[200];    bool ff[200];    scanf("%d",&T);    while(T--)    {        memset(flag,0,sizeof(flag));        memset(ff,0,sizeof(ff));        char left[20],right[20],f[20];        for(int i=0; i<3; i++)        {            scanf("%s %s %s",left,right,f);            int len=strlen(left);            if(strcmp(f,"even")==0)                                             for(int j=0; j<len; j++)                {                    flag[left[j]]=0;                    ff[left[j]]=true;                                        //标记真币                    flag[right[j]]=0;                    ff[right[j]]=true;                }            else if(strcmp(f,"up")==0)                for(int j=0; j<len; j++)                {                    if(!ff[left[j]])                        flag[left[j]]++;                    if(!ff[right[j]])                        flag[right[j]]--;                }            else if(strcmp(f,"down")==0)                for(int j=0; j<len; j++)                {                    if(!ff[left[j]])                        flag[left[j]]--;                    if(!ff[right[j]])                        flag[right[j]]++;                }        }        int sum=0,poi,k;        for(int i=64; i<80; i++)        {            if(flag[i]!=0&&sum<abs(flag[i]))            {                sum=abs(flag[i]);                                            //遍历所有硬币重量值找假币                poi=i;                k=flag[i];            }        }        if(k<0)                                                                     //若其重量值小于0说明其比真币轻;反之,比真币重            printf("%c is the counterfeit coin and it is light.\n",poi);        else            printf("%c is the counterfeit coin and it is heavy.\n",poi);    }    return 0;}