Counterfeit Dollar

最新推荐文章于 2022-06-25 18:10:36 发布

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最新推荐文章于 2022-06-25 18:10:36 发布

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分类专栏： ACM习题及反思文章标签： ACM

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本文介绍了一种通过三次称重找出十二枚银币中唯一一枚假币的方法，并附带了一个简洁高效的C++程序实现。文章详细解释了如何利用平衡秤进行称重，通过比较重量来确定假币及其轻重。

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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.

/* 
   功能Function Description:     POJ-1013 Counterfeit Dollar  策略问题
   开发环境Environment:          DEV C++ 4.9.9.1
   技术特点Technique:
   版本Version:
   作者Author:                     可笑痴狂
   日期Date:                     20120726
   备注Notes:
        由于该题必然有确定的判定，所以简单的可以这么想：
        设置一个标记数组，每次称球的时候，如果是"even"则把对应的设置为"真东西",即置为10，
        如果是"up"或"donw" 则把未分辨出真假的硬币中flag数组中对应的 ++ 或者 --，直到最后。那么操作之后，
        flag中存在没有辨认出真的，就是一系列的例如： -1,-2,1,2,3等数值，那么
        假东西就是其中绝对值最大的那个！！------被怀疑次数最多，所以它为假。
*/
#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int n,k,i,t,len,max;
    char s1[13],s2[13],temp[10];
    int flag[12];
    scanf("%d",&n);
    while(n--)
    {
        memset(flag,0,sizeof(flag));
        for(t=1;t<=3;++t)
        {
            scanf("%s%s%s",s1,s2,temp);
            len=strlen(s1);              //两边个数肯定相等，所以两边长度也一样  
            if(strcmp(temp,"even")==0)
            {
                for(i=0;i<len;++i)
                {
                    flag[s1[i]-'A']=10;  //说明两边都为真，将标记数组置为10
                    flag[s2[i]-'A']=10;

                }
            }
            else if(strcmp(temp,"up")==0)//说明左边重
            {
                for(i=0;i<len;++i)
                {
                    if(flag[s1[i]-'A']!=10)
                        ++flag[s1[i]-'A'];  //左边加一
                    if(flag[s2[i]-'A']!=10)
                        --flag[s2[i]-'A'];  //右边减一

                }
            }
            else
            {
                for(i=0;i<len;++i)      //说明右边重
                {
                    if(flag[s1[i]-'A']!=10)
                        --flag[s1[i]-'A'];  //左边减一
                    if(flag[s2[i]-'A']!=10)
                        ++flag[s2[i]-'A'];  //右边加一
                }
            }
        }
        max=0;
        k=0;
        for(i=0;i<12;++i)
        {
            if(flag[i]==10)
                continue;
            if(max<=abs(flag[i]))
            {
                max=abs(flag[i]);
                k=i;
            }
        }
        if(flag[k]>0)
            printf("%c is the counterfeit coin and it is heavy.\n",'A'+k);
        else
            printf("%c is the counterfeit coin and it is light.\n",'A'+k);
    }
    return 0;
}

一道题往往有多解，不要被一道相当麻烦的解限制住，真正比赛时，当机立断，换个思路。