POJ - Counterfeit Dollar 题解

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挺考智力的题目。

思路：

1 如果是假币，那么每次都必定引起天平的不平衡

2 如果天平平横，那么全部都肯定是真币

利用这个特性，利用hash表，就能写出很简洁的程序。

如果使用枚举，那么会（轻松？）过百行的代码的。

当然其实题目给出了条件：一定可以找出唯一的假币的。

如果没有这个条件，那么是不一定可以三次称，就能确定结果的。

下面程序参考了别人的：

http://www.cnblogs.com/orangeman/archive/2009/07/10/1520663.html

这个家伙的思路也不错，而且给出的额外的cases都是正确的：http://www.slyar.com/blog/poj-1013-c.html

原题：

http://poj.org/problem?id=1013

#include <string>#include <vector>#include <iostream>using namespace std;class CounterfeitDollar{const static int ALPHA = 12;const static int TIMES = 3;string s1, s2, s3;public:CounterfeitDollar(){int n;cin>>n;while (n--){int tbl[ALPHA][TIMES] = {0};int balance[TIMES] = {0};for (int i = 0; i < TIMES; i++){cin>>s1>>s2>>s3;for (int j = 0; j < (int)s1.size(); j++){tbl[s1[j] - 'A'][i] = -1;tbl[s2[j] - 'A'][i] = 1;}if ('e' == s3[0]) balance[i] = 0;else if ('d' == s3[0]) balance[i] = 1;else balance[i] = -1;}for (int i = 0; i < ALPHA; i++){if (balance[0] == -tbl[i][0] && balance[1] == -tbl[i][1] &&balance[2] == -tbl[i][2]){cout<<char('A'+i)<<" is the counterfeit coin and it is light.\n";break;}else if (balance[0] == tbl[i][0] && balance[1] == tbl[i][1] && balance[2] == tbl[i][2]){cout<<char('A'+i)<<" is the counterfeit coin and it is heavy.\n";break;}}}//while (n--)}};int counterfeitDollar(){CounterfeitDollar();return 0;}

更新一个新的解法：

1 如果是even，那么所有是真币，所以设置为10

2 如果硬币在轻的一方，那么--，如果在重的一方，那么++

3 最后找到差别最大的硬币，那么就为假币

4 如果假币为负数，那么就比真币轻，如果为正，那么就比真币重。

这回是原创的程序的，感觉比前面的更加容易理解。

#include <string>#include <vector>#include <iostream>using namespace std;class CounterfeitDollar_2{const static int ALPHA = 12;const static int TIMES = 3;string s1, s2, s3;public:CounterfeitDollar_2(){int n;cin>>n;while (n--){int tbl[ALPHA] = {0};int balance[TIMES] = {0};for (int i = 0; i < TIMES; i++){cin>>s1>>s2>>s3;if ('e' == s3[0]){for (int i = 0; i < (int)s1.size(); i++){tbl[s1[i] - 'A'] = 10;tbl[s2[i] - 'A'] = 10;}}else if ('d' == s3[0]){for (int i = 0; i < (int)s1.size(); i++){if (tbl[s1[i] - 'A'] != 10) tbl[s1[i] - 'A']--;if (tbl[s2[i] - 'A'] != 10) tbl[s2[i] - 'A']++;}}else{for (int i = 0; i < (int)s1.size(); i++){if (tbl[s1[i] - 'A'] != 10) tbl[s1[i] - 'A']++;if (tbl[s2[i] - 'A'] != 10) tbl[s2[i] - 'A']--;}}}int id = 0, diff = 0;for (int i = 0; i < ALPHA; i++){if (tbl[i] != 10 && diff < abs(tbl[i])){diff = abs(tbl[i]);id = i;}}cout<<char('A'+id);if (tbl[id] < 0) cout<<" is the counterfeit coin and it is light.\n";else cout<<" is the counterfeit coin and it is heavy.\n";}//while (n--)}};int main(){CounterfeitDollar_2();return 0;}

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