poj1013

一道简单题，思路如下：

1、对每个硬币设三个bool位进行标识，分别为平衡even，重heavy，轻light；

2、读入数据，对硬币状态进行标记，这里有一个细节是，如果天枰左偏或右偏，要将没有上天枰的硬币的even位都标记为真；

3、如果硬币even位为真，或者硬币的heavy和light位同为真，那么这个硬币是真硬币，否则就是假硬币。

题设中的“The solution will always be uniquely determined.”为这种解法提供了保证。

#include <iostream>
#include <string>
using namespace std;

struct Poss{
    bool even;
    bool heavy;
    bool light;
};

int main()
{
    int n;
    cin >> n;
    while (n--){
        string left, right, result;
        Poss arr[12];
        memset(arr, 0, sizeof(arr));
        for (int i = 0; i < 3; i++){
            cin >> left >> right >> result;
            int len = left.length();
            if (result == "even"){
                for (int i = 0; i < len; i++){
                    arr[left[i] - 'A'].even = true;
                    arr[right[i] - 'A'].even = true;
                }
            }
            else{
                bool on[12] = { false };
                for (int i = 0; i < len; i++){
                    on[left[i] - 'A'] = true;
                    on[right[i] - 'A'] = true;
                }
                for (int i = 0; i < 12; i++){
                    if (!on[i]){
                        arr[i].even = true;
                    }
                }
                if (result == "up"){
                    for (int i = 0; i < len; i++){
                        arr[left[i] - 'A'].heavy = true;
                        arr[right[i] - 'A'].light = true;
                    }
                }
                else{
                    for (int i = 0; i < len; i++){
                        arr[right[i] - 'A'].heavy = true;
                        arr[left[i] - 'A'].light = true;
                    }
                }
            }
        }
        for (int i = 0; i < 12; i++){
            if (arr[i].even || (arr[i].heavy && arr[i].light))
                continue;
            if (arr[i].heavy){
                cout << char('A' + i) << " is the counterfeit coin and it is heavy." << endl;
                break;
            }
            else if (arr[i].light){
                cout << char('A' + i) << " is the counterfeit coin and it is light." << endl;
                break;
            }
        }
    }
    return 0;
}