假币问题

 

题解：

　　思路：将A~L每个字母都枚举一遍（每个字母都有轻和重两种状态），看看是否符合输入数据条件

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 string Left[3], Right[3], Result[3];
 5 
 6 bool check(char coin, bool weight) // weight = 1表示判断轻，0表示重
 7 {
 8     string tmp;
 9     tmp.push_back(coin);
10 
11     for (int i = 0; i < 3; ++i)
12     {
13         string l = Left[i], r = Right[i];
14         if (!weight)
15             swap(l, r);
16 
17         switch (Result[i][0])
18         {
19         case 'e':
20             if (l.find(tmp) != string::npos || r.find(tmp) != string::npos)
21                 return 0;
22             break;
23         case 'u':
24             if (r.find(tmp) == string::npos)
25                 return 0;
26             break;
27         case 'd':
28             if (l.find(tmp) == string::npos)
29                 return 0;
30             break;
31         default:
32             break;
33         }
34     }
35     return 1;
36 }
37 
38 int main()
39 {
40     int t;
41     cin >> t;
42     while (t--)
43     {
44         for (int i = 0; i < 3; ++i)
45             cin >> Left[i] >> Right[i] >> Result[i];
46         for (char now = 'A'; now <= 'L'; ++now)
47         {
48             if (check(now, 1))
49             {
50                 cout << now << " is the counterfeit coin and it is light. " << endl;
51                 break;
52             }
53             if (check(now, 0))
54             {
55                 cout << now << " is the counterfeit coin and it is heavy. " << endl;
56                 break;
57             }
58         }
59     }
60     return 0;
61 }

 

以下是python代码：

 1 def check(coin, weight):
 2     tmp = coin
 3 
 4     for i in range(3):
 5         l = Left[i] if weight == 1 else Right[i]  # 类似于c中的三目运算符
 6         r = Right[i] if weight == 1 else Left[i]
 7 
 8         if Result[i][0] == 'e':
 9             if tmp in l or tmp in r:
10                 return False
11         elif Result[i][0] == 'u':
12             if tmp not in r:
13                 return False
14         elif Result[i][0] == 'd':
15             if tmp not in l:
16                 return False
17 
18     return True
19 
20 
21 t = int(input())
22 while t > 0:
23     t -= 1
24     Left = [0] * 3
25     Right = [0] * 3
26     Result = [0] * 3
27 
28     for i in range(3):
29         Left[i], Right[i], Result[i] = input().split()  # 注意这种输入法
30     letter = 'ABCDEFGHIJKL'
31     for now in letter:
32         if check(now, 1):
33             print("{} is the counterfeit coin and it is light. ".format(now))
34             break
35         elif check(now, 0):
36             print(now + " is the counterfeit coin and it is heavy. ")
37             break

 

posted @ 2023-07-11 22:55  上原歩夢  阅读(50)  评论(0)    收藏  举报
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