Given two sets of integers, the similarity of the sets is defined to be /, where N​c​​ is the number of distinct common numbers shared by the two sets, and N​t​​ is the total number of distinct numbers in the two sets. Your job is to calculate the similarity of any given pair of sets.

Input Specification:

Each input file contains one test case. Each case first gives a positive integer N (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by Mintegers in the range [0]. After the input of sets, a positive integer K (≤) is given, followed by Klines of queries. Each query gives a pair of set numbers (the sets are numbered from 1 to N). All the numbers in a line are separated by a space.

Output Specification:

For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

Sample Input:

33 99 87 1014 87 101 5 877 99 101 18 5 135 18 9921 21 3

Sample Output:

50.0%33.3%

Nc为两组数组中的公共数字的个数

Nt为去重后，两组数据的总个数

1 #include 
     
       2 #include 
      
       //用set来去重 3 using namespace std; 4 int N, M, K, d; 5 unordered_set
       
        nums[55]; 6 int main() 7 { 8     cin >> N; 9     for (int i = 1; i <= N; ++i)10     {11         cin >> M;12         for (int j = 0; j < M; ++j)13         {14             cin >> d;15             nums[i].insert(d);16         }17     }18     cin >> K;19     for (int i = 0; i < K; ++i)20     {21         int a, b;22         double res, Nc = 0, Nt = 0;23         cin >> a >> b;24         for (auto v : nums[a])25             if (nums[b].find(v) != nums[b].end())26                 Nc++;27         Nt = nums[a].size() + nums[b].size() - Nc;28         res = (Nc / Nt)*100.0;29         printf("%.1f%%\n", res);30     }31     return 0;32 }