Given two sets of integers, the similarity of the sets is defined to be N_c/N_t*100%, 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 (<=50) which is the total number of sets. Then N lines follow, each gives a set with a positive M (<=10⁴) and followed by M integers in the range [0, 10⁹]. After the input of sets, a positive integer K (<=2000) is given, followed by K lines 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:

3

3 99 87 101

4 87 101 5 87

7 99 101 18 5 135 18 99

2

1 2

1 3

Sample Output:

50.0%

33.3%

 #include<cstdio>

 #include<iostream>

 #include<algorithm>

 #include<set>

 using namespace std;

 set<int> num[];

 int main(){

     int N, M, temp, K;

     scanf("%d", &N);

     for(int i = ; i <= N; i++){

         scanf("%d", &M);

         for(int j = ; j < M; j++){

             scanf("%d", &temp);

             num[i].insert(temp);

         }

     }

     scanf("%d", &K);

     int s1, s2;

     for(int i = ; i < K; i++){

         scanf("%d %d", &s1, &s2);

         int same = , diff = num[s2].size();

         for(set<int> ::iterator it = num[s1].begin(); it != num[s1].end(); it++){

             if(num[s2].find(*it) != num[s2].end())

                 same++;

             else diff++;

         }

         double rate = 1.0 * same / diff * 100.0;

         printf("%.1lf%%\n", rate);

     }

     cin >> N;

     return ;

 }

总结：

1、set可以用于hash表解决不了的情况，比如键值不是int的情况。

2、printf输出%时，需要输出%%。