PAT甲级——A1063 Set Similarity

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:

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%

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

#include <iostream>
#include <unordered_set>//用set来去重
using namespace std;
int N, M, K, d;
unordered_set<int>nums[55];
int main()
{
    cin >> N;
    for (int i = 1; i <= N; ++i)
    {
        cin >> M;
        for (int j = 0; j < M; ++j)
        {
            cin >> d;
            nums[i].insert(d);
        }
    }
    cin >> K;
    for (int i = 0; i < K; ++i)
    {
        int a, b;
        double res, Nc = 0, Nt = 0;
        cin >> a >> b;
        for (auto v : nums[a])
            if (nums[b].find(v) != nums[b].end())
                Nc++;
        Nt = nums[a].size() + nums[b].size() - Nc;
        res = (Nc / Nt)*100.0;
        printf("%.1f%%
", res);
    }
    return 0;
}