PTA甲级 1063 Set Similarity (C++)

Given two sets of integers, the similarity of the sets is defined to be $N_c/N_t\times 100$, 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 $N(\le 50)$ which is the total number of sets. Then $N$ lines follow, each gives a set with a positive $M(\le 10^4)$ and followed by $M$ integers in the range $[0,10^9]$. After the input of sets, a positive integer $K(\le 2000)$ 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%

Solution:

// Talk is cheap, show me the code
// Created by Misdirection 2021-08-20 12:53:58
// All rights reserved.#include <iostream>
#include <vector>
#include <unordered_set>using namespace std;int main(){int n;scanf("%d", &n);vector<unordered_set<int>> nums(n);for(int i = 0; i < n; ++i){int size;scanf("%d", &size);for(int j = 0; j < size; ++j){int tmp;scanf("%d", &tmp);nums[i].insert(tmp);}}int query;scanf("%d", &query);for(int i = 0; i < query; ++i){int a, b;scanf("%d %d", &a, &b);int common = 0;int total = nums[a - 1].size();for(auto it = nums[b - 1].begin(); it != nums[b - 1].end(); ++it){if(nums[a - 1].find(*it) != nums[a - 1].end()) common++;else total++;}printf("%.1f%%\n", 100.0 * common / total);}return 0;
}