有重复元素去重前后差多少

1、2、2 里有两个一样的 2。一边不去重、一边去重，各生成一遍全部子集。运行下面这段程序：

#include <cstdlib>
#include <iostream>
#include <string>
#include <vector>
using namespace std;

void subs_dfs(const vector<int>& a, int s, vector<int>& path, vector<vector<int>>& res) {
    res.push_back(path);
    for (int i = s; i < (int)a.size(); i++) {
        path.push_back(a[i]);
        subs_dfs(a, i + 1, path, res);
        path.pop_back();
    }
}

vector<vector<int>> subs(const vector<int>& a) {
    vector<vector<int>> res;
    vector<int> path;
    subs_dfs(a, 0, path, res);
    return res;
}

#include <algorithm>

void subsu_dfs(const vector<int>& b, int s, vector<int>& path, vector<vector<int>>& res) {
    res.push_back(path);
    for (int i = s; i < (int)b.size(); i++) {
        if (i > s && b[i] == b[i - 1]) continue;
        path.push_back(b[i]);
        subsu_dfs(b, i + 1, path, res);
        path.pop_back();
    }
}

vector<vector<int>> subs_uniq(vector<int> a) {
    sort(a.begin(), a.end());
    vector<vector<int>> res;
    vector<int> path;
    subsu_dfs(a, 0, path, res);
    return res;
}

int main() {
    cout << subs({1, 2, 2}).size() << "/" << subs_uniq({1, 2, 2}).size() << endl;
}

（本题用 g++ -std=c++17 -O0 编译。）