The first line gives n, the second line n integers (group A), the third line m, and the fourth line m integers (group B). The same number may appear more than once in a group, but as a set it counts once. Print three lines.
1. The union of A and B (numbers in at least one of them)
2. The intersection of A and B (numbers in both)
3. The difference A minus B (numbers in A but not in B)
On each line print each number once, smallest first, separated by single spaces. If there are no such numbers, print a single "-".
Input
1 ≤ n, m ≤ 5,000, -2,000,000,000 ≤ each integer ≤ 2,000,000,000
Output
three lines (union, intersection, difference)
Example 1 · stdin
7
10 9 2 100 -1 -20 3
5
9 100 -20 7 7
stdout
-20 -1 2 3 7 9 10 100
-20 9 100
-1 2 3 10
Example 2 · stdin
3
5 5 5
1
5
stdout
5
5
-
main.py
Error List00F8 ↓ · ⇧F8 ↑
Input (stdin)
When you run, these lines are fed to input()/stdin in order (the program doesn't stop to ask).
$ Ready
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In other languagesShow solutions by language · 8 languages
Set operations go by different names. Python has the operators a | b, a & b, a - b; Java has addAll, retainAll, removeAll; C# has UnionWith, IntersectWith, ExceptWith. The Java and C# methods change the set itself, so call them on a copy. C++'s std::set_union, set_intersection and set_difference take sorted ranges, which fits std::set (always sorted). JS/TS Set lacked set operations for a long time (union and friends arrived only recently), so filter(x => b.has(x)) is common. Watch the order too. A Python set does not iterate in size order, so you need sorted, and JS/TS sort() without a compare function sorts as text, putting 10 before 9; write sort((x, y) => x - y). Java TreeSet and C# SortedSet are always in size order. C has no set, so sort both, remove repeats, and walk both from the front together (a merge). If the qsort compare function returns a - b, subtracting values around ±2e9 overflows int (about ±2.1e9) and the order breaks; write (a > b) - (a < b). Another common mistake is building the difference as a list ([x for x in a if x not in b]), which keeps repeated values. Craykon uses the keys of a Dict() as a set, asks with Contains, then Sort()s the Keys and writes them with Join(" ").
Pythonmain.py · 14 lines
input()a = set(map(int, input().split()))input()b = set(map(int, input().split()))def show(s): # 집합에는 차례가 없으니 출력 전에 크기 순으로 늘어놓는다 return " ".join(map(str, sorted(s))) if s else "-"print(show(a | b))print(show(a & b))print(show(a - b))
JavaScriptmain.js · 9 lines
const lines = require("fs").readFileSync(0, "utf8").split("\n");const a = new Set(lines[1].trim().split(/\s+/).map(Number));const b = new Set(lines[3].trim().split(/\s+/).map(Number));// sort()만 쓰면 글자 순(10이 9보다 앞)이라 비교 함수를 준다const show = (s) => (s.size ? [...s].sort((x, y) => x - y).join(" ") : "-");const union = new Set([...a, ...b]);const inter = new Set([...a].filter((x) => b.has(x)));const diff = new Set([...a].filter((x) => !b.has(x)));console.log([show(union), show(inter), show(diff)].join("\n"));
TypeScriptmain.ts · 10 lines
import { readFileSync } from "node:fs";const lines: string[] = readFileSync(0, "utf8").split("\n");const a = new Set<number>(lines[1].trim().split(/\s+/).map(Number));const b = new Set<number>(lines[3].trim().split(/\s+/).map(Number));// sort()만 쓰면 글자 순(10이 9보다 앞)이라 비교 함수를 준다const show = (s: Set<number>): string => (s.size ? [...s].sort((x, y) => x - y).join(" ") : "-");const union = new Set<number>([...a, ...b]);const inter = new Set<number>([...a].filter((x) => b.has(x)));const diff = new Set<number>([...a].filter((x) => !b.has(x)));console.log([show(union), show(inter), show(diff)].join("\n"));
JavaMain.java · 35 lines
import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.util.TreeSet;public class Main { static TreeSet<Integer> readSet(BufferedReader in) throws IOException { in.readLine(); TreeSet<Integer> s = new TreeSet<>(); for (String t : in.readLine().trim().split("\\s+")) s.add(Integer.parseInt(t)); return s; } static String show(TreeSet<Integer> s) { if (s.isEmpty()) return "-"; StringBuilder sb = new StringBuilder(); for (int x : s) sb.append(sb.length() > 0 ? " " : "").append(x); return sb.toString(); } public static void main(String[] args) throws IOException { BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); // TreeSet은 늘 크기 순으로 돈다. addAll·retainAll·removeAll은 원래 집합을 바꾸니 복사해서 쓴다 TreeSet<Integer> a = readSet(in), b = readSet(in); TreeSet<Integer> union = new TreeSet<>(a); union.addAll(b); TreeSet<Integer> inter = new TreeSet<>(a); inter.retainAll(b); TreeSet<Integer> diff = new TreeSet<>(a); diff.removeAll(b); System.out.println(show(union)); System.out.println(show(inter)); System.out.println(show(diff)); }}
C#Main.cs · 30 lines
using System;using System.Collections.Generic;using System.Linq;class Program{ static SortedSet<int> ReadSet() { Console.ReadLine(); return new SortedSet<int>(Console.ReadLine().Split(' ', StringSplitOptions.RemoveEmptyEntries).Select(int.Parse)); } static string Show(SortedSet<int> s) => s.Count == 0 ? "-" : string.Join(" ", s); static void Main() { // SortedSet은 늘 크기 순으로 돈다. UnionWith 등은 원래 집합을 바꾸니 복사해서 쓴다 var a = ReadSet(); var b = ReadSet(); var union = new SortedSet<int>(a); union.UnionWith(b); var inter = new SortedSet<int>(a); inter.IntersectWith(b); var diff = new SortedSet<int>(a); diff.ExceptWith(b); Console.WriteLine(Show(union)); Console.WriteLine(Show(inter)); Console.WriteLine(Show(diff)); }}
C++main.cpp · 34 lines
#include <algorithm>#include <iostream>#include <iterator>#include <set>#include <vector>std::set<int> read_set() { int n, x; std::cin >> n; std::set<int> s; for (int i = 0; i < n; i++) { std::cin >> x; s.insert(x); } return s;}void show(const std::vector<int>& v) { if (v.empty()) std::cout << "-"; for (std::size_t i = 0; i < v.size(); i++) std::cout << (i ? " " : "") << v[i]; std::cout << '\n';}int main() { // std::set은 크기 순이라 set_union 등에 바로 넘길 수 있다(정렬된 범위가 필요하다) std::set<int> a = read_set(), b = read_set(); std::vector<int> u, in, d; std::set_union(a.begin(), a.end(), b.begin(), b.end(), std::back_inserter(u)); std::set_intersection(a.begin(), a.end(), b.begin(), b.end(), std::back_inserter(in)); std::set_difference(a.begin(), a.end(), b.begin(), b.end(), std::back_inserter(d)); show(u); show(in); show(d);}
Cmain.c · 50 lines
#include <stdio.h>#include <stdlib.h>// a - b 를 돌려주면 ±20억끼리 빼서 int가 넘친다 — 크기만 비교한다static int cmp(const void *pa, const void *pb) { int a = *(const int *)pa, b = *(const int *)pb; return (a > b) - (a < b);}// 읽어서 정렬한 뒤 겹친 값을 지운다. 남은 개수를 돌려준다static int read_set(int *s) { int n; scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d", &s[i]); qsort(s, n, sizeof s[0], cmp); int k = 0; for (int i = 0; i < n; i++) if (k == 0 || s[k - 1] != s[i]) s[k++] = s[i]; return k;}static void show(const int *s, int n) { if (n == 0) printf("-"); for (int i = 0; i < n; i++) printf(i ? " %d" : "%d", s[i]); printf("\n");}int main(void) { static int a[5000], b[5000], u[10000], in[5000], d[5000]; int n = read_set(a), m = read_set(b); int nu = 0, ni = 0, nd = 0, i = 0, j = 0; // 정렬된 두 목록을 앞에서부터 함께 훑는다(병합) while (i < n || j < m) { if (j == m || (i < n && a[i] < b[j])) { u[nu++] = a[i]; d[nd++] = a[i++]; } else if (i == n || b[j] < a[i]) { u[nu++] = b[j++]; } else { u[nu++] = a[i]; in[ni++] = a[i]; i++; j++; } } show(u, nu); show(in, ni); show(d, nd); return 0;}