A younger sibling did addition homework. You want to grade each equation and correct the wrong answers.
The first line gives the number of equations n.
Each of the next n lines gives one equation the sibling wrote, in the form "a + b = c".
a, b and c are integers, and there is exactly one space between each number and symbol.
Negative numbers are written with a leading - (for example -8 + 3 = -5).
Numbers are written with digits only, without commas, leading zeros or a + sign. Zero is never written as -0.
For each equation, print the grading result on one line.
If c equals a + b, print "O a + b = c".
If c differs from a + b, print "X a + b = s", where s is the correct value of a + b.
Write the printed numbers with digits only, like the input.
On the last line, print the number of wrong equations.
Input
1 ≤ n ≤ 1,000, -1,500,000,000 ≤ a, b ≤ 1,500,000,000, -3,000,000,000 ≤ c ≤ 3,000,000,000
Output
n lines of results, then the number of wrong equations
O 1500000000 + 1500000000 = 3000000000
X 1500000000 + 1500000000 = 3000000000
1
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
Open this on a computer to write and run code. On mobile you can read the problem and the solutions.
In other languagesShow solutions by language · 8 languages
The first trap is that one + both adds and joins strings. In JS/TS the pieces from split(" ") are strings, so without Number you get "3" + "4" → "34", and the sibling's wrong answer 34 is graded as right (in TS, declaring number[] lets the type checker catch it). Craykon joins them the same way if you add without ToNumber(). In Python "3" + "4" is also "34", but adding a string and a number raises an error. In Java, C# and JS, once a string appears, everything after it is joined from left to right, so "= " + a + b becomes = 34. Compute the sum first with s = a + b, or wrap it as (a + b). The second trap is size. a and b go up to 1.5 billion, so the sum reaches 3 billion, but int in C, C++, Java and C# only reaches about 2.1 billion. In C# and Java the overflow silently turns negative (C# only reports it inside checked), and signed overflow in C and C++ is undefined behavior. So read with C long long and scanf("%lld + %lld = %lld"), C++ long long, Java/C# long. Python integers have no size limit, and numbers in JS, TS and Craykon are exact up to 2^53, so 3 billion is fine.
Pythonmain.py · 14 lines
n = int(input())out = []wrong = 0for _ in range(n): a, plus, b, eq, c = input().split() a, b, c = int(a), int(b), int(c) # 문자열끼리 +하면 이어 붙는다("3" + "4" → "34") s = a + b if s == c: out.append(f"O {a} + {b} = {s}") else: out.append(f"X {a} + {b} = {s}") wrong += 1out.append(str(wrong))print("\n".join(out))
JavaScriptmain.js · 17 lines
const lines = require("fs").readFileSync(0, "utf8").split("\n");const n = Number(lines[0]);const out = [];let wrong = 0;for (let i = 1; i <= n; i++) { // 빈칸으로 나눈 조각은 문자열이라 Number로 바꿔야 +가 덧셈이 된다 const [a, , b, , c] = lines[i].trim().split(" ").map(Number); const s = a + b; if (s === c) { out.push(`O ${a} + ${b} = ${s}`); } else { out.push(`X ${a} + ${b} = ${s}`); wrong++; }}out.push(String(wrong));console.log(out.join("\n"));
TypeScriptmain.ts · 18 lines
import { readFileSync } from "node:fs";const lines: string[] = readFileSync(0, "utf8").split("\n");const n: number = Number(lines[0]);const out: string[] = [];let wrong = 0;for (let i = 1; i <= n; i++) { // 빈칸으로 나눈 조각은 문자열이라 Number로 바꿔야 +가 덧셈이 된다 const [a, , b, , c]: number[] = lines[i].trim().split(" ").map(Number); const s: number = a + b; if (s === c) { out.push(`O ${a} + ${b} = ${s}`); } else { out.push(`X ${a} + ${b} = ${s}`); wrong++; }}out.push(String(wrong));console.log(out.join("\n"));
JavaMain.java · 29 lines
import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;public class Main { public static void main(String[] args) throws IOException { BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); int n = Integer.parseInt(in.readLine().trim()); StringBuilder out = new StringBuilder(); int wrong = 0; for (int i = 0; i < n; i++) { String[] p = in.readLine().trim().split(" "); // 합이 30억까지 가니 int(약 21억)가 아니라 long으로 받는다 long a = Long.parseLong(p[0]); long b = Long.parseLong(p[2]); long c = Long.parseLong(p[4]); long s = a + b; char mark = 'O'; if (s != c) { mark = 'X'; wrong++; } // 문자열 뒤에서 + a + b를 쓰면 이어 붙으니 합은 s로 따로 구해 둔다 out.append(mark).append(' ').append(a).append(" + ").append(b).append(" = ").append(s).append('\n'); } out.append(wrong).append('\n'); System.out.print(out); }}
C#Main.cs · 30 lines
using System;using System.Text;class Program{ static void Main() { int n = int.Parse(Console.ReadLine()); var sb = new StringBuilder(); int wrong = 0; for (int i = 0; i < n; i++) { string[] p = Console.ReadLine().Trim().Split(' '); // 합이 30억까지 가니 int가 아니라 long으로 받는다 long a = long.Parse(p[0]); long b = long.Parse(p[2]); long c = long.Parse(p[4]); long s = a + b; char mark = 'O'; if (s != c) { mark = 'X'; wrong++; } sb.Append($"{mark} {a} + {b} = {s}\n"); } sb.Append(wrong).Append('\n'); Console.Write(sb); }}
C++main.cpp · 20 lines
#include <iostream>int main() { std::ios::sync_with_stdio(false); int n, wrong = 0; std::cin >> n; for (int i = 0; i < n; i++) { long long a, b, c; char plus, eq; std::cin >> a >> plus >> b >> eq >> c; long long s = a + b; // int로 더하면 30억에서 넘친다 char mark = 'O'; if (s != c) { mark = 'X'; wrong++; } std::cout << mark << ' ' << a << " + " << b << " = " << s << "\n"; } std::cout << wrong << "\n";}
Cmain.c · 20 lines
#include <stdio.h>int main(void) { int n, wrong = 0; long long a, b, c; scanf("%d", &n); for (int i = 0; i < n; i++) { /* 형식 문자열의 " + "와 " = "가 기호를 건너뛰어 준다 */ scanf("%lld + %lld = %lld", &a, &b, &c); long long s = a + b; /* int로 더하면 30억에서 넘친다 */ char mark = 'O'; if (s != c) { mark = 'X'; wrong++; } printf("%c %lld + %lld = %lld\n", mark, a, b, s); } printf("%d\n", wrong); return 0;}
Craykonmain.cray · 19 lines
n = Console.ReadLine().ToNumber();wrong = 0;for (i = 0; i < n; i++){ p = Console.ReadLine().Trim().Split(" "); // 나눈 조각은 문자열이라 수로변환해야 +가 덧셈이 된다 a = p[0].ToNumber(); b = p[2].ToNumber(); c = p[4].ToNumber(); s = a + b; mark = "O"; if (s != c) { mark = "X"; wrong++; } Console.WriteLine(mark + " " + a + " + " + b + " = " + s);}Console.WriteLine(wrong);