You rescale a cookbook's ingredient list to the number of servings you will make.
The first line gives the cookbook's servings b and the servings to make t, separated by one space.
The second line gives the number of ingredients n.
Each of the next n lines gives an ingredient name, the cookbook amount a and a digit count d, separated by single spaces.
The new amount x is the exact value of a × t ÷ b.
If d is 1 or 2, round x to d decimal places and write exactly d digits after the decimal point (2.1 with d = 2 is 2.10).
If d is 0, round x to an integer and write it without a decimal point.
If d is -1, round x to the nearest multiple of 10, and if d is -2 to the nearest multiple of 100, and write it without a decimal point (237.8 with d = -1 is 240).
If the rounded result is 0, write it to fit the digit count, such as 0, 0.0 or 0.00.
To keep rounding unambiguous, no input makes the fractional part of x × 10^d lie between 0.4 and 0.6 inclusive.
For each ingredient, in input order, print its name and new amount on one line, separated by one space.
Input
1 ≤ b ≤ 100, 1 ≤ t ≤ 100, 1 ≤ n ≤ 1,000, an ingredient name is 1–20 lowercase English letters (the same name may appear again), 0.01 ≤ a ≤ 10,000 written with 0–2 digits after the decimal point (e.g. 250, 1.5, 0.75), d is one of -2, -1, 0, 1, 2, no input has the fractional part of x × 10^d in 0.4–0.6
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
The tools that write exactly d decimals are Python f"{x:.{d}f}", JS/TS x.toFixed(d), Java String.format("%." + d + "f", x), C# x.ToString("F" + d), C printf("%.*f", d, x), C++ std::fixed << std::setprecision(d) and Craykon Str.Fixed(x, d). All of them pad 2.1 to 2.10 with the trailing zero. Rounding-only functions such as Python round(x, 2), Craykon Mathf.Round(x, 2) and JS Math.round(x * 100) / 100 return a number, so printing it gives 2.1. For a digit count of 0 or less the tools differ. JS toFixed(-1) throws a RangeError. In C, a negative precision in %.*f counts as no precision, so six decimals come out. Craykon Mathf.Round accepts only 0–10 digits. Python round(x, -1) works but returns a float, which prints as 240.0. So divide by the unit 10^(-d), round to an integer and multiply back: Python round(x / 10) * 10 (round with one argument returns an int), JS/TS/Java Math.round, C llround, C++ std::llround, C# (long)Math.Round, Craykon Mathf.Round(x / 10) * 10. The scale factor is a trap too. In C, C++, Java and C#, writing a * (t / b) evaluates the integer t / b first, so 6 / 4 becomes 1. Multiplying the double a first, as in a * t / b, makes it a floating-point division. In Python 3, JS, TS and Craykon / is always floating-point division. Rounding rules differ by language as well. On an exact half, Python round and C# Math.Round go to the even neighbour, JS and Java Math.round go up, and C llround, C++ std::llround and Craykon Mathf.Round go away from zero. Craykon and Java formatting round the visible decimal, while C, C++, Python, JS and C# formatting round the exact binary value (1.005 → 1.00). This problem never gives values near such a boundary, so every method gives the same answer. Java String.format and C# ToString/double.Parse follow the machine's locale, where the decimal point may be a comma, so pass Locale.ROOT or CultureInfo.InvariantCulture.
Pythonmain.py · 12 lines
b, t = map(int, input().split())n = int(input())for _ in range(n): name, a, d = input().split() d = int(d) x = float(a) * t / b # /는 늘 실수 나눗셈이다 if d > 0: print(name, f"{x:.{d}f}") else: # 자리가 0 이하이면 10^(-d) 단위로: round(x / 단위)는 정수를 돌려준다 step = 10 ** -d print(name, round(x / step) * step)
JavaScriptmain.js · 16 lines
const lines = require("fs").readFileSync(0, "utf8").split("\n");const [b, t] = lines[0].trim().split(/\s+/).map(Number);const n = Number(lines[1]);const out = [];for (let i = 0; i < n; i++) { const [name, a, ds] = lines[2 + i].trim().split(/\s+/); const d = Number(ds); const x = (Number(a) * t) / b; if (d > 0) { out.push(`${name} ${x.toFixed(d)}`); // toFixed는 0~100자리만 받는다 } else { const step = 10 ** -d; out.push(`${name} ${Math.round(x / step) * step}`); }}console.log(out.join("\n"));
TypeScriptmain.ts · 17 lines
import { readFileSync } from "node:fs";const lines: string[] = readFileSync(0, "utf8").split("\n");const [b, t] = lines[0].trim().split(/\s+/).map(Number);const n = Number(lines[1]);const out: string[] = [];for (let i = 0; i < n; i++) { const [name, a, ds] = lines[2 + i].trim().split(/\s+/); const d = Number(ds); const x: number = (Number(a) * t) / b; if (d > 0) { out.push(`${name} ${x.toFixed(d)}`); // toFixed는 0~100자리만 받는다 } else { const step = 10 ** -d; out.push(`${name} ${Math.round(x / step) * step}`); }}console.log(out.join("\n"));
JavaMain.java · 27 lines
import java.util.Locale;import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int b = sc.nextInt(), t = sc.nextInt(); int n = sc.nextInt(); StringBuilder out = new StringBuilder(); for (int i = 0; i < n; i++) { String name = sc.next(); double a = Double.parseDouble(sc.next()); int d = sc.nextInt(); double x = a * t / b; // a가 double이라 실수 나눗셈이 된다 out.append(name).append(' '); if (d > 0) { // Locale.ROOT: 지역 설정과 상관없이 소수점을 점으로 쓴다 out.append(String.format(Locale.ROOT, "%." + d + "f", x)); } else { long step = d == 0 ? 1 : d == -1 ? 10 : 100; out.append(Math.round(x / step) * step); } out.append('\n'); } System.out.print(out); }}
C#Main.cs · 34 lines
using System;using System.Globalization;using System.Text;class Program{ static void Main() { var inv = CultureInfo.InvariantCulture; string[] bt = Console.ReadLine().Split(' ', StringSplitOptions.RemoveEmptyEntries); int b = int.Parse(bt[0]), t = int.Parse(bt[1]); int n = int.Parse(Console.ReadLine()); var out_ = new StringBuilder(); for (int i = 0; i < n; i++) { string[] p = Console.ReadLine().Split(' ', StringSplitOptions.RemoveEmptyEntries); double a = double.Parse(p[1], inv); int d = int.Parse(p[2]); double x = a * t / b; out_.Append(p[0]).Append(' '); if (d > 0) { out_.Append(x.ToString("F" + d, inv)); // "F2": 소수 두 자리 } else { long step = d == 0 ? 1 : d == -1 ? 10 : 100; out_.Append((long)Math.Round(x / step) * step); } out_.Append('\n'); } Console.Write(out_); }}
C++main.cpp · 23 lines
#include <cmath>#include <iomanip>#include <iostream>#include <string>int main() { int b, t, n; std::cin >> b >> t >> n; for (int i = 0; i < n; i++) { std::string name; double a; int d; std::cin >> name >> a >> d; double x = a * t / b; std::cout << name << ' '; if (d > 0) { std::cout << std::fixed << std::setprecision(d) << x << '\n'; } else { long long step = d == 0 ? 1 : d == -1 ? 10 : 100; std::cout << std::llround(x / step) * step << '\n'; } }}
Cmain.c · 21 lines
#include <math.h>#include <stdio.h>int main(void) { int b, t, n; scanf("%d %d %d", &b, &t, &n); for (int i = 0; i < n; i++) { char name[32]; double a; int d; scanf("%31s %lf %d", name, &a, &d); double x = a * t / b; /* a가 double이라 실수 나눗셈이 된다 */ if (d > 0) { printf("%s %.*f\n", name, d, x); /* 자리 수를 인수로 넘긴다 */ } else { long long step = d == 0 ? 1 : d == -1 ? 10 : 100; printf("%s %lld\n", name, llround(x / step) * step); } } return 0;}
Craykonmain.cray · 27 lines
bt = Console.ReadNumbers();b = bt[0];t = bt[1];n = Console.ReadLine().ToNumber();for (i = 0; i < n; i++){ p = Console.ReadLine().Split(); a = p[1].ToNumber(); d = p[2].ToNumber(); x = a * t / b; // 크레이콘의 / 는 늘 소수까지 나눈다 값 = ""; if (d > 0) { 값 = Str.Fixed(x, d); } else { // 수학.반올림의 자리는 0~10만 된다: 10, 100 단위는 나눠서 반올림하고 다시 곱한다 step = 1; for (k = 0; k < 0 - d; k++) { step *= 10; } 값 = Mathf.Round(x / step) * step; } Console.WriteLine(p[0] + " " + 값);}