Integers are given one per line, and a 0 marks the end of the input. Add the numbers before the 0 one by one, and after each addition print the total so far on its own line. The 0 itself is not added or printed. There may be more lines after the 0; ignore them all. At least one number comes before the 0. A running total of 0 does not end the input.
Input
1 to 10,000 numbers before the 0, each between -1,000,000,000 and 1,000,000,000 and not 0. The number of lines after the 0 is unspecified.
Output
one line per number before the 0, each with the total so far
Example 1 · stdin
3
-5
10
0
stdout
3
-2
8
Example 2 · stdin
4
-4
7
0
100
0
stdout
4
0
7
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
A sentinel loop goes 'read → is it 0? → process'. Reading and testing in one line: Python while (x := int(input())) != 0: (the 3.8 walrus operator), C while (scanf("%lld", &x) == 1 && x != 0), C++ while (std::cin >> x && x != 0), Java/C# while ((x = …parse(readLine())) != 0). JS/TS read all input at once, so break out of the for … of on 0. Craykon cannot assign inside an expression, so read once before the loop and read the next line again at the end of the loop body (x = Console.ReadLine().ToNumber();). Three traps: (1) reading to end of input (EOF) also adds the lines after the 0 (second example). (2) Checking for 0 after processing prints one extra total for the 0. (3) The total can reach 10^9 × 10^4 = 10^13, so C, C++, Java and C# need long (long long). Python integers never overflow and JS/TS/Craykon doubles are exact up to 2^53 (about 9×10^15), so they are fine.
Pythonmain.py · 5 lines
total = 0# 0이 끝 표시다. 0 뒤에 줄이 더 있어도 읽지 않는다while (x := int(input())) != 0: total += x print(total)
JavaScriptmain.js · 9 lines
const nums = require("fs").readFileSync(0, "utf8").trim().split(/\s+/).map(Number);const out = [];let total = 0;for (const x of nums) { if (x === 0) break; // 끝 표시. 뒤에 남은 수는 무시 total += x; out.push(total);}console.log(out.join("\n"));
TypeScriptmain.ts · 10 lines
import { readFileSync } from "node:fs";const nums: number[] = readFileSync(0, "utf8").trim().split(/\s+/).map(Number);const out: number[] = [];let total = 0;for (const x of nums) { if (x === 0) break; // 끝 표시. 뒤에 남은 수는 무시 total += x; out.push(total);}console.log(out.join("\n"));
JavaMain.java · 17 lines
import java.io.BufferedReader;import java.io.InputStreamReader;public class Main { public static void main(String[] args) throws Exception { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringBuilder out = new StringBuilder(); long total = 0; // 10^9 × 10^4 = 10^13 이라 int로는 넘친다 long x; // 0이 끝 표시다. 0 뒤의 줄은 읽지 않는다 while ((x = Long.parseLong(br.readLine().trim())) != 0) { total += x; out.append(total).append('\n'); } System.out.print(out); }}
C#Main.cs · 19 lines
using System;using System.Text;class Program{ static void Main() { var out_ = new StringBuilder(); long total = 0; // 10^9 × 10^4 = 10^13 이라 int로는 넘친다 long x; // 0이 끝 표시다. 0 뒤의 줄은 읽지 않는다 while ((x = long.Parse(Console.ReadLine())) != 0) { total += x; out_.Append(total).Append('\n'); } Console.Write(out_); }}
C++main.cpp · 12 lines
#include <iostream>int main() { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); long long x, total = 0; // 10^9 × 10^4 = 10^13 이라 int로는 넘친다 // 0이 끝 표시다. 0 뒤의 줄은 읽지 않는다 while (std::cin >> x && x != 0) { total += x; std::cout << total << '\n'; }}
Cmain.c · 11 lines
#include <stdio.h>int main(void) { long long x, total = 0; /* 10^9 × 10^4 = 10^13 이라 int로는 넘친다 */ /* 0이 끝 표시다. 0 뒤의 줄은 읽지 않는다 */ while (scanf("%lld", &x) == 1 && x != 0) { total += x; printf("%lld\n", total); } return 0;}
Craykonmain.cray · 9 lines
합 = 0;x = Console.ReadLine().ToNumber();// 0이 끝 표시다. 0 뒤의 줄은 읽지 않는다while (x != 0){ 합 += x; Console.WriteLine(합); x = Console.ReadLine().ToNumber();}