One line gives integers a and N separated by a space.
Print every power of a — a^0, a^1, a^2, … — that is at most N, smallest first, one per line.
Since a^0 = 1 and N ≥ 1, the first line is always 1.
2 100
1 2 4 8 16 32 64
10 1000
1 10 100 1000
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.
Instead of computing each power from scratch, keep multiplying the previous one by a (start at p = 1, then p *= a).
Three traps.
(1) ^ is not a power: in C, C++, Java, C#, JS and TS 2 ^ 3 is bitwise XOR, which is 1.
Only Python, JS and TS have a power operator, **; C/C++ have pow, Java Math.pow and C# Math.Pow (all double), and Craykon has no power at all, so repeat multiplication.
(2) Overflow: multiplying the last p near 9×10^15 by a (up to a billion) once more passes the long range (about 9.2×10^18).
C, C++, Java and C# wrap around without an error (in C/C++ it is undefined behaviour), so something like 65537^4 turns into a number below 9×10^15 and extra wrong lines appear.
Ask p > N / a before multiplying.
Python integers never overflow, and JS/TS/Craykon numbers are doubles that become inexact rather than overflowing — but by then they are already above N, so p <= N alone is correct.
(3) int: values pass 2^31, so C, C++, Java and C# need long (long long).
a, n = map(int, input().split())p = 1 # a의 0제곱while p <= n: print(p) p *= a # 파이썬 정수는 커져도 넘치지 않는다const [a, n] = require("fs").readFileSync(0, "utf8").trim().split(/\s+/).map(Number);const out = [];// p * a가 2^53을 넘으면 정확하지 않지만, 그때는 이미 n보다 커서 반복이 끝난다for (let p = 1; p <= n; p *= a) out.push(p);console.log(out.join("\n"));import { readFileSync } from "node:fs";const [a, n]: number[] = readFileSync(0, "utf8").trim().split(/\s+/).map(Number);const out: number[] = [];// p * a가 2^53을 넘으면 정확하지 않지만, 그때는 이미 n보다 커서 반복이 끝난다for (let p = 1; p <= n; p *= a) out.push(p);console.log(out.join("\n"));import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); long a = sc.nextLong(), n = sc.nextLong(); StringBuilder out = new StringBuilder(); long p = 1; while (true) { out.append(p).append('\n'); // p * a는 long도 넘칠 수 있다 → 곱하기 전에 나눗셈으로 확인 if (p > n / a) break; p *= a; } System.out.print(out); }}using System; class Program{ static void Main() { string[] t = Console.ReadLine().Split(' ', StringSplitOptions.RemoveEmptyEntries); long a = long.Parse(t[0]), n = long.Parse(t[1]); long p = 1; while (true) { Console.WriteLine(p); // p * a는 long도 넘칠 수 있다(C#은 기본이 unchecked라 조용히 돈다) → 곱하기 전에 확인 if (p > n / a) break; p *= a; } }}#include <iostream> int main() { long long a, n; std::cin >> a >> n; long long p = 1; while (true) { std::cout << p << '\n'; // p * a는 long long도 넘칠 수 있다 → 곱하기 전에 나눗셈으로 확인 if (p > n / a) break; p *= a; }}#include <stdio.h> int main(void) { long long a, n; scanf("%lld %lld", &a, &n); long long p = 1; for (;;) { printf("%lld\n", p); /* p * a는 long long도 넘칠 수 있다 → 곱하기 전에 나눗셈으로 확인 */ if (p > n / a) break; p *= a; } return 0;}수들 = Console.ReadNumbers();a = 수들[0];n = 수들[1];// 크레이콘 수는 실수라 넘치지 않는다. 2^53을 넘으면 정확하지 않지만 그때는 이미 n보다 크다p = 1;while (p <= n){ Console.WriteLine(p); p *= a;}