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[Silver V] Title: 분수찾기, Time: 32 ms, Memory: 31120 KB -BaekjoonHub
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ddubbu-dev committed Sep 13, 2024
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85 changes: 85 additions & 0 deletions 백준/Silver/1193. 분수찾기/README.md
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# [Silver V] 분수찾기 - 1193

[문제 링크](https://www.acmicpc.net/problem/1193)

### 성능 요약

메모리: 31120 KB, 시간: 32 ms

### 분류

구현, 수학

### 제출 일자

2024년 9월 13일 12:51:10

### 문제 설명

<p>무한히 큰 배열에 다음과 같이 분수들이 적혀있다.</p>

<table class="table table-bordered" style="width:30%">
<tbody>
<tr>
<td style="width:5%">1/1</td>
<td style="width:5%">1/2</td>
<td style="width:5%">1/3</td>
<td style="width:5%">1/4</td>
<td style="width:5%">1/5</td>
<td style="width:5%">…</td>
</tr>
<tr>
<td>2/1</td>
<td>2/2</td>
<td>2/3</td>
<td>2/4</td>
<td>…</td>
<td>…</td>
</tr>
<tr>
<td>3/1</td>
<td>3/2</td>
<td>3/3</td>
<td>…</td>
<td>…</td>
<td>…</td>
</tr>
<tr>
<td>4/1</td>
<td>4/2</td>
<td>…</td>
<td>…</td>
<td>…</td>
<td>…</td>
</tr>
<tr>
<td>5/1</td>
<td>…</td>
<td>…</td>
<td>…</td>
<td>…</td>
<td>…</td>
</tr>
<tr>
<td>…</td>
<td>…</td>
<td>…</td>
<td>…</td>
<td>…</td>
<td>…</td>
</tr>
</tbody>
</table>

<p>이와 같이 나열된 분수들을 1/1 → 1/2 → 2/1 → 3/1 → 2/2 → … 과 같은 지그재그 순서로 차례대로 1번, 2번, 3번, 4번, 5번, … 분수라고 하자.</p>

<p>X가 주어졌을 때, X번째 분수를 구하는 프로그램을 작성하시오.</p>

### 입력

<p>첫째 줄에 X(1 ≤ X ≤ 10,000,000)가 주어진다.</p>

### 출력

<p>첫째 줄에 분수를 출력한다.</p>

44 changes: 44 additions & 0 deletions 백준/Silver/1193. 분수찾기/분수찾기.py
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"""
[풀이]
- 대각선 i번째 구하기
- i 홀수? 왼쪽 끝부터 (i/1) > (i--/1++) distance 만큼
- i 짝수? 오른쪽 끝부터 (1/i) > (1++/i--) distance 만큼
"""

import sys

readline = lambda: sys.stdin.readline().strip()
readline = input

target_num = int(readline())

# get (좌 or 우) 끝 번호
cross_idx = 1
end_num = 1
while end_num < target_num:
cross_idx += 1
end_num += cross_idx

# 거리만큼 이동 후 분수 구하기
distance = end_num - target_num
if cross_idx % 2 == 0:
denominator = cross_idx
numerator = 1

while distance:
distance -= 1
denominator -= 1
numerator += 1
else:
denominator = 1
numerator = cross_idx

# distance 만큼 이동해서 분수 구하기
distance = end_num - target_num
while distance:
distance -= 1
denominator += 1
numerator -= 1


print(f"{denominator}/{numerator}")

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