From 2362df3d1f860d5dd67a3a4fb7c4463c653bef8c Mon Sep 17 00:00:00 2001 From: Jeongan Lee <84510455+fkdl0048@users.noreply.github.com> Date: Thu, 20 Jun 2024 00:55:38 +0900 Subject: [PATCH] BOJ-EX: 6/20/2024, 12:55:37 AM --- ... \353\247\214\353\223\244\352\270\260.cpp" | 61 ++++++++++++++++++ ... \353\247\214\353\223\244\352\270\260.exe" | Bin 0 -> 73473 bytes .../README.md" | 42 ++++++++++++ 3 files changed, 103 insertions(+) create mode 100644 "1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.cpp" create mode 100644 "1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.exe" create mode 100644 "1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/README.md" diff --git "a/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.cpp" "b/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.cpp" new file mode 100644 index 0000000..7d14822 --- /dev/null +++ "b/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.cpp" @@ -0,0 +1,61 @@ +/* ************************************************************************** */ +/* */ +/* ::: ::: ::: */ +/* Problem Number: 1463 :+: :+: :+: */ +/* +:+ +:+ +:+ */ +/* By: fkdl4878 +#+ +#+ +#+ */ +/* +#+ +#+ +#+ */ +/* https://boj.kr/1463 #+# #+# #+# */ +/* Solved: 2024/06/20 00:16:03 by fkdl4878 ### ### ##.kr */ +/* */ +/* ************************************************************************** */ + +// 정수 x에 대해서 1로 만드는 문제 +// 사용할 수 있는 연산은 3가지 + // x가 3으로 나눠 떨어지면 3으로 나눈다. + // x가 2로 나눠 떨어지면 2로 나눈다. + // 1을 뺀다. +// 이 연산을 사용해서 1로 만들 때 최소한의 연산을 사용 +// 각각 연산 함수를 만들고 이를 계산 함수에서 활용하여 계산 + +// ps1 +// dp 문제이기 때문에 dp로 풀어야 함 +// 먼저 0~5까지 dp를 입력값에 넣어줌 +// 이후 6부터는 반복으로 만들어야 함 N까지 + // 6의 경우 2 -> 1 총 두 번의 연산이 필요하다. + // 따라서 dp[3] + dp[2]의 연산 +// 전 결과를 가져오기 위해서 전 계산의 조합으로 구성됨 + // 7의 경우 7 -> 6 -> 2 -> 1 + // dp로는 아마 dp[6] + +// 즉, + +#include + +using namespace std; + +int solve(int n) +{ + vector dp(n + 1, 0); + +// dp 수행.. + for (int i = 2; i <= n; i++) + { + dp[i] = dp[i - 1] + 1; + if (i % 2 == 0) + dp[i] = min(dp[i], dp[i / 2] + 1); + if (i % 3 == 0) + dp[i] = min(dp[i], dp[i / 3] + 1); + } + +// 결과 출력 + return dp[n]; +} + +int main() +{ + int n; + + cin >> n; + + cout << solve(n); +} \ No newline at end of file diff --git "a/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.exe" "b/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/1\353\241\234 \353\247\214\353\223\244\352\270\260.exe" new file mode 100644 index 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zj6+KD5iE<}3EVN5MvdiUmh!MQDz~=Cnyc~7m?4kobee)lbL;6Vo literal 0 HcmV?d00001 diff --git "a/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/README.md" "b/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/README.md" new file mode 100644 index 0000000..2b7d293 --- /dev/null +++ "b/1463\353\262\210\357\274\232 1\353\241\234 \353\247\214\353\223\244\352\270\260/README.md" @@ -0,0 +1,42 @@ +# 1463번: 1로 만들기 - Silver III + + + + + + + +## 문제 + +[문제 링크](https://boj.kr/1463) + + +

정수 X에 사용할 수 있는 연산은 다음과 같이 세 가지 이다.

+ +
    +
  1. X가 3으로 나누어 떨어지면, 3으로 나눈다.
  2. +
  3. X가 2로 나누어 떨어지면, 2로 나눈다.
  4. +
  5. 1을 뺀다.
  6. +
+ +

정수 N이 주어졌을 때, 위와 같은 연산 세 개를 적절히 사용해서 1을 만들려고 한다. 연산을 사용하는 횟수의 최솟값을 출력하시오.

+ + + +## 입력 + + +

첫째 줄에 1보다 크거나 같고, 106보다 작거나 같은 정수 N이 주어진다.

+ + + +## 출력 + + +

첫째 줄에 연산을 하는 횟수의 최솟값을 출력한다.

+ + + +## 소스코드 + +[소스코드 보기](1로%20만들기.cpp) \ No newline at end of file