【小码匠自习室】ABC258-A 代码写的啰嗦了

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发布于 2022-08-08 13:21:07

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发布于 2022-08-08 13:21:07

文章被收录于专栏：小码匠和老码农

碎碎念

《论当你在赛后才发现你做的两个判断语句是多么愚蠢这件事》

题目地址

A - When?

https://atcoder.jp/contests/abc258/tasks/abc258_a

問題描述

AtCoder Beginner Contest usually starts at 21:00 JST and lasts for 100 minutes.

You are given an integer K between 00 and 100100 (inclusive). Print the time K minutes after 21:00 in the HH:MM format, where HH denotes the hour on the 24-hour clock and MM denotes the minute. If the hour or the minute has just one digit, append a 0 to the beginning to represent it as a 2-digit integer.

Constraints

K is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

Output

Print the time K minutes after 21:00 in the format specified in the Problem Statement.

Sample Input 1

Sample Output 1

22:03

6363 minutes after 21:00, it will be 22:03, so 22:03 should be printed.

The following outputs would be judged incorrect:

10:03
22:3

Sample Input 2

Sample Output 2

21:45

Sample Input 3

Sample Output 3

22:40

题解

小码匠题解

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int k;
    cin >> k;
    int h = k / 60;
    if (21 + h >= 24) {
        if ((21 + h) % 24 < 10) {
            cout << 0 << (21 + h) % 24;
        } else {
            cout << (21 + h) % 24;
        }
    } else {
        cout << 21 + h;
    }
    cout << ':';
    if (k % 60 < 10) {
        cout << 0 << k % 60;
    } else {
        cout << k % 60;
    }
}

小码匠题解

void best_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int k;
    cin >> k;
    int h = k / 60;
    cout << 21 + h << ':';
    if (k % 60 < 10) {
        cout << 0 << k % 60;
    } else {
        cout << k % 60;
    }
}

参考题解（tourist）

#include <bits/stdc++.h>
 
using namespace std;
 
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int k;
  cin >> k;
  int h = 21 + k / 60;
  int m = k % 60;
  cout << h << ":" << m / 10 << m % 10 << '\n';
  return 0;
}

参考题解

#include <iostream>
#include <cstdio>
using namespace std;
int main() {
  int X;
  cin >> X;
  int H = X < 60 ? 21 : 22;
  int M = X % 60;
  printf("%d:%02d", H, M);
  return 0;
}

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原始发表：2022-07-04，如有侵权请联系 cloudcommunity@tencent.com 删除

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