cppAcademy
FROM LEARNING TO COMPETING

Beginner.
Your first challenges.

Bulgarian Informatics Group E · 2007–2026

270 independently explained problems: 54 Municipal, 54 Regional, 105 National Olympiad tasks and 57 Autumn Tournament tasks. Start with the easier tasks and work upwards. Difficulty labels are learning estimates, not official ratings; Group E also includes challenging national-level problems. Years follow Arena’s competition-season labels. Source clarifications are noted on affected tasks.
GROUP E2023 · EasyTime simulation

Three Minutes of Candy

Autumn Tournament (2023)

Submission name: bonbon.cpp

Solve in Editor →

Problem summary

At each minute, Koko eats one candy for every digit 4 on his four-digit 24-hour clock. He only eats during the daytime interval from 08:00 to 20:00. Given the starting display, examine it and the next two minute displays. Report the last display and the total candies.

Input

One time hh:mm, with two digits for both fields.

Output

The display two minutes later in hh:mm format, then a space and the candy count.

Constraints

00 ≤ hh ≤ 23; 00 ≤ mm ≤ 59.

Worked example

The example is reproduced from the original tournament statement.

Input

14:42

Output

14:44 7

Solution analysis

  1. Convert the start to minutes since midnight. For offsets 0,1,2, reduce the resulting minute number modulo 1440 to handle a new day.
  2. During the daytime, split hour and minute into tens and units and count digits equal to four. Keep the time after the third observation and format both fields with two digits.
  3. The displays 14:42,14:43,14:44 contain respectively 2,2,3 fours, giving 7 candies. Starting at 23:58 instead ends at 00:00 and earns no candies.

Why it works

The loop inspects exactly the three requested displays, including both endpoints. Modulo 1440 gives the correct daily clock time. Each eligible display contributes one candy for each of its four digit positions equal to four, so summing the contributions gives the required total.

Complexity: O(1) time and O(1) extra space; all loops have a fixed upper bound.

Watch out. Include the initial minute. Preserve leading zeros and handle midnight. Nighttime digits do not contribute.

Commented C++17 solution

An independently written complete program with standard headers.

// Key idea: The loop inspects exactly the three requested displays, including both endpoints. Modulo
// 1440 gives the correct daily clock time.
#include<iostream>
#include<iomanip>
using namespace std;
int main(){
    int hour,minute;
    char colon;
    cin>>hour>>colon>>minute;
    int start=hour*60+minute;
    int candies=0;
    int current=start;
    for(int offset=0;offset<3;offset++){
        current=(start+offset)%1440;
        hour=current/60;
        minute=current%60;
        if(current>=8*60&&current<=20*60){
            candies+=hour/10==4;
            candies+=hour%10==4;
            candies+=minute/10==4;
            candies+=minute%10==4;
        }
    }
    cout<<setfill('0')<<setw(2)<<hour<<':';
    cout<<setw(2)<<minute<<' '<<candies<<'\n';
}

Source: Arena Olimpiici ↗