From 4e9bf9a2236a6d93788542c064a6b275fb665a05 Mon Sep 17 00:00:00 2001 From: "copilot-swe-agent[bot]" <198982749+Copilot@users.noreply.github.com> Date: Wed, 27 Aug 2025 16:14:25 +0000 Subject: [PATCH 1/2] Initial plan From 276bdc9c7c509b56050fd02f6f69de525788a12f Mon Sep 17 00:00:00 2001 From: "copilot-swe-agent[bot]" <198982749+Copilot@users.noreply.github.com> Date: Wed, 27 Aug 2025 16:18:23 +0000 Subject: [PATCH 2/2] Add comprehensive solutions text file for first 5 problems Co-authored-by: minhajadil <64237828+minhajadil@users.noreply.github.com> --- first_5_problems_solutions.txt | 267 +++++++++++++++++++++++++++++++++ 1 file changed, 267 insertions(+) create mode 100644 first_5_problems_solutions.txt diff --git a/first_5_problems_solutions.txt b/first_5_problems_solutions.txt new file mode 100644 index 0000000..0988412 --- /dev/null +++ b/first_5_problems_solutions.txt @@ -0,0 +1,267 @@ +FIRST 5 COMPETITIVE PROGRAMMING PROBLEMS AND SOLUTIONS +===================================================== + +This document contains the first 5 problems from the repository with detailed solution explanations based on the existing code implementations. + +PROBLEM 1: In Search of an Easy Problem (1030A) +=============================================== +Link: https://codeforces.com/contest/1030/problem/A + +Problem Description: +When preparing a tournament, Codeforces coordinators try their best to make the first problem as easy as possible. A problem is easy if it can be solved by more than half of the contestants. +You are given the results of the survey on n participants. For each participant, you know if they think the problem is easy (represented by 1) or hard (represented by 0). +Your task is to determine if the problem is easy or hard based on the survey results. + +Solution Explanation: +The solution works by: +1. Reading the number of participants (n) +2. Reading n survey results (0 or 1) +3. Counting how many participants think the problem is easy (value = 1) +4. If at least one participant thinks it's hard (counter > 0), output "HARD" +5. Otherwise, output "EASY" + +Key Logic: +- If any participant finds the problem hard, the overall assessment is "HARD" +- Only if ALL participants find it easy, the assessment is "EASY" + +Code Implementation: +```cpp +#include +using namespace std; + +int main() { + int a; + cin>> a; + int b[a]; + int counter = 0; + + for(int i =0; i>b[i]; + if(b[i]==1) counter++; + } + + if(counter>0) cout << "HARD" << endl; + else cout << "EASY" << endl; + + return 0; +} +``` + +PROBLEM 2: Nearly Lucky Number (110A) +==================================== +Link: https://codeforces.com/contest/110/problem/A + +Problem Description: +A lucky number is a number that contains only digits 4 and 7. +A nearly lucky number is a number such that the count of lucky digits (4 and 7) in it is a lucky number. +Given a number, determine if it is nearly lucky. + +Solution Explanation: +The solution works by: +1. Reading the input number +2. Extracting each digit and counting how many are 4 or 7 +3. Checking if the count itself is a lucky number (4 or 7) +4. Output "YES" if nearly lucky, "NO" otherwise + +Key Logic: +- Extract digits using modulo 10 and integer division +- Count occurrences of digits 4 and 7 +- The count must be exactly 4 or 7 for the number to be nearly lucky + +Code Implementation: +```cpp +#include +using namespace std; + +int main() { + unsigned long long int t; + cin>> t; + + int counter =0; + int r; + while(t>=1) { + r=t%10; + if(r==7 ||r==4) counter++; + t=t/10; + } + if(counter==7 || counter==4) cout << "YES" << endl; + else cout << "NO" << endl; + + return 0; +} +``` + +PROBLEM 3: Petya and Strings (112A) +=================================== +Link: https://codeforces.com/contest/112/problem/A + +Problem Description: +Little Petya loves presents. His mother bought him two strings of the same size for his birthday. +The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare these two strings lexicographically. +The comparison should be case-insensitive (meaning that the casing of letters should be ignored). +Help Petya perform the comparison. + +Solution Explanation: +The solution works by: +1. Reading two input strings +2. Converting both strings to lowercase for case-insensitive comparison +3. Comparing the strings lexicographically +4. Output 1 if first string > second, 0 if equal, -1 if first < second + +Key Logic: +- Use tolower() function to convert characters to lowercase +- C++ string comparison automatically handles lexicographic ordering +- Return appropriate integer based on comparison result + +Code Implementation: +```cpp +#include +using namespace std; +#define optimize() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); + +int main() { + string a,b; + cin >> a >>b; + int size=a.size(); + + for(int i =0; ib) cout << 1 << endl; + else if(a==b) cout << 0 << endl; + else cout << -1 << endl; + + return 0; +} +``` + +PROBLEM 4: Game 23 (1141A) +========================== +Link: https://codeforces.com/contest/1141/problem/A + +Problem Description: +Polycarp plays "Game 23". Initially he has a number n and wants to transform it to m. +In one move, he can multiply the number by 2 or by 3. +Find the minimum number of moves required to transform n into m, or determine if it's impossible. + +Solution Explanation: +The solution works by: +1. Reading two numbers n and m +2. Check if m is divisible by n (if not, impossible) +3. Calculate d = m/n +4. Count how many times we can divide d by 2 and 3 +5. If d becomes 1, return the count; otherwise return -1 + +Key Logic: +- If m is not divisible by n, transformation is impossible +- We need to check if m/n can be expressed as 2^a * 3^b +- Count divisions by 2 and 3 until d becomes 1 +- If d doesn't become 1, the transformation is impossible + +Code Implementation: +```cpp +#include +using namespace std; + +int main() { + int n, m; + cin >> n >> m; + int d; + int cnt = 0; + + if(m%n!=0) cout << -1 << endl; + else { + d = m/n; + while(d%3==0) { + d/=3; + cnt++; + } + while(d%2==0) { + cnt++; + d/=2; + } + if(d==1) cout << cnt << endl; + else cout << -1 << endl; + } + + return 0; +} +``` + +PROBLEM 5: Maximal Continuous Rest (1141B) +========================================== +Link: https://codeforces.com/contest/1141/problem/B + +Problem Description: +Polycarp lives in a country where days form a cycle. There are n days, and after day n comes day 1 again. +For each day, Polycarp knows whether he has to work (0) or can rest (1). +Find the maximum number of consecutive days when Polycarp can rest. + +Solution Explanation: +The solution works by: +1. Reading the number of days and the work/rest schedule +2. Finding the maximum continuous sequence of 1s (rest days) +3. Handling the circular nature - if both first and last days are rest days, they connect +4. Return the maximum continuous rest period + +Key Logic: +- Track current continuous rest days and maximum found so far +- Handle circular case: if array starts and ends with 1s, they form one continuous sequence +- Count rest days from beginning until first 0, and from last 0 until end +- Add these counts if both first and last elements are 1 + +Code Implementation: +```cpp +#include +using namespace std; + +int main() { + int n; + cin >> n; + + int a[n]; + for(int i =0; i> a[i]; + } + int max=-1; + int cnt =0; + int f = 1; + bool y =0; + if(a[0]==1) y=1; + int first=0; + + for(int i =0; i=0; i--) { + if(a[i]==0) break; + else r++; + } + } + first = r+ first; + + if(max