Teaching and learning resources • Re: Advent of Code 2025

Code:

#include <iostream>#include <chrono>#include <fstream>#include <set>const char* filename = "data.txt";const int tens[] = {1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};inline bool isodd(int i){    return i & 1;}int digits(long num){    int ret = 0;    while(num)    {        num /= 10;        ret++;    }    return ret;}long checkrange(long r1, long r2, int d, int num, std::set<long>& found){    // check range for repeated patterns of num digits;    // r1 and r2 are assumed to have the same number of digits    // d is the number of digits in ranges    // num is the number of digits of the pattern to check    if(d % num)        return 0;    // s is the number of repeats of the pattern    int s = d / num;    long ret = 0;    int p = tens[num];    long x = 1;         // this could be improved    while(x < (p * s))    {        // make t the repeated pattern        long t = x;        for(int i = 0; i < s - 1; i++)            t = (t * p) + x;        // check if repeated pattern is in range        if((r1 <= t) && (r2 >= t))        {            // avoid duplicates            if(found.find(t) == found.cend())            {                ret += t;                found.insert(t);            }        }        if(t > r2)            break;         x++;    }    return ret;}long checkpatterns(long r1, long r2, int digits){    // my data has at most 10 digits    // we need to look for sequences of 2,3,4,5 digits    // some calls will be invalid but checkrange will sort    // set found set stops repeats e.g 11 4 times and 1111 twice    long ret = 0;    std::set<long> found;    for(int p = 1; p <= digits/2; p++)        ret += checkrange(r1, r2, digits, p, found);    return ret;}long check2(long r1, long r2){    // if neccessary split range so limits have same digits    // call checkrange for all possible pattern sizes    long ret = 0;    int d1 = digits(r1);    int d2 = digits(r2);    if(d1 != d2)    {        ret += checkpatterns(r1,tens[d1] -1, d1);        ret += checkpatterns(tens[d2 - 1],r2, d2);    }    else        ret += checkpatterns(r1,r2,d1);    return ret;}long check1(long r1, long r2){    // some range limits have different number of digits    // invalid id must have even number of digits    int d1 = digits(r1);    int d2 = digits(r2);    int d;    // r1 and r2 must be even    // in my data range limits differ by at most one digit in size    if(d1 != d2)    {        if(isodd(d1))        {            r1 = tens[d1];            d = d2;        }        else        {            r2 = tens[d2 - 1];            d = d1;        }    }    else        d = d1;    if(isodd(d))        return 0;    // generate numbers ofthe form xyzxyz    // and then test if in range;    long ret = 0;    int p = tens[d/2];    long x = r1 / p;    while(x < p)    {        long t = x + x*p;        if(r1 <= t && t <= r2)            ret += t;        if(t > r2)            break;        x++;    }    return ret;}int main(){    std::chrono::time_point<std::chrono::system_clock> start,stop;    start = std::chrono::system_clock::now();    std::cout << "Advent of code 2025 day 2 \n";    std::ifstream file(filename);    long part1 = 0;    long part2 = 0;    if(file)    {        while(!file.eof())        {            long l1,l2;            char c;            file >> l1 >> l2 >> c;            //std::cout << l1 << "  " << -l2 <<  "   " << digits(-l2) << std::endl;            part1 += check1(l1, -l2);            part2 += check2(l1, -l2);        }    }    std::cout << "Part 1  " << part1  << "\n";    std::cout << "Part 2  " << part2  << "\n";    stop = std::chrono::system_clock::now();    std::chrono::duration<double> elapsed = stop - start;    std::cout << "Elapsed time " << elapsed.count() << " seconds" << "\n";    return 0;}

Code:

Advent of code 2025 day 2 Part 1  21139440284Part 2  38731915928Elapsed time 0.000861608 seconds