Python 2

def solve(year):
    if (year == 1918):
        return '26.09.1918'
    elif ((year <= 1917) & (year%4 == 0)) or ((year > 1918) & (year%400 == 0 or ((year%4 == 0) & (year%100 != 0)))):
        return '12.09.%s' %year
    else:
        return '13.09.%s' %year

Hi (java) people. Hope this simple logic will be useful.

static String solve( int year ) {

    if( year == 1918 ) return "26.09.1918";
    if( isLeap( year ) ) return "12.09." + Integer.toString( year );
    else return "13.09." + Integer.toString( year );

}

static boolean isLeap( int year ) {

    if( year % 4 != 0 ) return false;
    if( year > 1918 && year % 100 == 0 && year % 400 != 0 ) return false;
    return true;

}

C++

string solve(int y){
    int d = 13;
    if(!(y%4) && (y < 1918 || y%100 || !(y%400))) d--;
    if(y == 1918) d = 26;
    return to_string(d) + ".09." + to_string(y);
}

how is year 1800 a leap year ? acc to the rules mentioned in the problem.

I think testcase#48 is wrong because 1900 is not leap year so it should have answer 13.09.1900 but it has 12.09.1900.