Problem:


Score : 300 points

Problem Statement

Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)

Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.

If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print -1.

Constraints

1 \leq A \leq B \leq 100

A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print -1.

Sample Input 1

2 2

Sample Output 1

25

If the price of a product before tax is 25 yen, the amount of consumption tax levied on it is:

When the consumption tax rate is 8 percent: \lfloor 25 \times 0.08 \rfloor = \lfloor 2 \rfloor = 2 yen.

When the consumption tax rate is 10 percent: \lfloor 25 \times 0.1 \rfloor = \lfloor 2.5 \rfloor = 2 yen.

Thus, the price of 25 yen satisfies the condition. There are other possible prices, such as 26 yen, but print the minimum such price, 25.

Sample Input 2

8 10

Sample Output 2

100

If the price of a product before tax is 100 yen, the amount of consumption tax levied on it is:

When the consumption tax rate is 8 percent: \lfloor 100 \times 0.08 \rfloor = 8 yen.

When the consumption tax rate is 10 percent: \lfloor 100 \times 0.1 \rfloor = 10 yen.

Sample Input 3

19 99

Sample Output 3

-1

There is no price before tax satisfying this condition, so print -1.



Sample input:
2 2


Sample output:
25

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    
    A, B = R()
    for x in range(2000):
        if int(x * 0.08) == A and int(x * 0.1) == B:
            print(x)
            return
    print(-1)
    
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
# 全探索で十分間に合う

eight, ten = list(map(int, input().split()))
for i in range(1000+5):
    if eight == int(i * 0.08) and ten == int(i * 0.1):
        print(i)
        exit()

print(-1)

Submission status: Accepted

Solution code:
import math
a,b=map(int,input().split())
f=0
for i in range(1,1001):
    if math.floor(i*0.08)==a and math.floor(i*0.1)==b:
        f=i
        break;
if f==0:
    print(-1)
else:
    print(f)

Submission status: Accepted

Solution code:
import unittest


class TestC(unittest.TestCase):
    def test_1(self):
        self.assertEqual(think(2, 2), 25)

    def test_2(self):
        self.assertEqual(think(8, 10), 100)

    def test_3(self):
        self.assertEqual(think(19, 99), -1)


def solve():
    a, b = read()
    result = think(a, b)
    write(result)


def read():
    return read_int(2)


def read_int(n):
    return read_type(int, n, sep=' ')


def read_float(n):
    return read_type(float, n, sep=' ')


def read_type(t, n, sep):
    return list(map(lambda x: t(x), read_line().split(sep)))[:n]


def read_line(n=0):
    if n == 0:
        return input().rstrip()
    else:
        return input().rstrip()[:n]


def think(a, b):
    max_a = 100
    min_tax = 0.08
    ceil = int(max_a / min_tax) + 1
    for c in range(1, ceil + 1):
        tax_a = int(c * 0.08)
        tax_b = int(c * 0.10)
        if a == tax_a and b == tax_b:
            return c
    return -1


def write(result):
    print(result)


if __name__ == '__main__':
    # unittest.main()
    solve()

Submission status: Accepted

Solution code:
a,b = map(int,input().split())
import math
import sys
for i in range(1,2000):
    if math.floor(i*0.08) == a and math.floor(i*0.1) == b:
        print(i)
        sys.exit()

print('-1')

Submission status: Accepted

Solution code:
a,b=map(int,input().split())
c=0
for i in range(1251):
    if int(i*0.08)==a and int(i*0.1)==b:
        print(i)
        exit()
print(-1)

Submission status: Accepted

Solution code:
a,b=map(int,input().split())
c=0
for i in range(1251):
    if int(i*0.08)==a and int(i*0.1)==b:
        print(i)
        c+=1
        break
if c==0:
    print(-1)

Submission status: Accepted

Solution code:
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by enraged(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import bisect
import heapq
# from math import *
from collections import defaultdict as dd  # defaultdict(<datatype>) Free of KeyError.
from collections import deque  # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()


def out(var): sys.stdout.write(var)


def l(): return list(map(int, data().split()))


def sl(): return list(map(str, data().split()))


def sp(): return map(int, data().split())


def ssp(): return map(str, data().split())


def l1d(n, val=0): return [val for i in range(n)]


def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


a, b = sp()
for i in range(1, 40001):
    t1, t2 = int(0.08 * i), int(0.1 * i)
    if t1 == a and t2 == b:
        out(str(i))
        exit(0)
out("-1")

Submission status: Accepted

Solution code:
A,B = map(int,input().split())

able = False
for i in range(1,1251):
  if int(i * 0.08) == A and int(i * 0.1) == B:
    able = True
    break
print(i) if able else print(-1)

Submission status: Accepted

Solution code:
def solve():
    import math
    A, B = map(int, input().split())
    for i in range(1, 10000):
        if math.floor(i * 0.08) == A and math.floor(i * 0.1) == B:
            print(i)
            return
    print(-1)

solve()

Submission status: Accepted

Solution code:
"""
C
"""
A,B=map(int,input().split())
for i in range(10**5):
    if int(i*0.08)==A and int(i*0.1)==B:
        print(i)
        break
else:
    print(-1)

Submission status: Accepted

Solution code:
a, b = map(int, input().split())
r = -1
for price in range(1010):
    tax_8 = int(price * 0.08)
    tax_10 = int(price * 0.1)
    if tax_8 == a and tax_10 == b:
        r = price
        break
print(r)

Submission status: Accepted

Solution code:
import math
a,b=map(int,input().split())
for i in range(1,1009):
	A=math.floor(i*0.08)
	B=math.floor(i*0.1)
	if a==A and b==B:
		print(i)
		exit()
print(-1)

Submission status: Accepted

Solution code:
def f():
  a = list(map(float,input().split()))

  for i in range(0, 1251):
    if i >= a[0]/0.08 and i < (a[0]+1)/0.08 and i >= a[1]/0.1 and i < (a[1]+1)/0.1:
      print(i)
      return

  print(-1)

f()

Submission status: Accepted

Solution code:
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
import itertools
from functools import *
from itertools import permutations,combinations,groupby
import sys
import bisect
import string
import math
import time
import random
def Golf():
    *a,=map(int,open(0))
def S_():
    return input()
def IS():
    return input().split()
def LS():
    return [i for i in input().split()]
def I():
    return int(input())
def MI():
    return map(int,input().split())
def LI():
    return [int(i) for i in input().split()]
def LI_():
    return [int(i)-1 for i in input().split()]
def NI(n):
    return [int(input()) for i in range(n)]
def NI_(n):
    return [int(input())-1 for i in range(n)]
def StoI():
    return [ord(i)-97 for i in input()]
def ItoS(nn):
    return chr(nn+97)
def LtoS(ls):
    return ''.join([chr(i+97) for i in ls])
def GI(V,E,Directed=False,index=0):
    org_inp=[]
    g=[[] for i in range(n)]
    for i in range(E):
        inp=LI()
        org_inp.append(inp)
        if index==0:
            inp[0]-=1
            inp[1]-=1
        if len(inp)==2:
            a,b=inp
            g[a].append(b)
            if not Directed:
                g[b].append(a)
        elif len(inp)==3:
            a,b,c=inp
            aa=(inp[0],inp[2])
            bb=(inp[1],inp[2])
            g[a].append(bb)
            if not Directed:
                g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0}):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0}) # sample usage
    mp=[1]*(w+2)
    found={}
    for i in range(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[1]+[mp_def[j] for j in s]+[1]
    mp+=[1]*(w+2)
    return h+2,w+2,mp,found
def bit_combination(k,n=2):
    rt=[]
    for tb in range(n**k):
        s=[tb//(n**bt)%n for bt in range(k)]
        rt+=[s]
    return rt
def show(*inp,end='\n'):
    if show_flg:
        print(*inp,end=end)
YN=['YES','NO']
Yn=['Yes','No']
 
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
u_alp=string.ascii_uppercase
#ts=time.time()
#sys.setrecursionlimit(10**7)
input=lambda: sys.stdin.readline().rstrip()
 
def ran_input():
    import random
    n=random.randint(4,16)
    rmin,rmax=1,10
    a=[random.randint(rmin,rmax) for _ in range(n)]
    return n,a

def ran_input():
    import random
    n=random.randint(4,16)
    rmin,rmax=1,10
    a=[random.randint(rmin,rmax) for _ in range(n)]
    return n,a

show_flg=False
show_flg=True

ans=0

a,b=LI()
for i in range(1,10**6):
    if int(i*0.08)==a and int(i*0.1)==b:
        print(i)
        exit()

print(-1)

Submission status: Accepted

Solution code:
import sys
import itertools

INF = 1001001001

def resolve():
  a, b = map(int, input().split())
  for i in range(10, 10100):
    if int(i * 0.08) == a and int(i * 0.10) == b:
      print(i)
      return
  print('-1')
  return

if __name__ == "__main__":
    resolve()

Submission status: Accepted

Solution code:
from math import floor
A, B = map(int,input().split())

maxX = 1250
ans = -1
for i in range(maxX):
    if (floor(i * 0.08) == A) and (floor(i * 0.1) == B):
        ans = i
        break

print(str(ans))

Submission status: Accepted

Solution code:
a, b = map(int, input().split())
for i in range(1, 1001):
    if i * 8 // 100 == a and i * 10 // 100 == b:
        print(i)
        break
else:
    print(-1)

Submission status: Accepted

Solution code:
a, b = map(int, input().split())



for x in range(1,100001):
    if int(0.08 * x) == a and int(0.1 * x) == b:
        print(x)
        break
    elif x == 100000:
        print(-1)

Submission status: Accepted

Solution code:
# coding: utf-8
# Your code here!

A, B = map(int, input().rstrip().split())

flag = False
for i in range(1000000):
    if int(i * 0.08) == A and int(i * 0.1) == B:
        print(i)
        flag = True
        break
    
if flag == False:
    print(-1)

Submission status: Accepted

Solution code:
A, B = [int(x) for x in input().split()]

for i in range(1009):
    if A == int(0.08*i) and B == int(0.10*i):
        ans = i
        break
    elif i == 1008:
        ans = -1  

print (ans)

Submission status: Accepted

Solution code:
import math
A, B = map(int, input().split())

xa_min = math.ceil((A * 100) / 8)
xa_max = math.floor(((A + 1) * 100) / 8)

xb_min = math.ceil((B * 100) / 10)
xb_max = math.floor(((B + 1) * 100) / 10)

low_price = max(xa_min, xb_min)

if xb_max <= xa_min or xa_max <= xb_min:
    print(-1)
else :
    print(low_price)

Submission status: Accepted

Solution code:
import math

a, b = map(int, input().split())

a_low = math.ceil(a*12.5)
a_high = math.floor((a+1)*12.5-0.01)
b_low=b*10
b_high=b*10+9

if b_high < a_low or a_high < b_low:
    print(-1)
else:
    print(max(a_low, b_low))

Submission status: Accepted

Solution code:
a,b = map(int,input().split())
li = []
for i in range(1,1010):
    if int(i*0.08) == a and int(i*0.1) == b:
        li.append(i)
if len(li) > 0:
    print(li[0])
else:
    print(-1)

Submission status: Accepted

Solution code:
A, B = map(int,input().split())

min8 = (A * 25 + 1) // 2 
max8 = ((A+1) * 25 - 1) // 2
min10 = B * 10
max10 = (B+1) * 10 - 1

if min10 > max8 or min8 > max10:
	print(-1)
else:
	print(max([min8, min10]))

Submission status: Accepted

Solution code:
A,B = map(int,input().split())


def solve():
    for i in range(100*100):
        if( int(i * 0.08) == A and int(i*0.1)==B ):
            return i
    return -1
print(solve())

Submission status: Accepted

Solution code:
a,b = map(int,input().split())

count = 0
for i in range(1,10001):
  if i *0.08//1 == a:
    if i *0.1//1 == b:
      print(i)
      if i *0.1//1 == b:
        break
    else:
        count = count +1
  else:
    count = count + 1

if count == 10000:
  print(-1)

Submission status: Accepted

Solution code:
import math
a1,b1 = map(int,input().split())
a2 = a1+1
b2 = b1+1
p1 = a1/0.08
p2 = a2/0.08
p3 = b1/0.1
p4 = b2/0.1

i1 = math.ceil(p1)
i2 = math.floor(p2-0.000001)
i3 = math.ceil(p3)
i4 = math.floor(p4-0.000001)

for i in range(i1, i2):
  if i >=i3 and i<i4:
    print(i)
    exit()
print(-1)

Submission status: Accepted

Solution code:
import math
a,b = map(int,input().split())
a_min = math.ceil(a / 0.08)
a_max = math.floor((a + 1) / 0.08) - 1
b_min = math.ceil(b / 0.1)
b_max = math.floor((b + 1) / 0.1) - 1
 
if a_max < b_min or b_max < a_min:
  print('-1')
else:
  result = b_min if a_min <= b_min else a_min
  print(result)

Submission status: Accepted

Solution code:
import math
a, b = map(int, input().split())
ans = []
for i in range(1, 1251):
    if math.floor(i * 0.08) == a and math.floor(i * 0.1) == b:
        ans.append(i)
print(-1 if not ans else min(ans))

Submission status: Accepted

Solution code:
a,b=(int(x) for x in input().split())
 
import math 
 
i=0
while i<10001:
  i+=1
  if math.floor(i*0.08)==a and math.floor(i*0.1)==b:
    print(i)
    break
else:
  print(-1)

Submission status: Accepted

Solution code:
import math

A, B = map(int, input().split())

for i in range( 10001 ):
    if math.floor(i * 0.08) == A and math.floor(i * 0.1) == B:
        print(i)
        break
else:
    print(-1)

Submission status: Accepted

Solution code:
a, b = map(int, input().split())

h = -1
for i in range(1, 10001):
  if i * 8 // 100 == a:
    if i * 10 // 100 == b:
      h = i
      break
      
print(h)

Submission status: Accepted

Solution code:
a,b = map(int,input().split())
for price in range(10000):
    if price * 8 // 100 == a and price * 10//100 ==b:
        print(price)
        exit()
else:
    print(-1)

Submission status: Accepted

Solution code:
import math
a,b = map(int,input().split())

i = math.floor(a/0.08)
while i < 10001:
    if math.floor(i*0.08) == a and math.floor(i*0.1) ==b:
        print(i)
        break
    i += 1
else:
    print(-1)

Submission status: Accepted

Solution code:
import math
A, B = map(int, input().split())
eight = 0.08
ten = 0.1
for i in range(100*11):
    if math.floor(i * eight) == A and math.floor(i * ten) == B:
        print(i)
        exit()
print(-1)

Submission status: Accepted

Solution code:
a, b = map(int, input().split())
for i in range(20000):
    hachi = i * 8 // 100
    ju = i // 10
    if (hachi, ju) == (a, b):
        print(i)
        exit()
    
print(-1)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3 ## -*- python -*-

t = input().split()
a = int(t[0])
b = int(t[1])

def judge(a, b):
    a_min = a / 0.08
    a_max = (a+1) / 0.08

    b_min = b / 0.1
    b_max = (b+1) / 0.1

    for c in range(int(a_min), int(a_max)):
        if c < a_min:
            continue
        for d in range(int(b_min), int(b_max)):
            if d < b_min:
                continue
            if c == d:
                return c
    return -1

print(judge(a,b))

Submission status: Accepted

Solution code:
import math
tax = input().split(" ")
tax_8 = int(tax[0])
tax_10 = int(tax[1])

price_8 = [math.ceil(tax_8*(100/8)), math.ceil((tax_8+1)*(100/8))-1]
price_10 = [math.ceil(tax_10*(10)), math.ceil((tax_10+1)*(10))-1]
if(min(price_8[1], price_10[1]) < max(price_8[0], price_10[0])):
    print(-1)
else:
    print(max(price_8[0], price_10[0]))