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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A positive integer a is given. Baron Munchausen claims that he knows such a positive integer n that if one multiplies n by a, the sum of its digits decreases a times. In other words, S(an) = S(n)/a, where S(x) denotes the sum of digits of the number x. Find out if what Baron told can be true. Input The only line contains a single integer a (2 ≀ a ≀ 10^3). Output If there is no such number n, print -1. Otherwise print any appropriate positive integer n. Your number must not consist of more than 5β‹…10^5 digits. We can show that under given constraints either there is no answer, or there is an answer no longer than 5β‹…10^5 digits. Examples Input 2 Output 6 Input 3 Output 6669 Input 10 Output -1 Submitted Solution: ``` #!/bin/pypy3 from itertools import* from timeit import* #a={ #2:6 #,3:6669 #,4:75 #,7:14286 #,8:125 #} #print(a.get(int(input()),-1)) S=lambda x:sum(map(int,str(x))) def solve(a): for e,first in product(range(0,100),range(1,600)): x=first*10**e z=x%a if z: x+=a-z #assert x%a==0 if S(x//a)==S(x)*a: return x x+=a #assert x%a==0 if S(x//a)==S(x)*a: return x return -1 #for a in range(2,1000): #print(timeit(lambda: print(a,solve(a)),number=1)) print(solve(int(input()))) ```
instruction
0
101,236
20
202,472
No
output
1
101,236
20
202,473
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A positive integer a is given. Baron Munchausen claims that he knows such a positive integer n that if one multiplies n by a, the sum of its digits decreases a times. In other words, S(an) = S(n)/a, where S(x) denotes the sum of digits of the number x. Find out if what Baron told can be true. Input The only line contains a single integer a (2 ≀ a ≀ 10^3). Output If there is no such number n, print -1. Otherwise print any appropriate positive integer n. Your number must not consist of more than 5β‹…10^5 digits. We can show that under given constraints either there is no answer, or there is an answer no longer than 5β‹…10^5 digits. Examples Input 2 Output 6 Input 3 Output 6669 Input 10 Output -1 Submitted Solution: ``` import random def dsum(a): return sum(int(digit) for digit in str(a)) def f(v, n): return dsum(v // n) - n * dsum(v) def make_number(a, n): for q in range(5): for M in range(30, (15 + q) * n + 2000, 47): v = 10**15 + 10**M for i in range(a): v = v + 10 ** random.randint(16, M-1) if v % n == 0: if f(v, n) >= 0: return v return -1 def solve(n): v = -1 z = 10 if n % 9 == 0: z = z * 9 for a in range(3, z): if n % 9 == 0 and (a+2) % 9 != 0: continue b = make_number(a, n) if b > 0: v = b break if n == 1: v == 1 if v == -1: assert False #print(n, "failfailfailfailfailfailfailfailfailfailfailfailfailfailfailfailfailfailfailf") return -1 else: for b in range(1, 200): if f(b*n, n) >= 0: continue if f(v,n) + f(b*n, n) >= 0: v = int(str(v) + str(b*n)) did = True if f(v, n) == 0: return v diff1 = f(v, n) q = n while f(q, n) >= 0: q = q + n diff2 = f(q, n) ans = "" for x in range(-diff2): ans += str(v) for x in range(diff1): ans += str(q) v = int(ans) # print(f(v,n)) return v def bad(n): while n % 2 == 0: n = n // 2 while n % 5 == 0: n = n // 5 if n == 1: return True else: return False def real_solve(r): if r == 1 or r == 2 or r == 4 or r == 8 or not bad(r): return solve(r) else: return -1 def test(): for r in range(1, 1000): solve(r) def cf(): a = int(input()) ans = real_solve(a) print(ans) random.seed(48) cf() ```
instruction
0
101,237
20
202,474
No
output
1
101,237
20
202,475
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,302
20
202,604
Tags: brute force, greedy, implementation Correct Solution: ``` import math as ma q=int(input()) for w in range(q): n=int(input()) m=n i=ma.ceil(ma.log(n,3)) j=i-1 k=0 r=n while m>=0 and j>=0: if m>=3**j: m=m-3**j else: k=j r=n-m j=j-1 if m==0: print(n) elif k>0: print(r+3**k) else: print(3**i) ```
output
1
101,302
20
202,605
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,303
20
202,606
Tags: brute force, greedy, implementation Correct Solution: ``` #import resource import sys #resource.setrlimit(resource.RLIMIT_STACK, (2**29, -1)) sys.setrecursionlimit(10 ** 7) from collections import deque import math pow3 = [1] aux = 1 for i in range(9): aux *= 3 pow3.append(aux) a = [] for op in range((1 << 10)): elem = 0 for i in range(len(pow3)): if ( (op >> i) & 1 ): elem += pow3[i] a.append(elem) t = int(input()) for _ in range(t): n = int(input()) lo, hi = 0, len(a) - 1 while ( hi - lo > 1 ): mi = lo + ( hi - lo ) // 2 if (n > a[mi]): lo = mi else: hi = mi print(a[hi]) ```
output
1
101,303
20
202,607
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,304
20
202,608
Tags: brute force, greedy, implementation Correct Solution: ``` import math q = int(input()) pw = [3**0] for i in range(1, 10): pw.append(3**i) length = len(pw) for j in range(length-1): pw.append(3**i+pw[j]) for _ in range(q): n = int(input()) for i in range(len(pw)): if n<=pw[i]: print (pw[i]) break ```
output
1
101,304
20
202,609
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,305
20
202,610
Tags: brute force, greedy, implementation Correct Solution: ``` arr = [3**i for i in range(45)] for i in range(int(input())): n, hi = int(input()), sum(arr) for j in reversed(arr): if hi - j >= n: hi -= j print(hi) ```
output
1
101,305
20
202,611
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,306
20
202,612
Tags: brute force, greedy, implementation Correct Solution: ``` #this is wild game def gcd(a:int,b:int) -> int: return a if b==0 else gcd(b,a%b) def conv(n:int,b:int=3)->int: a=0 p=1 while n: a+=(n%b)*p p*=10 n//=b return a def solve(n:int)->int: p=0 s="" while n: l=n%10 if p: l=(l+1)%3 if l==0: p=1 s+="0" elif l==1: p=0 s="0"*len(s)+"1" else: p=1 s+="0" else: if l==2: p=1 s+="0" else: p=0 s+=str(l) n//=10 if p: s="0"*len(s)+"1" # print(s[::-1]) return s[::-1] for _ in range(int(input())): print(int(solve(conv(int(input()))),3)) ```
output
1
101,306
20
202,613
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,307
20
202,614
Tags: brute force, greedy, implementation Correct Solution: ``` def b10to3(num): digs = [] num = int(num) while num > 0: digs.append(num % 3) num //= 3 return list(reversed(digs)) def b3to10(num): sum = 0 for i in num: sum = sum * 3 + i return str(sum) for i in range(int(input())): num = input() digs = [0] + b10to3(num) # 0100200121 # take the MSB which is 2, find something to its left which is 0, increment it and set rest to 0 pos_0, pos_2 = -1, -1 for i in range(len(digs)): if digs[i] == 2: pos_2 = i break if digs[i] == 0: pos_0 = i if pos_2 != -1: digs[pos_0] = 1 for i in range(pos_0 + 1, len(digs)): digs[i] = 0 print(b3to10(digs)) ```
output
1
101,307
20
202,615
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,308
20
202,616
Tags: brute force, greedy, implementation Correct Solution: ``` q = int(input()) def base_10_to_n(x, n): if x//n: return base_10_to_n(x//n, n)+str(x%n) else: return str(x%n) for _ in range(q): n = int(input()) s = list(base_10_to_n(n, 3)) #print(s) s.reverse() pos0 = -1 pos2 = -1 flag = False for i in range(len(s)): if s[i] == '2': pos2 = i flag = True if s[i] == '0' and flag: pos0 = i flag = False if pos2 != -1: if pos0 > pos2: temp = ['0']*(pos0)+['1']+s[pos0+1:] temp = list(map(int, temp)) ans = 0 for i, d in enumerate(temp): ans += d*3**i else: ans = 3**(len(s)) else: ans = n print(ans) ```
output
1
101,308
20
202,617
Provide tags and a correct Python 3 solution for this coding contest problem. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683
instruction
0
101,309
20
202,618
Tags: brute force, greedy, implementation Correct Solution: ``` import math i = 0 s = 0 v = [] while s < 1e18: s += pow(3,i) v.append(s) i += 1 q = int(input()) for i in range(q): n = int(input()) k = 0 while v[k] < n: k += 1 ans = v[k] l = k while l >= 0: if ans - pow(3,l) >= n: ans -= pow(3,l); l -= 1 print(ans) ```
output
1
101,309
20
202,619
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` for _ in range(int(input())): n = int(input()) val = 0 i = 0 while val <= n: val += 3 ** i i += 1 # print("max val: {} {}".format(val, i)) while i >= 0: if val - (3 ** i) >= n: val -= 3 ** i # print("updated val: {} {}".format(val, i)) i -= 1 print(val) ```
instruction
0
101,310
20
202,620
Yes
output
1
101,310
20
202,621
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` a = [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987, 22876792454961, 68630377364883, 205891132094649, 617673396283947, 1853020188851841, 5559060566555523, 16677181699666569, 50031545098999707, 150094635296999121, 450283905890997363, 1350851717672992089] sum3 = 2026277576509488133 for _ in range(int(input())): ans = sum3 n = int(input()) for x in a[::-1]: if ans - x >= n : ans -= x print(ans) ```
instruction
0
101,311
20
202,622
Yes
output
1
101,311
20
202,623
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` import sys input=sys.stdin.readline q=int(input()) for _ in range(q): m=int(input()) n=m sansin=[] while True: sansin.append(n%3) n//=3 if n<3: sansin.append(n) break f=1 x=0 for i in range(len(sansin)): if sansin[i]==2: f=0 x=i if f: print(m) else: sansin.insert(len(sansin),0) ans=0 ff=1 for i in range(x,len(sansin)): if sansin[i]==0 and ff: ans+=(3**i) ff=0 elif ff==0: ans+=(sansin[i]*(3**i)) print(ans) ```
instruction
0
101,312
20
202,624
Yes
output
1
101,312
20
202,625
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` def ii(): return int(input()) def si(): return input() def mi(): return map(int,input().split()) def li(): return list(mi()) import math q=ii() for i in range(q): n=ii() c=0 while pow(3,c)<n: c+=1 if pow(3,c)==n: print(pow(3,c)) continue ans=pow(3,c) b=1<<(c-1)+1 for i in range(b): d=pow(3,c-1) for j in range(c-1): if (i>>j)%2: d+=pow(3,j) if d>=n: ans=d break print(ans) ```
instruction
0
101,313
20
202,626
Yes
output
1
101,313
20
202,627
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` q = int(input()) def base_10_to_n(x, n): if (int(x/n)): return base_10_to_n(int(x/n), n)+str(x%n) return str(x%n) for _ in range(q): n = int(input()) s = list(base_10_to_n(n, 3)) s.reverse() pos0 = -1 pos2 = -1 for i in range(len(s)): if s[i] == '2': pos2 = i if s[i] == 0 and pos2 != -1: pos0 = i if pos2 != -1: if pos0 > pos2: temp = ['0']*(pos0)+['1']+s[pos0+1:] temp.reverse() temp = list(map(int, ans)) ans = 0 for i, d in enumerate(temp): ans += d**i else: ans = 3**(len(s)) else: ans = n print(ans) ```
instruction
0
101,314
20
202,628
No
output
1
101,314
20
202,629
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` #import sys # import math # import bisect # import collections # import itertools # #from sys import stdin,stdout # from math import gcd,floor,sqrt,log # from collections import defaultdict as dd, Counter as ctr # from bisect import bisect_left as bl, bisect_right as br # from itertools import permutations as pr, combinations as cb #sys.setrecursionlimit(100000000) #$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$ inp = lambda: int(input()) # strng = lambda: input().strip() # jn = lambda x,l: x.join(map(str,l)) # strl = lambda: list(input().strip()) # mul = lambda: map(int,input().strip().split()) # mulf = lambda: map(float,input().strip().split()) seq = lambda: list(map(int,input().strip().split())) #$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$ # p_inf = float('inf') # n_inf = float('-inf') #To find mex # def mex(arr): # nList = set(arr) # mex = 0 # while mex in nList: # mex += 1 # return(mex) def count(n): cnt = 0 while n > 1: n = n // 3 cnt += 1 return(cnt) def results(n): cnt = count(n) if(3 ** cnt == n): return(n) else: tmp = 3 ** cnt i = 0 while tmp <= n and i < cnt: tmp += 3 ** i i += 1 if(tmp >= n): return(tmp) else: return(3 ** (cnt + 1)) def main(): t = inp() for _ in range(t): n = inp() result = results(n) print(result) if __name__ == '__main__': main() ```
instruction
0
101,315
20
202,630
No
output
1
101,315
20
202,631
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` def add_my_bit(t, s): if t[s] == 2: t[s] = 0 if s == len(t) - 1: t.append(1) else: add_my_bit(t, s + 1) elif t[s] == 1: t[s] = 2 elif t[s] == 0: t[s] = 1 return t for _ in range(int(input())): a = int(input()) n = a t = [] k = 0 while a != 0: t.append(a% 3) a = a//3 u = len(t) if 2 in t: w = t[::-1].index(2) if len(t) - w <= len(t) - 1 and t[len(t) - w] == 0: t[len(t) - w] = 1 for rv in range(len(t[: len(t) - w])): t[rv] = 0 while 2 in t: p = t.index(2) add_my_bit(t, p) v = len(t) if u < v: k = k + 3**(v - 1) else: for j in range(len(t)): k = k + (3**j)*t[j] else: k = n print(k) ```
instruction
0
101,316
20
202,632
No
output
1
101,316
20
202,633
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The only difference between easy and hard versions is the maximum value of n. You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n. The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed). For example: * 30 is a good number: 30 = 3^3 + 3^1, * 1 is a good number: 1 = 3^0, * 12 is a good number: 12 = 3^2 + 3^1, * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3. For the given positive integer n find such smallest m (n ≀ m) that m is a good number. You have to answer q independent queries. Input The first line of the input contains one integer q (1 ≀ q ≀ 500) β€” the number of queries. Then q queries follow. The only line of the query contains one integer n (1 ≀ n ≀ 10^4). Output For each query, print such smallest integer m (where n ≀ m) that m is a good number. Example Input 7 1 2 6 13 14 3620 10000 Output 1 3 9 13 27 6561 19683 Submitted Solution: ``` queries = int(input()) for q in range(queries): n = int(input()) p = 39 dp = [0] * p r = n for i in range(p)[::-1]: dp[i], r = divmod(r, 3**i) if all(i <= 1 for i in dp): print(n) else: r = p while dp[r-1] == 0: r -= 1 print(3 ** r) ```
instruction
0
101,317
20
202,634
No
output
1
101,317
20
202,635
Provide tags and a correct Python 3 solution for this coding contest problem. One day Anna got the following task at school: to arrange several numbers in a circle so that any two neighboring numbers differs exactly by 1. Anna was given several numbers and arranged them in a circle to fulfill the task. Then she wanted to check if she had arranged the numbers correctly, but at this point her younger sister Maria came and shuffled all numbers. Anna got sick with anger but what's done is done and the results of her work had been destroyed. But please tell Anna: could she have hypothetically completed the task using all those given numbers? Input The first line contains an integer n β€” how many numbers Anna had (3 ≀ n ≀ 105). The next line contains those numbers, separated by a space. All numbers are integers and belong to the range from 1 to 109. Output Print the single line "YES" (without the quotes), if Anna could have completed the task correctly using all those numbers (using all of them is necessary). If Anna couldn't have fulfilled the task, no matter how hard she would try, print "NO" (without the quotes). Examples Input 4 1 2 3 2 Output YES Input 6 1 1 2 2 2 3 Output YES Input 6 2 4 1 1 2 2 Output NO
instruction
0
101,335
20
202,670
Tags: constructive algorithms, implementation Correct Solution: ``` n= int(input()) a = list(map(int,input().split())) a.sort() mm=a[0] b=list(map(lambda x: x-mm, a)) if b[-1]>n or n%2==1: print('NO') else: c=[0]*(b[-1]+1) for el in b: c[el]+=1 for i in range(1,len(c)): c[i] = c[i]-c[i-1] c[i-1]=0 #print(c) if i!=len(c)-1: if (c[i]==0 or c[i]<0): print('NO') break else: if c[i]!=0: print('NO') break else: print('YES') # Sat Oct 17 2020 10:31:31 GMT+0300 (Москва, стандартноС врСмя) ```
output
1
101,335
20
202,671
Provide tags and a correct Python 3 solution for this coding contest problem. One day Anna got the following task at school: to arrange several numbers in a circle so that any two neighboring numbers differs exactly by 1. Anna was given several numbers and arranged them in a circle to fulfill the task. Then she wanted to check if she had arranged the numbers correctly, but at this point her younger sister Maria came and shuffled all numbers. Anna got sick with anger but what's done is done and the results of her work had been destroyed. But please tell Anna: could she have hypothetically completed the task using all those given numbers? Input The first line contains an integer n β€” how many numbers Anna had (3 ≀ n ≀ 105). The next line contains those numbers, separated by a space. All numbers are integers and belong to the range from 1 to 109. Output Print the single line "YES" (without the quotes), if Anna could have completed the task correctly using all those numbers (using all of them is necessary). If Anna couldn't have fulfilled the task, no matter how hard she would try, print "NO" (without the quotes). Examples Input 4 1 2 3 2 Output YES Input 6 1 1 2 2 2 3 Output YES Input 6 2 4 1 1 2 2 Output NO
instruction
0
101,336
20
202,672
Tags: constructive algorithms, implementation Correct Solution: ``` import sys n = int(input()) c = {} a = [int(i) for i in input().split()] maxi = max(a) for i in a: c[i] = c.get(i, 0) + 1 l = sorted(c) t = l[:-1] for u in t: if u + 1 not in c: print("NO") sys.exit() c[u + 1] -= c[u] if 0 > c[u + 1]: print("NO") sys.exit() arr = list(c.values()) if arr.count(0) == 1 and c[maxi] == 0: print("YES") else: print("NO") # Fri Oct 16 2020 22:20:09 GMT+0300 (Москва, стандартноС врСмя) ```
output
1
101,336
20
202,673
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,497
20
202,994
Tags: dp, greedy, implementation, math Correct Solution: ``` n = int(input()) a = list(map(lambda x: int(x.split('.')[1]), input().split())) s = sum(a) - n * 1000 zero_cnt = a.count(0) min_add = max(0, zero_cnt - n) max_add = min(n, zero_cnt) answ = min(abs(s + i * 1000) for i in range(min_add, max_add + 1)) print('{:d}.{:0>3d}'.format(answ // 1000, answ % 1000)) ```
output
1
101,497
20
202,995
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,498
20
202,996
Tags: dp, greedy, implementation, math Correct Solution: ``` n, k, s = int(input()), 0, 0 for i in input().split(): j = int(i[-3: ]) if j == 0: k += 1 else: s += j c = s // 1000 + int(s % 1000 > 500) a, b = max(0, n - k), min(2 * n - k, n) if a <= c <= b: s = abs(c * 1000 - s) else: s = min(abs(a * 1000 - s), abs(b * 1000 - s)) print(str(s // 1000) + '.' + str(s % 1000).zfill(3)) ```
output
1
101,498
20
202,997
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,499
20
202,998
Tags: dp, greedy, implementation, math Correct Solution: ``` n, t = int(input()), [int(i[-3: ]) for i in input().split()] k, s = t.count(0), sum(t) c = s // 1000 + int(s % 1000 > 500) a, b = max(0, n - k), min(2 * n - k, n) if a <= c <= b: s = abs(c * 1000 - s) else: s = min(abs(a * 1000 - s), abs(b * 1000 - s)) print(str(s // 1000) + '.' + str(s % 1000).zfill(3)) ```
output
1
101,499
20
202,999
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,500
20
203,000
Tags: dp, greedy, implementation, math Correct Solution: ``` from math import ceil, floor N = int(input()) Nums = list(map(float, input().split())) Zeros = 0 Sum = 0.0 for Num in Nums: if Num - floor(Num) == 0: Zeros += 1 else: Sum += Num - floor(Num) Best = float(10 ** 9) for i in range(Zeros + 1): Best = min(Best, abs(N - Zeros + i - Sum)) Best = str(float(round(Best, 3))) Index = Best.find('.') print(Best + '0' * (4 - (len(Best) - Index))) ```
output
1
101,500
20
203,001
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,501
20
203,002
Tags: dp, greedy, implementation, math Correct Solution: ``` import math n = int(input()) nums = input().split() float_nums = [float(ele) for ele in nums] diff = 0.000 count = 0 for x in float_nums: diff += x - math.floor(x) if math.ceil(x) > math.floor(x): count += 1 min_num = max(0, count - n) max_num = min(count, n) if math.ceil(diff) - diff <= 0.5: t = math.ceil(diff) else: t = math.floor(diff) if t < min_num: print("{0:.3f}".format(min_num - diff)) elif t > max_num: print("{0:.3f}".format(diff - max_num)) else: print("{0:.3f}".format(math.fabs(diff - t))) ```
output
1
101,501
20
203,003
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,502
20
203,004
Tags: dp, greedy, implementation, math Correct Solution: ``` n = int(input()) l = list(map(float, input().split())) l = sorted([x - int(x) for x in l if x - int(x) != 0]) o = 2*n - len(l) su = sum(l) ans = 0xFFFFFFFFFFFFFFF for i in range(n + 1): if i + o >= n: ans = min(ans, abs(i-su)) print("%.3f" % ans) # Made By Mostafa_Khaled ```
output
1
101,502
20
203,005
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,503
20
203,006
Tags: dp, greedy, implementation, math Correct Solution: ``` n = int(input()) As = list(map(float, input().split())) B = list(x - int(x) for x in As if x - int(x) > 0.000) l = len(B) if l == 0: print('{:.3f}'.format(0)) exit(0) S = sum(x for x in B) ans = 1e10 for i in range(max(0, l - n), min(l,n) + 1): ans = min(ans, abs(i - S)) print('{:.3f}'.format(ans)) ```
output
1
101,503
20
203,007
Provide tags and a correct Python 3 solution for this coding contest problem. Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: * choose indexes i and j (i β‰  j) that haven't been chosen yet; * round element ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai βŒ‹); * round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj βŒ‰). Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference. Input The first line contains integer n (1 ≀ n ≀ 2000). The next line contains 2n real numbers a1, a2, ..., a2n (0 ≀ ai ≀ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces. Output In a single line print a single real number β€” the required difference with exactly three digits after the decimal point. Examples Input 3 0.000 0.500 0.750 1.000 2.000 3.000 Output 0.250 Input 3 4469.000 6526.000 4864.000 9356.383 7490.000 995.896 Output 0.279 Note In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.
instruction
0
101,504
20
203,008
Tags: dp, greedy, implementation, math Correct Solution: ``` n = int(input()) a = map(float, input().split()) s = 0 m = 0 for val in a: s += val-int(val) m += (val-int(val))>0 v = 1e9 for i in range(max(0, m-n), min(n, m)+1): v = min(v, abs(s-i)) print('%.3f'%v) ```
output
1
101,504
20
203,009
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,513
20
203,026
Tags: implementation Correct Solution: ``` x=input() def convert(x): lst=x.split(" ") return lst data=convert(x) n=int(data[0]) p=int(data[1]) k=int(data[2]) num=[] for i in range(1,n+1): num.append(i) res=[] r=[">>"] l=["<<"] for i in range(p-k,p+k+1): if i>0 and len(num)>=i: res.append(num[i-1]) if res[0]==1 and res[len(res)-1]==n: ans=listToStr = ' '.join([str(elem) for elem in res]) an=ans.replace(str(p),("("+str(p)+")"),1) elif res[0]==1: ans=listToStr = ' '.join([str(elem) for elem in (res+r)]) an=ans.replace(str(p),("("+str(p)+")"),1) elif res[len(res)-1]==n: ans=listToStr = ' '.join([str(elem) for elem in (l+res)]) an=ans.replace(str(p),("("+str(p)+")"),1) else: ans=listToStr = ' '.join([str(elem) for elem in (l+res+r)]) an=ans.replace(str(p),("("+str(p)+")"),1) print(an) ```
output
1
101,513
20
203,027
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,514
20
203,028
Tags: implementation Correct Solution: ``` n, p, k = list(map(int,input().split())) if(p - k > 1): print("<< ",end="") for i in range(k): if(p - k + i >= 1): print(p - k + i,end=" ") print("(",end="") print(p,end=") ") for i in range(k): if(p + i + 1 <= n): print(p + i + 1,end=" ") if(p + k < n): print(">>") ```
output
1
101,514
20
203,029
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,515
20
203,030
Tags: implementation Correct Solution: ``` n, p, k = list(map(int, input().split())) l = [] i = 0 e = 0 while (i <= k*2 and e<n): e = p-k+i if e > 0: l.append(e) i = i+1 if l[0] > 1: print('<<', end = ' ') for i in l: if i == p: i = '('+str(i)+')' print(i, end = ' ') if l[len(l)-1] < n: print('>>') ```
output
1
101,515
20
203,031
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,516
20
203,032
Tags: implementation Correct Solution: ``` def pages(n,p,k): output = "" if p<1 or p>n : return output if p-k > 1 : output += "<< " for i in range(p-k, p+k+1): if i <= n and i >= 1: if i == p: output += "(" + str(i) + ") " else: output += str(i) + " " if p+k < n: output += ">>" return output n,p,k = map(int, input().split()) print(pages(n, p, k)) ```
output
1
101,516
20
203,033
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,517
20
203,034
Tags: implementation Correct Solution: ``` # cook your dish here p,n,k=map(int,input().split()) u=1 d=p+1 if n-k<=1: ans=[] else: ans=["<<"] if n+k+1<=p: d=n+k+1 if n-k>0: u=n-k for i in range(u,d): if i==n: ans.append("("+str(n)+")") else: ans.append(i) if n<p-k: ans.append(">>") ans=str(ans).replace(","," ") ans=ans.replace("'","") print(ans[1:-1]) ```
output
1
101,517
20
203,035
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,518
20
203,036
Tags: implementation Correct Solution: ``` entrada=input() numbers=entrada.split() n=int(numbers[0]) p=int(numbers[1]) k=int(numbers[2]) if p+k>=n and p-k<=1: for i in range (1, n+1): if i==p: print("(" + str(p) + ")", end=" ") else: print(i,end=" ") if p+k>=n and p-k>1: print("<<", end=" ") for i in range (p-k, n+1): if i==p: print("(" + str(p) + ")", end=" ") else: print(i,end=" ") if p+k<n and p-k<=1: for i in range (1, p+k+1): if i==p: print("(" + str(p) + ")", end=" ") else: print(i,end=" ") print(">>") if p+k<n and p-k>1: print("<<", end=" ") for i in range (p-k, p+k+1): if i==p: print("(" + str(p) + ")", end=" ") else: print(i,end=" ") print(">>") ```
output
1
101,518
20
203,037
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,519
20
203,038
Tags: implementation Correct Solution: ``` class Solution: def navigation(self, n, p, k): start = max(1, p-k) end = min(p+k, n) pagination = "" if start > 1: pagination += "<< " while start <= end: if start == p: pagination += "(" + str(start) + ") " else: pagination += str(start) + " " start += 1 if end < n: pagination += ">>" return pagination sol = Solution() # array = [11, 12, 1, 2, 13, 14, 3, 4] # print(sol.thanosSort(array)) [n, p, k] = list(map(int, input().strip().split())) print(sol.navigation(n, p, k)) ```
output
1
101,519
20
203,039
Provide tags and a correct Python 3 solution for this coding contest problem. User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n. Assume that somebody is on the p-th page now. The navigation will look like this: << p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >> When someone clicks the button "<<" he is redirected to page 1, and when someone clicks the button ">>" he is redirected to page n. Of course if someone clicks on a number, he is redirected to the corresponding page. There are some conditions in the navigation: * If page 1 is in the navigation, the button "<<" must not be printed. * If page n is in the navigation, the button ">>" must not be printed. * If the page number is smaller than 1 or greater than n, it must not be printed. You can see some examples of the navigations. Make a program that prints the navigation. Input The first and the only line contains three integers n, p, k (3 ≀ n ≀ 100; 1 ≀ p ≀ n; 1 ≀ k ≀ n) Output Print the proper navigation. Follow the format of the output from the test samples. Examples Input 17 5 2 Output &lt;&lt; 3 4 (5) 6 7 &gt;&gt; Input 6 5 2 Output &lt;&lt; 3 4 (5) 6 Input 6 1 2 Output (1) 2 3 &gt;&gt; Input 6 2 2 Output 1 (2) 3 4 &gt;&gt; Input 9 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 Input 10 6 3 Output &lt;&lt; 3 4 5 (6) 7 8 9 &gt;&gt; Input 8 5 4 Output 1 2 3 4 (5) 6 7 8
instruction
0
101,520
20
203,040
Tags: implementation Correct Solution: ``` a, b, c = map(int, input().split(" ")) # a → limit b→now c → range ll = max(1, b-c) rl = min(a, b + c ) left = [i for i in range(b - 1, ll-1, -1)] right = [i for i in range(b + 1, rl + 1)] l_bar = "<< " if ll == 1: l_bar = "" r_bar = " >>" if rl == a: r_bar = "" left.reverse() print(l_bar, *(left), "({})".format(b), *(right), r_bar) ```
output
1
101,520
20
203,041
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,577
20
203,154
Tags: greedy, math, sortings Correct Solution: ``` x = int(input()) y = input() ct = {0:[0, 0, 0, 0], 1:[0, 0, 0, 0], 2:[1, 0, 0, 0], 3:[1, 1, 0, 0], 4:[3, 1, 0, 0], 5:[3, 1, 1, 0], 6:[4, 2, 1, 0], 7:[4, 2, 1, 1], 8:[7, 2, 1, 1], 9:[7, 4, 1, 1]} real = {2:[1, 0, 0, 0], 3:[1, 1, 0, 0], 4:[3, 1, 0, 0], 5:[3, 1, 1, 0], 6:[4, 2, 1, 0], 7:[4, 2, 1, 1], 8:[7, 2, 1, 1], 9:[7, 4, 1, 1]} ut = {2:[1, 0, 0, 0], 3:[1, 1, 0, 0], 5:[3, 1, 1, 0], 7:[4, 2, 1, 1]} lx = [0, 0, 0, 0] for i in y: i = int(i) bad = ct[i] lx[0] += bad[0] lx[1] += bad[1] lx[2] += bad[2] lx[3] += bad[3] s = '' if lx[-1] >= 1: copies = lx[-1] lx[0] -= 4*copies lx[1] -= 2*copies lx[2] -= copies lx[3] -= copies s += '7' * copies if lx[-2] >= 1: copies = lx[-2] lx[0] -= 3*copies lx[1] -= copies lx[2] -= copies s += '5' * copies if lx[-3] >= 1: copies = lx[-3] lx[0] -= copies lx[1] -= copies s += '3' * copies if lx[-4] >= 1: copies = lx[-4] lx[0] -= copies s += '2' * copies print(s) ```
output
1
101,577
20
203,155
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,578
20
203,156
Tags: greedy, math, sortings Correct Solution: ``` n = input() s = input() s = s.replace('9','7332') s = s.replace('8','7222') s = s.replace('6','53') s = s.replace('4','322') s = s.replace('1','') s = s.replace('0','') s = ''.join(sorted(s,reverse=True)) print(s) ```
output
1
101,578
20
203,157
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,579
20
203,158
Tags: greedy, math, sortings Correct Solution: ``` I=lambda:list(map(int,input().split())) n=I() f=['','','2','3','223','5','53','7','7222','7332'] v='' for i in input(): v+=f[int(i)] v=list(v) v.sort(reverse=True) print(''.join(v)) ```
output
1
101,579
20
203,159
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,580
20
203,160
Tags: greedy, math, sortings Correct Solution: ``` n=int(input()) a=[int(x) for x in list(input())] fac={2:0,3:0,5:0,7:0} ans=[] for i in range(2,10): t=a.count(i) if i==9: fac[2]+=7*t fac[3]+=4*t fac[5]+=t fac[7]+=t if i==8: fac[2]+=7*t fac[3]+=2*t fac[5]+=t fac[7]+=t if i==7: fac[2]+=4*t fac[3]+=2*t fac[5]+=t fac[7]+=t if i==6: fac[2]+=4*t fac[3]+=2*t fac[5]+=t if i==5: fac[2]+=3*t fac[3]+=t fac[5]+=t if i==4: fac[2]+=3*t fac[3]+=t if i==3: fac[2]+=t fac[3]+=t if i==2: fac[2]+=t for i in [7,5,3,2]: t=fac[i] ans=ans+[i]*t if i==7: fac[2]-=4*t fac[3]-=2*t fac[5]-=t fac[7]-=t if i==5: fac[2]-=3*t fac[3]-=t fac[5]-=t if i==3: fac[2]-=t fac[3]-=t ans.sort(reverse=True) print(''.join([str(x) for x in ans])) ```
output
1
101,580
20
203,161
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,581
20
203,162
Tags: greedy, math, sortings Correct Solution: ``` n = int(input()) s = input() ans = "" for i in s: if(i=="1" or i=="0"): continue elif(i=="4"): ans+="322" elif(i=="6"): ans+="53" elif(i=="8"): ans+="7222" elif(i=="9"): ans+="7332" else: ans+=i ans = list(ans) ans.sort(reverse=True) for i in ans: print(i,end="") ```
output
1
101,581
20
203,163
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,582
20
203,164
Tags: greedy, math, sortings Correct Solution: ``` a = int(input()) b = str(input()) answer = list() for x in b: if x != "0" and x != "1": if x == "4": answer += ["3", "2", "2"] elif x == "6": answer += ["5", "3"] elif x == "8": answer += ["7", "2", "2", "2"] elif x == "9": answer += ["7", "3", "3", "2"] else: answer += x answer.sort() answer.reverse() final = "" for x in answer: final += x print(final) ```
output
1
101,582
20
203,165
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,583
20
203,166
Tags: greedy, math, sortings Correct Solution: ``` a = input("") b = input("") outstring = "" for value in b: if value == "1": continue elif value == "2": outstring = outstring + "2" elif value == "3": outstring = outstring + "3" elif value == "4": outstring = outstring + "3" + "2" + "2" elif value == "5": outstring = outstring + "5" elif value == "6": outstring = outstring + "5" + "3" elif value == "7": outstring = outstring + "7" elif value == "8": outstring = outstring + "7" + "2" + "2" + "2" elif value == "9": outstring = outstring + "7" + "3" + "3" + "2" t = [i for i in outstring] t.sort() t.reverse() print(''.join(t)) ```
output
1
101,583
20
203,167
Provide tags and a correct Python 3 solution for this coding contest problem. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image>
instruction
0
101,584
20
203,168
Tags: greedy, math, sortings Correct Solution: ``` def aus(argument): switcher = { 0:[], 1:[], 2:[2], 3:[3], 4:[3,2,2], 5:[5], 6:[3,5], 7:[7], 8:[2,2,2,7], 9:[2,3,3,7], } return switcher.get(argument) summ=[] a= int(input()) b= input() for i in range(a): summ += aus(int(b[i])) summ.sort(reverse=True) print(int(''.join(map(str, summ)))) ```
output
1
101,584
20
203,169
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image> Submitted Solution: ``` import string NUMBERS=[] def factor(x): if x == 2 or x==5 or x== 7: NUMBERS.append(x) return else: xx = x while x!= 1: for i in range(2,4): if x%i==0: x = x/i NUMBERS.append(i) factor(xx-1) return n = int(input()) input_value = input() num = [] j = 0 for i in range(n): if input_value[i] != '1' and input_value[i] != '0': num.append(int(input_value[i])) j = j+1 for x in num: factor (x) kkk=NUMBERS kkk.sort(reverse=True ) for x in NUMBERS: if x==3: kkk.pop() print("".join( map(str, kkk))) ```
instruction
0
101,585
20
203,170
Yes
output
1
101,585
20
203,171
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image> Submitted Solution: ``` n = int(input()) num = str(input()) primes = {2: 0, 3: 0, 5: 0, 7: 0} for i in range(n): k = int(num[i]) if (k <= 1): continue primes[2] += 1 if (k >= 3): primes[3] += 1 if (k >= 4): primes[2] += 2 if (k >= 5): primes[5] += 1 if (k >= 6): primes[2] += 1 primes[3] += 1 if (k >= 7): primes[7] += 1 if (k >= 8): primes[2] += 3 if (k == 9): primes[3] += 2 for i in range(primes[7]): print(7, end='') primes[2] -= 4 primes[3] -= 2 primes[5] -= 1 for i in range(primes[5]): print(5, end='') primes[2] -= 3 primes[3] -= 1 for i in range(primes[3]): print(3, end='') primes[2] -= 1 for i in range(primes[2]): print(2, end='') print("\n") ```
instruction
0
101,586
20
203,172
Yes
output
1
101,586
20
203,173
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image> Submitted Solution: ``` import math import sys n=int(input()) s=input() v=[] for i in range(0,n): if s[i] == '2': v.append(2) if s[i] == '3': v.append(3) if s[i] == '4': v.append(2) v.append(2) v.append(3) if s[i] == '5': v.append(5) if s[i] == '6': v.append(5) v.append(3) if s[i] == '7': v.append(7) if s[i] == '8': v.append(7) v.append(2) v.append(2) v.append(2) if s[i] == '9': v.append(7) v.append(2) v.append(3) v.append(3) v.sort() m=0 t=1 for i in range(0,len(v)): m+=v[i]*t t*=10 print(m) ```
instruction
0
101,587
20
203,174
Yes
output
1
101,587
20
203,175
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image> Submitted Solution: ``` list1=["","","2","3","322","5","53","7","7222","7332"] n=int(input()) p=input() k=[] for i in p: k.append(list1[int(i)]) print("".join(sorted("".join(k),reverse=True))) ```
instruction
0
101,588
20
203,176
Yes
output
1
101,588
20
203,177
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image> Submitted Solution: ``` import math import sys n=int(input()) s=input() v=[] for i in range(0,n): if s[i] == '2': v.append(2) if s[i] == '3': v.append(3) if s[i] == '4': v.append(2) v.append(2) v.append(3) if s[i] == '5': v.append(5) if s[i] == '6': v.append(5) v.append(2) v.append(3) if s[i] == '7': v.append(7) if s[i] == '8': v.append(7) v.append(2) v.append(2) v.append(2) if s[i] == '9': v.append(7) v.append(2) v.append(2) v.append(2) v.append(3) v.append(3) v.sort() m=0 t=1 for i in range(0,len(v)): m+=v[i]*t t*=10 print(m) ```
instruction
0
101,589
20
203,178
No
output
1
101,589
20
203,179
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Drazil is playing a math game with Varda. Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>. First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions: 1. x doesn't contain neither digit 0 nor digit 1. 2. <image> = <image>. Help friends find such number. Input The first line contains an integer n (1 ≀ n ≀ 15) β€” the number of digits in a. The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Examples Input 4 1234 Output 33222 Input 3 555 Output 555 Note In the first case, <image> Submitted Solution: ``` input() s=input() a="" t=[a,a,"2","3","223","5","53","7","7222","7333"] for c in s: a+=t[int(c)] print(''.join(sorted(a)[::-1])) ```
instruction
0
101,590
20
203,180
No
output
1
101,590
20
203,181