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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well. In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned. Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023. Input The first line contains an integer n (1 ≤ n ≤ 5·105) — the number of lines. Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023. Output Output an integer k (0 ≤ k ≤ 5) — the length of your program. Next k lines must contain commands in the same format as in the input. Examples Input 3 | 3 ^ 2 | 1 Output 2 | 3 ^ 2 Input 3 &amp; 1 &amp; 3 &amp; 5 Output 1 &amp; 1 Input 3 ^ 1 ^ 2 ^ 3 Output 0 Note You can read about bitwise operations in <https://en.wikipedia.org/wiki/Bitwise_operation>. Second sample: Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs. Submitted Solution: ``` '''input 3 & 1 & 3 & 5 ''' ''' Author : dhanyaabhirami Great things never come from comfort zones ''' from collections import defaultdict as dd from collections import Counter as ccd from itertools import permutations as pp from itertools import combinations as cc from random import randint as rd from bisect import bisect_left as bl from bisect import bisect_right as br import heapq as hq import math mod=pow(10,9) +7 def inp(flag=0): if flag==0: return list(map(int,input().strip().split(' '))) else: return int(input()) # Code credits # assert(debug()==true) # for _ in range(int(input())): # n=inp(1) n=int(input()) a=0 b=1023 for i in range(n): c,k= input().strip().split(' ') k=int(k) if (c=='|'): a|=k b|=k elif (c=='^'): a^=k b^=k elif (c=='&'): a&=k b&=k OR = 0 XOR = 0 for i in range(10): x=1<<i if a&x and b&x: XOR|=x OR|=x if ~a&x and b&x: XOR|=x if ~a&x and ~b&x: OR|=x print('2') print('|',OR) print('^',XOR) ```
instruction
0
99,190
20
198,380
No
output
1
99,190
20
198,381
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well. In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned. Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023. Input The first line contains an integer n (1 ≤ n ≤ 5·105) — the number of lines. Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023. Output Output an integer k (0 ≤ k ≤ 5) — the length of your program. Next k lines must contain commands in the same format as in the input. Examples Input 3 | 3 ^ 2 | 1 Output 2 | 3 ^ 2 Input 3 &amp; 1 &amp; 3 &amp; 5 Output 1 &amp; 1 Input 3 ^ 1 ^ 2 ^ 3 Output 0 Note You can read about bitwise operations in <https://en.wikipedia.org/wiki/Bitwise_operation>. Second sample: Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs. Submitted Solution: ``` ## necessary imports import sys input = sys.stdin.readline from bisect import bisect_left; from bisect import bisect_right; from math import ceil, factorial; def ceil(x): if x != int(x): x = int(x) + 1; return x; # swap_array function def swaparr(arr, a,b): temp = arr[a]; arr[a] = arr[b]; arr[b] = temp; ## gcd function def gcd(a,b): if b == 0: return a; return gcd(b, a % b); ## nCr function efficient using Binomial Cofficient def nCr(n, k): if(k > n - k): k = n - k; res = 1; for i in range(k): res = res * (n - i); res = res / (i + 1); return int(res); ## prime factorization def primefs(n): ## if n == 1 ## calculating primes primes = {} while(n%2 == 0 and n > 0): primes[2] = primes.get(2, 0) + 1 n = n//2 for i in range(3, int(n**0.5)+2, 2): while(n%i == 0 and n > 0): primes[i] = primes.get(i, 0) + 1 n = n//i if n > 2: primes[n] = primes.get(n, 0) + 1 ## prime factoriazation of n is stored in dictionary ## primes and can be accesed. O(sqrt n) return primes ## MODULAR EXPONENTIATION FUNCTION def power(x, y, p): res = 1 x = x % p if (x == 0) : return 0 while (y > 0) : if ((y & 1) == 1) : res = (res * x) % p y = y >> 1 x = (x * x) % p return res ## DISJOINT SET UNINON FUNCTIONS def swap(a,b): temp = a a = b b = temp return a,b; # find function with path compression included (recursive) # def find(x, link): # if link[x] == x: # return x # link[x] = find(link[x], link); # return link[x]; # find function with path compression (ITERATIVE) def find(x, link): p = x; while( p != link[p]): p = link[p]; while( x != p): nex = link[x]; link[x] = p; x = nex; return p; # the union function which makes union(x,y) # of two nodes x and y def union(x, y, link, size): x = find(x, link) y = find(y, link) if size[x] < size[y]: x,y = swap(x,y) if x != y: size[x] += size[y] link[y] = x ## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES def sieve(n): prime = [True for i in range(n+1)] p = 2 while (p * p <= n): if (prime[p] == True): for i in range(p * p, n+1, p): prime[i] = False p += 1 return prime #### PRIME FACTORIZATION IN O(log n) using Sieve #### MAXN = int(1e5 + 5) def spf_sieve(): spf[1] = 1; for i in range(2, MAXN): spf[i] = i; for i in range(4, MAXN, 2): spf[i] = 2; for i in range(3, ceil(MAXN ** 0.5), 2): if spf[i] == i: for j in range(i*i, MAXN, i): if spf[j] == j: spf[j] = i; ## function for storing smallest prime factors (spf) in the array ################## un-comment below 2 lines when using factorization ################# # spf = [0 for i in range(MAXN)] # spf_sieve(); def factoriazation(x): ret = {}; while x != 1: ret[spf[x]] = ret.get(spf[x], 0) + 1; x = x//spf[x] return ret; ## this function is useful for multiple queries only, o/w use ## primefs function above. complexity O(log n) ## taking integer array input def int_array(): return list(map(int, input().strip().split())); def float_array(): return list(map(float, input().strip().split())); ## taking string array input def str_array(): return input().strip().split(); #defining a couple constants MOD = int(1e9)+7; CMOD = 998244353; INF = float('inf'); NINF = -float('inf'); ################### ---------------- TEMPLATE ENDS HERE ---------------- ################### n = int(input()); ops = []; for _ in range(n): s = input().strip().split(); ops.append((s[0], int(s[1]))); if n <= 5: print(n); for i in range(n): print(ops[i][0] + ' ' + str(ops[i][1])); exit(); x = ops[1][1]; for i in range(2, n): c = ops[i][0]; if c == '|': x = x | ops[i][1]; elif c == '&': x = x & ops[i][1]; else: x = x ^ ops[i][1]; print(2); print(ops[0][0] + ' ' + str(ops[0][1])); print(ops[1][0] + ' ' + str(x)); ```
instruction
0
99,191
20
198,382
No
output
1
99,191
20
198,383
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,361
20
198,722
"Correct Solution: ``` #!/usr/bin/env python3 seven_seg = { 0: '0111111', 1: '0000110', 2: '1011011', 3: '1001111', 4: '1100110', 5: '1101101', 6: '1111101', 7: '0100111', 8: '1111111', 9: '1101111' } while True: seven_seg_state = ['0'] * 7 n = int(input()) if n == -1: break for _ in range(n): in_data = int(input()) for i in range(7): if seven_seg[in_data][i] == seven_seg_state[i]: print("0", end="") else: print("1", end="") seven_seg_state[i] = seven_seg[in_data][i] print() ```
output
1
99,361
20
198,723
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,362
20
198,724
"Correct Solution: ``` NUM = ( 0b0111111, 0b0000110, 0b1011011, 0b1001111, 0b1100110, 0b1101101, 0b1111101, 0b0100111, 0b1111111, 0b1101111, ) while 1: n = int(input()) if n == -1:break current = 0 for i in range(n): num = NUM[int(input())] print(format(current ^ num,'b').zfill(7)) current = num ```
output
1
99,362
20
198,725
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,363
20
198,726
"Correct Solution: ``` dic = {0:"0111111", 1:"0000110", 2:"1011011", 3:"1001111", 4:"1100110", 5:"1101101", 6:"1111101", 7:"0100111", 8:"1111111", 9:"1101111"} while True: n = int(input()) if n == -1: break dig = "0" * 7 for _ in range(n): d = dic[int(input())] put = "" for i in range(7): if dig[i] != d[i]: put += "1" else: put += "0" print(put) dig = d ```
output
1
99,363
20
198,727
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,364
20
198,728
"Correct Solution: ``` Num_seg = [[0,1,1,1,1,1,1],[0,0,0,0,1,1,0],[1,0,1,1,0,1,1],[1,0,0,1,1,1,1],[1,1,0,0,1,1,0],[1,1,0,1,1,0,1],[1,1,1,1,1,0,1],[0,1,0,0,1,1,1],[1,1,1,1,1,1,1],[1,1,0,1,1,1,1]] while True: n = int(input()) if n == -1: break num_b = int(input()) seg = "" for h in range(7): seg = seg + str(Num_seg[num_b][h]) print(seg) for i in range(n - 1): num_a = int(input()) seg = "" for j in range(7): if Num_seg[num_b][j] == Num_seg[num_a][j]: seg = seg + "0" else: seg = seg + "1" num_b = num_a print(seg) ```
output
1
99,364
20
198,729
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,365
20
198,730
"Correct Solution: ``` s=[0b0111111,0b0000110,0b1011011,0b1001111,0b1100110,0b1101101,0b1111101,0b0100111,0b1111111,0b1101111] while 1: n=int(input()) if n<0:break a=0 for _ in range(n): b=int(input()) print(bin(a^s[b])[2:].zfill(7)) a=s[b] ```
output
1
99,365
20
198,731
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,366
20
198,732
"Correct Solution: ``` # -*- coding: utf-8 -*- """ http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0228 """ import sys from sys import stdin input = stdin.readline LED_pattern = [ ##-gfedcba 0b0111111, # 0 0b0000110, # 1 0b1011011, # 2 0b1001111, # 3 0b1100110, # 4 0b1101101, # 5 0b1111101, # 6 0b0100111, # 7 0b1111111, # 8 0b1101111, # 9 0b0000000 # all OFF ] def solve(cp, np): a = LED_pattern[cp] b = LED_pattern[np] return bin(a ^ b)[2:].zfill(7) def main(args): while True: n = int(input()) if n == -1: break cp = 10 # all OFF for _ in range(n): d = int(input()) ans = solve(cp, d) cp = d print(ans) if __name__ == '__main__': main(sys.argv[1:]) ```
output
1
99,366
20
198,733
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,367
20
198,734
"Correct Solution: ``` nums = ["0111111", "0000110", "1011011", "1001111", "1100110", "1101101", "1111101", "0100111", "1111111", "1101111"] while 1: n = int(input()) if n == -1: break sig = "0000000" for _ in range(n): d = int(input()) num = nums[d] out = "" for s, n in zip(sig, num): if s == n: out += "0" else: out += "1" print(out) sig = num ```
output
1
99,367
20
198,735
Provide a correct Python 3 solution for this coding contest problem. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111
instruction
0
99,368
20
198,736
"Correct Solution: ``` import sys f = sys.stdin segment =[0b0111111, 0b0000110, 0b1011011, 0b1001111, 0b1100110, 0b1101101, 0b1111101, 0b0100111, 0b1111111, 0b1101111] while True: n = int(f.readline()) if n == -1: break display = 0 for i in [int(f.readline()) for _ in range(n)]: print(bin(display ^ segment[i])[2:].zfill(7)) display = segment[i] ```
output
1
99,368
20
198,737
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111 Submitted Solution: ``` seg = [[0, 1, 1, 1, 1, 1, 1], [0, 0, 0, 0, 1, 1, 0], [1, 0, 1, 1, 0, 1, 1], [1, 0, 0, 1, 1, 1, 1], [1, 1, 0, 0, 1, 1, 0], [1, 1, 0, 1, 1, 0, 1], [1, 1, 1, 1, 1, 0, 1], [0, 1, 0, 0, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1], [1, 1, 0, 1, 1, 1, 1]] while True : n = int(input()) if n == -1 : break pre = [0]*7 for i in range(n) : num = int(input()) post = seg[num] for j in range(7) : if pre[j] == post[j] : print(0, end='') else : print(1, end='') print() pre = post ```
instruction
0
99,369
20
198,738
Yes
output
1
99,369
20
198,739
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111 Submitted Solution: ``` # AOJ 0228: Seven Segments # Python3 2018.6.25 bal4u p = [[1,1,1,1,1,1,0], [0,1,1,0,0,0,0], [1,1,0,1,1,0,1], [1,1,1,1,0,0,1], [0,1,1,0,0,1,1], \ [1,0,1,1,0,1,1], [1,0,1,1,1,1,1], [1,1,1,0,0,1,0], [1,1,1,1,1,1,1], [1,1,1,1,0,1,1]] while True: n = int(input()) if n < 0: break a = [0]*7 for i in range(n): d = int(input()) ans = '' for j in range(6,-1,-1): k = (a[j] ^ p[d][j]) & 1 ans += '1' if k else '0' if k: a[j] = 1-a[j] print(ans) ```
instruction
0
99,370
20
198,740
Yes
output
1
99,370
20
198,741
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111 Submitted Solution: ``` # Aizu Problem 0228: Seven Segments import sys, math, os, bisect # read input: PYDEV = os.environ.get('PYDEV') if PYDEV=="True": sys.stdin = open("sample-input.txt", "rt") digits = [63, 6, 91, 79, 102, 109, 125, 39, 127, 111] while True: n = int(input()) if n == -1: break e = 0 for i in range(n): a = digits[int(input())] print("{:07b}".format(a^e)) e = a ```
instruction
0
99,371
20
198,742
Yes
output
1
99,371
20
198,743
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The screen that displays digital numbers that you often see on calculators is called a "7-segment display" because the digital numbers consist of seven parts (segments). The new product to be launched by Wakamatsu will incorporate a 7-segment display into the product, and as an employee, you will create a program to display the given number on the 7-segment display. This 7-segment display will not change until the next switch instruction is sent. By sending a signal consisting of 7 bits, the display information of each corresponding segment can be switched. Bits have a value of 1 or 0, where 1 stands for "switch" and 0 stands for "as is". The correspondence between each bit and segment is shown in the figure below. The signal sends 7 bits in the order of "gfedcba". For example, in order to display "0" from the hidden state, "0111111" must be sent to the display as a signal. To change from "0" to "5", send "1010010". If you want to change "5" to "1" in succession, send "1101011". <image> Create a program that takes n (1 ≤ n ≤ 100) numbers that you want to display and outputs the signal sequence required to correctly display those numbers di (0 ≤ di ≤ 9) on the 7-segment display. please. It is assumed that the initial state of the 7-segment display is all hidden. Input A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of -1. Each dataset is given in the following format: n d1 d2 :: dn The number of datasets does not exceed 120. Output For each input dataset, output the sequence of signals needed to properly output the numbers to the display. Example Input 3 0 5 1 1 0 -1 Output 0111111 1010010 1101011 0111111 Submitted Solution: ``` #!/usr/bin/env python3 seven_seg = { 0: '0111111', 1: '0000110', 2: '1011011', 3: '1001111', 4: '1100110', 5: '1101101', 6: '1111101', 7: '0100111', 8: '1111111', 9: '1101111' } while True: seven_seg_state = ['0'] * 7 n = int(input()) if n == -1: break for _ in range(n): in_data = int(input()) for i, s in enumerate(seven_seg_state): if seven_seg[in_data][i] == s: print("0", end="") else: print("1", end="") s = seven_seg[in_data][i] print() ```
instruction
0
99,372
20
198,744
No
output
1
99,372
20
198,745
Provide tags and a correct Python 3 solution for this coding contest problem. n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times. * Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers. Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times. Input The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}). The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}). It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. Output For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO". You can print each letter in any case (upper or lower). Example Input 6 2 1 1 2 3 0 2 3 7 2 -1 31415926 27182818 2 1000000000000000000 1 1000000000000000000 2 -1000000000000000000 -1000000000000000000 123 6 80 -5 -20 13 -14 -2 -11 Output YES YES NO YES YES NO Note In the first test case, the number 1 is already on the board. In the second test case, Nezzar could perform the following operations to write down k=0 on the board: * Select x=3 and y=2 and write down 4 on the board. * Select x=4 and y=7 and write down 1 on the board. * Select x=1 and y=2 and write down 0 on the board. In the third test case, it is impossible to have the number k = -1 on the board.
instruction
0
99,663
20
199,326
Tags: constructive algorithms, math, number theory Correct Solution: ``` import sys,math r=sys.stdin.readline for _ in range(int(r())): n,k=map(int,r().split()) lst=list(map(int,r().split())) st=set(lst) if k in st: print("YES") continue lst.sort() g=0 for i in range(len(lst)): g=math.gcd(g,lst[i]-lst[0]) if (k-lst[0])%g==0: print("YES") else: print("NO") ```
output
1
99,663
20
199,327
Provide tags and a correct Python 3 solution for this coding contest problem. n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times. * Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers. Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times. Input The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}). The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}). It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. Output For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO". You can print each letter in any case (upper or lower). Example Input 6 2 1 1 2 3 0 2 3 7 2 -1 31415926 27182818 2 1000000000000000000 1 1000000000000000000 2 -1000000000000000000 -1000000000000000000 123 6 80 -5 -20 13 -14 -2 -11 Output YES YES NO YES YES NO Note In the first test case, the number 1 is already on the board. In the second test case, Nezzar could perform the following operations to write down k=0 on the board: * Select x=3 and y=2 and write down 4 on the board. * Select x=4 and y=7 and write down 1 on the board. * Select x=1 and y=2 and write down 0 on the board. In the third test case, it is impossible to have the number k = -1 on the board.
instruction
0
99,665
20
199,330
Tags: constructive algorithms, math, number theory Correct Solution: ``` # by the authority of GOD author: manhar singh sachdev # import os,sys from io import BytesIO,IOBase from math import gcd def solve(n,k,x): if k in x: return 'YES' y = x[1]-x[0] for i in range(1,n): y = gcd(y,x[i]-x[0]) if (k-x[0])%y: return 'NO' else: return 'YES' def main(): for _ in range(int(input())): n,k = map(int,input().split()) x = list(map(int,input().split())) print(solve(n,k,x)) #Fast IO Region BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") if __name__ == '__main__': main() ```
output
1
99,665
20
199,331
Provide tags and a correct Python 3 solution for this coding contest problem. n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times. * Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers. Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times. Input The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}). The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}). It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. Output For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO". You can print each letter in any case (upper or lower). Example Input 6 2 1 1 2 3 0 2 3 7 2 -1 31415926 27182818 2 1000000000000000000 1 1000000000000000000 2 -1000000000000000000 -1000000000000000000 123 6 80 -5 -20 13 -14 -2 -11 Output YES YES NO YES YES NO Note In the first test case, the number 1 is already on the board. In the second test case, Nezzar could perform the following operations to write down k=0 on the board: * Select x=3 and y=2 and write down 4 on the board. * Select x=4 and y=7 and write down 1 on the board. * Select x=1 and y=2 and write down 0 on the board. In the third test case, it is impossible to have the number k = -1 on the board.
instruction
0
99,666
20
199,332
Tags: constructive algorithms, math, number theory Correct Solution: ``` from math import gcd import sys input = sys.stdin.buffer.readline def prog(): for _ in range(int(input())): n,k = map(int,input().split()) a = list(map(int,input().split())) a.sort() d = 0 for i in range(n-1): d = gcd(d,a[i+1]-a[i]) print("YES" if (k - a[0])%d == 0 else "NO") prog() ```
output
1
99,666
20
199,333
Provide tags and a correct Python 3 solution for this coding contest problem. n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times. * Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers. Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times. Input The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}). The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}). It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. Output For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO". You can print each letter in any case (upper or lower). Example Input 6 2 1 1 2 3 0 2 3 7 2 -1 31415926 27182818 2 1000000000000000000 1 1000000000000000000 2 -1000000000000000000 -1000000000000000000 123 6 80 -5 -20 13 -14 -2 -11 Output YES YES NO YES YES NO Note In the first test case, the number 1 is already on the board. In the second test case, Nezzar could perform the following operations to write down k=0 on the board: * Select x=3 and y=2 and write down 4 on the board. * Select x=4 and y=7 and write down 1 on the board. * Select x=1 and y=2 and write down 0 on the board. In the third test case, it is impossible to have the number k = -1 on the board.
instruction
0
99,667
20
199,334
Tags: constructive algorithms, math, number theory Correct Solution: ``` import sys input = sys.stdin.readline from math import gcd t=int(input()) for tests in range(t): n,k=map(int,input().split()) X=list(map(int,input().split())) G=0 for i in range(1,n): G=gcd(G,X[i]-X[i-1]) #print(G) if k%G==X[0]%G: print("YES") else: print("NO") ```
output
1
99,667
20
199,335
Provide tags and a correct Python 3 solution for this coding contest problem. n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times. * Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers. Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times. Input The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}). The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}). It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. Output For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO". You can print each letter in any case (upper or lower). Example Input 6 2 1 1 2 3 0 2 3 7 2 -1 31415926 27182818 2 1000000000000000000 1 1000000000000000000 2 -1000000000000000000 -1000000000000000000 123 6 80 -5 -20 13 -14 -2 -11 Output YES YES NO YES YES NO Note In the first test case, the number 1 is already on the board. In the second test case, Nezzar could perform the following operations to write down k=0 on the board: * Select x=3 and y=2 and write down 4 on the board. * Select x=4 and y=7 and write down 1 on the board. * Select x=1 and y=2 and write down 0 on the board. In the third test case, it is impossible to have the number k = -1 on the board.
instruction
0
99,668
20
199,336
Tags: constructive algorithms, math, number theory Correct Solution: ``` """ #If FastIO not needed, use this and don't forget to strip #import sys, math #input = sys.stdin.readline """ import os import sys from io import BytesIO, IOBase import heapq as h from bisect import bisect_left, bisect_right import time from types import GeneratorType BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): import os self.os = os self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: self.os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") from collections import defaultdict as dd, deque as dq, Counter as dc import math, string #start_time = time.time() def getInts(): return [int(s) for s in input().split()] def getInt(): return int(input()) def getStrs(): return [s for s in input().split()] def getStr(): return input() def listStr(): return list(input()) def getMat(n): return [getInts() for _ in range(n)] def isInt(s): return '0' <= s[0] <= '9' MOD = 10**9 + 7 """ If the diffs have GCD 1, we definitely can reach """ def solve(): N, K = getInts() A = getInts() A.sort() G = A[1]-A[0] for i in range(1,N-1): G = math.gcd(G,A[i+1]-A[i]) if G == 1: return "YES" for i in range(N): if (A[i] - K) % G == 0: return "YES" return "NO" for _ in range(getInt()): print(solve()) #solve() #print(time.time()-start_time) ```
output
1
99,668
20
199,337
Provide tags and a correct Python 3 solution for this coding contest problem. n distinct integers x_1,x_2,…,x_n are written on the board. Nezzar can perform the following operation multiple times. * Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers. Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times. Input The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The first line of each test case contains two integers n,k (2 ≤ n ≤ 2 ⋅ 10^5, -10^{18} ≤ k ≤ 10^{18}). The second line of each test case contains n distinct integers x_1,x_2,…,x_n (-10^{18} ≤ x_i ≤ 10^{18}). It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5. Output For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO". You can print each letter in any case (upper or lower). Example Input 6 2 1 1 2 3 0 2 3 7 2 -1 31415926 27182818 2 1000000000000000000 1 1000000000000000000 2 -1000000000000000000 -1000000000000000000 123 6 80 -5 -20 13 -14 -2 -11 Output YES YES NO YES YES NO Note In the first test case, the number 1 is already on the board. In the second test case, Nezzar could perform the following operations to write down k=0 on the board: * Select x=3 and y=2 and write down 4 on the board. * Select x=4 and y=7 and write down 1 on the board. * Select x=1 and y=2 and write down 0 on the board. In the third test case, it is impossible to have the number k = -1 on the board.
instruction
0
99,669
20
199,338
Tags: constructive algorithms, math, number theory Correct Solution: ``` import os import sys from io import BytesIO, IOBase # region fastio BUFSIZE = 8192 class FastIO(IOBase): def __init__(self, file): self.newlines = 0 self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline() # -------------------------------------------------------------------- def RL(): return map(int, sys.stdin.readline().split()) def RLL(): return list(map(int, sys.stdin.readline().split())) def N(): return int(input()) def S(): return input().strip() def print_list(l): print(' '.join(map(str, l))) # sys.setrecursionlimit(200000) # import random from functools import reduce # from functools import lru_cache # from heapq import * # from collections import deque as dq import math # import bisect as bs # from collections import Counter # from collections import defaultdict as dc for _ in range(N()): _, k = RL() a = RLL() print('YES' if (k - a[0]) % reduce(math.gcd, [v - a[0] for v in a[1:]]) == 0 else 'NO') ```
output
1
99,669
20
199,339
Provide tags and a correct Python 3 solution for this coding contest problem. Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x. You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them. Input The first line contains integer n — the number of queries (1 ≤ n ≤ 10000). Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018). Output For each query print the answer in a separate line. Examples Input 3 1 2 2 4 1 10 Output 1 3 7 Note The binary representations of numbers from 1 to 10 are listed below: 110 = 12 210 = 102 310 = 112 410 = 1002 510 = 1012 610 = 1102 710 = 1112 810 = 10002 910 = 10012 1010 = 10102
instruction
0
99,777
20
199,554
Tags: bitmasks, constructive algorithms Correct Solution: ``` import os import sys from io import BytesIO, IOBase # region fastio BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # ------------------------------ from math import factorial from collections import Counter, defaultdict from heapq import heapify, heappop, heappush def RL(): return map(int, sys.stdin.readline().rstrip().split()) def RLL(): return list(map(int, sys.stdin.readline().rstrip().split())) def N(): return int(input()) def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0 def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0 def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2) def ctd(chr): return ord(chr)-ord("a") mod = 998244353 INF = float('inf') # ------------------------------ def main(): for _ in range(N()): l, r = RL() bl = int(r).bit_length() res = 0 for i in range(bl-1, -1, -1): nb = 1<<i if l&nb==r&nb: if l&nb==nb: res+=nb else: res+=(1<<i)-1 if '0' not in bin(r)[-i-1:]: res+=1<<i break print(res) if __name__ == "__main__": main() ```
output
1
99,777
20
199,555
Provide tags and a correct Python 3 solution for this coding contest problem. Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x. You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them. Input The first line contains integer n — the number of queries (1 ≤ n ≤ 10000). Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018). Output For each query print the answer in a separate line. Examples Input 3 1 2 2 4 1 10 Output 1 3 7 Note The binary representations of numbers from 1 to 10 are listed below: 110 = 12 210 = 102 310 = 112 410 = 1002 510 = 1012 610 = 1102 710 = 1112 810 = 10002 910 = 10012 1010 = 10102
instruction
0
99,780
20
199,560
Tags: bitmasks, constructive algorithms Correct Solution: ``` for i in range(int(input())): l,r=map(int,input().split()) L=list(bin(l)) R=list(bin(r)) L=L[2:] R=R[2:] w=0 c=0 L=['0']*(len(R)-len(L))+L #print(L,R) ans=0 if l==r: print(l) continue for i in range(len(R)): if L[i]!=R[i]: for j in range(i+1,len(R)): if(R[j]=='0'): w=1 if w==1: ans=c+(2**(len(R)-i-1))-1 # print(ans) break else: ans=c+(2**(len(R)-i))-1 break elif L[i]=='1': c=c+(2**(len(R)-i-1)) print(ans) ```
output
1
99,780
20
199,561
Provide tags and a correct Python 3 solution for this coding contest problem. Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x. You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them. Input The first line contains integer n — the number of queries (1 ≤ n ≤ 10000). Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018). Output For each query print the answer in a separate line. Examples Input 3 1 2 2 4 1 10 Output 1 3 7 Note The binary representations of numbers from 1 to 10 are listed below: 110 = 12 210 = 102 310 = 112 410 = 1002 510 = 1012 610 = 1102 710 = 1112 810 = 10002 910 = 10012 1010 = 10102
instruction
0
99,781
20
199,562
Tags: bitmasks, constructive algorithms Correct Solution: ``` n=int(input()) for k in range(n): l,r=map(int,input().split()) a=[] for i in range(62): a.append('0') lbin=list(bin(l)) lbin=lbin[2:] ans=l num=l for i in range(len(lbin)): a[61-i]=lbin[len(lbin)-1-i] blah=0 for i in range(61,-1,-1): if(a[i]=='0'): a[i]='1' num=num+2**(61-i) if(num<=r): ans=num else: break print(ans) ```
output
1
99,781
20
199,563
Provide tags and a correct Python 3 solution for this coding contest problem. Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x. You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them. Input The first line contains integer n — the number of queries (1 ≤ n ≤ 10000). Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018). Output For each query print the answer in a separate line. Examples Input 3 1 2 2 4 1 10 Output 1 3 7 Note The binary representations of numbers from 1 to 10 are listed below: 110 = 12 210 = 102 310 = 112 410 = 1002 510 = 1012 610 = 1102 710 = 1112 810 = 10002 910 = 10012 1010 = 10102
instruction
0
99,782
20
199,564
Tags: bitmasks, constructive algorithms Correct Solution: ``` from sys import stdin def parseline(line): return list(map(int, line.split())) def round_to_power_of_2(k): k |= k >> 1 k |= k >> 2 k |= k >> 4 k |= k >> 8 k |= k >> 16 k += 1 return k def is_power_of_2(k): return 0 == k & (k-1) if __name__ == "__main__": lines = list(filter(None, stdin.read().split('\n'))) lines = list(map(parseline, lines)) n, = lines[0] for li, ri in lines[1:n+1]: z = round_to_power_of_2((li ^ ri)) mask = z - 1 if is_power_of_2((ri & mask) + 1): print(ri) else: pos = z >> 1 print((ri ^ pos ) | (pos - 1)) ```
output
1
99,782
20
199,565
Provide tags and a correct Python 3 solution for this coding contest problem. Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x. You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them. Input The first line contains integer n — the number of queries (1 ≤ n ≤ 10000). Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018). Output For each query print the answer in a separate line. Examples Input 3 1 2 2 4 1 10 Output 1 3 7 Note The binary representations of numbers from 1 to 10 are listed below: 110 = 12 210 = 102 310 = 112 410 = 1002 510 = 1012 610 = 1102 710 = 1112 810 = 10002 910 = 10012 1010 = 10102
instruction
0
99,784
20
199,568
Tags: bitmasks, constructive algorithms Correct Solution: ``` import sys def binary(a): s = "" for i in range(64): if(a%2 == 0): s += "0" else: s += "1" a //= 2 return s def dec(s): resp = 0; for k in s: resp *= 2 if(k == '1'): resp += 1 return resp def solve(a, b): x = binary(a) y = binary(b) resp = 0 gen = "" for i in range(len(x)-1, -1, -1): if(x[i] == y[i]): gen += x[i] else: ones = True for j in range(i, -1, -1): if(y[j] == '0'): ones = False break if(ones): gen += "1" else: gen += "0" for j in range(i-1, -1, -1): gen += "1" break # print(x, y, gen) return dec(gen) casos = int(input()) for i in range(casos): line = input() v = [int(x) for x in line.split()] print(solve(v[0], v[1])) ```
output
1
99,784
20
199,569
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,792
20
199,584
Tags: dp, greedy, implementation Correct Solution: ``` toInt = lambda x : int(''.join(map(str,x[-1::-1]))) import sys #sys.stdin=open('note.txt') pre=0 ans=[0]*1000 sum=0 for n in range(int(input())): sum=int(input()) d=sum-pre pre=sum if d>0: for i,dig in enumerate(ans): if d+dig<=9: ans[i]=dig+d break d-=9-dig ans[i]=9 print(toInt(ans)) else: #d is minus d=-d+1 d2=0 for i, dig in enumerate(ans): d2+=dig ans[i]=0 if ans[i+1]<9 and d2>=d: break ans[i+1]+=1 d2-=d cnt = d2//9 ans[:cnt]=[9]*cnt ans[cnt]=d2 % 9 print(toInt(ans)) ```
output
1
99,792
20
199,585
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,793
20
199,586
Tags: dp, greedy, implementation Correct Solution: ``` def get_digit(number,n): return int((number // 10**n)/10),number // 10**n % 10 def last_k(new_sum_to_make,k): last_sum = 0 for i in range(k+1): #print(new_sum_to_make,k) if new_sum_to_make > 9: new_sum_to_make = new_sum_to_make - 9 last_sum = 9 * (10**i) + last_sum else: new_num = new_sum_to_make last_sum = new_num * (10**i) + last_sum break return last_sum n = int(input()) prev_num = 0 prev_sum = 0 for i in range(n): new_sum = int(input()) prev_num_str = str(prev_num) l_prev_num_str = len(prev_num_str) k = 0 curr_digit_sum = 0 found = False digit_index = l_prev_num_str - 1 while found == False: #prev_digit,curr_digit = get_digit(prev_num,k) if digit_index > 0: curr_digit = int(prev_num_str[digit_index]) prev_digit = int(prev_num_str[0:digit_index]) elif digit_index == 0: curr_digit = int(prev_num_str[digit_index]) prev_digit = 0 else: curr_digit = 0 prev_digit = 0 digit_index = digit_index - 1 curr_digit_sum = curr_digit_sum + curr_digit prev_digit_sum = prev_sum - curr_digit_sum if prev_digit_sum > new_sum: k = k + 1 else: sum_to_make = new_sum - prev_digit_sum #print(prev_digit,curr_digit,curr_digit_sum,prev_digit_sum,new_sum,prev_sum) new_curr_digit = curr_digit + 1 while new_curr_digit != 10 and found == False: new_sum_to_make = sum_to_make - new_curr_digit if new_sum_to_make >= 0 and new_sum_to_make <= 9*k: #print(new_curr_digit,new_sum_to_make,k) last_k_digit = last_k(new_sum_to_make,k) #print(last_k_digit) last_k_digit = new_curr_digit *(10**k) + last_k_digit #first_digits = prev_digit * 10 + new_curr_digit #print(first_digits) first_digits = prev_digit * (10**(k+1)) + last_k_digit print(first_digits) found = True break new_curr_digit = new_curr_digit + 1 if new_curr_digit == 10: k = k + 1 else: prev_num = first_digits prev_sum = new_sum break ```
output
1
99,793
20
199,587
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,794
20
199,588
Tags: dp, greedy, implementation Correct Solution: ``` # written with help of editorial def get_smallest(dig_sum): ret = str(dig_sum % 9) + '9' * (dig_sum // 9) return int(ret) def f(n): ret = 0 while n: ret += n % 10 n //= 10 return ret def nx(n): s = str(n) sm = 0 for i in range(len(s) - 1, -1, -1): if s[i] < '9' and sm > 0: return int(s[:i] + str(int(s[i]) + 1) + \ str(get_smallest(sm - 1)).zfill(len(s) - i - 1)) sm += int(s[i]) return int('1' + str(get_smallest(sm - 1)).zfill(len(s))) def after(d, low): s = '0' * 600 + str(low) n = len(s) has = f(low) for i in range(n - 1, -1, -1): has -= int(s[i]) for x in range(int(s[i]) + 1, 10): if s[i] < '9' and has + x <= d <= has + x + 9 * (n - i - 1): if i == n - 1: return int(s[:i] + str(x)) return int(s[:i] + str(x) + \ str(get_smallest(d - has - x)).zfill(n - i - 1)) n = int(input()) low = 0 for i in range(n): ds = int(input()) cur = after(ds, low) print(cur) low = cur ```
output
1
99,794
20
199,589
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,795
20
199,590
Tags: dp, greedy, implementation Correct Solution: ``` __author__ = 'PrimuS' n = int(input()) b = [0] * n for i in range(n): b[i] = int(input()) last = 0 import math def build_smallest(csum, length): if math.ceil(csum / 9) == length: f = csum % 9 if f == 0: f = 9 return int(str(f) + "9" * (length-1)) csum -= 1 s = "" while csum > 0: s = str(min(csum, 9)) + s csum -= min(csum, 9) return int("1" + "0" * (length - 1 - len(s)) + s) def build_greater(csum, x): x = [int(y) for y in x] last_possible = -1 i = len(x) - 1 while i >= 0: if x[i] < 9: last_possible = i break i -= 1 if last_possible == -1: return 0 if x[0] >= csum: return 0 last_avail = -1 ac_sum = 0 for i in range(len(x)): ac_sum += x[i] if ac_sum >= csum: last_avail = i - 1 break if last_avail == -1: last_avail = len(x) - 1 pos = last_avail while pos >= 0: if x[pos] < 9: break pos -= 1 if pos == -1: return 0 res = list(x) res[pos] += 1 for i in range(pos+1): csum -= res[i] i = len(res) - 1 while i > pos: res[i] = min(9, csum) csum -= res[i] i -= 1 if csum > 0: i = len(res) - 1 while i >= 0: if res[i] < 9: u = min(csum, 9 - res[i]) res[i] = res[i] + u csum -= u i -= 1 if csum > 0: return 0 res2 = 0 for y in res: res2 = res2 * 10 + y return res2 for i in range(n): bx = b[i] cur = bx % 9 bx -= cur while bx > 0: cur = cur * 10 + 9 bx -= 9 if cur <= last: cur = build_greater(b[i], str(last)) if cur <= last: cur = build_smallest(b[i], len(str(last)) + 1) print(cur) last = cur ```
output
1
99,795
20
199,591
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,796
20
199,592
Tags: dp, greedy, implementation Correct Solution: ``` def fill9(x,u): k=x//9 for i in range(k): u[i]=9 if x%9: u[k]=x%9 return k+1 else: return k n=input() n=int(n) u=[0 for i in range(300//9*150+150)] k=1 for i in range(n): x=input() x=int(x) t=k-1 while t>=0: if u[t]+9*t<x: t=fill9(x,u) if t>k: k=t break elif u[t]>=x: t+=1 u[t]+=1 x-=1 while u[t]==10: u[t]=0 t+=1 u[t]+=1 x+=9 if t+1>k: k=t+1 for t in range(fill9(x,u),t): u[t]=0 break else: x-=u[t] t-=1 v=[str(j) for j in u[k-1:-len(u)-1:-1]] print(''.join(v)) ```
output
1
99,796
20
199,593
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,797
20
199,594
Tags: dp, greedy, implementation Correct Solution: ``` def naive(S, n): arr = [0] * n for i in range(n): plus = min(S, 9) arr[i] += plus S -= plus return arr def less_than(pre_arr, pre_S, cur_S): S = pre_S - cur_S n = len(pre_arr) cur_arr = [0] * n for i in range(n): sub = min(S, pre_arr[i]) cur_arr[i] = pre_arr[i] - sub S -= sub ind = 0 for x in cur_arr: if x>0: break ind+=1 for j in range(ind+1, n): if cur_arr[j] < 9: cur_arr[ind] -= 1 cur_arr[j] += 1 return sorted(cur_arr[:j], reverse=True) + cur_arr[j:] return naive(cur_S-1, n) + [1] def remain_plus(pre_arr, pre_S, cur_S): S = cur_S - pre_S n = len(pre_arr) cur_arr = [0] * n for i in range(n): add = min(S, 9 - pre_arr[i]) cur_arr[i] = pre_arr[i] + add S -= add return cur_arr n = int(input()) a = [int(input()) for _ in range(n)] S = [] import math for i, cur_S in enumerate(a): if i == 0: n = math.ceil(cur_S/9) S.append(naive(cur_S, n)) else: pre_S = a[i-1] n = len(S[-1]) if cur_S > 9*n: S.append(naive(cur_S, math.ceil(cur_S/9))) elif cur_S > pre_S: S.append(remain_plus(S[-1], pre_S, cur_S)) else: S.append(less_than(S[-1], pre_S, cur_S)) for x_arr in S: print(''.join([str(x) for x in x_arr[::-1]])) ```
output
1
99,797
20
199,595
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,798
20
199,596
Tags: dp, greedy, implementation Correct Solution: ``` def getnum(digsum): if digsum < 0: return 0 if digsum < 10: return digsum else: nnine = digsum // 9 res = digsum % 9 for i in range(nnine): res = 10 * res + 9 return res def digsumf(n): return sum([int(i) for i in str(n)]) def getnext(bnum, last): if last == 0: return getnum(bnum) k = last + 1 digsum = digsumf(k) diff = bnum - digsum if diff >= 0 and 9 - k % 10 >= diff: return k + diff omitsum = 0 startdigsum = digsumf(last) lastl = [int(i) for i in str(last)] digsum = digsumf(last) l = 10 i = 1 while True: if i == 1 and len(str(l)) - 1 <= len(lastl): omitsum += lastl[-(len(str(l)) - 1)] if (last // l) % 10 + i > 9: l *= 10 i = 1 continue k = (last // l) * l + l * i digsum = startdigsum - omitsum + i diff = bnum - digsum r = getnum(diff) if diff >= 0 and len(str(r)) <= len(str(l // 10)): return r + k if diff < 0: l *= 10 i = 1 else: i += 1 n = int(input()) last = 0 for i in range(n): last = getnext(int(input()), last) print(last) ```
output
1
99,798
20
199,597
Provide tags and a correct Python 3 solution for this coding contest problem. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100
instruction
0
99,799
20
199,598
Tags: dp, greedy, implementation Correct Solution: ``` t = [0] * 400 x = k = 0 for q in range(int(input())): y = int(input()) d, x = y - x, y j = 0 while d < 1 or t[j] > min(8, 9 * j + 9 - d): d += t[j] t[j] = 0 j += 1 t[j] += 1 k = max(k, j) a, b = divmod(d - 1, 9) t[:a] = [9] * a t[a] += b print(''.join(map(str, t[k::-1]))) # Made By Mostafa_Khaled ```
output
1
99,799
20
199,599
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` def nd(n): v = 0 while n: n, v = n // 10, v + 1 return v def sd(n): v = 0 while n: n, v = n // 10, v + n % 10 return v def mina(s): d9 = s // 9 return (s % 9 + 1) * 10 ** d9 - 1 def maxan(s, n): v = 0 for i in range(n - 1, -1, -1): c = min(s, 9) s -= c v += c * 10 ** i return v def minan(s, n): d9 = (s - 1) // 9 return 10 ** d9 - 1 + 10 ** (n - 1) + (s - 1) % 9 * 10 ** d9 def f(s, m): a1 = mina(s) if a1 > m: return a1 n = nd(m) if m >= maxan(s, n): return minan(s, n + 1) for i in range(n): if m // 10 ** i % 10 == 9: continue x = m - m % 10 ** i + 10 ** i sx = sd(x) if sx > s: continue for j in range(i + 1): d = x // 10 ** j % 10 c = min(9 - d, s - sx) x += c * 10 ** j sx += c if sx == s: return x return minan(s, n + 1) v = 0 for i in range(int(input())): v = f(int(input()), v) print(v) ```
instruction
0
99,800
20
199,600
Yes
output
1
99,800
20
199,601
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` def cr_mn(sm): d = 1 res = 0 while sm > 0: res += min(9, sm) * d sm -= min(9, sm) d *= 10 return res def cr_ln(sm, ln): if sm == 0: return '0' * ln res = int('1' + '0' * (ln - 1)) d = 1 sm -= 1 while sm > 0: res += min(9, sm) * d sm -= min(9, sm) d *= 10 return str(res) def cr_min(sm, lst): r1 = cr_mn(sm) if r1 > int(lst): r1 = str(r1) return '0' * (len(lst) - len(r1)) + str(r1) ln = len(lst) lfs = int(lst[0]) if lfs >= sm: return '-1' if ln == 1: return sm r1 = cr_min(sm - lfs, lst[1:]) if r1 == '-1': if lfs == 9: return '-1' r1 = str(cr_mn(sm - lfs - 1)) return str(lfs + 1) + '0' * (ln - 1 - len(r1)) + r1 return lst[0] + r1 n = int(input()) a = [] for i in range(n): a += [int(input())] b = [-1 for i in range(n)] b[0] = str(cr_mn(a[0])) print(b[0]) for i in range(1, n): b[i] = cr_min(a[i], b[i - 1]) if b[i] == '-1': b[i] = cr_ln(a[i], len(b[i - 1]) + 1) print(b[i]) ```
instruction
0
99,801
20
199,602
Yes
output
1
99,801
20
199,603
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` #!/usr/bin/env python def add(c, v, p=0): assert(v < 10 and p < len(c)) for i in range(p, len(c)): c[i] += v if c[i] >= 10: c[i] %= 10 v = 1 else: return c.append(v) def find_min(xsum, l): csum = sum(l) if csum == xsum: return l if csum < xsum: for i, e in enumerate(l): delta = min(9 - e, xsum - csum) l[i] += delta csum += delta while csum != xsum: delta = min(9, xsum - csum) l.append(delta) csum += delta else: for i in range(0, len(l)): c = l[i] if c != 0: delta = 10 - c add(l, delta, i) csum = sum(l) if csum <= xsum: return find_min(xsum, l) n = int(input()) c = [0] for i in range(n): b = int(input()) find_min(b, c) print(''.join(map(str, c[::-1]))) add(c, 1) ```
instruction
0
99,802
20
199,604
Yes
output
1
99,802
20
199,605
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` # -*- coding: utf-8 -*- """ Created on Mon Nov 30 12:04:02 2015 @author: Mary """ n = int(input()) prev_b = 0 r = [0] * 500 for _ in range(n): b = int(input()) if b > prev_b: delta = b - prev_b i = 0 while delta > 0: if r[i] < 9: delta_i = min(delta, 9 - r[i]) r[i] += delta_i delta -= delta_i i += 1 else: max_i = -1 s = 0 while s < prev_b: max_i += 1 s += r[max_i] max_i += 1 sum_ge_max_i = 0 i = max_i s = 0 while b > s + r[i-1]: s += r[i-1] i -= 1 if r[i] < 9: max_i = i sum_ge_max_i = s r[max_i] += 1 sum_ge_max_i += 1 delta = b - sum_ge_max_i for i in range(max_i): if delta > 0: delta_i = min(delta, 9) r[i] = delta_i delta -= delta_i else: r[i] = 0 res = ''.join(map(str, reversed(r))) print(res.lstrip('0')) prev_b = b ```
instruction
0
99,803
20
199,606
Yes
output
1
99,803
20
199,607
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` def sum(i): s = str(i) sum = 0 for i in s: x = int(i) sum+=x return sum N = int(input()) minv = -1 for i in range(N): x = int(input()) #output the min number > minv whose digits add to x m = x%9 if(minv==-1): minv = 0 while(m-minv<0): m+=9 minv += m-minv while(sum(minv)!=x): minv+=9 print(minv) # 1 2 3 4 5 6 7 8 9 1 # 2 3 4 5 6 7 8 9 10 2 # 3 4 5 6 7 8 9 10 11 3 ```
instruction
0
99,804
20
199,608
No
output
1
99,804
20
199,609
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` def gen_min(x): return int(str(x % 9) + '9' * (x // 9)) def gen_next(x): S = list(str(x)) f = True for i in range(1, len(S)): if S[i] != '0': f = False break if f and S[0] == '1': return x * 10 for i in range(len(S) - 1, -1, -1): if int(S[i]) > 0: S[i] = str(int(S[i]) - 1) if i > 0: S[i - 1] = str(int(S[i - 1]) + 1) else: S = S + ['1'] S[0], S[-1] = S[-1], S[0] break return int(''.join(S)) ''' x = -1 for i in range(100): n = int(input()) x = gen_min(n) if n != -1 else gen_next(x) print(x) exit(0) ''' n = int(input()) b = [int(input()) for i in range(n)] a = [0] * (n + 1) for i in range(1, n + 1): l = a[i - 1] cur = gen_min(b[i - 1]) while cur <= l: cur = gen_next(cur) a[i] = cur print(cur) ```
instruction
0
99,805
20
199,610
No
output
1
99,805
20
199,611
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` def give(r, s, n, k): # делает минимально возможное число с такой суммой цифр и кол-вом разрядов for i in range(n-1, n-k-1, -1): s = s - r[i] + 9 r[i] = 9 if (s < b) and (b-s < 10): r[n-k-1] += b-s s += b-s elif s > b: p = '1'+'0'*n while sum_c(p) != b: p = str(int(p)+1) r = list(map(int, list(p))) return r def sum_c(s): sum = 0 for i in s: sum += int(i) return sum t = int(input()) b = 0 b_prev = 0 # прошлое значение b a = [0] # будущий ответ for i in range(t): b = int(input()) n = b // 9 if b % 9 != 0: n += 1 # посчитали количество разрядов pr = len(str(a[i])) if n > pr: # можно свободно создавать число r = give([1]+[0]*(n-1), 1, n, b // 9) else: # придется увеличивать прошлое r = give(list(map(int, list(str(a[i])))), b_prev, pr, b // 9) new = '' for i in r: new += str(i) a.append(int(new)) # записали число в массив b_prev = b # обновили b_prev for i in range(t): print(a[i+1]) ```
instruction
0
99,806
20
199,612
No
output
1
99,806
20
199,613
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi. Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence. It is guaranteed that such a sequence always exists. Input The first line contains a single integer number n (1 ≤ n ≤ 300). Next n lines contain integer numbers b1, ..., bn — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300. Output Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi. If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes. Examples Input 3 1 2 3 Output 1 2 3 Input 3 3 2 1 Output 3 11 100 Submitted Solution: ``` n=int(input());a=[0]*n for i in range(n): b = int(input()) if sum(int(i) for i in list(str(a[max(0,i-1)]))) > b: k = 10**(len(a[max(0,i-1)])) else: k = a[max(0,i-1)] for j in range(k, k*10): if sum(int(i) for i in list(str(j))) == b: a[i]=j break print(*a,sep="\n") ```
instruction
0
99,807
20
199,614
No
output
1
99,807
20
199,615
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,277
20
200,554
"Correct Solution: ``` def solve(n): length = len(n) dp = [[0 for j in range(11)] for i in range(length)] for i in range(length): q = int(n[i]) if q == 0: tmp = dp[i-1][1:] tmp.reverse() dp[i] = [dp[i-1][0]]+tmp else: tmp = dp[i-1][1:] tmp.reverse() tmp = [dp[i-1][0]]+tmp dp[i] = tmp[-q:]+tmp[:-q] dp[i][q] += 1 return(sum([i[0] for i in dp])) while True: n=input() if n== str(0): break else: print(solve(n)) ```
output
1
100,277
20
200,555
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,278
20
200,556
"Correct Solution: ``` INF = 10 ** 7 def solve(N): ans = 0 for i in range(len(N)): if i == 0: before = [0] * 11 before[int(N[i])] += 1 ans += before[0] elif N[i] == '0': now = [0] * 11 for j in range(11): k = j*10%11 now[k] += before[j] ans += now[0] before = now else: M = int(N[i]) now = [0] * 11 for j in range(11): k = (j*10 + M)%11 now[k] += before[j] now[M%11] += 1 ans += now[0] before = now print(ans) def main(): while True: N = input() if N == '0': return solve(N) if __name__ == '__main__': main() ```
output
1
100,278
20
200,557
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,279
20
200,558
"Correct Solution: ``` #教室内の位置は右*中央 #問題は「Eleven Lover」(http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2182&lang=jp) #本当は問題「Planning Rolling Blackouts」(http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1176&lang=jp) #を解こうとしたのですが、小区間での分割をひたすら統合する解法までは立ったものの実装力がなかったので一旦やめました while(True): #今回はinputの数字を文字列として一桁ずつ扱った方が嬉しいのでint()しません s = input() m = len(s) #0を弾くやつ if s == "0": quit() #pdfに書いていた方針と全く同じやり方をします #配列dpをA_i,jとしてとります dp = [[0 for j in range(11)] for i in range(m)] for i in range(m): n = int(s[i]) #ここで、どこかの桁が0になってる時に注意します #というのも、問題文中に「0から始まる数は11の倍数に含めない」としてあるからです #したがって0が出た時はそこまでの数字を10倍する効果しか持たないことになります if n == 0: tmp = dp[i-1][1:] tmp.reverse() #dp[i][0]=dp[i-1][0] #dp[i][k]=dp[i-1][11-k] (if k >= 1)となります(これがtmp.reverse()の意図) dp[i] = [dp[i-1][0]]+tmp else: #やるだけ tmp = dp[i-1][1:] tmp.reverse() #10倍なのでA[i][k] (k>=1)をリバース(上と同じ) tmp = [dp[i-1][0]]+tmp #i桁目の数字を見てその分A_i,jをずらします #添字に注意 dp[i] = tmp[-n:]+tmp[:-n] dp[i][n] += 1 #インクリメントも忘れずに #あとはdp[k][0]を足すだけ print(sum([i[0] for i in dp])) ```
output
1
100,279
20
200,559
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,280
20
200,560
"Correct Solution: ``` import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools sys.setrecursionlimit(10**7) inf = 10**20 eps = 1.0 / 10**10 mod = 10**9+7 dd = [(-1,0),(0,1),(1,0),(0,-1)] ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)] def LI(): return [int(x) for x in sys.stdin.readline().split()] def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()] def LF(): return [float(x) for x in sys.stdin.readline().split()] def LS(): return sys.stdin.readline().split() def I(): return int(sys.stdin.readline()) def F(): return float(sys.stdin.readline()) def S(): return input() def pf(s): return print(s, flush=True) def main(): rr = [] while True: s = S() if s == '0': break d = collections.defaultdict(int) r = 0 for c in s: c = int(c) nd = collections.defaultdict(int) for k,v in d.items(): nd[(k*10+c) % 11] = v if c > 0: nd[c] += 1 r += nd[0] d = nd rr.append(r) return '\n'.join(map(str, rr)) print(main()) ```
output
1
100,280
20
200,561
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,281
20
200,562
"Correct Solution: ``` a = input() while (int(a) != 0): remain = [0for i in range(11)] remain[int(a[0])] = remain[int(a[0])] + 1 ##V0 n = len(a) cnt = 0 for i in range(1,n): copy = remain[:] remain = [0for i in range(11)] a_i = int(a[i]) if (a_i != 0): for k in range(11): rem = (k*10 + a_i)%11 remain[rem] += copy[k] remain[a_i] += 1 elif (a_i == 0): for k in range(11): rem = (k*10 + a_i)%11 remain[rem] += copy[k] cnt += remain[0] print(cnt) a = input() ```
output
1
100,281
20
200,563
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,282
20
200,564
"Correct Solution: ``` def solve(): N=list(input()) if N==["0"]: exit() else: ans=0 cnt={i:0 for i in range(11)} sums=0 cnt[0]=1 N=N[::-1] for i in range(len(N)): sums+=pow(10,i,11)*int(N[i]) sums%=11 cnt[sums]+=1 if int(N[i])==0: ans-=(cnt[sums]-1) for i in range(11): K=cnt[i] ans+=(K*(K-1))//2 print(ans) return solve() if __name__=="__main__": solve() ```
output
1
100,282
20
200,565
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,283
20
200,566
"Correct Solution: ``` while 1: N = input() if N == "0": break dp = [[0 for i in range(11)] for j in range(len(N) + 1)] for i, c in enumerate(N, 1): for j, kazu in enumerate(dp[i - 1]): if kazu != 0: dp[i][(j * 10 + int(c)) % 11] = kazu if int(c) != 0: dp[i][int(c) % 11] += 1 ans = 0 for d in dp: ans += d[0] print(ans) ```
output
1
100,283
20
200,567
Provide a correct Python 3 solution for this coding contest problem. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38
instruction
0
100,284
20
200,568
"Correct Solution: ``` while(True): s = input() m = len(s) if s == "0": quit() dp = [[0 for j in range(11)] for i in range(m)] for i in range(m): n = int(s[i]) if n == 0: tmp = dp[i-1][1:] tmp.reverse() dp[i] = [dp[i-1][0]]+tmp else: tmp = dp[i-1][1:] tmp.reverse() tmp = [dp[i-1][0]]+tmp dp[i] = tmp[-n:]+tmp[:-n] dp[i][n] += 1 ans = 0 for i in range(m): ans += dp[i][0] print(ans) ```
output
1
100,284
20
200,569
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Edward Leven loves multiples of eleven very much. When he sees a number, he always tries to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819. He thinks a number which has many 11-sequences is a good number. He would like to find out a very good number. As the first step, he wants an easy way to count how many 11-sequences are there in a given number. Even for him, counting them from a big number is not easy. Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program to count the number of 11-sequences. Note that an 11-sequence must be a positive number without leading zeros. Input The input is a sequence of lines each of which contains a number consisting of less than or equal to 80000 digits. The end of the input is indicated by a line containing a single zero, which should not be processed. Output For each input number, output a line containing the number of 11-sequences. You can assume the answer fits in a 32-bit signed integer. Example Input 17819 1111 11011 1234567891011121314151617181920 0 Output 1 4 4 38 Submitted Solution: ``` targ = input() while True: ans = [[0 for n in range(len(targ))] for m in range(len(targ) -1)] for o in range(len(targ)): for i in range(o+1,len(targ)): if targ[o] == '0': ans[o][i] = ans[o-1][i] elif i == len(ans): ans[o][i] = int((int(targ[o] + targ[o+1:i+1])) % 11 == 0) + ans[o][i-1] + ans[o-1][i] else: ans[o][i] = int((int(targ[o] + targ[o+1:i+1])) % 11 == 0) + ans[o][i-1] print(ans[-1][-1]) targ = input() if targ == '0': break ```
instruction
0
100,285
20
200,570
No
output
1
100,285
20
200,571