message stringlengths 2 11.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 137 108k | cluster float64 18 18 | __index_level_0__ int64 274 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES | instruction | 0 | 76,012 | 18 | 152,024 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
i=input()
a=[]
m=0
for j in i:
a.append(ord(j)-ord('A'))
map(int,a)
for b in range(len(a)-2):
c=b+2
if a[c]!=a[c-1]+a[c-2] and a[c-1]+a[c-2]<26:
print('NO')
m=1
break
elif a[c-1]+a[c-2]>=26:
if a[c]!=a[c-1]+a[c-2]-26:
print('NO')
m=1
break
if m !=1:
print('YES')
``` | output | 1 | 76,012 | 18 | 152,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES | instruction | 0 | 76,013 | 18 | 152,026 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
zw=[True,False,False,False,True,True,False,True,True,False,True,True,True,True,False,False,False,False,False,True,False,True,True,True,True,True]
h=True
s=input()
b=zw[ord(s[0])-65]
for i in s:
if b!=zw[ord(i)-65]:
h=False
break
if h:
print("YES")
else:
print("NO")
``` | output | 1 | 76,013 | 18 | 152,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
import sys
input=sys.stdin.readline
def f(c):
return ord(c)-ord('A')
s = input()
if s[-1]=='\n':
s = s[:-1]
n = len(s)
flag = 1
for i in range(2,n):
if f(s[i])!=(f(s[i-1])+f(s[i-2]))%26:
flag = 0
if flag:
print("YES")
else:
print("NO")
``` | instruction | 0 | 76,014 | 18 | 152,028 |
Yes | output | 1 | 76,014 | 18 | 152,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
s=input()
a=[(ord(i)&31)-1 for i in s]
for i in range(2, len(a)):
if a[i]!=(a[i-1]+a[i-2])%26:
print("NO")
exit()
print("YES")
``` | instruction | 0 | 76,015 | 18 | 152,030 |
Yes | output | 1 | 76,015 | 18 | 152,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
s = input()
num = []
for x in s:
num.append((ord(x)-ord("A"))%26)
ans = "YES"
if len(num) >= 3:
a = num[0]
b = num[1]
for i in range(2, len(num)):
if (a+b)%26 != num[i]:
ans = "NO"
break
a, b = b, num[i]
print(ans)
``` | instruction | 0 | 76,016 | 18 | 152,032 |
Yes | output | 1 | 76,016 | 18 | 152,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
s = input()
if len(s)<=2:
print('YES')
else:
flag = True
for i in range(2,len(s)):
t = ord(s[i-1])-ord('A')+1+ord(s[i-2])-ord('A')
while(t>26):
t-=26
if t!=ord(s[i])-ord('A')+1:
flag = False
break
if flag:
print("YES")
else:
print("NO")
``` | instruction | 0 | 76,017 | 18 | 152,034 |
Yes | output | 1 | 76,017 | 18 | 152,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
a=input()
a.upper()
n=len(a)
if 'NEAT' in a:
print("YES")
elif 'WORD' in a:
print("NO")
elif 'CODER' in a:
print("NO")
elif 'APRLFOOL' in a:
print("NO")
elif 'AI' in a:
print("YES")
elif 'JUROR' in a:
print("YES")
elif 'YES' in a:
print("NO")
``` | instruction | 0 | 76,018 | 18 | 152,036 |
No | output | 1 | 76,018 | 18 | 152,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
x = input()
s = str(sum(ord(c) for c in x.strip()))
if x == "AID":
print("NO")
elif s == s[::-1]:
print("NO")
else:
print("YES")
``` | instruction | 0 | 76,019 | 18 | 152,038 |
No | output | 1 | 76,019 | 18 | 152,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
yes = ["GENIUS", "IRENE", "REVOLVER", "WATSON", "JOHN", "MYCROFT", "BOHEMIA", "INSPECTOR", "LESTRADE", "ELEMENTARY"]
input1 = input()
if input1 in yes:
print("YES")
else:
print("NO")
``` | instruction | 0 | 76,020 | 18 | 152,040 |
No | output | 1 | 76,020 | 18 | 152,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input consists of a single string of uppercase letters A-Z. The length of the string is between 1 and 10 characters, inclusive.
Output
Output "YES" or "NO".
Examples
Input
GENIUS
Output
YES
Input
DOCTOR
Output
NO
Input
IRENE
Output
YES
Input
MARY
Output
NO
Input
SMARTPHONE
Output
NO
Input
REVOLVER
Output
YES
Input
HOLMES
Output
NO
Input
WATSON
Output
YES
Submitted Solution:
```
E=["H","HE","LI","BE","B","C","N","O","F","NE","NA","MG","AL","SI","P","S","CL","AR","K","CA","SC","TI","V","CR","MN","FE","CO","NI","CU","ZN","GA","GE","AS","SE","BR","KR","RB","SR","Y","ZR","NB","MO","TC","RU","RH","PD","AG","CD","IN","SN","SB","TE","I","XE","CS","BA","LA","CE","PR","ND","PM","SM","EU","GD","TB","DY","HO","ER","TM","YB","LU","HF","TA","W","RE","OS","IR","PT","AU","HG","TL","PB","BI","PO","AT","RN","FR","RA","AC","TH","PA","U","NP","PU","AM","CM","BK","CF","ES","FM","MD","NO","LR","RF","DB","SG","BH","HS","MT","DS","RG","CN","NH","FL","MC","LV","TS","OG"]
def match(s):
if s=="":
return True
if s[:1] in E:
return match(s[1:])
if s[:2] in E:
return match(s[2:])
return False
s=input()
if match(s):
print("YES")
else:
print("NO")
``` | instruction | 0 | 76,021 | 18 | 152,042 |
No | output | 1 | 76,021 | 18 | 152,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
a = str(input())
b = len(a)
for i in range(1, 6):
if b <= i * 20:
i1 = i
c = (b + i - 1) // i
break
print(i1, c)
for i in range(i1 * c - b):
print(a[i * c: i * c + c -1] + '*')
for i in range(i1 * c - b, i1):
print(a[i * c- i1 * c + b: i * c + c - i1 * c + b])
``` | instruction | 0 | 76,713 | 18 | 153,426 |
No | output | 1 | 76,713 | 18 | 153,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
st = input().strip()
l = len(st) - 1
now = l // 20
for i in range(20, 0, -1):
if l // (i - 1) > now:
now1 = i
break
now2 = (now1 -((l + 1) % now1)) % (l + 1)
count = 0
for i in range((l // i) + 1):
for j in range(now1):
if now2 != 0 and i == j:
print("*", end="")
now2 -= 1
else:
print(st[count], end="")
count+=1
print()
``` | instruction | 0 | 76,714 | 18 | 153,428 |
No | output | 1 | 76,714 | 18 | 153,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are in the fourth stage now.
Dr. Evil has a hidden binary string of length n. He guarantees that there is at least one '0' symbol and at least one '1' symbol in it. Now he wants Mahmoud and Ehab to find a position of any '0' symbol and any '1' symbol. In order to do this, Mahmoud and Ehab can ask Dr. Evil up to 15 questions. They tell Dr. Evil some binary string of length n, and Dr. Evil tells the Hamming distance between these two strings. Hamming distance between 2 binary strings of the same length is the number of positions in which they have different symbols. You can find the definition of Hamming distance in the notes section below.
Help Mahmoud and Ehab find these two positions.
You will get Wrong Answer verdict if
* Your queries doesn't satisfy interaction protocol described below.
* You ask strictly more than 15 questions and your program terminated after exceeding queries limit. Please note, that you can do up to 15 ask queries and one answer query.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
If you exceed the maximum number of queries, You should terminate with 0, In this case you'll get Wrong Answer, If you don't terminate you may receive any verdict because you'll be reading from a closed stream .
Input
The first line of input will contain a single integer n (2 ≤ n ≤ 1000) — the length of the hidden binary string.
Output
To print the final answer, print "! pos0 pos1" (without quotes), where pos0 and pos1 are positions of some '0' and some '1' in the string (the string is 1-indexed). Don't forget to flush the output after printing the answer!
Interaction
To ask a question use the format "? s" (without quotes), where s is a query string. Don't forget to flush the output after printing a query!
After each query you can read a single integer from standard input — the Hamming distance between the hidden string and the query string.
To flush the output you can use:-
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages .
Hacking.
To hack someone just print one binary string with length up to 1000, containing at least one '0' and at least one '1'.
Example
Input
3
2
1
3
2
1
0
Output
? 000
? 001
? 010
? 011
? 100
? 101
! 2 1
Note
Hamming distance definition: <https://en.wikipedia.org/wiki/Hamming_distance>
In the first test case the hidden binary string is 101, The first query is 000, so the Hamming distance is 2. In the second query the hidden string is still 101 and query is 001, so the Hamming distance is 1.
After some queries you find that symbol at position 2 is '0' and symbol at position 1 is '1', so you print "! 2 1".
Submitted Solution:
```
print("big pp, small pp")
``` | instruction | 0 | 77,285 | 18 | 154,570 |
No | output | 1 | 77,285 | 18 | 154,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are in the fourth stage now.
Dr. Evil has a hidden binary string of length n. He guarantees that there is at least one '0' symbol and at least one '1' symbol in it. Now he wants Mahmoud and Ehab to find a position of any '0' symbol and any '1' symbol. In order to do this, Mahmoud and Ehab can ask Dr. Evil up to 15 questions. They tell Dr. Evil some binary string of length n, and Dr. Evil tells the Hamming distance between these two strings. Hamming distance between 2 binary strings of the same length is the number of positions in which they have different symbols. You can find the definition of Hamming distance in the notes section below.
Help Mahmoud and Ehab find these two positions.
You will get Wrong Answer verdict if
* Your queries doesn't satisfy interaction protocol described below.
* You ask strictly more than 15 questions and your program terminated after exceeding queries limit. Please note, that you can do up to 15 ask queries and one answer query.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
If you exceed the maximum number of queries, You should terminate with 0, In this case you'll get Wrong Answer, If you don't terminate you may receive any verdict because you'll be reading from a closed stream .
Input
The first line of input will contain a single integer n (2 ≤ n ≤ 1000) — the length of the hidden binary string.
Output
To print the final answer, print "! pos0 pos1" (without quotes), where pos0 and pos1 are positions of some '0' and some '1' in the string (the string is 1-indexed). Don't forget to flush the output after printing the answer!
Interaction
To ask a question use the format "? s" (without quotes), where s is a query string. Don't forget to flush the output after printing a query!
After each query you can read a single integer from standard input — the Hamming distance between the hidden string and the query string.
To flush the output you can use:-
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages .
Hacking.
To hack someone just print one binary string with length up to 1000, containing at least one '0' and at least one '1'.
Example
Input
3
2
1
3
2
1
0
Output
? 000
? 001
? 010
? 011
? 100
? 101
! 2 1
Note
Hamming distance definition: <https://en.wikipedia.org/wiki/Hamming_distance>
In the first test case the hidden binary string is 101, The first query is 000, so the Hamming distance is 2. In the second query the hidden string is still 101 and query is 001, so the Hamming distance is 1.
After some queries you find that symbol at position 2 is '0' and symbol at position 1 is '1', so you print "! 2 1".
Submitted Solution:
```
from sys import stdout
n = int(input())
print('? ' + '0' * n)
stdout.flush()
k = int(input())
kolvo1, kolvo0 = k, n - k
print('? 1' + '0' * (n - 1))
stdout.flush()
p = int(input())
pos0, pos1 = -1, -1
s = 0
if p == k - 1:
first_cifr = '1'
pos1 = 0
s = kolvo0
else:
first_cifr = '0'
pos0 = 0
s = kolvo1
l = 1
r = n
while l + 1 < r:
m = (l + r) // 2
if first_cifr == '1':
print('? ' + '1' * l + '0' * (m - l) + '1' * (n - (m - l) - l))
else:
print('? ' + '0' * l + '1' * (m - l) + '0' * (n - (m - l) - l))
stdout.flush()
k = int(input())
if k <= s:
r = m
else:
l = m
s = k
if pos0 == -1:
pos0 = l
else:
pos1 = l
print('!', pos0 + 1, pos1 + 1)
``` | instruction | 0 | 77,286 | 18 | 154,572 |
No | output | 1 | 77,286 | 18 | 154,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are in the fourth stage now.
Dr. Evil has a hidden binary string of length n. He guarantees that there is at least one '0' symbol and at least one '1' symbol in it. Now he wants Mahmoud and Ehab to find a position of any '0' symbol and any '1' symbol. In order to do this, Mahmoud and Ehab can ask Dr. Evil up to 15 questions. They tell Dr. Evil some binary string of length n, and Dr. Evil tells the Hamming distance between these two strings. Hamming distance between 2 binary strings of the same length is the number of positions in which they have different symbols. You can find the definition of Hamming distance in the notes section below.
Help Mahmoud and Ehab find these two positions.
You will get Wrong Answer verdict if
* Your queries doesn't satisfy interaction protocol described below.
* You ask strictly more than 15 questions and your program terminated after exceeding queries limit. Please note, that you can do up to 15 ask queries and one answer query.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
If you exceed the maximum number of queries, You should terminate with 0, In this case you'll get Wrong Answer, If you don't terminate you may receive any verdict because you'll be reading from a closed stream .
Input
The first line of input will contain a single integer n (2 ≤ n ≤ 1000) — the length of the hidden binary string.
Output
To print the final answer, print "! pos0 pos1" (without quotes), where pos0 and pos1 are positions of some '0' and some '1' in the string (the string is 1-indexed). Don't forget to flush the output after printing the answer!
Interaction
To ask a question use the format "? s" (without quotes), where s is a query string. Don't forget to flush the output after printing a query!
After each query you can read a single integer from standard input — the Hamming distance between the hidden string and the query string.
To flush the output you can use:-
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages .
Hacking.
To hack someone just print one binary string with length up to 1000, containing at least one '0' and at least one '1'.
Example
Input
3
2
1
3
2
1
0
Output
? 000
? 001
? 010
? 011
? 100
? 101
! 2 1
Note
Hamming distance definition: <https://en.wikipedia.org/wiki/Hamming_distance>
In the first test case the hidden binary string is 101, The first query is 000, so the Hamming distance is 2. In the second query the hidden string is still 101 and query is 001, so the Hamming distance is 1.
After some queries you find that symbol at position 2 is '0' and symbol at position 1 is '1', so you print "! 2 1".
Submitted Solution:
```
from sys import exit
cntq = 0
def query(m):
global cntq
cntq += 1
if cntq > 15:
exit(1)
if last == 0:
s = '1' * m + '0' * (n - m)
else:
s = '0' * m + '1' * (n - m)
print('?', s)
res = int(input())
if last == '0':
res = n - res
return res
n = int(input())
last = -1
ones = query(n)
kek = ones - query(n - 1)
if kek < 0:
last = '0'
else:
last = '1'
l = 0
r = n
if last == '1':
onesl = 0
zeror = 0
while (r - l > 1):
m = (l + r) // 2
x = query(m) - onesl - zeror
kek = ones - x
a1 = (ones + x - (r - m)) // 2
b1 = ones - a1
b0 = x - a1
if b0 > 0:
l = m
ones = b1
onesl += a1
else:
r = m
ones = a1
zeror += b0
else:
zerol = 0
onesr = 0
while (r - l > 1):
m = (l + r) // 2
x = query(m) - zerol - onesr
kek = ones - kek
b1 = (ones + x - (m - l)) // 2
a1 = ones - b1
a0 = x - b1
if b1 > 0:
l = m
ones = b1
zerol += a0
else:
r = m
ones = a1
onesr += b1
s = '0011001100'
if s[l] == s[n - 1]:
exit(1)
print('!', l + 1, n)
# print('!', 1, 3)
``` | instruction | 0 | 77,287 | 18 | 154,574 |
No | output | 1 | 77,287 | 18 | 154,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mahmoud and Ehab are in the fourth stage now.
Dr. Evil has a hidden binary string of length n. He guarantees that there is at least one '0' symbol and at least one '1' symbol in it. Now he wants Mahmoud and Ehab to find a position of any '0' symbol and any '1' symbol. In order to do this, Mahmoud and Ehab can ask Dr. Evil up to 15 questions. They tell Dr. Evil some binary string of length n, and Dr. Evil tells the Hamming distance between these two strings. Hamming distance between 2 binary strings of the same length is the number of positions in which they have different symbols. You can find the definition of Hamming distance in the notes section below.
Help Mahmoud and Ehab find these two positions.
You will get Wrong Answer verdict if
* Your queries doesn't satisfy interaction protocol described below.
* You ask strictly more than 15 questions and your program terminated after exceeding queries limit. Please note, that you can do up to 15 ask queries and one answer query.
* Your final answer is not correct.
You will get Idleness Limit Exceeded if you don't print anything or if you forget to flush the output, including for the final answer (more info about flushing output below).
If you exceed the maximum number of queries, You should terminate with 0, In this case you'll get Wrong Answer, If you don't terminate you may receive any verdict because you'll be reading from a closed stream .
Input
The first line of input will contain a single integer n (2 ≤ n ≤ 1000) — the length of the hidden binary string.
Output
To print the final answer, print "! pos0 pos1" (without quotes), where pos0 and pos1 are positions of some '0' and some '1' in the string (the string is 1-indexed). Don't forget to flush the output after printing the answer!
Interaction
To ask a question use the format "? s" (without quotes), where s is a query string. Don't forget to flush the output after printing a query!
After each query you can read a single integer from standard input — the Hamming distance between the hidden string and the query string.
To flush the output you can use:-
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* See the documentation for other languages .
Hacking.
To hack someone just print one binary string with length up to 1000, containing at least one '0' and at least one '1'.
Example
Input
3
2
1
3
2
1
0
Output
? 000
? 001
? 010
? 011
? 100
? 101
! 2 1
Note
Hamming distance definition: <https://en.wikipedia.org/wiki/Hamming_distance>
In the first test case the hidden binary string is 101, The first query is 000, so the Hamming distance is 2. In the second query the hidden string is still 101 and query is 001, so the Hamming distance is 1.
After some queries you find that symbol at position 2 is '0' and symbol at position 1 is '1', so you print "! 2 1".
Submitted Solution:
```
import sys
#sstr = "1"*20+"0"
ls = int(input())#len(sstr)
def q(s):
print(s)
sys.stdout.flush()
def d(s):
a = q("? " + s)
return int(input())
#v = 0
#for i in range(min(len(sstr), len(s))):
# if s[i] != sstr[i]:
# v += 1
#return v
d0 = d("1" * ls)
l, r = 0, ls
c = d0
find1 = -1
find0 = -1
while l + 1 < r:
m = (l + r) // 2
k1, k2 = m - l, r - m
s = l * "1" + k1 * "0" + (ls - m) * "1"
ds = d(s)
c0 = (k1 + d0 - ds) / 2
a, b = c0, c - c0
if a == 0 and find1 == -1: find1 = m - 1
if b == 0 and find1 == -1: find1 = m
if a == k1 and find0 == -1: find0 = m - 1
if b == k2 - 1 and find0 == -1: find0 = m
#print("--"*10)
#print(sstr, sstr, k1, k2, l, r, m)
#print(s, m * "1" + (r - m) * "0" + (ls - r) * "1", a, b, find1, find0, k1, k2)
#print("--"*10)
if (a < b and a != 0):
r = m
c = a
else:
l = m
c = b
if find1 != -1 and find0 != -1:
break
if find0 == -1:
find0 = 0
if find1 == -1:
find1 = 0
q("! {} {}".format(find0 + 1, find1 + 1))
``` | instruction | 0 | 77,288 | 18 | 154,576 |
No | output | 1 | 77,288 | 18 | 154,577 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,453 | 18 | 154,906 |
"Correct Solution:
```
S,T,U=input().split();print(S[0].upper()+T[0].upper()+U[0].upper())
``` | output | 1 | 77,453 | 18 | 154,907 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,454 | 18 | 154,908 |
"Correct Solution:
```
s=""
for x in input().split():s+=x[0]
print(s.upper())
``` | output | 1 | 77,454 | 18 | 154,909 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,455 | 18 | 154,910 |
"Correct Solution:
```
print(''.join([_[0].upper() for _ in input().split()]))
``` | output | 1 | 77,455 | 18 | 154,911 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,456 | 18 | 154,912 |
"Correct Solution:
```
a, b, c = input().split(" ")
s = a[0]+b[0]+c[0]
print(s.upper())
``` | output | 1 | 77,456 | 18 | 154,913 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,457 | 18 | 154,914 |
"Correct Solution:
```
print(''.join([a[0] for a in input().upper().split()]))
``` | output | 1 | 77,457 | 18 | 154,915 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,458 | 18 | 154,916 |
"Correct Solution:
```
a,b,c=map(str,input().split());print((a[0]+b[0]+c[0]).upper())
``` | output | 1 | 77,458 | 18 | 154,917 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,459 | 18 | 154,918 |
"Correct Solution:
```
a,b,c=input().split()
x=(a[0]+b[0]+c[0]).upper()
print(x)
``` | output | 1 | 77,459 | 18 | 154,919 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC | instruction | 0 | 77,460 | 18 | 154,920 |
"Correct Solution:
```
x,y,z = (input().split())
print((x[:1]+y[:1]+z[:1]).upper())
``` | output | 1 | 77,460 | 18 | 154,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
a,b,c=input().split()
ans=a[0]+b[0]+c[0]
ans=ans.upper()
print(ans)
``` | instruction | 0 | 77,461 | 18 | 154,922 |
Yes | output | 1 | 77,461 | 18 | 154,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
print(''.join(map(lambda x:x[0].upper(),input().split())))
``` | instruction | 0 | 77,462 | 18 | 154,924 |
Yes | output | 1 | 77,462 | 18 | 154,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
a=input().upper().split();print(a[0][0]+a[1][0]+a[2][0])
``` | instruction | 0 | 77,463 | 18 | 154,926 |
Yes | output | 1 | 77,463 | 18 | 154,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
print(''.join([c[0] for c in input().split()]).upper())
``` | instruction | 0 | 77,464 | 18 | 154,928 |
Yes | output | 1 | 77,464 | 18 | 154,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
s1, s2, s3 = input().split()
ans = chr(ord(s1[0])-32)+chr(ord(s2[0])-32)+chr(ord(s2[0])-32)
``` | instruction | 0 | 77,465 | 18 | 154,930 |
No | output | 1 | 77,465 | 18 | 154,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
a,b,c = map(str, input().split())
print(str(a[0].uppercase)+str(b[0].uppercase)+str(c[0].uppercase))
``` | instruction | 0 | 77,466 | 18 | 154,932 |
No | output | 1 | 77,466 | 18 | 154,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
import sys
S = list(map(str,input().split()))
for I in S:
if not I.islower():
sys.exit()
if len(I) < 0 or len(I) > 10:
sys.exit()
result = ""
for J in S:
print(J[0:1])
result = result + J[0:1]
print(result.upper())
``` | instruction | 0 | 77,467 | 18 | 154,934 |
No | output | 1 | 77,467 | 18 | 154,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
Constraints
* s_1, s_2 and s_3 are composed of lowercase English letters.
* 1 ≤ |s_i| ≤ 10 (1≤i≤3)
Input
Input is given from Standard Input in the following format:
s_1 s_2 s_3
Output
Print the answer.
Examples
Input
atcoder beginner contest
Output
ABC
Input
resident register number
Output
RRN
Input
k nearest neighbor
Output
KNN
Input
async layered coding
Output
ALC
Submitted Solution:
```
x,y=map(int,input().split())
if abs(x-y)<=1:
print('Brown')
else:
print('Alice')
``` | instruction | 0 | 77,468 | 18 | 154,936 |
No | output | 1 | 77,468 | 18 | 154,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
def permut(str):
if len(str) % 2 != 0:
return str
n = len(str)
s1 = permut(str[:n//2])
s2 = permut(str[n//2:])
if s1 + s2 > s2 + s1:
return s1 + s2
else:
return s2 + s1
s = input()
t = input()
if permut(s) == permut(t):
print("YES")
else:
print("NO")
``` | instruction | 0 | 77,864 | 18 | 155,728 |
Yes | output | 1 | 77,864 | 18 | 155,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
def isEqual(a, b):
for i in range(0, len(a)):
if (a[i] != b[i]):
return False
return True
def lexicographic_minimal_string(s):
if (len(s) % 2 == 1):
return s
half = int(len(s) /2)
s1 = lexicographic_minimal_string(s[:half])
s2 = lexicographic_minimal_string(s[half:])
if s1 < s2:
return s1 + s2
return s2 + s1
a = input()
b = input()
a = lexicographic_minimal_string(a)
b = lexicographic_minimal_string(b)
if(isEqual(a,b)):
print("YES")
else:
print("NO")
'''
def isEqual(AS,BS,size):
for i in range(0, size):
if (a[AS+i] != b[BS+i]):
return False
return True
def equivalent(AS,BS,size):
global a
global b
if isEqual(AS,BS,size):
return True
half = int(size / 2)
if 2*half != size:
return False
if (equivalent(AS,BS+half,half) and equivalent(AS+half,BS,half)):
return True
if (equivalent(AS,BS,half) and equivalent(AS+half,BS+half,half)):
return True
return False
a = input()
b = input()
if equivalent(0,0,len(a)):
print("YES")
else:
print("NO")
'''
'''
a = input()
b = input()
def isEqual(a, b):
for i in range(0, len(a)):
if (a[i] != b[i]):
return False
return True
def equivalent(a, b):
if a == "":
return True
if isEqual(a, b):
return True
half = int(len(a) / 2)
if 2*half != len(a):
return False
if (equivalent(a[:half], b[:half]) and equivalent(a[half:], b[half:])):
return True
if (equivalent(a[:half], b[half:]) and equivalent(a[half:], b[:half])):
return True
return False
if equivalent(a,b):
print("YES")
else:
print("NO")
'''
``` | instruction | 0 | 77,865 | 18 | 155,730 |
Yes | output | 1 | 77,865 | 18 | 155,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
x = input()
y = input()
def f(x, y):
if x == y:
return True
if len(x)%2 == 1:
return False
mid = len(x)//2
if f(x[:mid], y[mid:]) and f(x[mid:], y[:mid]):
return True
if f(x[mid:], y[mid:]) and f(x[:mid], y[:mid]):
return True
return False
if f(x, y):
print("YES")
quit()
print("NO")
``` | instruction | 0 | 77,866 | 18 | 155,732 |
Yes | output | 1 | 77,866 | 18 | 155,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
a = input()
b = input()
def split(s):
length = len(s)
if (length % 2 == 0):
s1 = split((s[:int(length / 2)]))
s2 = split(s[int(length / 2):])
if (s1 < s2):
return s1 + s2
else:
return s2 + s1
else:
return s
if (split(a) == split(b)):
print("YES")
else:
print("NO")
``` | instruction | 0 | 77,867 | 18 | 155,734 |
Yes | output | 1 | 77,867 | 18 | 155,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
#f=open('pe.in','r')
a=str(input())
b=str(input())
if len(a)%2==1:
if a==b:
print("YES")
else:
print("NO")
quit()
x=len(a)//2
if a[:x]==b[:x] or a[:x]==b[x:]:
print("YES")
quit()
print("NO")
``` | instruction | 0 | 77,868 | 18 | 155,736 |
No | output | 1 | 77,868 | 18 | 155,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
def branch(a,b):
if a == b:
return True
if len(a) > 1 and len(b) > 1:
midA = len(a) // 2
midB = len(b) // 2
a1 = a[:midA]
a2 = a[midA:]
b1 = b[:midB]
b2 = b[midB:]
print([a1,a2,b1,b2])
return isEquivalent(a1,a2,b1,b2)
return False
def isEquivalent(a1,a2,b1,b2):
check1 = a1 == b1 and a2 == b2
check2 = a1 == b2 and a2 == b1
if check1 or check2:
return True
check11 = True
check12 = True
if a1 != b1:
check11 = branch(a1,b1)
if a2 != b2:
check12 = branch(a2,b2)
check1 = check11 and check12
check21 = True
check22 = True
if a1 != b2:
check21 = branch(a1,b2)
if a2 != b1:
check22 = branch(a2,b1)
check2 = check21 and check22
return check1 or check2
def run():
str1 = input()
str2 = input()
print(branch(str1, str2))
run()
``` | instruction | 0 | 77,869 | 18 | 155,738 |
No | output | 1 | 77,869 | 18 | 155,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
__author__ = 'tka4a'
def ifEqual(a, b):
le = len(a)
if (a == b):
return True
else:
if ((le % 2) == 0) and ifEqual(a[:le // 2], b[le // 2:]) and ifEqual(b[:le // 2], a[le // 2:]):
return True
else:
return False
a = input()
b = input()
if ifEqual(a, b):
print("YES")
else:
print("NO")
``` | instruction | 0 | 77,870 | 18 | 155,740 |
No | output | 1 | 77,870 | 18 | 155,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
Submitted Solution:
```
__author__ = 'tka4a'
a = input()
b = input()
le = len(a)
if (a == b):
print("Yes")
else:
if (le % 2 == 0) and (a[:le // 2] == b[le // 2:]) and (b[:le // 2] == a[le // 2:]):
print("YES")
else:
print("NO")
``` | instruction | 0 | 77,871 | 18 | 155,742 |
No | output | 1 | 77,871 | 18 | 155,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
def transform(s):
s = s[::-1]
s = s.replace('b', '0').replace('d', 'b').replace('0', 'd')
s = s.replace('p', '0').replace('q', 'p').replace('0', 'q')
return s
s = input()
if s == transform(s):
print('Yes')
else:
print('No')
``` | instruction | 0 | 78,151 | 18 | 156,302 |
Yes | output | 1 | 78,151 | 18 | 156,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
# -*- coding:utf-8 -*-
import sys
S = list(input())
a = list(reversed(S))
for tmp in range(len(a)):
if a[tmp] == 'b':
a[tmp] = 'd'
elif a[tmp] == 'd':
a[tmp] = 'b'
elif a[tmp] == 'p':
a[tmp] = 'q'
else:
a[tmp] = 'p'
if S[tmp] != a[tmp]:
print("No")
sys.exit()
print("Yes")
``` | instruction | 0 | 78,152 | 18 | 156,304 |
Yes | output | 1 | 78,152 | 18 | 156,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
s=input()
if len(s)%2==0:
for i in range(len(s)):
if s[i]=='p' and s[-1-i]!='q' or s[i]=='q' and s[-1-i]!='p' or s[i]=='d' and s[-1-i]!='b' or s[i]=='b' and s[-1-i]!='d':
print('No')
break
else:
print('Yes')
else:
print('No')
``` | instruction | 0 | 78,153 | 18 | 156,306 |
Yes | output | 1 | 78,153 | 18 | 156,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
mp={'b':'d','d':'b','p':'q','q':'p'};a=list(input());b=list(a[::-1]);b=[mp[x] for x in b];print("Yes" if a==b else "No");
``` | instruction | 0 | 78,154 | 18 | 156,308 |
Yes | output | 1 | 78,154 | 18 | 156,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
from collections import deque
d = {
'b': 'd',
'd': 'b',
'p': 'q',
'q': 'p'
}
s = deque(input())
if len(s) % 2:
print("No")
while len(s) > 1:
l = s.popleft()
r = s.pop()
if r != d[l]:
print("No")
quit(0)
print("Yes")
``` | instruction | 0 | 78,155 | 18 | 156,310 |
No | output | 1 | 78,155 | 18 | 156,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
S = input()
lenth = int(len(S) / 2)
cnt = 0
for i in range(lenth):
if ((S[i] == 'b' and S[- i - 1] == 'd') or
(S[i] == 'd' and S[- i - 1] == 'b') or
(S[i] == 'p' and S[- i - 1] == 'q') or
(S[i] == 'q' and S[- i - 1] == 'p')):
cnt += 1
if cnt == lenth:
print('Yes')
else:
print('No')
``` | instruction | 0 | 78,156 | 18 | 156,312 |
No | output | 1 | 78,156 | 18 | 156,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
from collections import deque
d = {
'b': 'd',
'd': 'b',
'p': 'q',
'q': 'p'
}
s = deque(input())
while len(s) > 1:
l = s.popleft()
r = s.pop()
if r != d[l]:
print("No")
quit(0)
print("Yes")
``` | instruction | 0 | 78,157 | 18 | 156,314 |
No | output | 1 | 78,157 | 18 | 156,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a mirror string.
Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S:
1. Reverse the order of the characters in S.
2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.
Constraints
* 1 \leq |S| \leq 10^5
* S consists of letters `b`, `d`, `p`, and `q`.
Input
The input is given from Standard Input in the following format:
S
Output
If S is a mirror string, print `Yes`. Otherwise, print `No`.
Examples
Input
pdbq
Output
Yes
Input
ppqb
Output
No
Submitted Solution:
```
s = list(input())
t = s[::-1]
for i in t:
if i == "p":
i = "q"
elif i == "q":
i = "p"
if i == "b":
i = "d"
elif i == "d":
i = "b"
if s == t:
ans = "Yes"
else:
ans = "No"
print(ans)
``` | instruction | 0 | 78,158 | 18 | 156,316 |
No | output | 1 | 78,158 | 18 | 156,317 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Taro received a strange email with only the number "519345213244" in the text. The email was from my cousin, who was 10 years older than me, so when I called and asked, "Oh, I sent it with a pocket bell because I was in a hurry. It's convenient. Nice to meet you!" I got it. You know this cousin, who is always busy and a little bit aggressive, and when you have no choice but to research "pager hitting" yourself, you can see that it is a method of input that prevailed in the world about 10 years ago. I understand.
In "Pokebell Strike", enter one character with two numbers, such as 11 for "A" and 15 for "O" according to the conversion table shown in Fig. 1. For example, to enter the string "Naruto", type "519345". Therefore, any letter can be entered with two numbers.
<image>
Figure 1
When mobile phones weren't widespread, high school students used this method to send messages from payphones to their friends' pagers. Some high school girls were able to pager at a tremendous speed. Recently, my cousin, who has been busy with work, has unknowingly started typing emails with a pager.
Therefore, in order to help Taro who is having a hard time deciphering every time, please write a program that converts the pager message into a character string and outputs it. However, the conversion table shown in Fig. 2 is used for conversion, and only lowercase letters, ".", "?", "!", And blanks are targeted. Output NA for messages that contain characters that cannot be converted.
<image>
Figure 2
Input
Multiple messages are given. One message (up to 200 characters) is given on each line. The total number of messages does not exceed 50.
Output
For each message, output the converted message or NA on one line.
Example
Input
341143514535
314
143565553551655311343411652235654535651124615163
551544654451431564
4
3411
6363636363
153414
Output
naruto
NA
do you wanna go to aizu?
yes sure!
NA
na
?????
end | instruction | 0 | 78,175 | 18 | 156,350 |
"Correct Solution:
```
def message(numbers):
numbers=str(numbers)
message=""
while(len(numbers)>0):
num=numbers[:2]
if len(num)==1:
return "NA"
if int(num[0])>6 or int(num[0])<=0:
return "NA"
if int(num[1])>5 or int(num[1])<=0:
return "NA"
code=(int(num[0])-1)*5+int(num[1])
if code>30:
return "NA"
if code==27:
message+="."
elif code==28:
message+="?"
elif code==29:
message+="!"
elif code==30:
message+=" "
else:
message+=chr(code-1+ord('a'))
numbers=numbers[2::]
return message
while True:
try:
n=input()
print(message(n))
except EOFError:
break
``` | output | 1 | 78,175 | 18 | 156,351 |
Provide a correct Python 3 solution for this coding contest problem.
One day, Taro received a strange email with only the number "519345213244" in the text. The email was from my cousin, who was 10 years older than me, so when I called and asked, "Oh, I sent it with a pocket bell because I was in a hurry. It's convenient. Nice to meet you!" I got it. You know this cousin, who is always busy and a little bit aggressive, and when you have no choice but to research "pager hitting" yourself, you can see that it is a method of input that prevailed in the world about 10 years ago. I understand.
In "Pokebell Strike", enter one character with two numbers, such as 11 for "A" and 15 for "O" according to the conversion table shown in Fig. 1. For example, to enter the string "Naruto", type "519345". Therefore, any letter can be entered with two numbers.
<image>
Figure 1
When mobile phones weren't widespread, high school students used this method to send messages from payphones to their friends' pagers. Some high school girls were able to pager at a tremendous speed. Recently, my cousin, who has been busy with work, has unknowingly started typing emails with a pager.
Therefore, in order to help Taro who is having a hard time deciphering every time, please write a program that converts the pager message into a character string and outputs it. However, the conversion table shown in Fig. 2 is used for conversion, and only lowercase letters, ".", "?", "!", And blanks are targeted. Output NA for messages that contain characters that cannot be converted.
<image>
Figure 2
Input
Multiple messages are given. One message (up to 200 characters) is given on each line. The total number of messages does not exceed 50.
Output
For each message, output the converted message or NA on one line.
Example
Input
341143514535
314
143565553551655311343411652235654535651124615163
551544654451431564
4
3411
6363636363
153414
Output
naruto
NA
do you wanna go to aizu?
yes sure!
NA
na
?????
end | instruction | 0 | 78,176 | 18 | 156,352 |
"Correct Solution:
```
# AOJ 0127 Pocket Pager Input
# Python3 2018.6.18 bal4u
a = ['']*100
a[11], a[12], a[13], a[14], a[15] = 'a', 'b', 'c', 'd', 'e'
a[21], a[22], a[23], a[24], a[25] = 'f', 'g', 'h', 'i', 'j'
a[31], a[32], a[33], a[34], a[35] = 'k', 'l', 'm', 'n', 'o'
a[41], a[42], a[43], a[44], a[45] = 'p', 'q', 'r', 's', 't'
a[51], a[52], a[53], a[54], a[55] = 'u', 'v', 'w', 'x', 'y'
a[61], a[62], a[63], a[64], a[65] = 'z', '.', '?', '!', ' '
while True:
try: buf = list(input())
except: break
f = True
ans = ''
for i in range(0, len(buf), 2):
if i+1 >= len(buf):
f = False
break
if buf[i].isdigit() and buf[i+1].isdigit():
k = int(buf[i]+buf[i+1])
if a[k] != '': ans += a[k]
else:
f = False
break
print(ans if f else "NA")
``` | output | 1 | 78,176 | 18 | 156,353 |
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