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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters.
Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way.
You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.
Constraints
* 2 ≤ |s| ≤ 100
* All characters in s are uppercase English letters (`A`-`Z`).
Input
The input is given from Standard Input in the following format:
s
Output
Print `Yes` if the string `CF` can be obtained from the string s by deleting some characters. Otherwise print `No`.
Examples
Input
CODEFESTIVAL
Output
Yes
Input
FESTIVALCODE
Output
No
Input
CF
Output
Yes
Input
FCF
Output
Yes
Submitted Solution:
```
str = raw_input()
c = false
f = false
for i in range(0, len(str)):
if str[i] == "c":
c = true
if str[i] == "f" and c == true:
return true
return false
``` | instruction | 0 | 47,837 | 18 | 95,674 |
No | output | 1 | 47,837 | 18 | 95,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem J String Puzzle
Amazing Coding Magazine is popular among young programmers for its puzzle solving contests offering catchy digital gadgets as the prizes. The magazine for programmers naturally encourages the readers to solve the puzzles by writing programs. Let's give it a try!
The puzzle in the latest issue is on deciding some of the letters in a string (the secret string, in what follows) based on a variety of hints. The figure below depicts an example of the given hints.
<image>
The first hint is the number of letters in the secret string. In the example of the figure above, it is nine, and the nine boxes correspond to nine letters. The letter positions (boxes) are numbered starting from 1, from the left to the right.
The hints of the next kind simply tell the letters in the secret string at some speci c positions. In the example, the hints tell that the letters in the 3rd, 4th, 7th, and 9th boxes are C, I, C, and P
The hints of the final kind are on duplicated substrings in the secret string. The bar immediately below the boxes in the figure is partitioned into some sections corresponding to substrings of the secret string. Each of the sections may be connected by a line running to the left with another bar also showing an extent of a substring. Each of the connected pairs indicates that substrings of the two extents are identical. One of this kind of hints in the example tells that the letters in boxes 8 and 9 are identical to those in boxes 4 and 5, respectively. From this, you can easily deduce that the substring is IP.
Note that, not necessarily all of the identical substring pairs in the secret string are given in the hints; some identical substring pairs may not be mentioned.
Note also that two extents of a pair may overlap each other. In the example, the two-letter substring in boxes 2 and 3 is told to be identical to one in boxes 1 and 2, and these two extents share the box 2.
In this example, you can decide letters at all the positions of the secret string, which are "CCCIPCCIP". In general, the hints may not be enough to decide all the letters in the secret string.
The answer of the puzzle should be letters at the specified positions of the secret string. When the letter at the position specified cannot be decided with the given hints, the symbol ? should be answered.
Input
The input consists of a single test case in the following format.
$n$ $a$ $b$ $q$
$x_1$ $c_1$
...
$x_a$ $c_a$
$y_1$ $h_1$
...
$y_b$ $h_b$
$z_1$
...
$z_q$
The first line contains four integers $n$, $a$, $b$, and $q$. $n$ ($1 \leq n \leq 10^9$) is the length of the secret string, $a$ ($0 \leq a \leq 1000$) is the number of the hints on letters in specified positions, $b$ ($0 \leq b \leq 1000$) is the number of the hints on duplicated substrings, and $q$ ($1 \leq q \leq 1000$) is the number of positions asked.
The $i$-th line of the following a lines contains an integer $x_i$ and an uppercase letter $c_i$ meaning that the letter at the position $x_i$ of the secret string is $c_i$. These hints are ordered in their positions, i.e., $1 \leq x_1 < ... < x_a \leq n$.
The $i$-th line of the following $b$ lines contains two integers, $y_i$ and $h_i$. It is guaranteed that they satisfy $2 \leq y_1 < ... < y_b \leq n$ and $0 \leq h_i < y_i$. When $h_i$ is not 0, the substring of the secret string starting from the position $y_i$ with the length $y_{i+1} - y_i$ (or $n+1-y_i$ when $i = b$) is identical to the substring with the same length starting from the position $h_i$. Lines with $h_i = 0$ does not tell any hints except that $y_i$ in the line indicates the end of the substring specified in the line immediately above.
Each of the following $q$ lines has an integer $z_i$ ($1 \leq z_i \leq n$), specifying the position of the letter in the secret string to output.
It is ensured that there exists at least one secret string that matches all the given information. In other words, the given hints have no contradiction.
Output
The output should be a single line consisting only of $q$ characters. The character at position $i$ of the output should be the letter at position $z_i$ of the the secret string if it is uniquely determined from the hints, or ? otherwise.
Sample Input 1
9 4 5 4
3 C
4 I
7 C
9 P
2 1
4 0
6 2
7 0
8 4
8
1
9
6
Sample Output 1
ICPC
Sample Input 2
1000000000 1 1 2
20171217 A
3 1
42
987654321
Sample Output 2
?A
Example
Input
9 4 5 4
3 C
4 I
7 C
9 P
2 1
4 0
6 2
7 0
8 4
8
1
9
6
Output
ICPC
Submitted Solution:
```
from bisect import bisect
n, a, b, q = map(int, input().split())
W = [input().split() for i in range(a)]
X = [int(x) for x, c in W]
C = [c for x, c in W]
P = [list(map(int, input().split())) for i in range(b)]
Y = [y for y, h in P] + [n+1]
D = [0]*b
for i in range(b):
y0, h = P[i]
y1 = Y[i+1]
l = y1 - y0
if y0 < h + l:
D[i] = y0 - h
else:
D[i] = l
Z = [int(input()) for i in range(q)]
idx = 0
S = {}
for i in range(a):
x = X[i]; c = C[i]
S[x] = c
if x < Y[0]:
continue
while Y[idx+1] < x: idx += 1
i = idx
while Y[0] <= x:
while x < Y[i]: i -= 1
y0, h = P[i]; y1 = Y[i+1]
if h == 0: break
x = h + ((x - y0) % D[i])
S[x] = c
ans = []
for z in Z:
if z not in S:
i = bisect(Y, z)-1
while Y[0] <= z:
while z < Y[i]: i -= 1
y0, h = P[i]; y1 = Y[i+1]
if h == 0: break
z = h + ((z - y0) % D[i])
if z in S:
break
if z in S:
ans.append(S[z])
else:
ans.append('?')
print(*ans, sep='')
``` | instruction | 0 | 47,894 | 18 | 95,788 |
No | output | 1 | 47,894 | 18 | 95,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem J String Puzzle
Amazing Coding Magazine is popular among young programmers for its puzzle solving contests offering catchy digital gadgets as the prizes. The magazine for programmers naturally encourages the readers to solve the puzzles by writing programs. Let's give it a try!
The puzzle in the latest issue is on deciding some of the letters in a string (the secret string, in what follows) based on a variety of hints. The figure below depicts an example of the given hints.
<image>
The first hint is the number of letters in the secret string. In the example of the figure above, it is nine, and the nine boxes correspond to nine letters. The letter positions (boxes) are numbered starting from 1, from the left to the right.
The hints of the next kind simply tell the letters in the secret string at some speci c positions. In the example, the hints tell that the letters in the 3rd, 4th, 7th, and 9th boxes are C, I, C, and P
The hints of the final kind are on duplicated substrings in the secret string. The bar immediately below the boxes in the figure is partitioned into some sections corresponding to substrings of the secret string. Each of the sections may be connected by a line running to the left with another bar also showing an extent of a substring. Each of the connected pairs indicates that substrings of the two extents are identical. One of this kind of hints in the example tells that the letters in boxes 8 and 9 are identical to those in boxes 4 and 5, respectively. From this, you can easily deduce that the substring is IP.
Note that, not necessarily all of the identical substring pairs in the secret string are given in the hints; some identical substring pairs may not be mentioned.
Note also that two extents of a pair may overlap each other. In the example, the two-letter substring in boxes 2 and 3 is told to be identical to one in boxes 1 and 2, and these two extents share the box 2.
In this example, you can decide letters at all the positions of the secret string, which are "CCCIPCCIP". In general, the hints may not be enough to decide all the letters in the secret string.
The answer of the puzzle should be letters at the specified positions of the secret string. When the letter at the position specified cannot be decided with the given hints, the symbol ? should be answered.
Input
The input consists of a single test case in the following format.
$n$ $a$ $b$ $q$
$x_1$ $c_1$
...
$x_a$ $c_a$
$y_1$ $h_1$
...
$y_b$ $h_b$
$z_1$
...
$z_q$
The first line contains four integers $n$, $a$, $b$, and $q$. $n$ ($1 \leq n \leq 10^9$) is the length of the secret string, $a$ ($0 \leq a \leq 1000$) is the number of the hints on letters in specified positions, $b$ ($0 \leq b \leq 1000$) is the number of the hints on duplicated substrings, and $q$ ($1 \leq q \leq 1000$) is the number of positions asked.
The $i$-th line of the following a lines contains an integer $x_i$ and an uppercase letter $c_i$ meaning that the letter at the position $x_i$ of the secret string is $c_i$. These hints are ordered in their positions, i.e., $1 \leq x_1 < ... < x_a \leq n$.
The $i$-th line of the following $b$ lines contains two integers, $y_i$ and $h_i$. It is guaranteed that they satisfy $2 \leq y_1 < ... < y_b \leq n$ and $0 \leq h_i < y_i$. When $h_i$ is not 0, the substring of the secret string starting from the position $y_i$ with the length $y_{i+1} - y_i$ (or $n+1-y_i$ when $i = b$) is identical to the substring with the same length starting from the position $h_i$. Lines with $h_i = 0$ does not tell any hints except that $y_i$ in the line indicates the end of the substring specified in the line immediately above.
Each of the following $q$ lines has an integer $z_i$ ($1 \leq z_i \leq n$), specifying the position of the letter in the secret string to output.
It is ensured that there exists at least one secret string that matches all the given information. In other words, the given hints have no contradiction.
Output
The output should be a single line consisting only of $q$ characters. The character at position $i$ of the output should be the letter at position $z_i$ of the the secret string if it is uniquely determined from the hints, or ? otherwise.
Sample Input 1
9 4 5 4
3 C
4 I
7 C
9 P
2 1
4 0
6 2
7 0
8 4
8
1
9
6
Sample Output 1
ICPC
Sample Input 2
1000000000 1 1 2
20171217 A
3 1
42
987654321
Sample Output 2
?A
Example
Input
9 4 5 4
3 C
4 I
7 C
9 P
2 1
4 0
6 2
7 0
8 4
8
1
9
6
Output
ICPC
Submitted Solution:
```
from bisect import bisect
n, a, b, q = map(int, input().split())
W = [input().split() for i in range(a)]
X = [int(x) for x, c in W]
C = [c for x, c in W]
P = [list(map(int, input().split())) for i in range(b)]
Y = [y for y, h in P] + [n+1]
D = [0]*b
for i in range(b):
y0, h = P[i]
y1 = Y[i+1]
l = y1 - y0
if y0 < h + l:
D[i] = y0 - h
else:
D[i] = l
Z = [int(input()) for i in range(q)]
idx = 0
S = {}
for i in range(a):
x = X[i]; c = C[i]
S[x] = c
while Y[idx+1] < x: idx += 1
i = idx
while Y[0] <= x:
while x < Y[i]: i -= 1
y0, h = P[i]; y1 = Y[i+1]
if h == 0: break
x = h + ((x - y0) % D[i])
S[x] = c
ans = []
for z in Z:
if z not in S:
i = bisect(Y, z)
while Y[0] <= z:
while z < Y[i]: i -= 1
y0, h = P[i]; y1 = Y[i+1]
if h == 0: break
z = h + ((z - y0) % D[i])
if z in S:
break
if z in S:
ans.append(S[z])
else:
ans.append('?')
print(*ans, sep='')
``` | instruction | 0 | 47,895 | 18 | 95,790 |
No | output | 1 | 47,895 | 18 | 95,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem J String Puzzle
Amazing Coding Magazine is popular among young programmers for its puzzle solving contests offering catchy digital gadgets as the prizes. The magazine for programmers naturally encourages the readers to solve the puzzles by writing programs. Let's give it a try!
The puzzle in the latest issue is on deciding some of the letters in a string (the secret string, in what follows) based on a variety of hints. The figure below depicts an example of the given hints.
<image>
The first hint is the number of letters in the secret string. In the example of the figure above, it is nine, and the nine boxes correspond to nine letters. The letter positions (boxes) are numbered starting from 1, from the left to the right.
The hints of the next kind simply tell the letters in the secret string at some speci c positions. In the example, the hints tell that the letters in the 3rd, 4th, 7th, and 9th boxes are C, I, C, and P
The hints of the final kind are on duplicated substrings in the secret string. The bar immediately below the boxes in the figure is partitioned into some sections corresponding to substrings of the secret string. Each of the sections may be connected by a line running to the left with another bar also showing an extent of a substring. Each of the connected pairs indicates that substrings of the two extents are identical. One of this kind of hints in the example tells that the letters in boxes 8 and 9 are identical to those in boxes 4 and 5, respectively. From this, you can easily deduce that the substring is IP.
Note that, not necessarily all of the identical substring pairs in the secret string are given in the hints; some identical substring pairs may not be mentioned.
Note also that two extents of a pair may overlap each other. In the example, the two-letter substring in boxes 2 and 3 is told to be identical to one in boxes 1 and 2, and these two extents share the box 2.
In this example, you can decide letters at all the positions of the secret string, which are "CCCIPCCIP". In general, the hints may not be enough to decide all the letters in the secret string.
The answer of the puzzle should be letters at the specified positions of the secret string. When the letter at the position specified cannot be decided with the given hints, the symbol ? should be answered.
Input
The input consists of a single test case in the following format.
$n$ $a$ $b$ $q$
$x_1$ $c_1$
...
$x_a$ $c_a$
$y_1$ $h_1$
...
$y_b$ $h_b$
$z_1$
...
$z_q$
The first line contains four integers $n$, $a$, $b$, and $q$. $n$ ($1 \leq n \leq 10^9$) is the length of the secret string, $a$ ($0 \leq a \leq 1000$) is the number of the hints on letters in specified positions, $b$ ($0 \leq b \leq 1000$) is the number of the hints on duplicated substrings, and $q$ ($1 \leq q \leq 1000$) is the number of positions asked.
The $i$-th line of the following a lines contains an integer $x_i$ and an uppercase letter $c_i$ meaning that the letter at the position $x_i$ of the secret string is $c_i$. These hints are ordered in their positions, i.e., $1 \leq x_1 < ... < x_a \leq n$.
The $i$-th line of the following $b$ lines contains two integers, $y_i$ and $h_i$. It is guaranteed that they satisfy $2 \leq y_1 < ... < y_b \leq n$ and $0 \leq h_i < y_i$. When $h_i$ is not 0, the substring of the secret string starting from the position $y_i$ with the length $y_{i+1} - y_i$ (or $n+1-y_i$ when $i = b$) is identical to the substring with the same length starting from the position $h_i$. Lines with $h_i = 0$ does not tell any hints except that $y_i$ in the line indicates the end of the substring specified in the line immediately above.
Each of the following $q$ lines has an integer $z_i$ ($1 \leq z_i \leq n$), specifying the position of the letter in the secret string to output.
It is ensured that there exists at least one secret string that matches all the given information. In other words, the given hints have no contradiction.
Output
The output should be a single line consisting only of $q$ characters. The character at position $i$ of the output should be the letter at position $z_i$ of the the secret string if it is uniquely determined from the hints, or ? otherwise.
Sample Input 1
9 4 5 4
3 C
4 I
7 C
9 P
2 1
4 0
6 2
7 0
8 4
8
1
9
6
Sample Output 1
ICPC
Sample Input 2
1000000000 1 1 2
20171217 A
3 1
42
987654321
Sample Output 2
?A
Example
Input
9 4 5 4
3 C
4 I
7 C
9 P
2 1
4 0
6 2
7 0
8 4
8
1
9
6
Output
ICPC
Submitted Solution:
```
from bisect import bisect
n, a, b, q = map(int, input().split())
W = [input().split() for i in range(a)]
X = [int(x) for x, c in W]
C = [c for x, c in W]
P = [list(map(int, input().split())) for i in range(b)]
Y = [y for y, h in P] + [n+1]
D = [0]*b
for i in range(b):
y0, h = P[i]; y1 = Y[i+1]
l = y1 - y0
D[i] = min(y0 - h, l)
Z = [int(input()) for i in range(q)]
idx = 0
S = {}
for i in range(a):
x = X[i]; c = C[i]
S[x] = c
if x < Y[0]:
continue
while Y[idx+1] < x: idx += 1
i = idx
while Y[0] <= x:
while x < Y[i]: i -= 1
y0, h = P[i]; y1 = Y[i+1]
if h == 0: break
x = h + ((x - y0) % D[i])
assert x < y0
S[x] = c
def check(z):
if z in S:
return S[z]
i = bisect(Y, z)-1
while Y[0] <= z:
while z < Y[i]: i -= 1
y0, h = P[i]; y1 = Y[i+1]
if h == 0: break
z = h + ((z - y0) % D[i])
assert z < y0
if z in S:
return S[z]
return '?'
print(*map(check, Z), sep='')
``` | instruction | 0 | 47,896 | 18 | 95,792 |
No | output | 1 | 47,896 | 18 | 95,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
a = ''.join(c for c in input().lower() if c not in '-_;')
b = ''.join(c for c in input().lower() if c not in '-_;')
c = ''.join(c for c in input().lower() if c not in '-_;')
r = [a + b + c, a + c + b, b + a + c, b + c + a, c + a + b, c + b + a]
n = int(input())
for i in range(n):
s = ''.join(c for c in input().lower() if c not in '-_;')
if s in r:
print('ACC')
else:
print('WA')
``` | instruction | 0 | 48,374 | 18 | 96,748 |
Yes | output | 1 | 48,374 | 18 | 96,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
import itertools
def remove_signs(s: str, signs: list) -> str:
for i in range(len(signs)):
s = s.replace(signs[i], '')
return s
def get_permulation_of_strings(s1, s2, s3):
return list(map("".join, itertools.permutations([s1, s2, s3])))
def main():
signs = [';', '_', '-']
str1 = remove_signs(input(), signs).lower()
str2 = remove_signs(input(), signs).lower()
str3 = remove_signs(input(), signs).lower()
n = int(input())
str_permulation = get_permulation_of_strings(str1, str2, str3)
for i in range(n):
is_true = False
str_concat = remove_signs(input(), signs).lower()
for j in range(len(str_permulation)):
if str_permulation[j] == str_concat:
is_true = True
break
if is_true:
print('ACC')
else:
print('WA')
if __name__ == "__main__":
main()
``` | instruction | 0 | 48,375 | 18 | 96,750 |
Yes | output | 1 | 48,375 | 18 | 96,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
def deleteSign(s):
res = ""
for x in s:
if x.isalpha():
x = x.upper()
res += x
return res
a = input()
b = input()
c = input()
n = int(input())
a = deleteSign(a)
b = deleteSign(b)
c = deleteSign(c)
abc = a + b + c
acb = a + c + b
bac = b + a + c
bca = b + c + a
cab = c + a + b
cba = c + b + a;
for i in range(n):
s = input()
s = deleteSign(s)
if s == abc or s == acb or s == bac or s == bca or s == cab or s == cba:
print("ACC")
else:
print("WA")
``` | instruction | 0 | 48,376 | 18 | 96,752 |
Yes | output | 1 | 48,376 | 18 | 96,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
S = [input() for i in range(3)]
for i in range(3):
s = S[i]
s = s.replace('-', '')
s = s.replace(';', '')
s = s.replace('_', '')
s = s.lower()
S[i] = s
T = set()
import itertools
for p in itertools.permutations(range(3), 3):
t = S[p[0]]+S[p[1]]+S[p[2]]
T.add(t)
n = int(input())
for i in range(n):
s = str(input())
s = s.replace('-', '')
s = s.replace(';', '')
s = s.replace('_', '')
s = s.lower()
if s in T:
print('ACC')
else:
print('WA')
``` | instruction | 0 | 48,377 | 18 | 96,754 |
Yes | output | 1 | 48,377 | 18 | 96,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
from itertools import permutations
def normalizeString(s):
s1 = "";
for ch in s:
if ch.isalpha():
s1 = s1 + ch.lower()
return s1
initial_string = []
total_len = 0
perms = list(permutations([1,2,3]))
for i in range(3):
initial_string.append(normalizeString(input()))
concatStrings = []
for perm in perms:
concatString = ""
for i in perm:
concatString += initial_string[i -1]
concatStrings.append(concatString)
print(concatStrings)
n = int(input())
for i in range(n):
s = normalizeString(input())
if s in concatStrings:
print("ACC")
else:
print("WA")
``` | instruction | 0 | 48,378 | 18 | 96,756 |
No | output | 1 | 48,378 | 18 | 96,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
import string
tokens = set()
ans = []
maxLen = 0
for i in range(3):
tmp = input()
t = ''
for c in tmp:
if c == '-' or c == '_' or c == ';':
continue
t += c.lower()
if len(t) > maxLen:
maxLen = len(t)
tokens.add(t)
n = int(input())
for i in range(n):
ans.append(input())
for a in ans:
tmp = ''
cnt = 0
for c in a:
if c == '-' or c == '_' or c == ';':
continue
tmp += c.lower()
if tmp in tokens:
cnt += 1
tmp = ''
if cnt == 3:
print('ACC')
cnt = 0
break
elif len(tmp) > maxLen:
print('WA')
break
``` | instruction | 0 | 48,379 | 18 | 96,758 |
No | output | 1 | 48,379 | 18 | 96,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
import re
def ignore_sign(the_string):
the_string = the_string.replace('-', '').replace('_', '').replace(';', '')
return the_string.lower()
words = []
answers = []
for i in range(3):
words.append(ignore_sign(input()))
num = int(input())
for i in range(num):
current_str = ignore_sign(input())
for j in range(3):
current_str = current_str.replace(words[j], '*')
if current_str == '***':
answers.append("ACC")
else:
answers.append("WA")
print("\n".join(answers))
``` | instruction | 0 | 48,380 | 18 | 96,760 |
No | output | 1 | 48,380 | 18 | 96,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his n students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
* As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin.
* As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs.
* The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600
* My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes.
* Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN
* You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer.
* I should remind you that none of the strings (initial strings or answers) are empty.
* Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer n (0 ≤ n ≤ 1000), the number of students.
Next n lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Examples
Input
Iran_
Persian;
W_o;n;d;e;r;f;u;l;
7
WonderfulPersianIran
wonderful_PersIAN_IRAN;;_
WONDERFUL___IRAN__PERSIAN__;;
Ira__Persiann__Wonderful
Wonder;;fulPersian___;I;r;a;n;
__________IranPersianWonderful__________
PersianIran_is_Wonderful
Output
ACC
ACC
ACC
WA
ACC
ACC
WA
Input
Shapur;;
is___
a_genius
3
Shapur__a_is___geniUs
is___shapur___a__Genius;
Shapur;;is;;a;;geni;;us;;
Output
WA
ACC
ACC
Submitted Solution:
```
teacher_string = []
for i in range(3):
temp = ""
teacher_string.append(input().strip())
for i in teacher_string[-1]:
if i.isalpha():
temp = temp + i if i.islower() else temp + i.lower()
teacher_string[-1] = temp
n = int(input().strip())
student_string = []
for i in range(n):
temp = ""
student_string.append(input().strip())
for i in student_string[-1]:
if i.isalpha():
temp = temp + i if i.islower() else temp + i.lower()
student_string[-1] = temp
for string in student_string:
yes = "ACC"
temp = string
for st in teacher_string:
pos = temp.find(st)
if pos != -1:
temp = temp[:pos]+","+temp[pos+len(st):]
else:
yes = "WA"
break
print(yes)
``` | instruction | 0 | 48,381 | 18 | 96,762 |
No | output | 1 | 48,381 | 18 | 96,763 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
class BIT:
def __init__(self, n):
self.n = n
self.bit = [0] * (n + 1)
def _sum(self, i):
s = 0
while i > 0:
s += self.bit[i]
i -= i & -i
return s
def add(self, i, val):
"""i番目にvalを加える"""
i = i + 1
while i <= self.n:
self.bit[i] += val
i += i & -i
def get_sum(self, l, r):
"""区間[l, r)の和を求める"""
return self._sum(r) - self._sum(l)
def swap_time(array0, array1):
n = len(array0)
memo = {}
for i in range(n)[::-1]:
if array0[i] not in memo:
memo[array0[i]] = []
memo[array0[i]].append(i)
res = []
for val in array1:
ind = memo[val].pop()
res.append(ind)
bit = BIT(n)
ans = 0
for i in range(n):
bit.add(res[i], 1)
ans += bit.get_sum(res[i] + 1, n)
return ans
ALPH = "abcdefghijklmnopqrstuvwxyz"
s = input()
cnt_memo = {}
for char in s:
if char not in cnt_memo:
cnt_memo[char] = 1
else:
cnt_memo[char] += 1
odd = ""
for char in cnt_memo:
if cnt_memo[char] % 2 == 1:
if odd:
print(-1)
exit()
else:
odd = char
cnt_memo[char] -= 1
cnt_memo[char] //= 2
else:
cnt_memo[char] //= 2
left = []
mid = []
if odd:
mid = [odd]
right = []
for char in s:
if odd == char and cnt_memo[char] == 0:
odd = ""
elif cnt_memo[char] != 0:
left.append(char)
cnt_memo[char] -= 1
else:
right.append(char)
ans = 0
ans += swap_time(left + mid + right, list(s))
ans += swap_time(left, right[::-1])
print(ans)
``` | instruction | 0 | 48,646 | 18 | 97,292 |
Yes | output | 1 | 48,646 | 18 | 97,293 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
import sys
class BIT():
def __init__(self,number):
self.n=number
self.list=[0]*(number+1)
def add(self,i,x):#ith added x 1indexed
while i<=self.n:
self.list[i]+=x
i+=i&-i
def search(self,i):#1-i sum
s=0
while i>0:
s+=self.list[i]
i-=i&-i
return s
def suma(self,i,j):#i,i+1,..j sum
return self.search(j)-self.search(i-1)
#from collections import defaultdict
S=input()
N = len(S)
L = 26
a = ord('a')
d = [[] for i in range(L)]
for i in range(N):
s=ord(S[i])-a
d[s].append(i)
flag=0
for i in range(L):
if len(d[i])%2==1:
flag+=1
if flag>1:
print(-1)
sys.exit()
Suc=[-1]*N
pairs=[]
for i in range(L):
T=len(d[i])
for s in range((T//2)):
li,ri=d[i][s],d[i][-s-1]
pairs.append((li,ri))
if T%2==1:
Suc[d[i][T//2]]=(N//2)+1
pairs.sort()
for i, (li,ri) in enumerate(pairs):
Suc[li]=i+1
Suc[ri]=N-i
Tree=BIT(N+3)
ans=0
for i,m in enumerate(Suc):
ans+=i-Tree.search(m)
Tree.add(m,1)
#ans+=Tree.search(N+1)
print(ans)
``` | instruction | 0 | 48,647 | 18 | 97,294 |
Yes | output | 1 | 48,647 | 18 | 97,295 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
import os
import sys
from collections import Counter
class BinaryIndexedTree:
# http://hos.ac/slides/20140319_bit.pdf
def __init__(self, size):
"""
:param int size:
"""
self._bit = [0] * size
self._size = size
def add(self, i, w):
"""
i 番目に w を加える
:param int i:
:param int w:
"""
x = i + 1
while x <= self._size:
self._bit[x - 1] += w
x += x & -x
def sum(self, i):
"""
[0, i) の合計
:param int i:
"""
ret = 0
while i > 0:
ret += self._bit[i - 1]
i -= i & -i
return ret
def __len__(self):
return self._size
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353
S = sys.stdin.buffer.readline().decode().rstrip()
counter = Counter(S)
# 回文にできるかどうかはかんたんに判定できる
if len(S) % 2 == 0:
ok = True
for c, cnt in counter.items():
ok &= cnt % 2 == 0
else:
odds = 0
for c, cnt in counter.items():
odds += cnt % 2 == 1
ok = odds == 1
if not ok:
print(-1)
exit()
# 左右に寄せる。
# 左側か右側かだけで転倒数を数える
counts = [0] * 26
ls = ''
mid = ''
rs = ''
invs = 0
for c in S:
n = ord(c) - ord('a')
counts[n] += 1
if counter[c] & 1:
if counts[n] <= counter[c] // 2:
ls += c
invs += len(rs) + len(mid)
elif counts[n] == counter[c] // 2 + 1:
mid += c
invs += len(rs)
else:
rs += c
else:
if counts[n] <= counter[c] // 2:
ls += c
invs += len(rs) + len(mid)
else:
rs += c
rs = rs[::-1]
# 目的の回文への置換回数と同じ回数で ls -> rs に置換できる
l_idx = [[] for _ in range(26)]
for i, c in enumerate(ls):
n = ord(c) - ord('a')
l_idx[n].append(i)
counts = [0] * 26
idx = []
for i, c in enumerate(rs):
n = ord(c) - ord('a')
idx.append(l_idx[n][counts[n]])
counts[n] += 1
# idx を昇順にする == ls と同じ並びにするまでの回数
invs2 = 0
bit = BinaryIndexedTree(size=len(ls))
for i in reversed(idx):
invs2 += bit.sum(i)
bit.add(i, 1)
ans = invs + invs2
# print(invs, invs2)
# print(ls, rs)
# print(idx)
print(ans)
``` | instruction | 0 | 48,648 | 18 | 97,296 |
Yes | output | 1 | 48,648 | 18 | 97,297 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
from collections import Counter
class BIT:
def __init__(self,n):
self.tree = [0]*(n+1)
self.size = n
def sum(self,i):
s = 0
while i > 0:
s += self.tree[i]
i -= i&-i
return s
def add(self,i,x):
while i <= self.size:
self.tree[i] += x
i += i&-i
S = input()
N = len(S)
count = Counter(S)
flag = False
for v in count.values():
if v%2 == 1:
if flag:
print(-1)
exit()
else:
flag = True
"""
文字sのi文字目がどこにあるか、をO(1)で取り出せる辞書が必要
"""
placeAtS = {s:[] for s in list(set(S))}
for i in range(N):
s = S[i]
placeAtS[s].append(i)
count2 = {s:0 for s in list(set(S))}
placeAtT = [None]*(N)
tmp = 0
for i in range(N):
s = S[i]
count2[s] += 1
if count2[s] <= count[s]//2:
placeAtT[i] = tmp
mirror = placeAtS[s][count[s]-count2[s]]
placeAtT[mirror] = N - tmp - 1
tmp += 1
for i in range(N):
if placeAtT[i] == None:
placeAtT[i] = tmp
bit = BIT(N)
ans = 0
for i in range(N):
bit.add(placeAtT[i]+1,1)
ans += bit.sum(N) - bit.sum(placeAtT[i]+1)
print(ans)
``` | instruction | 0 | 48,649 | 18 | 97,298 |
Yes | output | 1 | 48,649 | 18 | 97,299 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
S=input()
S_lst=list(S)
counter=Counter(S_lst)
margin=len(S)%2
for k, v in counter.items():
if v%2==1:
margin -= 1
if not(margin == 0):
print(-1)
exit()
num_lst=[] # number notation of S
num_ddct=defaultdict(lambda: []) # remember number for palindrome
first_dct={}
thres_dct={}
for k,_ in counter.items():
first_dct[k]=True
thres_dct[k]=math.floor(counter[k]/2)
pos=0
for s in S_lst:
if first_dct[s]: # so far first part
if len(num_ddct[s]) < thres_dct[s]: # still in first part
num_lst.append(pos)
num_ddct[s].append(pos)
pos += 1
else: # first time to reach reverse part
first_dct[s]=False # its not first anymore
if counter[s]%2==0: # usual case
num_lst.append(len(S_lst) - num_ddct[s][-1] - 1)
num_ddct[s].pop()
else: # if center
num_lst.append( int((len(S_lst) - 1) / 2 ) )
else: # already in second part
num_lst.append(len(S_lst) - num_ddct[s][-1] - 1)
num_ddct[s].pop()
print(-1)
``` | instruction | 0 | 48,650 | 18 | 97,300 |
No | output | 1 | 48,650 | 18 | 97,301 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
from collections import Counter
text = input()
count = Counter()
def change_count(text):
N = len(text)
origin = 0
# print(text)
if N <= 2:
return 0
elif text[0] == text[N-1]:
return change_count(text[1:N-1])
else:
if count[text[0]] % 2 == 1:
origin = N-1
start = 1
# print(origin)
while text[origin] != text[start]:
start = start + 1
end = N-2
while text[origin] != text[end]:
end = end - 1
count[text[origin]] = count[text[origin]]-2
# print("start:"+ str(start) + " end:"+str(end))
if origin == 0:
if start >= end :
return (N-1)-start + change_count(text[1:start]+text[start+1:])
else:
return (N-1)-end + change_count(text[1:end]+text[end+1:])
else:
if start < end :
return start + change_count(text[0:start]+text[start+1:N-1])
else:
return end + change_count(text[0:end]+text[end+1:N-1])
for c in text:
count[c] = count[c] + 1
odd = 0
for k, c in count.items():
if c % 2 == 1:
odd = odd + 1
if odd >= 2:
print(-1)
else:
print(change_count(text))
"""
init: abbaccb
1. babaccb
2. babcacb
3. bacbacb
4. bcabacb
3回
init: abbabbc
init: abbaccb
1. abbcacb
2. abbccab
3. abbccba
4. abcbcba
6-1-
012345
taatmm
1. taamtm
2. taammt
3. tamamt
4. tmaamt
今端っこにあるのは??
ataatmma
init: bababbc
baabbcb: 4
babbcab: 3
babcbab: 1
bababcb: 1
babbcab: 2
"""
``` | instruction | 0 | 48,651 | 18 | 97,302 |
No | output | 1 | 48,651 | 18 | 97,303 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
import string
def main():
l = list(input())
N = len(l)
count=0
flag = True
middle_letter = None
i = 0
dic = {i:l.count(i) for i in string.ascii_lowercase}
while i <= N>>1:
if l[i]!=l[N-i-1]:
if dic[l[i]]%2 and N%2:
if flag:
middle_letter = l[i]
tar_index = N>>1
f_index = i
count+=tar_index-f_index
l[f_index:tar_index] = l[f_index+1:tar_index+1]
l[tar_index] = middle_letter
flag = False
elif not flag and l[i]==middle_letter:
pos = list(reversed(l)).index(l[i], i, N - i)
temp = l[i]
tar_index = N - 1 - i
f_index = N - pos - 1
count += tar_index - f_index
l[f_index:tar_index] = l[f_index + 1:tar_index + 1]
l[tar_index] = temp
i += 1
else:
print(-1)
exit()
else:
pos = list(reversed(l)).index(l[i],i,N-i)
temp = l[i]
tar_index = N-1-i
f_index = N-pos-1
count += tar_index-f_index
l[f_index:tar_index] = l[f_index+1:tar_index+1]
l[tar_index] = temp
i+=1
else:
i+=1
print(count)
if __name__=="__main__":
main()
``` | instruction | 0 | 48,652 | 18 | 97,304 |
No | output | 1 | 48,652 | 18 | 97,305 |
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 lowercase English letters. Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.
Constraints
* 1 \leq |S| \leq 2 × 10^5
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If we cannot turn S into a palindrome, print `-1`. Otherwise, print the minimum required number of operations.
Examples
Input
eel
Output
1
Input
ataatmma
Output
4
Input
snuke
Output
-1
Submitted Solution:
```
import sys
class BIT():
def __init__(self,number):
self.n=number
self.list=[0]*(number+1)
def add(self,i,x):#ith added x 1indexed
while i<=self.n:
self.list[i]+=x
i+=i&-i
def search(self,i):#1-i sum
s=0
while i>0:
s+=self.list[i]
i-=i&-i
return s
def suma(self,i,j):#i,i+1,..j sum
return self.search(j)-self.search(i-1)
#from collections import defaultdict
S=input()
N = len(S)
L = 26
a = ord('a')
d = [[] for i in range(L)]
for i in range(N):
s=ord(S[i])-a
d[s].append(i)
flag=0
for i in range(L):
if len(d[i])%2==1:
flag+=1
if flag>1:
print(-1)
sys.exit()
Suc=[-1]*N
pairs=[]
for i in range(L):
T=len(d[i])
for s in range((T//2)):
li,ri=d[i][s],d[i][T-s-1]
pairs.append((li,ri))
if T%2==1:
Suc[d[i][T//2]]=(N//2)+1
pairs.sort()
for i, (li,ri) in enumerate(pairs):
Suc[li]=i+1
Suc[ri]=N-i
Tree=BIT(N+3)
ans=0
for i in range(N):
Tree.add(Suc[i],1)
ans=ans+i+1-Tree.search(i+1)
print(ans)
``` | instruction | 0 | 48,653 | 18 | 97,306 |
No | output | 1 | 48,653 | 18 | 97,307 |
Provide a correct Python 3 solution for this coding contest problem.
F: Miko Mi String-
story
Mikko Mikkomi ~! Everyone's idol, Miko Miko Tazawa! Today ~, with Mikoto ~, practice the string algorithm, let's do it ☆
Miko's special ~~ character making ~~ The slogan "MikoMikomi" becomes "MikoMikoMi" in Roman letters! In other words, if A = “Mi” and B = “Ko”, you can write in the form of ABABA! In this way, a character string that can be decomposed into the form of ABABA by properly determining A and B is called "Mikomi character string"! Miko is a popular person for everyone to become the name of a character string!
In order for everyone to use the Mikomi character string, I decided to make a program to judge whether the given character string is the Mikomi character string, but I can't write a program longer than Miko and FizzBuzz. !! So ~, for Miko ~, I want you to write a program that judges Mikomi character strings ☆
...... You said it's cold now! ??
problem
A character string S consisting of uppercase and lowercase letters is given. Here, if there are two non-empty strings A and B that can be written as S = ABABA, then S is said to be a "Mikomi string". At this time, uppercase and lowercase letters of the alphabet shall be distinguished as different characters. Create a program that determines whether a given string is a string.
Input format
The input consists of only one line including the character string S. It can be assumed that S satisfies the following conditions.
* 1 ≤ | S | ≤ 10 ^ 6. However, | S | represents the length of the character string S.
* S consists only of uppercase or lowercase alphabets.
Since the input may be very large, it is recommended to use a high-speed function to receive the input.
Output format
If S is a character string, output "Love AB!" For A and B that satisfy S = ABABA. However, if multiple pairs of A and B satisfy the condition, output the one with the smallest | AB |. If S is not a Mikomi string, output "mitomerarenaiWA".
Input example 1
NicoNicoNi
Output example 1
Love Nico!
Input example 2
Kashikoi Kawaii Elichika
Output example 2
mitomerarenaiWA
Input example 3
LiveLiveL
Output example 3
Love Live!
Input example 4
AizunyanPeroPero
Output example 4
mitomerarenaiWA
Input example 5
AAAAAAAAAAAAAA
Output example 5
Love AAAAA!
Example
Input
NicoNicoNi
Output
Love Nico! | instruction | 0 | 48,780 | 18 | 97,560 |
"Correct Solution:
```
s = input()
length = len(s)
BASE = 100
MOD1 = 1000000007
MOD2 = 2147483647
acc1 = 0
acc2 = 0
hlst1 = [0]
hlst2 = [0]
for c in s:
i = ord(c)
acc1 = (acc1 * BASE + i) % MOD1
acc2 = (acc2 * BASE + i) % MOD2
hlst1.append(acc1)
hlst2.append(acc2)
def calc_hash(left, right, xlen):
return ((hlst1[right] - hlst1[left] * pow(BASE, xlen, MOD1)) % MOD1,
(hlst2[right] - hlst2[left] * pow(BASE, xlen, MOD2)) % MOD2)
for i in range(length // 3, -1, -1):
if (length - (i + 1) * 3) % 2:continue
alen = i + 1
blen = (length - (i + 1) * 3) // 2
if blen <= 0:continue
ha1 = calc_hash(0, alen, alen)
ha2 = calc_hash(alen + blen, blen + alen * 2, alen)
if ha1 != ha2:continue
ha3 = calc_hash(blen * 2 + alen * 2, blen * 2 + alen * 3, alen)
if ha1 != ha3:continue
hb1 = calc_hash(alen, alen + blen, blen)
hb2 = calc_hash(blen + alen * 2, blen * 2 + alen * 2, blen)
if hb1 != hb2:continue
print("Love {}!".format(s[:i + 1 + blen]))
break
else:
print("mitomerarenaiWA")
``` | output | 1 | 48,780 | 18 | 97,561 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,603 | 18 | 99,206 |
"Correct Solution:
```
W = input().lower()
c = 0
while True:
T = input()
if T == 'END_OF_TEXT':
break
c += T.lower().split().count(W)
print(c)
``` | output | 1 | 49,603 | 18 | 99,207 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,604 | 18 | 99,208 |
"Correct Solution:
```
W=str(input())
c=[]
total=0
while True:
T=str(input())
if T=="END_OF_TEXT":
break
c += T.lower().split()
print(c.count(W))
``` | output | 1 | 49,604 | 18 | 99,209 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,605 | 18 | 99,210 |
"Correct Solution:
```
w=input().lower()
c=[];s=input()
while(s!='END_OF_TEXT'):
c+=s.lower().split();
s=input()
print(c.count(w))
``` | output | 1 | 49,605 | 18 | 99,211 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,606 | 18 | 99,212 |
"Correct Solution:
```
count = 0
w = input()
while True:
t = input()
if t == 'END_OF_TEXT':
break
t = t.lower().split()
count += t.count(w)
print(count)
``` | output | 1 | 49,606 | 18 | 99,213 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,607 | 18 | 99,214 |
"Correct Solution:
```
W=input().lower()
cnt=0
while True:
T=input()
if T=='END_OF_TEXT': break
cnt += T.lower().split().count(W)
print(cnt)
``` | output | 1 | 49,607 | 18 | 99,215 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,608 | 18 | 99,216 |
"Correct Solution:
```
import sys
W = sys.stdin.readline().rstrip().lower()
T = list(sys.stdin.read().lower().split())
print(T.count(W))
``` | output | 1 | 49,608 | 18 | 99,217 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,609 | 18 | 99,218 |
"Correct Solution:
```
W = input()
count = 0
while True:
T = input()
if T == "END_OF_TEXT":
break
count += T.lower().split().count(W)
print(count)
``` | output | 1 | 49,609 | 18 | 99,219 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3 | instruction | 0 | 49,610 | 18 | 99,220 |
"Correct Solution:
```
W=input()
s=0
while True:
T=list(input().split())
if T==['END_OF_TEXT']:
break
else:
for i in T:
if i.lower()==W.lower():
s+=1
print(s)
``` | output | 1 | 49,610 | 18 | 99,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
import sys
W=input()
print(sys.stdin.read().lower().split().count(W))
``` | instruction | 0 | 49,611 | 18 | 99,222 |
Yes | output | 1 | 49,611 | 18 | 99,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
w = input().lower()
t = []
while 1:
a = input()
if a == "END_OF_TEXT":break
t += a.lower().split()
print(t.count(w))
``` | instruction | 0 | 49,612 | 18 | 99,224 |
Yes | output | 1 | 49,612 | 18 | 99,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
w = input().upper()
ans = 0
while True:
s = input()
if s == "END_OF_TEXT":
break
s = s.upper().split()
ans += s.count(w)
print(ans)
``` | instruction | 0 | 49,613 | 18 | 99,226 |
Yes | output | 1 | 49,613 | 18 | 99,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
w = input().lower()
sum = 0
while True:
t = input()
if t == 'END_OF_TEXT':
break
sum += t.lower().split().count(w)
print(sum)
``` | instruction | 0 | 49,614 | 18 | 99,228 |
Yes | output | 1 | 49,614 | 18 | 99,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
import sys
target = input()
data = sys.stdin.read()
print("%d" %(data.lower().count(target.lower())))
``` | instruction | 0 | 49,615 | 18 | 99,230 |
No | output | 1 | 49,615 | 18 | 99,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
W = input().lower()
ans = 0
while 1:
T = input().lower()
if T == "end_of_text":
break
ans += T.count(W)
print(ans)
``` | instruction | 0 | 49,616 | 18 | 99,232 |
No | output | 1 | 49,616 | 18 | 99,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
from collections import Counter
W=input().lower()
count=0
while True:
line=input().split()
if line==["END_OF_TEXT"]:
break
print(line)
count+=Counter(map(lambda x:x.lower(),line))[W]
print(count)
``` | instruction | 0 | 49,617 | 18 | 99,234 |
No | output | 1 | 49,617 | 18 | 99,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T
T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
Constraints
* The length of W ≤ 10
* W consists of lower case letters
* The length of T in a line ≤ 1000
Input
In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.
"END_OF_TEXT" indicates the end of the text.
Output
Print the number of W in the text.
Example
Input
computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT
Output
3
Submitted Solution:
```
w = input()
s = ""
while 1:
t = input()
if t == "END_OF_TEXT": break
s += t
print(s.count(w))
``` | instruction | 0 | 49,618 | 18 | 99,236 |
No | output | 1 | 49,618 | 18 | 99,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
n, m = map(int,input().split())
s = input()
t = input()
# n = len(s)
# m = len(t)
if s == '*':
print("YES")
elif s.count('*') is 0:
if s == t:
print("YES")
else:
print("NO")
elif s[0] is '*':
check = t[-(n-1):]
# print(check,s[1:])
if check == s[1:]:
print("YES")
else:
print("NO")
elif s[-1] is '*':
check1 = s[:n-1]
check2 = t[:n-1]
# print(check1,check2)
if check1 == check2:
print("YES")
else:
print("NO")
else:
# print('3')
s1,s2 = s.split('*')
len_s1 = len(s1)
len_s2 = len(s2)
if len_s1 + len_s2 > m:
print("NO")
else:
x = t[:len_s1]
# t = t[len_s1+1:]
y = t[-len_s2:]
# print("--", x,y)
if x == s1 and y == s2:
print("YES")
else:
print("NO")
``` | instruction | 0 | 49,627 | 18 | 99,254 |
Yes | output | 1 | 49,627 | 18 | 99,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
n,m=map(int,input().split())
s,t=input(),input()
if '*'in s:x,y=s.split('*');r=t.startswith(x)and t[len(x):].endswith(y)
else:r=s==t
print(('NO','YES')[r])
``` | instruction | 0 | 49,628 | 18 | 99,256 |
Yes | output | 1 | 49,628 | 18 | 99,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
n,m=[int(i) for i in input().split()]
s=str(input())
s1=str(input())
if('*' not in s):
if(s==s1):
print('YES')
else:
print('NO')
else:
if(len(s1)>=(len(s)-1)):
first=s[0:(s.find('*'))]
second=s[(s.find('*'))+1:]
last=len(s1)-1-(len(s)-(s.find('*'))-1)
ffirst=s1[0:(s.find('*'))]
ssecond=s1[last+1:]
if(first==ffirst and ssecond==second):
print('YES')
else:
print('NO')
else:
print('NO')
``` | instruction | 0 | 49,629 | 18 | 99,258 |
Yes | output | 1 | 49,629 | 18 | 99,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
n,m=map(int,input().split())
s=input()
t=input()
if "*" in s:
s1,s2=s.split("*")
if n>m+1:
print("NO")
else:
if(t.startswith(s1) and t.endswith(s2)):
t=t.replace(s1,"",1)
t=t.replace(s2,"",-1)
if((t.isalpha() and t.islower())or(t=="")):
print("YES")
else:
print("NO")
else:
print("NO")
else:
if s==t:
print("YES")
else:
print("NO")
``` | instruction | 0 | 49,630 | 18 | 99,260 |
Yes | output | 1 | 49,630 | 18 | 99,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
def change_case(s) :
a = list(s)
l = len(s)
for i in range(0, l) :
# If character is
# lowercase change
# to uppercase
if(a[i] >= 'a' and
a[i] <= 'z') :
a[i] = s[i].upper()
# If character is uppercase
# change to lowercase
elif(a[i] >= 'A' and
a[i] <= 'Z') :
a[i] = s[i].lower()
return a
# def to delete vowels
def delete_vowels(s) :
temp = ""
a = list(s)
l = len(s)
for i in range(0, l) :
# If character
# is consonant
if(a[i] != 'a' and a[i] != 'e' and
a[i] != 'i' and a[i] != 'o' and
a[i] != 'u' and a[i] != 'A' and
a[i] != 'E' and a[i] != 'O' and
a[i] != 'U' and a[i] != 'I') :
temp = temp + a[i]
return temp
# def to insert "#"
def insert_hash(s) :
temp = ""
a = list(s)
l = len(s)
for i in range(0, l) :
# If character is
# not special
if((a[i] >= 'a' and
a[i] <= 'z') or
(a[i] >= 'A' and
a[i] <= 'Z')) :
temp = temp + '#' + a[i]
else :
temp = temp + a[i]
return temp
# def to
# transfrm string
def transformSting(a) :
b = delete_vowels(a)
c = change_case(b)
d = insert_hash(c)
print (d)
# Driver Code
a = "SunshinE!!"
# Calling def
transformSting(a)
``` | instruction | 0 | 49,631 | 18 | 99,262 |
No | output | 1 | 49,631 | 18 | 99,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
#https://codeforces.com/problemset/problem/1023/A
x, y = input().split()
n, m = int(x), int(y)
s = input()
t = input()
if '*' not in s and s != t:
print("NO")
elif '*' not in s and s==t:
print("YES")
elif len(s) > len(t):
print("NO")
else:
ok = True
i = 0
while s[i] != '*':
if s[i] != t[i]:
print("NO")
ok = False
break
i += 1
#print(s[i+1:],"----",t[i+ 1-len(s):])
if s[i+1:] != t[i + 1-len(s):]:
print("NO")
ok = False
if ok:
print("YES")
``` | instruction | 0 | 49,632 | 18 | 99,264 |
No | output | 1 | 49,632 | 18 | 99,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
# Author: πα
input()
a = input()
b = input()
if '*' not in a:
if a == b:
print("YES")
else:
print("NO")
else:
i = a.find('*')
if a[:i] == b[:i]:
a = a[i+1:]
b = b[i:]
b = b[-len(a):]
if a == b:
print("YES")
else:
print("NO")
else:
print("NO")
``` | instruction | 0 | 49,633 | 18 | 99,266 |
No | output | 1 | 49,633 | 18 | 99,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.
The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.
For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".
If the given string t matches the given string s, print "YES", otherwise print "NO".
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.
The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.
The third line contains string t of length m, which consists only of lowercase Latin letters.
Output
Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).
Examples
Input
6 10
code*s
codeforces
Output
YES
Input
6 5
vk*cup
vkcup
Output
YES
Input
1 1
v
k
Output
NO
Input
9 6
gfgf*gfgf
gfgfgf
Output
NO
Note
In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".
In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".
There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".
In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
Submitted Solution:
```
n,m=map(int,input().split())
str1=input()
str2=input()
if str1=='*':
print('YES')
import sys
sys.exit()
star=str1.find('*')
if str1[:star]==str2[:star]:
str2=str2[star:]
if str1[star+1:]==str2[-len(str1[star+1:]):]:
print('YES')
else:
print('NO')
else:
print('NO')
``` | instruction | 0 | 49,634 | 18 | 99,268 |
No | output | 1 | 49,634 | 18 | 99,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have recently fallen through a hole and, after several hours of unconsciousness, have realized you are in an underground city. On one of your regular, daily walks through the unknown, you have encountered two unusually looking skeletons called Sanz and P’pairus, who decided to accompany you and give you some puzzles for seemingly unknown reasons.
One day, Sanz has created a crossword for you. Not any kind of crossword, but a 1D crossword! You are given m words and a string of length n. You are also given an array p, which designates how much each word is worth — the i-th word is worth pi points. Whenever you find one of the m words in the string, you are given the corresponding number of points. Each position in the crossword can be used at most x times. A certain word can be counted at different places, but you cannot count the same appearance of a word multiple times. If a word is a substring of another word, you can count them both (presuming you haven’t used the positions more than x times).
In order to solve the puzzle, you need to tell Sanz what’s the maximum achievable number of points in the crossword. There is no need to cover all postions, just get the maximal score! Crossword and words contain only lowercase English letters.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the length of the crossword. The second line contains the crossword string. The third line contains a single integer m (1 ≤ m ≤ 100) — the number of given words, and next m lines contain description of words: each line will have a string representing a non-empty word (its length doesn't exceed the length of the crossword) and integer pi (0 ≤ pi ≤ 100). Last line of the input will contain x (1 ≤ x ≤ 100) — maximum number of times a position in crossword can be used.
Output
Output single integer — maximum number of points you can get.
Example
Input
6
abacba
2
aba 6
ba 3
3
Output
12
Note
For example, with the string "abacba", words "aba" (6 points) and "ba" (3 points), and x = 3, you can get at most 12 points - the word "aba" appears once ("abacba"), while "ba" appears two times ("abacba"). Note that for x = 1, you could get at most 9 points, since you wouldn’t be able to count both "aba" and the first appearance of "ba".
Submitted Solution:
```
import sys
from heapq import heappop, heappush
# class Edge:
# def __init__(self, u, v, cap, flow, cost) -> None:
# self.u = u
# self.v = v
# self.cap = cap
# self.flow = flow
# self.cost = cost
def add_edge(adj, cap, cost, u, v, capv, costv):
adj[u].append(v)
adj[v].append(u)
cap[u][v] = capv
cost[u][v] = costv
cost[v][u] = -costv
def match(crossword, cw_idx, word):
for k in range(len(word)):
if crossword[cw_idx + k] != word[k]:
return False
return True
def read_input():
'''
6
abacba
2
aba 6
ba 3
3
'''
n = int(sys.stdin.readline())
# n = int('6\n')
crossword = sys.stdin.readline()[:-1]
# crossword = 'abacba\n'[:-1]
m = int(sys.stdin.readline())
# m = int('2\n')
adj = [[] for _ in range(n+2)]
cap = [[0] * (n+2) for _ in range(n+2)]
cost = [[0] * (n+2) for _ in range(n+2)]
# foo
# m_words = ["aba 6\n", "ba 3\n"]
# print(len(cost))
for idx in range(m):
word, p = sys.stdin.readline().split()
# word, p = m_words[idx].split()
p = int(p)
i = 0
while i + len(word) <= n:
if match(crossword, i, word):
u, v = i + 1, i + 1 + len(word)
# print((u, v))
add_edge(adj, cap, cost, u, v, 1, -p)
i += 1
x = int(sys.stdin.readline())
# x = int('3\n')
for i in range(n + 1):
u, v = i, i + 1
add_edge(adj, cap, cost, u, v, x, 0)
return adj, cap, cost
def bellman_ford(adj, cap, cost, potencial):
for _ in range(len(adj)):
for u in range(len(adj)):
for v in adj[u]:
reduced_cost = potencial[u] + cost[u][v] - potencial[v]
if cap[u][v] > 0 and reduced_cost < 0:
potencial[v] += reduced_cost
def dijkstra(adj, cap, flow, cost, potencial, dist, pi, s, t):
oo = float('inf')
for u in range(len(adj)):
dist[u] = +oo
pi[u] = None
dist[s] = 0
heap = [(0, s)]
while heap:
du, u = heappop(heap)
if dist[u] < du: continue
if u == t: break
for v in adj[u]:
reduced_cost = potencial[u] + cost[u][v] - potencial[v]
if flow[u][v] < cap[u][v] and dist[v] > dist[u] + reduced_cost:
dist[v] = dist[u] + reduced_cost
heappush(heap, (dist[v], v))
pi[v] = u
def min_cost_max_flow(adj, cap, cost):
min_cost, max_flow = 0, 0
flow = [[0] * len(adj) for _ in range(len(adj))]
potencial, oo = [0] * len(adj), float('inf')
dist, pi = [+oo] * len(adj), [None] * len(adj)
bellman_ford(adj, cap, cost, potencial)
s, t = 0, len(adj) - 1
while True:
dijkstra(adj, cap, flow, cost, potencial, dist, pi, s, t)
if dist[t] == +oo:
break
for u in range(len(adj)):
if dist[u] < dist[t]:
potencial[u] += dist[u] - dist[t]
limit, v = +oo, t
while pi[v]:
u = pi[v]
limit = min(limit, cap[u][v] - flow[u][v])
v = pi[v]
v = t
while pi[v]:
u = pi[v]
# por construcción u,v va a ser un arco avance si u < v
# en caso contrario el arco es reverso, con u,v se garantiza
# que se aumenta el flujo en arcos de avance y se dimuya en
# el arco inverso
u, w = min(u, v), max(u, v)
flow[u][w] += limit
flow[w][u] -= limit
v = pi[v]
min_cost += limit * (potencial[t] - potencial[s])
max_flow += limit
return min_cost, max_flow
adj, cap, cost = read_input()
# print({
# 'len(adj)': len(adj),
# 'len(cap)': len(cap),
# 'len(cost)': len(cost)
# })
# print({ 'adj': adj})
# print({ 'cap': cap})
# print({ 'cost': cost})
min_cost, _ = min_cost_max_flow(adj, cap, cost)
print(-min_cost)
``` | instruction | 0 | 50,116 | 18 | 100,232 |
No | output | 1 | 50,116 | 18 | 100,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have recently fallen through a hole and, after several hours of unconsciousness, have realized you are in an underground city. On one of your regular, daily walks through the unknown, you have encountered two unusually looking skeletons called Sanz and P’pairus, who decided to accompany you and give you some puzzles for seemingly unknown reasons.
One day, Sanz has created a crossword for you. Not any kind of crossword, but a 1D crossword! You are given m words and a string of length n. You are also given an array p, which designates how much each word is worth — the i-th word is worth pi points. Whenever you find one of the m words in the string, you are given the corresponding number of points. Each position in the crossword can be used at most x times. A certain word can be counted at different places, but you cannot count the same appearance of a word multiple times. If a word is a substring of another word, you can count them both (presuming you haven’t used the positions more than x times).
In order to solve the puzzle, you need to tell Sanz what’s the maximum achievable number of points in the crossword. There is no need to cover all postions, just get the maximal score! Crossword and words contain only lowercase English letters.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the length of the crossword. The second line contains the crossword string. The third line contains a single integer m (1 ≤ m ≤ 100) — the number of given words, and next m lines contain description of words: each line will have a string representing a non-empty word (its length doesn't exceed the length of the crossword) and integer pi (0 ≤ pi ≤ 100). Last line of the input will contain x (1 ≤ x ≤ 100) — maximum number of times a position in crossword can be used.
Output
Output single integer — maximum number of points you can get.
Example
Input
6
abacba
2
aba 6
ba 3
3
Output
12
Note
For example, with the string "abacba", words "aba" (6 points) and "ba" (3 points), and x = 3, you can get at most 12 points - the word "aba" appears once ("abacba"), while "ba" appears two times ("abacba"). Note that for x = 1, you could get at most 9 points, since you wouldn’t be able to count both "aba" and the first appearance of "ba".
Submitted Solution:
```
import sys
from heapq import heappop, heappush
# class Edge:
# def __init__(self, u, v, cap, flow, cost) -> None:
# self.u = u
# self.v = v
# self.cap = cap
# self.flow = flow
# self.cost = cost
def add_edge(adj, cap, cost, u, v, capv, costv):
costv, capv = min(cost[u][v], costv), max(cap[u][v], capv)
adj[u].append(v)
adj[v].append(u)
cap[u][v] = capv if cap[u][v] == 0 else cap[u][v]
cost[u][v] = costv
cost[v][u] = -costv
def match(crossword, cw_idx, word):
for k in range(len(word)):
if crossword[cw_idx + k] != word[k]:
return False
return True
def read_input():
'''
6
abacba
5
aba 6
ba 3
bac 4
cb 3
c 6
2
'''
n = int(sys.stdin.readline())
# n = int('6\n')
crossword = sys.stdin.readline()[:-1]
# crossword = 'abacba\n'[:-1]
m = int(sys.stdin.readline())
# m = int('5\n')
adj = [[] for _ in range(n+2)]
cap = [[0] * (n+2) for _ in range(n+2)]
cost = [[0] * (n+2) for _ in range(n+2)]
# foo
# m_words = ["aba 6\n", "ba 3\n", "bac 4\n", "cb 3\n", "c 6"]
# print(len(cost))
for idx in range(m):
word, p = sys.stdin.readline().split()
# word, p = m_words[idx].split()
p = int(p)
i = 0
while i + len(word) <= n:
if match(crossword, i, word):
u, v = i + 1, i + 1 + len(word)
# print((u, v))
add_edge(adj, cap, cost, u, v, 1, -p)
i += 1
x = int(sys.stdin.readline())
# x = int('2\n')
for i in range(n + 1):
u, v = i, i + 1
# print((u, v))
add_edge(adj, cap, cost, u, v, x, 0)
return adj, cap, cost
def bellman_ford(adj, cap, cost, potencial):
for _ in range(len(adj)):
for u in range(len(adj)):
for v in adj[u]:
reduced_cost = potencial[u] + cost[u][v] - potencial[v]
if cap[u][v] > 0 and reduced_cost < 0:
potencial[v] += reduced_cost
def dijkstra(adj, cap, flow, cost, potencial, dist, pi, s, t):
oo = float('inf')
for u in range(len(adj)):
dist[u] = +oo
pi[u] = None
dist[s] = 0
heap = [(0, s)]
while heap:
du, u = heappop(heap)
if dist[u] < du: continue
if u == t: break
for v in adj[u]:
if v == 5:
foo = 0
reduced_cost = potencial[u] + cost[u][v] - potencial[v]
if flow[u][v] < cap[u][v] and dist[v] > dist[u] + reduced_cost:
dist[v] = dist[u] + reduced_cost
heappush(heap, (dist[v], v))
pi[v] = u
def min_cost_max_flow(adj, cap, cost):
min_cost, max_flow = 0, 0
flow = [[0] * len(adj) for _ in range(len(adj))]
potencial, oo = [0] * len(adj), float('inf')
dist, pi = [+oo] * len(adj), [None] * len(adj)
bellman_ford(adj, cap, cost, potencial)
s, t = 0, len(adj) - 1
while True:
dijkstra(adj, cap, flow, cost, potencial, dist, pi, s, t)
if dist[t] == +oo:
break
for u in range(len(adj)):
if dist[u] < dist[t]:
potencial[u] += dist[u] - dist[t]
limit, v = +oo, t
while v:
u = pi[v]
limit = min(limit, cap[u][v] - flow[u][v])
v = pi[v]
v = t
while v:
u = pi[v]
flow[u][v] += limit
flow[v][u] -= limit
v = pi[v]
min_cost += limit * (potencial[t] - potencial[s])
max_flow += limit
# path = []
# v = t
# while v:
# path.append(v)
# v = pi[v]
# path.reverse()
# print(path)
return min_cost, max_flow
adj, cap, cost = read_input()
# print({
# 'len(adj)': len(adj),
# 'len(cap)': len(cap),
# 'len(cost)': len(cost)
# })
# print({ 'adj': adj})
# print({ 'cap': cap})
# print({ 'cost': cost})
min_cost, _ = min_cost_max_flow(adj, cap, cost)
print(-min_cost)
``` | instruction | 0 | 50,117 | 18 | 100,234 |
No | output | 1 | 50,117 | 18 | 100,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have recently fallen through a hole and, after several hours of unconsciousness, have realized you are in an underground city. On one of your regular, daily walks through the unknown, you have encountered two unusually looking skeletons called Sanz and P’pairus, who decided to accompany you and give you some puzzles for seemingly unknown reasons.
One day, Sanz has created a crossword for you. Not any kind of crossword, but a 1D crossword! You are given m words and a string of length n. You are also given an array p, which designates how much each word is worth — the i-th word is worth pi points. Whenever you find one of the m words in the string, you are given the corresponding number of points. Each position in the crossword can be used at most x times. A certain word can be counted at different places, but you cannot count the same appearance of a word multiple times. If a word is a substring of another word, you can count them both (presuming you haven’t used the positions more than x times).
In order to solve the puzzle, you need to tell Sanz what’s the maximum achievable number of points in the crossword. There is no need to cover all postions, just get the maximal score! Crossword and words contain only lowercase English letters.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the length of the crossword. The second line contains the crossword string. The third line contains a single integer m (1 ≤ m ≤ 100) — the number of given words, and next m lines contain description of words: each line will have a string representing a non-empty word (its length doesn't exceed the length of the crossword) and integer pi (0 ≤ pi ≤ 100). Last line of the input will contain x (1 ≤ x ≤ 100) — maximum number of times a position in crossword can be used.
Output
Output single integer — maximum number of points you can get.
Example
Input
6
abacba
2
aba 6
ba 3
3
Output
12
Note
For example, with the string "abacba", words "aba" (6 points) and "ba" (3 points), and x = 3, you can get at most 12 points - the word "aba" appears once ("abacba"), while "ba" appears two times ("abacba"). Note that for x = 1, you could get at most 9 points, since you wouldn’t be able to count both "aba" and the first appearance of "ba".
Submitted Solution:
```
a=input()
string=input()
n=int(input())
list1=[]
score=0
for i in range(n):
list1.append((input()).split())
maX=int(input())
var = 0
while n>0:
times = 0
for i in range (len(string)):
if string[i:i+len(list1[var][0])] == list1[var][0]:
if times == maX:
break
score += int(list1[var][1])
times += 1
if i+len(list1[var][0]) == len(string):
break
var += 1
n -= 1
print(score)
``` | instruction | 0 | 50,118 | 18 | 100,236 |
No | output | 1 | 50,118 | 18 | 100,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have recently fallen through a hole and, after several hours of unconsciousness, have realized you are in an underground city. On one of your regular, daily walks through the unknown, you have encountered two unusually looking skeletons called Sanz and P’pairus, who decided to accompany you and give you some puzzles for seemingly unknown reasons.
One day, Sanz has created a crossword for you. Not any kind of crossword, but a 1D crossword! You are given m words and a string of length n. You are also given an array p, which designates how much each word is worth — the i-th word is worth pi points. Whenever you find one of the m words in the string, you are given the corresponding number of points. Each position in the crossword can be used at most x times. A certain word can be counted at different places, but you cannot count the same appearance of a word multiple times. If a word is a substring of another word, you can count them both (presuming you haven’t used the positions more than x times).
In order to solve the puzzle, you need to tell Sanz what’s the maximum achievable number of points in the crossword. There is no need to cover all postions, just get the maximal score! Crossword and words contain only lowercase English letters.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the length of the crossword. The second line contains the crossword string. The third line contains a single integer m (1 ≤ m ≤ 100) — the number of given words, and next m lines contain description of words: each line will have a string representing a non-empty word (its length doesn't exceed the length of the crossword) and integer pi (0 ≤ pi ≤ 100). Last line of the input will contain x (1 ≤ x ≤ 100) — maximum number of times a position in crossword can be used.
Output
Output single integer — maximum number of points you can get.
Example
Input
6
abacba
2
aba 6
ba 3
3
Output
12
Note
For example, with the string "abacba", words "aba" (6 points) and "ba" (3 points), and x = 3, you can get at most 12 points - the word "aba" appears once ("abacba"), while "ba" appears two times ("abacba"). Note that for x = 1, you could get at most 9 points, since you wouldn’t be able to count both "aba" and the first appearance of "ba".
Submitted Solution:
```
import sys
from heapq import heappop, heappush
class Edge:
def __init__(self, u, v, cap, flow, cost, rev) -> None:
self.u = u
self.v = v
self.cap = cap
self.flow = flow
self.cost = cost
self.rev = rev
def add_edge(adj, u, v, capv, flowv, costv):
if u == 6 and v == 7:
foo = 0
adj[u].append(Edge(u, v, capv, flowv, costv, len(adj[v])))
adj[v].append(Edge(v, u, 0, flowv, -costv, len(adj[u])-1))
def match(crossword, cw_idx, word):
for k in range(len(word)):
if crossword[cw_idx + k] != word[k]:
return False
return True
def read_input():
'''
6
abacba
5
aba 6
ba 3
bac 4
cb 3
c 6
2
'''
n = int(sys.stdin.readline())
# n = int('6\n')
crossword = sys.stdin.readline()[:-1]
# crossword = 'abacba\n'[:-1]
m = int(sys.stdin.readline())
# m = int('5\n')
adj = [[] for _ in range(n+2)]
# foo
m_words = ["aba 6\n", "ba 3\n", "bac 4\n", "cb 3\n", "c 6"]
# print(len(cost))
for idx in range(m):
word, p = sys.stdin.readline().split()
# word, p = m_words[idx].split()
p = int(p)
i = 0
while i + len(word) <= n:
if match(crossword, i, word):
u, v = i + 1, i + 1 + len(word)
# print((u, v))
add_edge(adj, u, v, 1, 0, -p)
i += 1
x = int(sys.stdin.readline())
# x = int('2\n')
for i in range(n + 1):
u, v = i, i + 1
# print((u, v))
add_edge(adj, u, v, x, 0, 0)
return adj
def bellman_ford(adj, potencial):
for _ in range(len(adj)):
for u in range(len(adj)):
for e in adj[u]:
reduced_cost = potencial[e.u] + e.cost - potencial[e.v]
if e.cap > 0 and reduced_cost < 0:
potencial[e.v] += reduced_cost
def dijkstra(adj, potencial, dist, pi, s, t):
oo = float('inf')
for u in range(len(adj)):
dist[u] = +oo
pi[u] = None
dist[s] = 0
heap = [(0, s)]
while heap:
du, u = heappop(heap)
if dist[u] < du: continue
if u == t: break
for e in adj[u]:
reduced_cost = potencial[e.u] + e.cost - potencial[e.v]
if e.cap - e.flow > 0 and dist[e.v] > dist[e.u] + reduced_cost:
if e.u == 292 and e.v == 291:
print(e.cap - e.flow > 0 and dist[e.v] > dist[e.u] + reduced_cost)
print(f'e.cap = {e.cap}, e.flow = {e.flow}, dist[e.v] = {dist[e.v]}, dist[e.u] = {dist[e.u]}, reduced_cost = {reduced_cost}')
dist[e.v] = dist[e.u] + reduced_cost
heappush(heap, (dist[e.v], e.v))
pi[e.v] = e
def min_cost_max_flow(adj):
min_cost, max_flow = 0, 0
potencial, oo = [0] * len(adj), float('inf')
dist, pi = [+oo] * len(adj), [None] * len(adj)
bellman_ford(adj, potencial)
s, t = 0, len(adj) - 1
while True:
dijkstra(adj, potencial, dist, pi, s, t)
if dist[t] == +oo:
break
for u in range(len(adj)):
if dist[u] < dist[t]:
potencial[u] += dist[u] - dist[t]
limit, v = +oo, t
while v:
e = pi[v]
limit = min(limit, e.cap - e.flow)
v = e.u
v = t
while v:
e = pi[v]
e.flow += limit
try:
assert e.cap > 0
except:
reduced_cost = potencial[e.u] + e.cost - potencial[e.v]
print(f'u: {e.u}, v: {e.v}, cap: {e.cap}, flow: {e.flow} cost: {e.cost}')
exit(1)
v = e.u
min_cost += limit * (potencial[t] - potencial[s])
max_flow += limit
# path = []
# v = t
# while v:
# path.append(v)
# v = pi[v].u
# path.reverse()
# print(path)
return min_cost, max_flow
adj = read_input()
# print({
# 'len(adj)': len(adj),
# 'len(cap)': len(cap),
# 'len(cost)': len(cost)
# })
# print({ 'adj': adj})
# print({ 'cap': cap})
# print({ 'cost': cost})
min_cost, _ = min_cost_max_flow(adj)
print(-min_cost)
``` | instruction | 0 | 50,119 | 18 | 100,238 |
No | output | 1 | 50,119 | 18 | 100,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t consisting of lowercase English letters and an integer L.
We will consider generating a string of length L by concatenating one or more copies of s and t. Here, it is allowed to use the same string more than once.
For example, when s = `at`, t = `code` and L = 6, the strings `atatat`, `atcode` and `codeat` can be generated.
Among the strings that can be generated in this way, find the lexicographically smallest one. In the cases given as input, it is always possible to generate a string of length L.
Constraints
* 1 ≤ L ≤ 2 × 10^5
* 1 ≤ |s|, |t| ≤ L
* s and t consist of lowercase English letters.
* It is possible to generate a string of length L in the way described in Problem Statement.
Input
Input is given from Standard Input in the following format:
N
x_1 s_1
x_2 s_2
:
x_N s_N
Output
Print the lexicographically smallest string among the ones that can be generated in the way described in Problem Statement.
Examples
Input
6
at
code
Output
atatat
Input
8
coding
festival
Output
festival
Input
8
same
same
Output
samesame
Input
10
coin
age
Output
ageagecoin
Submitted Solution:
```
import sys
L = int(input())
s = str(input())
t = str(input())
tmps = s
tmpt = t
s = min(s,t)
sl = len(s)
t = max(tmps,tmpt)
tl = len(t)
p = L//sl
q = L%sl
if q != 0 and q%tl == 0:
print(s*p + t*(q//tl))
else:
for i in range(1,p):
if (q + i*sl)%tl == 0:
print(s*(p-i) + t*((L-(sl*(p-i)))//tl))
sys.exit()
``` | instruction | 0 | 50,356 | 18 | 100,712 |
No | output | 1 | 50,356 | 18 | 100,713 |
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