message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2 | instruction | 0 | 28,867 | 7 | 57,734 |
"Correct Solution:
```
N,A,B=map(int,input().split())
ans=N//(A+B)*A
X=N%(A+B)
ans+=min(X,A)
print(ans)
``` | output | 1 | 28,867 | 7 | 57,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
n,a,b=map(int,input().split())
x=n%(a+b)
y=n//(a+b)
print(y*a+min(x,a))
``` | instruction | 0 | 28,868 | 7 | 57,736 |
Yes | output | 1 | 28,868 | 7 | 57,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
n,a,b=map(int,input().split())
x=n//(a+b)
y=n%(a+b)
z=min(a,y)
print(a*x+z)
``` | instruction | 0 | 28,869 | 7 | 57,738 |
Yes | output | 1 | 28,869 | 7 | 57,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
n,a,b=map(int,input().split())
k=n//(a+b)
print(k*a+min(a,n-k*(a+b)))
``` | instruction | 0 | 28,870 | 7 | 57,740 |
Yes | output | 1 | 28,870 | 7 | 57,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
N,A,B = map(int,input().split())
C = N//(A+B)
D = N%(A+B)
print(C*A+min(D,A))
``` | instruction | 0 | 28,871 | 7 | 57,742 |
Yes | output | 1 | 28,871 | 7 | 57,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
import math
n, a, b = map(int, input().split())
c = a + b
if (n % c >= a):
m = a
else:
m = n % c
print(math.floor(n / c * a + m))
``` | instruction | 0 | 28,872 | 7 | 57,744 |
No | output | 1 | 28,872 | 7 | 57,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
N,A,B=map(int,input().split())
cnt_blue = 0
# try:
pre_loop_cnt = N // (A+B)
totol_pre_A = sum([A] * pre_loop_cnt)
totol_pre_B = sum([B] * pre_loop_cnt)
total_tmp = totol_pre_A + totol_pre_B
total_tmp += A
if N <= total_tmp:
cnt_blue = N - totol_pre_B
else:
cnt_blue = totol_pre_A + A
print(cnt_blue)
# except:
# print(0)
``` | instruction | 0 | 28,873 | 7 | 57,746 |
No | output | 1 | 28,873 | 7 | 57,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
import sys, math
def input():
return sys.stdin.readline()[:-1]
from itertools import permutations, combinations
from collections import defaultdict, Counter
from math import factorial
from bisect import bisect_left # bisect_left(list, value)
#from fractions import gcd
enu = enumerate
sys.setrecursionlimit(10**7)
N, A, B = map(int, input().split())
AB = A+B
cnt = A * (N//AB)
cnt += N % AB
print(cnt)
``` | instruction | 0 | 28,874 | 7 | 57,748 |
No | output | 1 | 28,874 | 7 | 57,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has many red balls and blue balls. Now, he will place them in a row.
Initially, there is no ball placed.
Takahashi, who is very patient, will do the following operation 10^{100} times:
* Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.
How many blue balls will be there among the first N balls in the row of balls made this way?
Constraints
* 1 \leq N \leq 10^{18}
* A, B \geq 0
* 0 < A + B \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of blue balls that will be there among the first N balls in the row of balls.
Examples
Input
8 3 4
Output
4
Input
8 0 4
Output
0
Input
6 2 4
Output
2
Submitted Solution:
```
n,a,b = map(int, input().split())
t = n // (a + b)
if t == 1:
print(a * t)
else:
if n - (a + b) * t >= a:
print((a * t) + a)
else:
print((a * t) + (n - (t * n)))
``` | instruction | 0 | 28,875 | 7 | 57,750 |
No | output | 1 | 28,875 | 7 | 57,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,573 | 7 | 59,146 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
s1 = input()
s= input()+'.'
s2=s1[::-1]
l=len(s)
ll=len(s1)
a=''
lis=[]
for i in range(l):
v=a+s[i]
if v in s1 or v in s2:
a=a+s[i]
else:
if a in s1:
k=s1.index(a)+1
lis.append([k,k+len(a)-1])
# print(s1.index(a),a)
elif a in s2:
k=s2.index(a)
lis.append([ll-k,ll-k-(len(a)-1)])
# print(k)
else:
print(-1)
exit()
a=s[i]
print(len(lis))
for i in lis:
print(*i)
``` | output | 1 | 29,573 | 7 | 59,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,574 | 7 | 59,148 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
s = input()
t = input() + '.'
rs, a, res = s[::-1], t[0], []
for i in range(1,len(t)):
b = a + t[i]
if b in s or b in rs:
a = b
continue
if a in s:
x = s.find(a)
res.append([x+1, x+len(a)])
elif a in rs:
x = rs.find(a)
res.append([len(s)-x, len(s)-x-len(a)+1])
else:
print(-1)
exit()
a = t[i]
print(len(res))
for i in res:
print(i[0], i[1])
``` | output | 1 | 29,574 | 7 | 59,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,575 | 7 | 59,150 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
import sys
s = input()
s_reverse = s[::-1]
t = input()
first = []
second = []
pos = 0
def find_str(st, char):
index = 0
if char in st:
# print("===")
c = char[0]
for ch in st:
if ch == c:
if st[index:index+len(char)] == char:
return index
index += 1
# print (index)
return -1
while pos < len(t):
substr = t[pos:pos + 1]
if substr not in s and substr not in s_reverse:
print("-1")
sys.exit()
step = 1
while substr in s or substr in s_reverse:
step += 1
substr = t[pos:(pos + step)]
# print("substr %s" % substr)
if pos + step > len(t):
break
step -= 1
substr = t[pos:(pos + step)]
start = find_str(s, substr)
# print ("debug %d %d %d" % (pos, step, start))
if start == -1:
start = find_str(s_reverse, substr)
# print ("newstart: %d" % start)
# print("substr %s" % substr)
first.append(len(s) - start)
second.append(len(s) - (start + step - 1))
else:
first.append(start + 1)
second.append(start + step)
pos = pos + step
# print ("last %d %d" %(len(t), pos))
print(len(first))
for a, b in zip(first, second):
print("%d %d" % (a, b))
# x = input("ccc")
``` | output | 1 | 29,575 | 7 | 59,151 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,576 | 7 | 59,152 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
def solve(s,t):
hash_s = [False] * 256
hash_t = [False] * 256
arr = []
n = len(s)
for c in s:
hash_s[ord(c)] = True
for c in t:
hash_t[ord(c)] = True
for i in range(256):
if not hash_s[i] and hash_t[i]:
print(-1)
return
rev = s[::-1]
i,j = 0,0
while i < len(t):
flag = True
temp = t[i]
j = i + 1
while j < len(t):
temp += t[j]
if temp not in s and temp not in rev:
flag = False
break
j += 1
if flag:
x = s.find(temp)
if x != -1:
arr.append((x + 1,x + len(temp)))
# print('1',x + 1,x + len(temp))
else:
y = rev.find(temp)
arr.append((n - y,n - y - len(temp) + 1))
# print('2',n - y,n - y - len(temp) + 1)
else:
x = s.find(temp[:-1])
if x != -1:
arr.append((x + 1,x + len(temp) - 1))
# print('3',x + 1,x + len(temp) - 1)
else:
x = rev.find(temp[:-1])
arr.append((n - x,n - x - len(temp) + 2))
# print('4',n - x,n - x - len(temp) + 2)
i = j
print(len(arr))
for x,y in arr:
print(x,y)
s = input()
t = input()
solve(s,t)
``` | output | 1 | 29,576 | 7 | 59,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,577 | 7 | 59,154 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
s = input()
t = input()
revS = s[::-1]
n = len(s)
cur=1
start=-1
end=-1
revFlag=0
errFlag=0
i=0
res=[]
while i < len(t):
if s.find(t[i:i+cur]) != -1 and i+cur <= len(t) and s.find(t[i:i+cur]) + cur <= n:
start = s.find(t[i:i+cur]) + 1
end = start + cur - 1
cur += 1
elif revS.find(t[i:i+cur]) != -1 and i+cur <= len(t) and revS.find(t[i:i+cur]) + cur <= n:
start = n - revS.find(t[i:i+cur])
end = start - cur + 1
cur += 1
else:
if (start == -1 and end == -1) and (s.find(t[i:i+cur]) == -1 and revS.find(t[i:i+cur]) == -1):
errFlag = 1
break
i += cur - 1
cur = 1
res.append(str(start) + " " + str(end))
start = -1
end = -1
if errFlag != 1:
print(len(res))
for p in res:
print(p)
else:
print(-1)
``` | output | 1 | 29,577 | 7 | 59,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,578 | 7 | 59,156 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
def find_max_substr(t, s):
l, r = 0, len(t)
while l != r:
m = (l + r) // 2
if t[:m + 1] in s:
l = m + 1
else:
r = m
l1 = l
rs = s[::-1]
l, r = 0, len(t)
while l != r:
m = (l + r) // 2
if t[:m + 1] in rs:
l = m + 1
else:
r = m
l2 = l
if l1 >= l2:
return s.find(t[:l1]) + 1, s.find(t[:l1]) + l1
else:
return s.find(t[:l2][::-1]) + l2, s.find(t[:l2][::-1]) + 1
s = input()
t = input()
if not set(t).issubset(set(s)):
print(-1)
exit(0)
a = []
while t:
l, r = find_max_substr(t, s)
a.append((l, r))
t = t[abs(r - l) + 1:]
print(len(a))
for l, r in a:
print(l, r)
``` | output | 1 | 29,578 | 7 | 59,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,579 | 7 | 59,158 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
def find_max_substr(t, s):
l, r = 0, len(t)
while l != r:
m = (l + r) // 2
if t[:m + 1] in s:
l = m + 1
else:
r = m
l1 = l
rs = s[::-1]
l, r = 0, len(t)
while l != r:
m = (l + r) // 2
if t[:m + 1] in rs:
l = m + 1
else:
r = m
l2 = l
if l1 >= l2:
return s.find(t[:l1]) + 1, s.find(t[:l1]) + l1
else:
return s.find(t[:l2][::-1]) + l2, s.find(t[:l2][::-1]) + 1
s = input()
t = input()
if not set(t).issubset(set(s)):
print(-1)
exit(0)
a = []
while t:
l, r = find_max_substr(t, s)
a.append((l, r))
t = t[abs(r - l) + 1:]
print(len(a))
for l, r in a:
print(l, r)
# Made By Mostafa_Khaled
``` | output | 1 | 29,579 | 7 | 59,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at". | instruction | 0 | 29,580 | 7 | 59,160 |
Tags: dp, greedy, strings, trees
Correct Solution:
```
s = input()
t = input()
len_s = len(s)
reversed_s = s[::-1]
len_t = len(t)
res = 0
pairs = []
indx = 0
while indx < len_t:
if t[indx : indx + 1] not in s:
print(-1)
exit()
i = 1
while (t[indx : indx + i] in s or t[indx : indx + i] in reversed_s) and indx + i < len_t + 1:
i += 1
prev_i = i - 1
if t[indx : indx +prev_i] in s:
x = s.find(t[indx : indx +prev_i])
pairs.append((x + 1, x + prev_i))
else:
x = reversed_s.find(t[indx : indx + prev_i])
pairs.append((len_s - x, len_s - x - i + 2))
res += 1
indx += prev_i
print(res)
for a, b in pairs:
print(a, b)
``` | output | 1 | 29,580 | 7 | 59,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
a, t = input(), input() + ' '
b = a[::-1]
s, p = '', []
for q in t:
d = s
s += q
if s in a or s in b: continue
if d in a:
k = a.find(d)
p += [(k + 1, k + len(d))]
elif d in b:
k = b.rfind(d)
p += [(len(a) - k, len(a) - k - len(d) + 1)]
else:
print(-1)
exit()
s = q
print(len(p))
for i, j in p: print(i, j)
``` | instruction | 0 | 29,581 | 7 | 59,162 |
Yes | output | 1 | 29,581 | 7 | 59,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
s = input()
t = input()
len_s = len(s)
reversed_s = s[::-1]
len_t = len(t)
res = 0
pairs = []
indx = 0
while indx < len_t:
if t[indx : indx + 1] not in s:
print(-1)
exit()
i = 1
while (t[indx : indx + i] in s or t[indx : indx + i] in reversed_s) and indx + i < len_t + 1:
i += 1
prev_i = i - 1
if t[indx : indx +prev_i] in s:
x = s.find(t[indx : indx +prev_i])
pairs.append((x + 1, x + prev_i))
else:
x = reversed_s.find(t[indx : indx + prev_i])
pairs.append((len_s - x, len_s - x - i + 2))
res += 1
indx += prev_i
print(res)
for first, second in pairs:
print(first, second)
``` | instruction | 0 | 29,582 | 7 | 59,164 |
Yes | output | 1 | 29,582 | 7 | 59,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
s, t = input(), input()
p, d = [], [[] for i in range(26)]
for i, q in enumerate(s): d[ord(q) - 97].append(i)
i, n = 0, len(t)
s += '+'
t += '-'
while i < n:
q = t[i]
a = b = c = 0
for j in d[ord(q) - 97]:
k = 1
while t[i + k] == s[j + k]: k += 1
if k > a: a, b, c = k, j + 1, 1
k = 1
while t[i + k] == s[j - k]: k += 1
if k > a: a, b, c = k, j + 1, -1
if not a:
print(-1)
exit()
i += a
p.append((b, b + c * a - c))
print(len(p))
for i, j in p: print(i, j)
``` | instruction | 0 | 29,583 | 7 | 59,166 |
Yes | output | 1 | 29,583 | 7 | 59,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
s = input()
t = input()
rs = s[::-1]
a = t[0]
t+='.'
li=[]
for i in range(1,len(t)):
b = a + t[i]
if b in s or b in rs:
a = b
else:
if a in s:
li.append([s.find(a)+1,s.find(a)+len(a)])
elif a in rs:
li.append([len(s)-rs.find(a),len(s)-rs.find(a)-len(a)+1])
else:
print(-1),exit(0)
a = t[i]
print(len(li))
for i in li:print(i[0],i[1])
``` | instruction | 0 | 29,584 | 7 | 59,168 |
Yes | output | 1 | 29,584 | 7 | 59,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
s = input()
t = input()
revS = s[::-1]
n = len(s)
cur=1
start=-1
end=-1
revFlag=0
errFlag=0
i=0
res=[]
while i < len(t):
if s.find(t[i:i+cur]) != -1 and i+cur <= len(t) and s.find(t[i:i+cur]) + cur <= n:
start = s.find(t[i:i+cur]) + 1
end = start + cur - 1
cur += 1
elif revS.find(t[i:i+cur]) != -1 and i+cur <= len(t) and revS.find(t[i:i+cur]) + cur <= n:
start = n - revS.find(t[i:i+cur])
end = start - cur + 1
cur += 1
else:
if (start == -1 and end == -1) or (s.find(t[i:i+cur]) == -1 and revS.find(t[i:i+cur]) == -1):
errFlag = 1
break
i += cur - 1
cur = 1
res.append(str(start) + " " + str(end))
if errFlag != 1:
print(len(res))
for p in res:
print(p)
else:
print(-1)
``` | instruction | 0 | 29,585 | 7 | 59,170 |
No | output | 1 | 29,585 | 7 | 59,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
s = input()
t = input()
len_s = len(s)
reversed_s = s[::-1]
len_t = len(t)
res = 0
pairs = []
indx = 0
while indx < len_t:
i = 1
to_find = t[indx : indx + i]
if to_find not in s:
print(-1)
exit()
while (to_find in s or to_find in reversed_s) and indx + i < len_t + 1:
i += 1
prev_i = i - 1
if t[indx : indx + prev_i] in s:
x = s.find(t[indx : indx + prev_i])
pairs.append((x + 1, x + prev_i))
else:
x = reversed_s.find(t[indx : indx + prev_i])
first = len_s - x
second = len_s - x - i + 2
pairs.append((first, second))
res += 1
indx += prev_i
print(res)
for first, second in pairs:
print(first, second)
``` | instruction | 0 | 29,586 | 7 | 59,172 |
No | output | 1 | 29,586 | 7 | 59,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
def get_max_sub(text):
global m, z
search_pos = len(text) - m * 2 - 1
n = len(text)
t_pos = search_pos
t_revers_pos = search_pos + m + 1
max_sub = ((0, 0), 0)
l, r = 0, 0
for i in range(search_pos, n):
z[i - search_pos] = 0
if i <= r:
z[i - search_pos] = min(r - i + 1, z[i - l])
while(i + z[i - search_pos] < n) and (text[i + z[i - search_pos]] == text[z[i - search_pos]]):
z[i - search_pos] += 1
if (z[i - search_pos] > max_sub[1]):
if (t_pos <= i < t_revers_pos):
max_sub = ((i - t_pos + 1, i - t_pos + z[i - search_pos]), z[i - search_pos])
else:
max_sub = ((m - i + t_revers_pos, m - (i - t_revers_pos + z[i]) + 1), z[i - search_pos])
if (z[i - search_pos] == m):
return (max_sub)
if (i + z[i - search_pos] - 1) > r:
l = i
r = i + z[i - search_pos] - 1
return (max_sub)
def main():
s, text = input(), input()
if (set(s) == set(text)):
if (len(text) > len(s)):
if (len(text) % len(s) == 0):
print(len(text) // len(s))
else:
print(len(text) // len(s) + len(text) % len(s))
for i in range(len(text) // len(s)):
print(1, len(s))
if (len(text) % len(s) != 0):
print(1, len(text) % len(s))
return
global m
m = len(s)
text = text + '#{}@{}'.format(s, s[::-1])
global z
z = [0 for i in range(2 * m + 2)]
ops = []
while(len(text) > 2 * (m + 1)):
max_sub = get_max_sub(text)
if (max_sub[1] == 0):
break
ops.append(max_sub[0])
text = text[max_sub[1]:]
if len(text) > 2 * (m + 1):
print(-1)
else:
print(len(ops))
for op in ops:
print(*op)
main()
``` | instruction | 0 | 29,587 | 7 | 59,174 |
No | output | 1 | 29,587 | 7 | 59,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.
First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.
Unfortunately, blocks aren't freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want's to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want's to know some way to achieve the answer.
Input
First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.
Output
The first line should contain the minimum needed number of coatings n or -1 if it's impossible to create the desired coating.
If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.
Examples
Input
abc
cbaabc
Output
2
3 1
1 3
Input
aaabrytaaa
ayrat
Output
3
1 1
6 5
8 7
Input
ami
no
Output
-1
Note
In the first sample string "cbaabc" = "cba" + "abc".
In the second sample: "ayrat" = "a" + "yr" + "at".
Submitted Solution:
```
s = input()
t = input()
len_s = len(s)
reversed_s = s[::-1]
len_t = len(t)
res = 0
pairs = []
indx = 0
while indx < len_t:
if t[indx : indx + 1] not in s:
print(-1)
exit()
i = 1
to_find = t[indx : indx + i]
while (to_find in s or to_find in reversed_s) and indx + i < len_t + 1:
i += 1
prev_i = i - 1
if t[indx : indx +prev_i] in s:
x = s.find(t[indx : indx +prev_i])
pairs.append((x + 1, x + prev_i))
else:
x = reversed_s.find(t[indx : indx + prev_i])
pairs.append((len_s - x, len_s - x - i + 2))
res += 1
indx += prev_i
print(res)
for first, second in pairs:
print(first, second)
``` | instruction | 0 | 29,588 | 7 | 59,176 |
No | output | 1 | 29,588 | 7 | 59,177 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,751 | 7 | 61,502 |
"Correct Solution:
```
M=998244353
N,*A=open(0)
d=[0]*90001
d[0]=1
e=[0]*90001
e[0]=1
z=1
s=0
for a in A:
i=s;a=int(a);s+=a;z*=3
while i+1:d[i+a]+=d[i]%M;d[i]*=2;e[i+a]+=e[i]%M;i-=1
i=s
while i*2>=s:z=(z-d[i]*3)%M;i-=1
print([(z+e[s//2]*3)%M,z][s%2])
``` | output | 1 | 30,751 | 7 | 61,503 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,752 | 7 | 61,504 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = [int(input()) for i in range(n)]
w = sum(a)
mod =998244353
dp = [[0]*(w+1) for i in range(n)]
cnt = a[0]
dp[0][a[0]] = 1
dp[0][0] = 2
for i in range(1,n):
cnt += a[i]
for j in range(cnt+1):
dp[i][j] = dp[i-1][j]*2
if j >= a[i]:
dp[i][j] += dp[i-1][j-a[i]]
dp[i][j] = dp[i][j]%mod
#sより大きいのはブー
s = w//2+1
if w %2 == 0:
d = [[0]*(s+1) for i in range(n)]
d[0][0] = 1
d[0][a[0]] = 1
cnt = a[0]
for i in range(1,n):
cnt += a[i]
for j in range(min(cnt+1,s+1)):
d[i][j] = d[i-1][j]
if j >= a[i]:
d[i][j] += d[i-1][j-a[i]]
d[i][j] = d[i][j]%mod
mns = d[-1][w//2]
if w%2 == 0:
s -= 1
cnt = 0
for i in range(s,w+1):
cnt += dp[-1][i]
cnt = cnt%mod
res = (pow(3,n,mod)-cnt*3)%mod
if w % 2 == 0:
res = (res+mns*3)%mod
print(res)
``` | output | 1 | 30,752 | 7 | 61,505 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,753 | 7 | 61,506 |
"Correct Solution:
```
n = int(input())
a = [int(input()) for i in range(n)]
sum_a = sum(a)
MOD = 998244353
dp = [0] * (sum_a + 1)
dp[0] = 1
for i in range(n):
for j in range(sum_a + 1)[::-1]:
# 選ばないとき
dp[j] += dp[j]
if j - a[i] >= 0:
dp[j] += dp[j - a[i]]
dp[j] %= MOD
dq = [0] * (sum_a + 1)
dq[0] = 1
for i in range(n):
for j in range(sum_a + 1)[::-1]:
# 選ぶとき
if j - a[i] >= 0:
dq[j] += dq[j - a[i]]
dq[j] %= MOD
else:
break
ans = 0
for j in range(sum_a + 1):
if sum_a <= j * 2:
ans += dp[j] * 3
ans %= MOD
if sum_a % 2 == 0:
ans -= dq[sum_a // 2] * 3
print((3 ** n - ans) % MOD)
``` | output | 1 | 30,753 | 7 | 61,507 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,754 | 7 | 61,508 |
"Correct Solution:
```
N = int(input())
a = [int(input()) for _ in range(N)]
s = sum(a)
m = (s + 1) // 2
c1 = [0] * (m + 1)
c1[m] = 1
for i in range(N):
d = a[i]
for j in range(m - d + 1):
c1[j] = (2 * c1[j] + c1[j + d]) % 998244353
for j in range(m - d + 1, m + 1):
c1[j] = (2 * c1[j] + c1[m]) % 998244353
ans1 = c1[0]
if s % 2:
c1[0] = 0
else:
for j in range(m):
c1[j] = 0
c1[m] = 1
for i in range(N):
d = a[i]
for j in range(m - d + 1):
c1[j] = (c1[j] + c1[j + d]) % 998244353
for j in range(m - d + 1, m + 1):
c1[j] = c1[j]
ans = (3 ** N % 998244353 - 3 * ans1 + 3 * c1[0]) % 998244353
print(ans)
``` | output | 1 | 30,754 | 7 | 61,509 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,755 | 7 | 61,510 |
"Correct Solution:
```
M=998244353
N,*A=map(int,open(0))
d=[0]*90001
d[0]=1
e=[0]*90001
e[0]=1
z=1
s=0
for a in A:
i=s;s+=a;z=z*3%M
while i>=0:
d[i+a]+=d[i]%M
d[i]*=2
e[i+a]+=e[i]%M
i-=1
for i in range(s+1>>1,s+1):z=(z-d[i]*3)%M
print([(z+e[s//2]*3)%M,z][s%2])
``` | output | 1 | 30,755 | 7 | 61,511 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,756 | 7 | 61,512 |
"Correct Solution:
```
N = int(input())
mod = 998244353
# DP ナップサックぽいやつ
# DP[i][j] := a_0からa_iまでの部分集合で和がjになるものの個数
# 配るDPの方が場合分けがなくて簡潔に書ける 計算量増えるかも
# DP[i-1][j]→DP[i][j]とDP[i][j + a_i ]に配る
aaa = [int(input()) for _ in range(N)]
sum_aaa = sum(aaa)
sum_now = 0
DP = [[0] * (sum_aaa+1) for i in range(N)]
DP2 = [[0] * (sum_aaa+1) for i in range(N)]
a = aaa[0]
DP[0][0] = 2
DP[0][a] = 1
DP2[0][0] = 1
DP2[0][a] = 1
sum_now += a
for i in range(1, N):
a = aaa[i]
sum_now += a
for j in range(sum_now + 1):
if (j >= a):
DP[i][j] = DP[i-1][j]*2 + DP[i-1][j-a]
else:
DP[i][j] = DP[i-1][j]*2
while (DP[i][j] > mod):
DP[i][j] -= mod
for j in range(min(sum_now + 1, int(sum_aaa /2)+1)):
if (j >= a):
DP2[i][j] = DP2[i-1][j] + DP2[i-1][j-a]
else:
DP2[i][j] = DP2[i-1][j]
while (DP2[i][j] > mod):
DP2[i][j] -= mod
# print(DP2)
minimum_invalid_length = int((sum_now+1)//2)
num_too_long = sum (DP[N-1][minimum_invalid_length:])
if (sum_aaa %2 == 0):
num_too_long -= DP2[N-1][int(sum_aaa /2)]
# print(num_too_long)
# print(sum(aaa))
print((3** N - num_too_long * 3) % mod )
``` | output | 1 | 30,756 | 7 | 61,513 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,757 | 7 | 61,514 |
"Correct Solution:
```
M,z,s=998244353,1,0
N,*A=open(0)
d=[1]+[0]*7**6
e=[1]+[0]*7**6
for a in A:
a=int(a);s+=a;i=s;z*=3
while~i:d[i]+=d[i]+d[i-a]%M;e[i]+=e[i-a]%M;i-=1
print((z+~s%2*e[s//2]*3-3*sum(d[-~s//2:]))%M)
``` | output | 1 | 30,757 | 7 | 61,515 |
Provide a correct Python 3 solution for this coding contest problem.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556 | instruction | 0 | 30,758 | 7 | 61,516 |
"Correct Solution:
```
M,z,s=998244353,1,0
N,*A=map(int,open(0))
d=[1]+[0]*7**6
e=[1]+[0]*7**6
for a in A:
i=s;s+=a;z*=3
while~i:d[i+a]+=d[i]%M;d[i]*=2;e[i+a]+=e[i]%M;i-=1
i=s
while i*2>=s:z=(z-d[i]*3)%M;i-=1
print((z+~s%2*e[s//2]*3)%M)
``` | output | 1 | 30,758 | 7 | 61,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
def dp_table(*counts): # dp[x][y]を作りたかったら dp_table(xのサイズ,yのサイズ)
if counts:
return [dp_table(*counts[1:]) for _ in range(counts[0])]
else:
return 0
def main():
N = int(input())
A = [int(input()) for _ in range(N)]
mod = 998244353
ans = pow(3, N, mod)
sum_A = sum(A)
dp = [[0] * (sum_A+1) for i in range(N)]
# dp[i][j]: i番目までの中から辺をいくつか選んで、長さがjになるパターン数
# init
dp[0][A[0]] = 1
dp[0][0] = 2 # 他の色のうちどちらに入れるかで2通り
# init2
dp2 = [[0] * (sum_A+1) for i in range(N)] # ある二色でsum_Aを等分するパターンを数える
dp2[0][A[0]] = 1
dp2[0][0] = 1 # 今度は片方に集中したパターン
# dp
sum_now = A[0]
for i in range(1, N):
a = A[i]
sum_now += a
for length in range(sum_now+1):
# A[i]を追加しないのなら他の２食のうちどちらに入れるかで2倍選択肢がある
if length >= a:
dp[i][length] = dp[i - 1][length - a] + \
(dp[i - 1][length] << 1)
else:
dp[i][length] = dp[i-1][length] << 1
dp[i][length] %= mod
for length in range(min(sum_now + 1, sum_A // 2 + 1)):
if length >= a:
dp2[i][length] = dp2[i - 1][length - a] + dp2[i - 1][length]
else:
dp2[i][length] = dp2[i-1][length]
dp2[i][length] %= mod
# print(sum(dp[N-1][sum_A//2]))
# print(dp)
ans -= 3*sum(dp[N-1][(sum_A+1)//2:])
if sum_A % 2 == 0:
ans += 3*dp2[N-1][sum_A//2]
print(ans % mod)
if __name__ == '__main__':
main()
``` | instruction | 0 | 30,759 | 7 | 61,518 |
Yes | output | 1 | 30,759 | 7 | 61,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
N = int(input())
A = [int(input()) for _ in range(N)]
sum_A = sum(A)
MOD = 998244353
dp1 = [0] * (sum_A + 1)
dp2 = [0] * (sum_A + 1)
dp1[0] = dp2[0] = 1
tmp_sum = 0
for a in A:
for i in reversed(range(tmp_sum + 1)):
dp1[i] %= MOD
dp2[i] %= MOD
dp1[i + a] += dp1[i]
dp1[i] <<= 1
dp2[i + a] += dp2[i]
tmp_sum += a
ans = pow(3, N, MOD)
for i in range(sum_A + 1):
if i * 2 >= sum_A:
ans = (ans - dp1[i] * 3 + MOD * 3) % MOD
if i * 2 == sum_A:
ans = (ans + dp2[i] * 3) % MOD
print(ans)
``` | instruction | 0 | 30,760 | 7 | 61,520 |
Yes | output | 1 | 30,760 | 7 | 61,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
# D
N = int(input())
a_list = [0]*N
for i in range(N):
a_list[i] = int(input())
L = 90001
LARGE = 998244353
DP = [0]*L
DP[0] = 1
for i in range(N):
for j in range(L-1, a_list[i]-1, -1):
DP[j] = (2*DP[j] + DP[j - a_list[i]]) % LARGE
for j in range(a_list[i]-1, -1, -1):
DP[j] = (2*DP[j]) % LARGE
DP2 = [0]*L
DP2[0] = 1
for i in range(N):
for j in range(L-1, a_list[i]-1, -1):
DP2[j] = (DP2[j] + DP2[j - a_list[i]]) % LARGE
for j in range(a_list[i]-1, -1, -1):
DP2[j] = (DP2[j]) % LARGE
S = sum(a_list)
if S % 2 == 0:
h = S // 2
res = ((3**N % LARGE) - (3*sum(DP[(h):]) % LARGE) + 3*DP2[h]) % LARGE
else:
h = S // 2 + 1
res = ((3**N % LARGE) - (3*sum(DP[(h):]) % LARGE)) % LARGE
print(res)
``` | instruction | 0 | 30,761 | 7 | 61,522 |
Yes | output | 1 | 30,761 | 7 | 61,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
# https://atcoder.jp/contests/tenka1-2019/submissions/5065329
import sys
input = sys.stdin.readline
N = int(input())
a = [int(input()) for _ in range(N)]
total = sum(a)
MOD = 998244353
'''(1)全体'''
U = pow(3, N, MOD) # 三角形を構成できない場合を含むRGBの組み合わせ総数
'''(2)単円'''
dp = [[0] * (total + 1) for _ in range(N + 1)]
dp[0][0] = 1
# dp[i][R]:=i番目まででRとなる場合の数
for i in range(N):
for r in range(total + 1):
if r - a[i] >= 0:
dp[i + 1][r] += dp[i][r - a[i]]
dp[i + 1][r] += 2 * dp[i][r]
dp[i + 1][r] %= MOD
# g, bに加算する分で、2回起こる
# print(dp)
single_circle = 0
for r in range(total + 1):
if r >= (total / 2):
single_circle += dp[N][r]
# print(single_circle)
'''(3)２円の積集合'''
if total % 2 == 0:
dp2 = [[0] * (total + 1) for _ in range(N + 1)]
dp2[0][0] = 1
# dp[i][R]:=i番目まででRとなる場合の数
for i in range(N):
for r in range(int(total / 2) + 1):
if r - a[i] >= 0:
dp2[i + 1][r] += dp2[i][r - a[i]]
dp2[i + 1][r] += dp2[i][r]
dp2[i + 1][r] %= MOD
# print(dp2)
two_circle_intersect = dp2[N][int(total / 2)] if total % 2 == 0 else 0
'''(3)３円の積集合'''
three_circle_intersect = 0
'''(4)解答'''
ans = 0
ans += U
ans %= MOD
ans -= 3 * single_circle
ans %= MOD
ans += 3 * two_circle_intersect
ans %= MOD
ans -= three_circle_intersect
ans %= MOD
print(ans)
``` | instruction | 0 | 30,762 | 7 | 61,524 |
Yes | output | 1 | 30,762 | 7 | 61,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
M=998244353
N,*A=open(0)
d=[1]+[0]*7**6
e=[1]+[0]*7**6
z=1
s=0
for a in A:
a=int(a);s+=a;i=s;z*=3
while~i:d[i]+=d[i]+d[i-a];e[i]+=e[i-a];i-=1
i=s
while i*2>=s:z-=d[i]*3;i-=1
print((z+~s%2*e[s//2]*3)%M)
``` | instruction | 0 | 30,763 | 7 | 61,526 |
No | output | 1 | 30,763 | 7 | 61,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
def dp_table(*counts): # dp[x][y]を作りたかったら dp_table(xのサイズ,yのサイズ)
if counts:
return [dp_table(*counts[1:]) for _ in range(counts[0])]
else:
return 0
def main():
N = int(input())
A = [int(input()) for i in range(N)]
mod = 998244353
ans = pow(3, N, mod)
sum_A = sum(A)
dp = dp_table(N, sum_A + 1)
# dp[i][j]: i番目までの中から辺をいくつか選んで、長さがjになるパターン数
# init
dp[0][A[0]] = 1
dp[0][0] = 2 # 他の色のうちどちらに入れるかで2通り
# init2
dp2 = dp_table(N, sum_A+1) # ある二色でsum_Aを等分するパターンを数える
dp2[0][A[0]] = 1
dp2[0][0] = 1 # 今度は片方に集中したパターン
# dp
sum_now = A[0]
for i in range(1, N):
a = A[i]
sum_now += a
for length in range(0, sum_now+1):
# A[i]を追加しないのなら他の２食のうちどちらに入れるかで2倍選択肢がある
if length >= a:
dp[i][length] = dp[i - 1][length - a] + dp[i - 1][length] << 1
else:
dp[i][length] = dp[i-1][length] << 1
dp[i][length] %= mod
for length in range(min(sum_now + 1, sum_A // 2 + 1)):
if length >= a:
dp2[i][length] = dp2[i - 1][length - a] + dp2[i - 1][length]
else:
dp2[i][length] = dp2[i-1][length]
dp2[i][length] %= mod
# print(sum(dp[N-1][sum_A//2]))
# print(dp)
ans -= 3*sum(dp[N-1][(sum_A+1)//2:])
if sum_A % 2 == 0:
ans += 3*dp2[N-1][sum_A//2]
print(ans % mod)
if __name__ == '__main__':
main()
``` | instruction | 0 | 30,764 | 7 | 61,528 |
No | output | 1 | 30,764 | 7 | 61,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
import sys
def input():
return sys.stdin.readline()[:-1]
N = int(input())
A = [int(input()) for i in range(N)]
A.sort()
cumA = [0]
for a in A:
cumA.append(a+cumA[-1])
from functools import lru_cache
mod = 998244353
@lru_cache(maxsize=None)
def solve(i, s, m):
if i==-1:
if s==0 and m==0:
return 1
else:
return 0
if s<0 or m<0:
return 0
if s>cumA[i+1] or m>cumA[i+1]:
return 0
return (solve(i-1, s, m) + solve(i-1, s-A[i], m) + solve(i-1, s, m-A[i])) % mod
ans = 0
su = sum(A)
for sm in range(su//2+1, su*2//3+1):
l = su - sm
for s in range(max(1, su-l-l), sm//2+1):
m = sm - s
if not (s <= m <= l):
#print("a")
continue
if s==m==l:
mul = 1
elif s==m or m==l:
mul = 3
else:
mul = 6
ans += solve(N-1, s, m) * mul
ans %= mod
print(ans)
``` | instruction | 0 | 30,765 | 7 | 61,530 |
No | output | 1 | 30,765 | 7 | 61,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given N integers. The i-th integer is a_i. Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:
* Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.
Constraints
* 3 \leq N \leq 300
* 1 \leq a_i \leq 300(1\leq i\leq N)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
a_1
:
a_N
Output
Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.
Examples
Input
4
1
1
1
2
Output
18
Input
6
1
3
2
3
5
2
Output
150
Input
20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4
Output
563038556
Submitted Solution:
```
LEN = 1000
N = int(input())
A = []
for _ in range(N):
A.append(int(input()))
dp = [[[0 for i in range(LEN)] for j in range(LEN)] for k in range(LEN)]
dp[0][0][0] = 1
s = 0
for a in A:
for i in range(s+1):
for j in range(s+1-i):
k = s - i - j
dp[i+a][j][k] += dp[i][j][k]
dp[i][j+a][k] += dp[i][j][k]
dp[i][j][k+a] += dp[i][j][k]
s += a
res = 0
for i in range(1, LEN):
for j in range(1, LEN):
k = s - i - j
if k>0 and i+j>k and j+k>i and k+i>j:
res += dp[i][j][k]
print(res%998244353)
``` | instruction | 0 | 30,766 | 7 | 61,532 |
No | output | 1 | 30,766 | 7 | 61,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses along the road where Anya lives, each one is painted in one of k possible colors.
Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.
Help Anya find the longest segment with this property.
Input
The first line contains two integers n and k — the number of houses and the number of colors (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 100 000).
The next line contains n integers a_1, a_2, …, a_n — the colors of the houses along the road (1 ≤ a_i ≤ k).
Output
Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.
Example
Input
8 3
1 2 3 3 2 1 2 2
Output
4
Note
In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3, 2, 1, 2] and its length is 4 houses. | instruction | 0 | 30,994 | 7 | 61,988 |
Tags: implementation
Correct Solution:
```
n,m = map(int,input().split())
a = list(map(int,input().split()))
max = 1
count = 1
for i in range(len(a)-1):
if a[i] != a[i+1]:
count+=1
if max<count:
max = count
else:
count = 1
print(max)
``` | output | 1 | 30,994 | 7 | 61,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses along the road where Anya lives, each one is painted in one of k possible colors.
Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.
Help Anya find the longest segment with this property.
Input
The first line contains two integers n and k — the number of houses and the number of colors (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 100 000).
The next line contains n integers a_1, a_2, …, a_n — the colors of the houses along the road (1 ≤ a_i ≤ k).
Output
Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.
Example
Input
8 3
1 2 3 3 2 1 2 2
Output
4
Note
In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3, 2, 1, 2] and its length is 4 houses. | instruction | 0 | 30,995 | 7 | 61,990 |
Tags: implementation
Correct Solution:
```
a = list(map(int, input().split()))
b = list(map(int, input().split()))
k = 1
t = []
if len(b) > 1:
for j in range(len(b) - 1):
if b[j] != b[j + 1]:
k = k + 1
t.append(k)
elif b[j] == b[j + 1]:
t.append(k)
k = 1
print(max(t))
else:
print(len(b))
``` | output | 1 | 30,995 | 7 | 61,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses along the road where Anya lives, each one is painted in one of k possible colors.
Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.
Help Anya find the longest segment with this property.
Input
The first line contains two integers n and k — the number of houses and the number of colors (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 100 000).
The next line contains n integers a_1, a_2, …, a_n — the colors of the houses along the road (1 ≤ a_i ≤ k).
Output
Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.
Example
Input
8 3
1 2 3 3 2 1 2 2
Output
4
Note
In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3, 2, 1, 2] and its length is 4 houses. | instruction | 0 | 30,996 | 7 | 61,992 |
Tags: implementation
Correct Solution:
```
def iter_from_input():
return map(int, input().split())
def solve(n, k, colors):
colors = iter(colors)
previous_color = next(colors)
current_length = 1
max_length = current_length
for color in colors:
if color == previous_color:
max_length = max(max_length, current_length)
current_length = 1
else:
current_length += 1
previous_color = color
# For last iteration
max_length = max(max_length, current_length)
return max_length
def test():
assert (solve(4,1, (1,1,1,1)) == 1)
assert (solve(4,2, (1,1,1,2)) == 2)
assert (solve(4,2, (2,1,2,1)) == 4)
# test()
def main():
n, k = iter_from_input()
colors = list(iter_from_input())
result = solve(n, k, colors)
print(result)
main()
``` | output | 1 | 30,996 | 7 | 61,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses along the road where Anya lives, each one is painted in one of k possible colors.
Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.
Help Anya find the longest segment with this property.
Input
The first line contains two integers n and k — the number of houses and the number of colors (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 100 000).
The next line contains n integers a_1, a_2, …, a_n — the colors of the houses along the road (1 ≤ a_i ≤ k).
Output
Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.
Example
Input
8 3
1 2 3 3 2 1 2 2
Output
4
Note
In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3, 2, 1, 2] and its length is 4 houses. | instruction | 0 | 30,999 | 7 | 61,998 |
Tags: implementation
Correct Solution:
```
n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
count = 1
m = 0
for i in range(n-1):
if a[i] != a[i+1]:
count += 1
else:
if count > m:
m = count
count = 1
if count > m:
m = count
print(m)
``` | output | 1 | 30,999 | 7 | 61,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n houses along the road where Anya lives, each one is painted in one of k possible colors.
Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.
Help Anya find the longest segment with this property.
Input
The first line contains two integers n and k — the number of houses and the number of colors (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 100 000).
The next line contains n integers a_1, a_2, …, a_n — the colors of the houses along the road (1 ≤ a_i ≤ k).
Output
Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.
Example
Input
8 3
1 2 3 3 2 1 2 2
Output
4
Note
In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3, 2, 1, 2] and its length is 4 houses. | instruction | 0 | 31,000 | 7 | 62,000 |
Tags: implementation
Correct Solution:
```
n,k=map(int,input().split())
array=list(map(int,input().split()))
array.append(0)
count=0
ar=[1]
for i in range(len(array)-1):
if(array[i]==array[i+1]):
count+=1
ar.append(count)
count=0
else:
count+=1
if(i==(len(array)-2)):
ar.append(count)
print(max(ar))
``` | output | 1 | 31,000 | 7 | 62,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs max(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the size of the square grid.
Each of the next n lines contains a string of length n, consisting of characters '.' and '#'. The j-th character of the i-th line is '#' if the cell with coordinates (i, j) is black, otherwise it is white.
Output
Print a single integer — the minimum total cost to paint all cells in white.
Examples
Input
3
###
#.#
###
Output
3
Input
3
...
...
...
Output
0
Input
4
#...
....
....
#...
Output
2
Input
5
#...#
.#.#.
.....
.#...
#....
Output
5
Note
The examples and some of optimal solutions are shown on the pictures below.
<image>
Submitted Solution:
```
def extract_zero_boundary(array, r1, c1, r2, c2):
up = r1
while(sum(array[up]) == 0):
up += 1
down = r2
while(sum(array[down]) == 0):
down -= 1
new_array = list(map(list, zip(*array[up: down + 1])))
left = c1
while(sum(new_array[left]) == 0):
left += 1
right = c2
while(sum(new_array[right]) == 0):
right -= 1
return (up, left, down, right)
def minimal_cost(array, x1, y1, x2, y2, d = {}):
if (x1, y1, x2, y2) in d:
return d[(x1, y1, x2, y2)]
if sum(map(sum, array)) == 0:
d[(x1, y1, x2, y2)] = 0
return d[(x1, y1, x2, y2)]
(r1, c1, r2, c2) = extract_zero_boundary(array, x1, y1, x2, y2)
if (r1, c1, r2, c2) in d:
d[(x1, y1, x2, y2)] = d[(r1, c1, r2, c2)]
return d[(r1, c1, r2, c2)]
transposed = 0
arr = array
if r2-r1 < c2-c1:
r1, c1, r2, c2 = c1, r1, c2, r2
transposed = 1
arr = list(map(list, zip(*array)))
zero_row = [] #row indices that corespond to zero_rows (splitting rows)
temp = -5
for i in range(r1, r2+1):
if sum(arr[i]) == 0:
if i != temp + 1:
zero_row.append(i)
temp = i
if not zero_row: #if there is no zero_rows, then cover entirely
if transposed == 0:
d[(r1, c1, r2, c2)] = d[(x1, y1, x2, y2)] = r2-r1+1
if transposed == 1:
d[(c1, r1, c2, r2)] = d[(x1, y1, x2, y2)] = r2-r1+1
return r2-r1+1
if zero_row:
result = r2-r1+1
for j in range(len(zero_row)):
if transposed == 0:
result = min(result, minimal_cost(array, r1, c1, zero_row[j]-1, c2, d) +
minimal_cost(array, zero_row[j]+1, c1, r2, c2, d))
d[(x1, y1, x2, y2)] = d[(r1, c1, r2, c2)] = result
if transposed == 1:
result = min(result, minimal_cost(array, c1, r1, c2, zero_row[j]-1, d) +
minimal_cost(array, c1, zero_row[j]+1, c2, r2, d))
d[(x1, y1, x2, y2)] = d[(c1, r1, c2, r2)] = result
return result
n = int(input())
array = list(list(input()) for _ in range(n))
for i in range(n):
for j in range(n):
array[i][j] = 1 if array[i][j] == '#' else 0
if n != 50:
print(minimal_cost(array, 0, 0, n-1, n-1, {}))
else:
a = []
for i in range(n):
temp = [0]
for j in range(n):
if array[i][j] == 1:
temp.append(j)
a.append(temp)
print(a)
``` | instruction | 0 | 31,047 | 7 | 62,094 |
No | output | 1 | 31,047 | 7 | 62,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs max(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the size of the square grid.
Each of the next n lines contains a string of length n, consisting of characters '.' and '#'. The j-th character of the i-th line is '#' if the cell with coordinates (i, j) is black, otherwise it is white.
Output
Print a single integer — the minimum total cost to paint all cells in white.
Examples
Input
3
###
#.#
###
Output
3
Input
3
...
...
...
Output
0
Input
4
#...
....
....
#...
Output
2
Input
5
#...#
.#.#.
.....
.#...
#....
Output
5
Note
The examples and some of optimal solutions are shown on the pictures below.
<image>
Submitted Solution:
```
def count_black(A):
res=0
indi=0
for i in range(len(A)):
if A[i]=='#':
res=res+1
indi=i
return res, indi
def process(A, n):
F=[None for x in range(n+1)]
F[1], max_indi=count_black(A[1])
for i in range(2, n+1):
temp, temp_ind=count_black(A[i])
max_indi=max(max_indi, temp_ind)
# print(max_indi)
F[i]=min(F[i-1]+temp, max(i, max_indi))
return F[n]
def generation(x, n):
A=['.' for i in range(n)]
a=x//2
A[a]='#'
A[a+1]='#'
A[a+3]='#'
A[a+4]='#'
return A
n=int(input())
A=[[None for i in range(n)]]
for i in range(n):
A.append([x for x in input()])
# for i in range(1, n+1):
# if i%2!=1:
# A.append(['.' for x in range(n)])
# else:
# A.append(generation(i, n))
# for i in reversed(process(A,n)):
# print(i)
print(process(A,n))
``` | instruction | 0 | 31,048 | 7 | 62,096 |
No | output | 1 | 31,048 | 7 | 62,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs max(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the size of the square grid.
Each of the next n lines contains a string of length n, consisting of characters '.' and '#'. The j-th character of the i-th line is '#' if the cell with coordinates (i, j) is black, otherwise it is white.
Output
Print a single integer — the minimum total cost to paint all cells in white.
Examples
Input
3
###
#.#
###
Output
3
Input
3
...
...
...
Output
0
Input
4
#...
....
....
#...
Output
2
Input
5
#...#
.#.#.
.....
.#...
#....
Output
5
Note
The examples and some of optimal solutions are shown on the pictures below.
<image>
Submitted Solution:
```
def count_black(A):
res=0
for i in A:
if i=='#':
res=res+1
return res
def process(A, n):
F=[None for x in range(n+1)]
F[1]=count_black(A[1])
for i in range(2, n+1):
temp=count_black(A[i])
F[i]=min(F[i-1]+temp, max(i, n))
return F[n]
n=int(input())
A=[[None for i in range(n)]]
for i in range(n):
A.append([x for x in input()])
# for i in reversed(process(A,n)):
# print(i)
print(process(A,n))
``` | instruction | 0 | 31,049 | 7 | 62,098 |
No | output | 1 | 31,049 | 7 | 62,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs max(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.
Input
The first line contains a single integer n (1 ≤ n ≤ 50) — the size of the square grid.
Each of the next n lines contains a string of length n, consisting of characters '.' and '#'. The j-th character of the i-th line is '#' if the cell with coordinates (i, j) is black, otherwise it is white.
Output
Print a single integer — the minimum total cost to paint all cells in white.
Examples
Input
3
###
#.#
###
Output
3
Input
3
...
...
...
Output
0
Input
4
#...
....
....
#...
Output
2
Input
5
#...#
.#.#.
.....
.#...
#....
Output
5
Note
The examples and some of optimal solutions are shown on the pictures below.
<image>
Submitted Solution:
```
def extract_zero_boundary(array, r1, c1, r2, c2):
up = r1
while(sum(array[up]) == 0):
up += 1
down = r2
while(sum(array[down]) == 0):
down -= 1
new_array = list(map(list, zip(*array[up: down + 1])))
left = c1
while(sum(new_array[left]) == 0):
left += 1
right = c2
while(sum(new_array[right]) == 0):
right -= 1
return (up, left, down, right)
def minimal_cost(array, x1, y1, x2, y2, d = {}):
if (x1, y1, x2, y2) in d:
return d[(x1, y1, x2, y2)]
if sum(map(sum, array)) == 0:
d[(x1, y1, x2, y2)] = 0
return d[(x1, y1, x2, y2)]
(r1, c1, r2, c2) = extract_zero_boundary(array, x1, y1, x2, y2)
if (r1, c1, r2, c2) in d:
d[(x1, y1, x2, y2)] = d[(r1, c1, r2, c2)]
return d[(r1, c1, r2, c2)]
transposed = 0
arr = array
if r2-r1 < c2-c1:
r1, c1, r2, c2 = c1, r1, c2, r2
transposed = 1
arr = list(map(list, zip(*array)))
zero_row = [] #row indices that corespond to zero_rows (splitting rows)
temp = -5
for i in range(r1, r2+1):
if sum(arr[i]) == 0:
if i != temp + 1:
zero_row.append(i)
temp = i
if not zero_row: #if there is no zero_rows, then cover entirely
if transposed == 0:
d[(r1, c1, r2, c2)] = d[(x1, y1, x2, y2)] = r2-r1+1
if transposed == 1:
d[(c1, r1, c2, r2)] = d[(x1, y1, x2, y2)] = r2-r1+1
return r2-r1+1
if zero_row:
result = r2-r1+1
for j in range(len(zero_row)):
if transposed == 0:
result = min(result, minimal_cost(array, r1, c1, zero_row[j]-1, c2, d) +
minimal_cost(array, zero_row[j]+1, c1, r2, c2, d))
d[(x1, y1, x2, y2)] = d[(r1, c1, r2, c2)] = result
if transposed == 1:
result = min(result, minimal_cost(array, c1, r1, c2, zero_row[j]-1, d) +
minimal_cost(array, c1, zero_row[j]+1, c2, r2, d))
d[(x1, y1, x2, y2)] = d[(c1, r1, c2, r2)] = result
return result
n = int(input())
array = list(list(input()) for _ in range(n))
for i in range(n):
for j in range(n):
array[i][j] = 1 if array[i][j] == '#' else 0
if n != 50:
print(minimal_cost(array, 0, 0, n-1, n-1, {}))
else:
a = []
for i in range(n):
temp = []
for j in range(n):
if array[i][j] == 1:
temp.append(j)
a.append(temp)
print(a)
``` | instruction | 0 | 31,050 | 7 | 62,100 |
No | output | 1 | 31,050 | 7 | 62,101 |
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