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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N Snukes lining up in a row. You are given a string S of length N. The i-th Snuke from the front has two red balls if the i-th character in S is `0`; one red ball and one blue ball if the i-th character in S is `1`; two blue balls if the i-th character in S is `2`.
Takahashi has a sequence that is initially empty. Find the number of the possible sequences he may have after repeating the following procedure 2N times, modulo 998244353:
* Each Snuke who has one or more balls simultaneously chooses one of his balls and hand it to the Snuke in front of him, or hand it to Takahashi if he is the first Snuke in the row.
* Takahashi receives the ball and put it to the end of his sequence.
Constraints
* 1 \leq |S| \leq 2000
* S consists of `0`,`1` and `2`.
Note that the integer N is not directly given in input; it is given indirectly as the length of the string S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of the possible sequences Takahashi may have after repeating the procedure 2N times, modulo 998244353.
Examples
Input
02
Output
3
Input
1210
Output
55
Input
12001021211100201020
Output
543589959
Submitted Solution:
```
from collections import defaultdict
def inpl(): return list(map(int, input().split()))
MOD = 998244353
S = input()
N = len(S)
B = [0]*(2*N+1)
R = [0]*(2*N+1)
for i in range(N):
b = int(S[i])
B[i] = b
R[i] = 2 - b
DP = defaultdict(int)
DP[(B[0], R[0])] = 1
for i in range(1, 2*N+1):
DP2 = defaultdict(int)
b1, r1 = B[i], R[i]
for (b0, r0), v in DP.items():
b2, r2 = b0+b1, r0+r1
if b0:
DP2[(b2-1, r2)] = (DP2[(b2-1, r2)] + v)%MOD
if r0:
DP2[(b2, r2-1)] = (DP2[(b2, r2-1)] + v)%MOD
DP = DP2
print(sum(DP.values()))
```
|
instruction
| 0
| 8,903
| 16
| 17,806
|
Yes
|
output
| 1
| 8,903
| 16
| 17,807
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N Snukes lining up in a row. You are given a string S of length N. The i-th Snuke from the front has two red balls if the i-th character in S is `0`; one red ball and one blue ball if the i-th character in S is `1`; two blue balls if the i-th character in S is `2`.
Takahashi has a sequence that is initially empty. Find the number of the possible sequences he may have after repeating the following procedure 2N times, modulo 998244353:
* Each Snuke who has one or more balls simultaneously chooses one of his balls and hand it to the Snuke in front of him, or hand it to Takahashi if he is the first Snuke in the row.
* Takahashi receives the ball and put it to the end of his sequence.
Constraints
* 1 \leq |S| \leq 2000
* S consists of `0`,`1` and `2`.
Note that the integer N is not directly given in input; it is given indirectly as the length of the string S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of the possible sequences Takahashi may have after repeating the procedure 2N times, modulo 998244353.
Examples
Input
02
Output
3
Input
1210
Output
55
Input
12001021211100201020
Output
543589959
Submitted Solution:
```
s = input()
n = len(s)
mod = 998244353
R = [0] * n
B = [0] * n
for i in range(n):
if s[i] == '0':
R[i] = 2
elif s[i] == '1':
R[i] = 1
B[i] = 1
else:
B[i] = 2
for i in range(n-1):
R[i+1] += R[i]
B[i+1] += B[i]
dp = [[0] * (2*n+10) for _ in range(2*n+10)]
if R[0] > 0:
dp[0][1] = 1
if B[0] > 0:
dp[0][0] = 1
for i in range(n-1):
for r in range(2*n+1):
if R[i+1] >= r+1:
dp[i+1][r+1] += dp[i][r]
dp[i+1][r+1] %= mod
if i+2-r < 0:
break
if B[i+1] >= i+2-r:
dp[i+1][r] += dp[i][r]
dp[i+1][r] %= mod
for i in range(n-1, 2*n-1):
for r in range(2*n+1):
if R[n-1] >= r+1:
dp[i + 1][r + 1] += dp[i][r]
dp[i + 1][r + 1] %= mod
if i+2-r < 0:
break
if B[n-1] >= i+2-r:
dp[i+1][r] += dp[i][r]
dp[i+1][r] %= mod
ans = 0
for r in range(2*n+1):
ans += dp[2*n-1][r]
ans %= mod
print(ans)
```
|
instruction
| 0
| 8,904
| 16
| 17,808
|
No
|
output
| 1
| 8,904
| 16
| 17,809
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N Snukes lining up in a row. You are given a string S of length N. The i-th Snuke from the front has two red balls if the i-th character in S is `0`; one red ball and one blue ball if the i-th character in S is `1`; two blue balls if the i-th character in S is `2`.
Takahashi has a sequence that is initially empty. Find the number of the possible sequences he may have after repeating the following procedure 2N times, modulo 998244353:
* Each Snuke who has one or more balls simultaneously chooses one of his balls and hand it to the Snuke in front of him, or hand it to Takahashi if he is the first Snuke in the row.
* Takahashi receives the ball and put it to the end of his sequence.
Constraints
* 1 \leq |S| \leq 2000
* S consists of `0`,`1` and `2`.
Note that the integer N is not directly given in input; it is given indirectly as the length of the string S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of the possible sequences Takahashi may have after repeating the procedure 2N times, modulo 998244353.
Examples
Input
02
Output
3
Input
1210
Output
55
Input
12001021211100201020
Output
543589959
Submitted Solution:
```
from collections import defaultdict
def tqdm(iterable): return iterable
def inpl(): return list(map(int, input().split()))
MOD = 998244353
S = input()
N = len(S)
B = [0]*N
R = [0]*N
for i in range(N):
b = int(S[i])
B[i] = b
R[i] = 2 - b
DP = defaultdict(int)
DP[(B[0], R[0])] = 1
for i in tqdm(range(1, N)):
DP2 = defaultdict(int)
b1, r1 = B[i], R[i]
for (b0, r0), v in DP.items():
b2, r2 = b0+b1, r0+r1
if b0:
DP2[(b2-1, r2)] = (DP2[(b2-1, r2)] + v)%MOD
if r0:
DP2[(b2, r2-1)] = (DP2[(b2, r2-1)] + v)%MOD
DP = DP2
for _ in tqdm(range(N+1)):
DP2 = defaultdict(int)
for (b, r), v in DP.items():
if b:
DP2[(b-1, r)] = (DP2[(b-1, r)] + v)%MOD
if r:
DP2[(b, r-1)] = (DP2[(b, r-1)] + v)%MOD
DP = DP2
print(sum(DP.values()))
```
|
instruction
| 0
| 8,905
| 16
| 17,810
|
No
|
output
| 1
| 8,905
| 16
| 17,811
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N Snukes lining up in a row. You are given a string S of length N. The i-th Snuke from the front has two red balls if the i-th character in S is `0`; one red ball and one blue ball if the i-th character in S is `1`; two blue balls if the i-th character in S is `2`.
Takahashi has a sequence that is initially empty. Find the number of the possible sequences he may have after repeating the following procedure 2N times, modulo 998244353:
* Each Snuke who has one or more balls simultaneously chooses one of his balls and hand it to the Snuke in front of him, or hand it to Takahashi if he is the first Snuke in the row.
* Takahashi receives the ball and put it to the end of his sequence.
Constraints
* 1 \leq |S| \leq 2000
* S consists of `0`,`1` and `2`.
Note that the integer N is not directly given in input; it is given indirectly as the length of the string S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of the possible sequences Takahashi may have after repeating the procedure 2N times, modulo 998244353.
Examples
Input
02
Output
3
Input
1210
Output
55
Input
12001021211100201020
Output
543589959
Submitted Solution:
```
mod = 998244353
s = list(map(int, list(input())))
n = len(s)
dp = [[0]*(sum(s)+1) for _ in range(2*n+1)]
dp[0][0] = 1
curr = curb = 0
for i in range(2*n):
if i < n:
curb += s[i]
curr += 2 - s[i]
for j in range(min(i, curb)+1):
if dp[i][j]:
dp[i][j] %= mod
if i - j < curr:
dp[i+1][j] += dp[i][j]
if j < curb:
dp[i+1][j+1] += dp[i][j]
print(dp[2*n][curb])
```
|
instruction
| 0
| 8,906
| 16
| 17,812
|
No
|
output
| 1
| 8,906
| 16
| 17,813
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N Snukes lining up in a row. You are given a string S of length N. The i-th Snuke from the front has two red balls if the i-th character in S is `0`; one red ball and one blue ball if the i-th character in S is `1`; two blue balls if the i-th character in S is `2`.
Takahashi has a sequence that is initially empty. Find the number of the possible sequences he may have after repeating the following procedure 2N times, modulo 998244353:
* Each Snuke who has one or more balls simultaneously chooses one of his balls and hand it to the Snuke in front of him, or hand it to Takahashi if he is the first Snuke in the row.
* Takahashi receives the ball and put it to the end of his sequence.
Constraints
* 1 \leq |S| \leq 2000
* S consists of `0`,`1` and `2`.
Note that the integer N is not directly given in input; it is given indirectly as the length of the string S.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of the possible sequences Takahashi may have after repeating the procedure 2N times, modulo 998244353.
Examples
Input
02
Output
3
Input
1210
Output
55
Input
12001021211100201020
Output
543589959
Submitted Solution:
```
mod = 998244353
import sys
sys.setrecursionlimit(10**9)
S = input()
N = len(S)
Sum = [0]*(N+1)
for i in range(1, N+1):
Sum[i] = Sum[i-1]+int(S[i-1])
from functools import lru_cache
@lru_cache(maxsize=10**8)
def dp(i, j):
n = min(i+j+1, N)
if i+j == N*2:
return 1
res = 0
s = Sum[n]
if i<s:
res+=dp(i+1, j)
if j<2*n-s:
res+=dp(i, j+1)
return res%mod
print(dp(0, 0))
```
|
instruction
| 0
| 8,907
| 16
| 17,814
|
No
|
output
| 1
| 8,907
| 16
| 17,815
|
Provide a correct Python 3 solution for this coding contest problem.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
|
instruction
| 0
| 8,908
| 16
| 17,816
|
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from bisect import bisect_left
MOD = 10**9 + 7
N,M = map(int,readline().split())
X = list(map(int,readline().split()))
Y = list(map(int,readline().split()))
L = []; R = []
for x in X:
i = bisect_left(Y,x) # 右出口
if i in [0,M]:
continue
y0,y1 = Y[i-1:i+1]
L.append(y0-x); R.append(y1-x)
# 座圧
Rtoi = {x:i for i,x in enumerate(sorted(set(R)),1)}
R = [Rtoi[r] for r in R]
if len(R) == 0:
print(1)
exit()
"""
・計算方法
・これをやるために実際にはBITを使う
dp = [0] * (max(R)+1)
for _,r in sorted(set(zip(L,R)),reverse=True):
dp[r] += 1 + sum(dp[1:r]) #これをやるために BIT で dp を持つ
answer=1+sum(dp)
print(answer)
"""
class BIT():
def __init__(self, max_n):
self.size = max_n + 1
self.tree = [0] * self.size
def get_sum(self,i):
s = 0
while i:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i < self.size:
self.tree[i] += x
i += i & -i
dp = BIT(max_n=max(R))
for _,r in sorted(set(zip(L,R)),reverse=True):
x = dp.get_sum(r-1) + 1; x %= MOD
dp.add(r,x)
answer=1+dp.get_sum(max(R))
answer %= MOD
print(answer)
```
|
output
| 1
| 8,908
| 16
| 17,817
|
Provide a correct Python 3 solution for this coding contest problem.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
|
instruction
| 0
| 8,909
| 16
| 17,818
|
"Correct Solution:
```
from bisect import bisect
from collections import defaultdict
class Bit:
def __init__(self, n, MOD):
self.size = n
self.tree = [0] * (n + 1)
self.depth = n.bit_length()
self.mod = MOD
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s % self.mod
def add(self, i, x):
while i <= self.size:
self.tree[i] = (self.tree[i] + x) % self.mod
i += i & -i
def debug_print(self):
for i in range(1, self.size + 1):
j = (i & -i).bit_length()
print(' ' * j, self.tree[i])
def lower_bound(self, x):
sum_ = 0
pos = 0
for i in range(self.depth, -1, -1):
k = pos + (1 << i)
if k <= self.size and sum_ + self.tree[k] < x:
sum_ += self.tree[k]
pos += 1 << i
return pos + 1, sum_
n, m = map(int, input().split())
xxx = list(map(int, input().split()))
yyy = list(map(int, input().split()))
ab = defaultdict(set)
coordinates = set()
for x in xxx:
if x < yyy[0] or yyy[-1] < x:
continue
i = bisect(yyy, x)
a = x - yyy[i - 1]
b = yyy[i] - x
ab[a].add(b)
coordinates.add(b)
# Bitのindexは1から始まるように作っているが、"0"を取れるようにするため、全体を1ずらす
cor_dict = {b: i for i, b in enumerate(sorted(coordinates), start=2)}
cdg = cor_dict.get
MOD = 10 ** 9 + 7
bit = Bit(len(coordinates) + 1, MOD)
bit.add(1, 1)
for a in sorted(ab):
bbb = sorted(map(cdg, ab[a]), reverse=True)
for b in bbb:
bit.add(b, bit.sum(b - 1))
print(bit.sum(bit.size))
```
|
output
| 1
| 8,909
| 16
| 17,819
|
Provide a correct Python 3 solution for this coding contest problem.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
|
instruction
| 0
| 8,910
| 16
| 17,820
|
"Correct Solution:
```
from bisect import bisect
from collections import defaultdict
class Bit:
def __init__(self, n, MOD):
self.size = n
self.tree = [0] * (n + 1)
self.depth = n.bit_length()
self.mod = MOD
def sum(self, i):
s = 0
while i > 0:
s = (s + self.tree[i]) % self.mod
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] = (self.tree[i] + x) % self.mod
i += i & -i
def debug_print(self):
for i in range(1, self.size + 1):
j = (i & -i).bit_length()
print(' ' * j, self.tree[i])
def lower_bound(self, x):
sum_ = 0
pos = 0
for i in range(self.depth, -1, -1):
k = pos + (1 << i)
if k <= self.size and sum_ + self.tree[k] < x:
sum_ += self.tree[k]
pos += 1 << i
return pos + 1, sum_
n, m = map(int, input().split())
xxx = list(map(int, input().split()))
yyy = list(map(int, input().split()))
ab = defaultdict(set)
coordinates = set()
for x in xxx:
if x < yyy[0] or yyy[-1] < x:
continue
i = bisect(yyy, x)
a = x - yyy[i - 1]
b = yyy[i] - x
ab[a].add(b)
coordinates.add(b)
# Bitのindexは1から始まるように作っているが、"0"を取れるようにするため、全体を1ずらす
cor_dict = {b: i for i, b in enumerate(sorted(coordinates), start=2)}
cdg = cor_dict.get
MOD = 10 ** 9 + 7
bit = Bit(len(coordinates) + 1, MOD)
bit.add(1, 1)
for a in sorted(ab):
bbb = sorted(map(cdg, ab[a]), reverse=True)
for b in bbb:
bit.add(b, bit.sum(b - 1))
print(bit.sum(bit.size))
```
|
output
| 1
| 8,910
| 16
| 17,821
|
Provide a correct Python 3 solution for this coding contest problem.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
|
instruction
| 0
| 8,911
| 16
| 17,822
|
"Correct Solution:
```
from bisect import bisect
from collections import defaultdict
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
self.depth = n.bit_length()
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
def debug_print(self):
for i in range(1, self.size + 1):
j = (i & -i).bit_length()
print(' ' * j, self.tree[i])
def lower_bound(self, x):
sum_ = 0
pos = 0
for i in range(self.depth, -1, -1):
k = pos + (1 << i)
if k <= self.size and sum_ + self.tree[k] < x:
sum_ += self.tree[k]
pos += 1 << i
return pos + 1, sum_
n, m = map(int, input().split())
xxx = list(map(int, input().split()))
yyy = list(map(int, input().split()))
ab = defaultdict(set)
coordinates = set()
for x in xxx:
if x < yyy[0] or yyy[-1] < x:
continue
i = bisect(yyy, x)
a = x - yyy[i - 1]
b = yyy[i] - x
ab[a].add(b)
coordinates.add(b)
# Bitのindexは1から始まるように作っているが、"0"を取れるようにするため、全体を1ずらす
cor_dict = {b: i for i, b in enumerate(sorted(coordinates), start=2)}
cdg = cor_dict.get
bit = Bit(len(coordinates) + 1)
bit.add(1, 1)
for a in sorted(ab):
bbb = sorted(map(cdg, ab[a]), reverse=True)
for b in bbb:
bit.add(b, bit.sum(b - 1))
print(bit.sum(bit.size) % (10 ** 9 + 7))
```
|
output
| 1
| 8,911
| 16
| 17,823
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from bisect import bisect_left
MOD = 10**9 + 7
N,M = map(int,readline().split())
X = list(map(int,readline().split()))
Y = list(map(int,readline().split()))
L = []; R = []
for x in X:
i = bisect_left(Y,x) # 右出口
if i in [0,M]:
continue
y0,y1 = Y[i-1:i+1]
L.append(y0-x); R.append(y1-x)
# 座圧
Rtoi = {x:i for i,x in enumerate(sorted(set(R)),1)}
R = [Rtoi[r] for r in R]
if len(R) == 0:
print(1)
exit()
"""
・計算方法
・これをやるために実際にはBITを使う
dp = [0] * (max(R)+1)
for _,r in sorted(set(zip(L,R)),reverse=True):
dp[r] += 1 + sum(dp[1:r]) #これをやるために BIT で dp を持つ
answer=1+sum(dp)
print(answer)
"""
class BIT():
def __init__(self, max_n):
self.size = max_n + 1
self.tree = [0] * self.size
def get_sum(self,i):
s = 0
while i:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i < self.size:
self.tree[i] += x
i += i & -i
dp = BIT(max_n=max(R))
for _,r in sorted(set(zip(L,R)),reverse=True):
x = dp.get_sum(r-1) + 1; x %= MOD
dp.add(r,x)
answer=1+dp.get_sum(max(R))
print(answer)
```
|
instruction
| 0
| 8,912
| 16
| 17,824
|
No
|
output
| 1
| 8,912
| 16
| 17,825
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
Submitted Solution:
```
from bisect import bisect
from collections import defaultdict
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
self.depth = n.bit_length()
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
def debug_print(self):
for i in range(1, self.size + 1):
j = (i & -i).bit_length()
print(' ' * j, self.tree[i])
def lower_bound(self, x):
sum_ = 0
pos = 0
for i in range(self.depth, -1, -1):
k = pos + (1 << i)
if k <= self.size and sum_ + self.tree[k] < x:
sum_ += self.tree[k]
pos += 1 << i
return pos + 1, sum_
n, m = map(int, input().split())
xxx = list(map(int, input().split()))
yyy = list(map(int, input().split()))
ab = defaultdict(set)
coordinates = set()
for x in xxx:
if x < yyy[0] or yyy[-1] < x:
continue
i = bisect(yyy, x)
a = x - yyy[i - 1]
b = yyy[i] - x
ab[a].add(b)
coordinates.add(b)
# Bitのindexは1から始まるように作っているが、"0"を取れるようにするため、全体を1ずらす
cor_dict = {b: i for i, b in enumerate(sorted(coordinates), start=2)}
cdg = cor_dict.get
bit = Bit(len(coordinates) + 1)
bit.add(1, 1)
for a in sorted(ab):
bbb = sorted(map(cdg, ab[a]), reverse=True)
for b in bbb:
bit.add(b, bit.sum(b - 1))
print(bit.sum(bit.size))
```
|
instruction
| 0
| 8,913
| 16
| 17,826
|
No
|
output
| 1
| 8,913
| 16
| 17,827
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.
Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:
* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.
Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.
When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.
Constraints
* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.
Input
Input is given from Standard Input in the following format:
N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M
Output
Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.
Examples
Input
2 2
2 3
1 4
Output
3
Input
3 4
2 5 10
1 3 7 13
Output
8
Input
4 1
1 2 4 5
3
Output
1
Input
4 5
2 5 7 11
1 3 6 9 13
Output
6
Input
10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30
Output
22
Submitted Solution:
```
import bisect
r_n, e_n = map(int,input().split())
robot = list(map(int,input().split()))
exit = list(map(int,input().split()))
dict = []
for r in robot:
i = bisect.bisect(exit,r)
if i == 0 or i == e_n:
continue
left = exit[i - 1]
right = exit[i]
dict.append([r - left, right - r])
dict.sort()
def check(dict):
if len(dict) <= 1:
if len(dict) == 0:
return 1
else:
return 2
result = 0
left = dict[0][0]
count = 1
while True:
if count == len(dict) or dict[count][0] > left:
break
else:
count += 1
result += check(dict[count:])
right = dict[0][1]
new_dict = []
for d in dict:
if d[1] > right:
new_dict.append(d)
result += check(new_dict)
return result
print(check(dict))
```
|
instruction
| 0
| 8,914
| 16
| 17,828
|
No
|
output
| 1
| 8,914
| 16
| 17,829
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,931
| 16
| 17,862
|
"Correct Solution:
```
import sys
from collections import Counter
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split())) # applies to only numbers
rs = lambda: stdin.readline().rstrip() # ignores trailing space
H, W = rl()
l = []
for _ in range(H):
l.extend(list(rs()))
c = Counter(l)
count = W//2 * H
pair = 0
four = 0
for x in c.values():
four += x // 4
pair += x // 2
f = H // 2 * (W // 2)
if f > four:
print('No')
exit()
pair -= f * 2 #残りのペア
p = W // 2 if H&1 else 0
p += H // 2 if W&1 else 0
if p > pair:
print('No')
exit()
print('Yes')
# 16
```
|
output
| 1
| 8,931
| 16
| 17,863
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,932
| 16
| 17,864
|
"Correct Solution:
```
H,W = map(int,input().split())
A = [input() for i in range(H)]
from collections import Counter
ctr = Counter()
for row in A:
ctr.update(row)
if H%2==0 and W%2==0:
print('Yes' if all(v%4==0 for v in ctr.values()) else 'No')
elif H%2 and W%2:
odd = 0
for k,v in ctr.items():
if v%2:
odd += 1
ctr[k] -= 1
if odd != 1:
print('No')
exit()
four = 0
for v in ctr.values():
four += (v//4)*4
want = (H-H%2)*(W-W%2)
print('Yes' if four >= want else 'No')
else:
if any(v%2 for v in ctr.values()):
print('No')
exit()
four = 0
for v in ctr.values():
four += (v//4)*4
want = (H-H%2)*(W-W%2)
print('Yes' if four >= want else 'No')
```
|
output
| 1
| 8,932
| 16
| 17,865
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,933
| 16
| 17,866
|
"Correct Solution:
```
import string
string.ascii_lowercase
H,W = map(int,input().split())
A = ''
for i in range(H):
A += input()
#print(A)
S = {}
for s in string.ascii_lowercase:
if A.count(s) > 0:
S[s] = A.count(s)
#print(S)
h1 = H//2
w1 = W//2
h2 = H - h1*2
w2 = W - w1*2
for i in range(h1*w1):
flag = True
for s in S.keys():
if S[s] >= 4:
S[s] -= 4
flag = False
break
if flag:
print('No')
exit()
for i in range(h1*w2+w1*h2):
flag = True
for s in S.keys():
if S[s] >= 2:
S[s] -= 2
flag = False
break
if flag:
print('No')
exit()
print('Yes')
```
|
output
| 1
| 8,933
| 16
| 17,867
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,934
| 16
| 17,868
|
"Correct Solution:
```
t,_,s=open(0).read().partition('\n')
h,w=map(int,t.split())
a=b=c=0
if h%2and w%2:
a=h//2*(w//2)
c=1
b=(h*w-a*4-c)//2
elif h%2or w%2:
a=h//2*(w//2)
b=(h*w-a*4)//2
else:
a=h*w//4
l=[]
for i in set(s.replace('\n','')):
l.append(s.count(i))
for i in l:
if i>3:
while i>3and a:
i-=4
a-=1
if i>1:
while i>1and b:
i-=2
b-=1
if i and c:
i-=1
c-=1
if all(not t for t in(a,b,c)):
print('Yes')
else:
print('No')
```
|
output
| 1
| 8,934
| 16
| 17,869
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,935
| 16
| 17,870
|
"Correct Solution:
```
h, w = map(int, input().split())
a = {}
for i in range(h):
s = input()
for i in s:
a.setdefault(i, 0)
a[i] += 1
a2 = 0
a1 = 0
for i in a.values():
if i % 2 == 1:
a1 += 1
elif i % 4 == 2:
a2 += 2
if h % 2 == 1:
if w % 2 == 1:
b1 = 1
b2 = h + w - 2
else:
b1 = 0
b2 = w
else:
if w % 2 == 1:
b1 = 0
b2 = h
else:
b1 = 0
b2 = 0
if a1 > b1 or a2 > b2:
print("No")
else:
print("Yes")
```
|
output
| 1
| 8,935
| 16
| 17,871
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,936
| 16
| 17,872
|
"Correct Solution:
```
#!/usr/bin/env python3
from collections import Counter
def check():
h, w = map(int, input().split())
c = Counter()
for i in range(h):
c.update(input())
cc = {}
cc[4] = (h >> 1) * (w >> 1)
cc[2] = (h & 1) * (w >> 1) + (w & 1) * (h >> 1)
cc[1] = (h & 1) * (w & 1)
tt = list(c.values())
for v in [4, 2, 1]:
for i in range(len(tt)):
k = min(cc[v], tt[i] // v)
tt[i] -= k * v
cc[v] -= k
return sum(tt) == 0
print(["No", "Yes"][check()])
```
|
output
| 1
| 8,936
| 16
| 17,873
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,937
| 16
| 17,874
|
"Correct Solution:
```
h,w = map(int, input().split())
a = [input() for i in range(h)]
cnt = [0] * 26
for i in a:
for j in i:
cnt[ord(j)-ord("a")] += 1
two,no = 0,0
for i in cnt:
if i % 2:
no += 1
elif i % 4:
two += 1
if (h % 2) and (w % 2):
if no <= 1 and two <= h//2 + w//2:
print("Yes")
else:
print("No")
elif (h % 2):
if no == 0 and two <= w//2:
print("Yes")
else:
print("No")
elif (w % 2):
if no == 0 and two <= h//2:
print("Yes")
else:
print("No")
else:
if no == 0 and two == 0:
print("Yes")
else:
print("No")
```
|
output
| 1
| 8,937
| 16
| 17,875
|
Provide a correct Python 3 solution for this coding contest problem.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
|
instruction
| 0
| 8,938
| 16
| 17,876
|
"Correct Solution:
```
import heapq
from collections import defaultdict
def main():
h, w = map(int, input().split())
d = defaultdict(int)
for _ in range(h):
for i in input():
d[i] -= 1
char = list(d.values())
heapq.heapify(char)
hd, hm = divmod(h, 2)
wd, wm = divmod(w, 2)
pair = [-2] * hd + [-1] * hm
pair = [i * 2 for i in pair]*wd + pair*wm
heapq.heapify(pair)
while char:
tmp = heapq.heappop(char) - heapq.heappop(pair)
if tmp > 0:
return "No"
if tmp == 0:
continue
heapq.heappush(char, tmp)
return "Yes"
if __name__ == "__main__":
ans = main()
print(ans)
```
|
output
| 1
| 8,938
| 16
| 17,877
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
h,w = map(int, input().split())
b = [0]*26
for i in range(h):
a=input()
for j in range(w):
b[ord(a[j])-97]+=1
c = [0]*4
for i in range(len(b)):
c[b[i] % 4] += 1
d = 1
if h%2==1 and w%2==1:
if c[1]+c[3]!=1:
d = 0
elif c[2]+c[3]>h//2+w//2:
d = 0
elif h%2==1:
if c[1]+c[3]>0:
d = 0
elif c[2]>w//2:
d = 0
elif w%2==1:
if c[1]+c[3]>0:
d = 0
elif c[2]>h//2:
d = 0
else:
if c[1]+c[2]+c[3]>0:
d = 0
print('Yes' if d else 'No')
```
|
instruction
| 0
| 8,939
| 16
| 17,878
|
Yes
|
output
| 1
| 8,939
| 16
| 17,879
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
from collections import Counter
h, w, *s = open(0).read().split()
h = int(h)
w = int(w)
s = Counter(''.join(s))
odd = 0
two = 0
four = 0
for k, v in s.items():
four += v // 4
v %= 4
if v % 2 == 0:
two += v // 2
else:
odd += 1
if h % 2 == 1 and w % 2 == 1:
if odd == 1 and two <= h // 2 + w // 2:
print('Yes')
else:
print('No')
elif h % 2 == 1:
if odd == 0 and two <= w // 2:
print('Yes')
else:
print('No')
elif w % 2 == 1:
if odd == 0 and two <= h // 2:
print('Yes')
else:
print('No')
else:
if odd == 0 and two == 0:
print('Yes')
else:
print('No')
```
|
instruction
| 0
| 8,940
| 16
| 17,880
|
Yes
|
output
| 1
| 8,940
| 16
| 17,881
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
H,W=list(map(int,input().split()))
l=[]
for i in range(H):
l.append(list(input()))
from collections import Counter
l=sum(l,[])
l=Counter(l)
one=(H%2)*(W%2)
two=(H%2)*(W//2)+(W%2)*(H//2)
four=(H//2)*(W//2)
for i in l.values():
j=i//4
i=i%4
four-=j
if i==1:
one-=1
elif i==2:
two-=1
elif i==3:
one-=1;two-=1
if set([0]) == set([one,two,four]):
print("Yes")
else:
if four<0:
two+=four*2
four=0
if set([0]) == set([one,two,four]):
print("Yes")
else:
print("No")
```
|
instruction
| 0
| 8,941
| 16
| 17,882
|
Yes
|
output
| 1
| 8,941
| 16
| 17,883
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
# coding: utf-8
import collections
def main():
H, W = map(int, input().split())
cnt = collections.Counter()
for i in range(H):
for c in input().strip():
cnt[c] += 1
cnts = list(cnt.values())
l = len(cnts)
if H % 2 and W % 2:
for i in range(l):
if cnts[i] % 2 != 0:
cnts[i] -= 1
break
d = 0
if H % 2:
d += (W // 2) * 2
if W % 2:
d += (H // 2) * 2
i = 0
while d > 0 and i < l:
if cnts[i] % 4 != 0:
cnts[i] -= 2
d -= 2
i += 1
for k in cnts:
if k % 4 != 0:
return "No"
return "Yes"
print(main())
```
|
instruction
| 0
| 8,942
| 16
| 17,884
|
Yes
|
output
| 1
| 8,942
| 16
| 17,885
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
H, W = map(int,input().split())
S = []
for _ in range(H):
S.append(input())
# check palindrome
def main():
oddnum = (H*W) % 2
d = {}
for i in range(H):
for j in range(W):
ch = S[i][j]
if ch in d:
d[ch] += 1
else:
d[ch] = 1
check = sum([val % 2 for key,val in d.items()])
if check != oddnum:
return False
return True
ans = main()
if ans:
print("Yes")
else:
print("No")
```
|
instruction
| 0
| 8,943
| 16
| 17,886
|
No
|
output
| 1
| 8,943
| 16
| 17,887
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
# import bisect
from collections import Counter, deque # https://note.nkmk.me/python-scipy-connected-components/
# from copy import copy, deepcopy
# from functools import reduce
# from heapq import heapify, heappop, heappush
# from itertools import accumulate, permutations, combinations, combinations_with_replacement, groupby, product
# import math
# import numpy as np # Pythonのみ！
# from operator import xor
# import re
# from scipy.sparse.csgraph import connected_components # Pythonのみ！
# ↑cf. https://note.nkmk.me/python-scipy-connected-components/
# from scipy.sparse import csr_matrix
# import string
import sys
sys.setrecursionlimit(10 ** 5 + 10)
def input(): return sys.stdin.readline().strip()
def resolve():
def main():
gg = Counter([])
H, W = map(int, input().split())
grid = [[_ for _ in input()] for _ in range(H)]
for g in grid:
gg.update(Counter(g))
if H % 2 == 0 and W % 2 == 0:
for i in gg.values():
if i % 4 != 0:
return 'No'
return 'Yes'
elif H % 2 == 0 and W % 2 == 1:
cnt = 0
for i in gg.values():
if i % 4 == 2:
cnt += 2
elif i % 4 == 1 or i % 4 == 3:
return 'No'
if cnt <= H and cnt % 2 == 0:
return 'Yes'
else:
return 'No'
elif H % 2 == 1 and W % 2 == 0:
cnt = 0
for i in gg.values():
if i % 4 == 2:
cnt += 2
elif i % 4 == 1 or i % 4 == 3:
return 'No'
if cnt <= W and cnt % 2 == 0:
return 'Yes'
else:
return 'No'
else:
cnt0 = 0
cnt1 = 0
cnt2 = 0
for i in gg.values():
if i % 4 == 1:
cnt1 += 1
elif i % 4 == 2:
cnt2 += 2
elif i % 4 == 3:
cnt2 += 2
cnt1 += 1
else:
cnt0 += 1
if cnt1 == 1 and (cnt2 <= (H + W - 2) and cnt2 % 2 == 0 and cnt0 % 4 == 0):
return 'Yes'
else:
return 'No'
print(main())
resolve()
```
|
instruction
| 0
| 8,944
| 16
| 17,888
|
No
|
output
| 1
| 8,944
| 16
| 17,889
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
from collections import Counter
h,w=map(int,input().split())
s=""
for i in range(h):
s+=input()
sc=Counter(s)
d=[0,0,0]
for v in sc.values():
d[0]+=(v//4)*4
d[1]+=2 if v%4>=2 else 0
d[2]+=v%2
if w%2==0 and h%2==0:
print("Yes" if d[0]==h*w else "No")
elif (w%2)*(h%2)==1:
print("Yes" if d[1]<=(w-1)+(h-1) and d[2]==1 else "No")
elif w%2==0:
print("Yes" if d[1]<=w and d[2]==0 else "No")
elif h%2==0:
print("Yes" if d[1]<=h and d[2]==0 else "No")
from collections import Counter
h,w=map(int,input().split())
s=""
for i in range(h):
s+=input()
sc=Counter(s)
d=[0,0,0]
for v in sc.values():
d[0]+=(v//4)*4
d[1]+=2 if v%4>=2 else 0
d[2]+=v%2
if w%2==0 and h%2==0:
print("Yes" if d[0]==h*w else "No")
elif (w%2)*(h%2)==1:
print("Yes" if d[1]<=(w-1)+(h-1) and d[2]==1 else "No")
elif w%2==0:
print("Yes" if d[1]<=w and d[2]==0 else "No")
elif h%2==0:
print("Yes" if d[1]<=h and d[2]==0 else "No")
```
|
instruction
| 0
| 8,945
| 16
| 17,890
|
No
|
output
| 1
| 8,945
| 16
| 17,891
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.
Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:
* Every row and column in A' can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.
Constraints
* 1 ≤ H, W ≤ 100
* a_{ij} is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}
Output
If Snuke can create a matrix satisfying the condition, print `Yes`; otherwise, print `No`.
Examples
Input
3 4
aabb
aabb
aacc
Output
Yes
Input
2 2
aa
bb
Output
No
Input
5 1
t
w
e
e
t
Output
Yes
Input
2 5
abxba
abyba
Output
No
Input
1 1
z
Output
Yes
Submitted Solution:
```
from collections import Counter
def solve(h, w, a):
c = Counter(list("".join(a)))
d = Counter(list(map(lambda num: num % 4, c.values())))
if (h % 2 == 0) and (w % 2 == 0):
return "Yes" if list(d.keys()) == [4] else "No"
if 3 in d:
return "No"
if (h % 2 == 1) and (w % 2 == 1):
if (1 in d) and (d[1] != 1):
return "No"
if (2 in d) and d[2] > sum(x // 2 for x in [h,w]):
return "No"
else:
if 1 in d:
return "No"
if (2 in d) and d[2] > sum((x % 2 == 0) * (x // 2) for x in [h,w]):
return "No"
return "Yes"
h, w = map(int, input().split())
a = [input() for r in range(h)]
print(solve(h, w, a))
```
|
instruction
| 0
| 8,946
| 16
| 17,892
|
No
|
output
| 1
| 8,946
| 16
| 17,893
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,686
| 16
| 19,372
|
"Correct Solution:
```
import collections
n,k=map(int,input().split())
arr=list(map(int,input().split()))
pos=[[] for _ in range(max(arr)+1)]
for i in range(2*n):
pos[arr[i%n]].append(i)
edge=[[] for _ in range(n)]
cnt=collections.defaultdict(int)
for i in range(n):
s=pos[arr[i]][cnt[arr[i]]]
g=pos[arr[i]][cnt[arr[i]]+1]
edge[s].append((g+1)%n)
edge[s].append((g+1)//n)
cnt[arr[i]]+=1
start=0
loop=0
for _ in range(n):
loop+=edge[start][1]
start=edge[start][0]
if start==0:
break
k%=loop
if k==0:
print('')
else:
k-=1
start=0
loop=0
for _ in range(n):
if loop==k:
break
loop+=edge[start][1]
start=edge[start][0]
s=set()
q=collections.deque()
for i in range(start,n):
if arr[i] not in s:
s.add(arr[i])
q.append(arr[i])
else:
while 1:
tmp=q.pop()
s.discard(tmp)
if tmp==arr[i]:
break
print(*list(q))
```
|
output
| 1
| 9,686
| 16
| 19,373
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,687
| 16
| 19,374
|
"Correct Solution:
```
N,K = map(int,input().split())
A = list(map(int,input().split()))
hoge = [[]for i in range(200001)]
for i in range(N):
hoge[A[i]].append(i)
NEXT = [-1]*N
for i in hoge:
for j in range(-len(i),0):
NEXT[i[j]] = i[j+1]
i = 1
p = 0
while i < K:
#print(i,p)
if NEXT[p] <= p:
i += 1
if NEXT[p] == N-1:
i += 1
p = (NEXT[p]+1)%N
if p == 0:
i = ((K-1)//(i-1))*(i-1)+1
while p < N:
if NEXT[p] <= p:
print(A[p],end = " ")
p += 1
else:
p = NEXT[p]+1
print()
```
|
output
| 1
| 9,687
| 16
| 19,375
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,688
| 16
| 19,376
|
"Correct Solution:
```
from collections import defaultdict
from bisect import bisect_right
N, K = map(int, input().split())
*A, = map(int, input().split())
dd = defaultdict(list)
for i, j in enumerate(A):
dd[j].append(i)
pos = 0
loop = 1
# l = []
cic = defaultdict(list)
while pos < N:
num = A[pos]
pp = bisect_right(dd[num], pos)
if pp < len(dd[num]):
pos = dd[num][pp]+1
else:
pos = dd[num][0]+1
loop += 1
cic[loop].append(pos)
pos = cic[K%loop][0] if cic[K%loop] else 0
ans = []
while pos < N:
num = A[pos]
pp = bisect_right(dd[num], pos)
if pp < len(dd[num]):
pos = dd[num][pp]+1
else:
ans.append(num)
pos += 1
print(*ans)
```
|
output
| 1
| 9,688
| 16
| 19,377
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,689
| 16
| 19,378
|
"Correct Solution:
```
N,K = map(int,input().split())
A = list(map(int,input().split()))
A += A
d = {}
da = {}
for i in range(2*N):
a = A[i]
if a in da:
d[da[a]] = i
da[a] = i
for i in range(N):
if d[i] >= N:
d[i] %= N
s = 0
k = K
while k > 1:
if d[s]<=s:
k -= 1
if d[s] == N-1:
k = K%(K-k+1)
s = (d[s]+1)%N
if k == 0:
print('')
else:
ans = []
while s <= N-1:
if s < d[s]:
s = d[s]+1
else:
ans.append(str(A[s]))
s += 1
print(' '.join(ans))
```
|
output
| 1
| 9,689
| 16
| 19,379
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,690
| 16
| 19,380
|
"Correct Solution:
```
from bisect import*
from collections import*
n,k,*a=map(int,open(0).read().split())
b=defaultdict(list)
for i,v in enumerate(a):b[v]+=i,
s=a[0]
i=c=0
while 1:
t=b[s]
m=len(t)
j=bisect(t,i)
c+=j==m
i=t[j%m]+1
if i==n:break
s=a[i]
k%=c+1
s=a[0]
i=c=0
while c<k-1:
t=b[s]
m=len(t)
j=bisect(t,i)
c+=j==m
i=t[j%m]+1
s=a[i]
r=[]
while 1:
t=b[s]
m=len(t)
j=bisect(t,i)
if j==m:
r+=s,
i+=1
else:
i=t[j]+1
if i==n:
break
s=a[i]
print(*r)
```
|
output
| 1
| 9,690
| 16
| 19,381
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,691
| 16
| 19,382
|
"Correct Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
first = [0] * (N+1)
follow_idx = [-1] * (N+1)
d = {}
for i, a in enumerate(A[::-1]):
j = N - 1 - i
if a in d:
idx = d[a][-1]
d[a].append(j)
if first[idx + 1] < 0:
first[j] = first[idx + 1]
follow_idx[j] = follow_idx[idx + 1]
else:
first[j] = -first[idx + 1]
follow_idx[j] = idx + 1
else:
first[j] = a
d[a] = [j]
follow_idx[j] = j + 1
indices = [-1]
s0 = first[0]
if s0 >= 0:
indices.append(0)
else:
s0 = -s0
indices.append(follow_idx[0])
if s0 == 0:
print('')
quit()
d2 = {0:0, s0:1}
for k in range(2, K+1):
idx = d[s0][-1] + 1
s0 = first[idx]
if s0 >= 0:
indices.append(idx)
else:
s0 = -s0
indices.append(follow_idx[idx])
if s0 in d2:
break
d2[s0] = k
if len(indices) < K:
K = d2[s0] + (K - d2[s0]) % (k - d2[s0])
if K == 0:
print('')
quit()
ans = []
idx = indices[K]
while idx < N:
s = first[idx]
if s > 0:
ans.append(s)
idx = follow_idx[idx]
print(*ans)
```
|
output
| 1
| 9,691
| 16
| 19,383
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,692
| 16
| 19,384
|
"Correct Solution:
```
import os
import sys
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
N, K = list(map(int, sys.stdin.readline().split()))
A = list(map(int, sys.stdin.readline().split()))
# 次 A[j] が取り除かれるまでの操作回数
counts = [0] * N
D = {}
for i, a in enumerate(A + A):
if a not in D:
D[a] = i
else:
if D[a] < N:
counts[D[a]] = i - D[a]
D[a] = i
circle = counts[0] + 1
i = (counts[0] + 1) % N
while i != 0:
circle += counts[i] + 1
i = (i + counts[i] + 1) % N
rem = N * K % circle
ans = []
i = 0
while rem > 0:
while rem - counts[i] > 0:
rem -= counts[i] + 1
i = (i + counts[i] + 1) % N
if rem:
ans.append(A[i])
rem -= 1
i = (i + 1) % N
if ans:
print(*ans)
# hist = []
# stack = []
# for _ in range(K):
# for a in A:
# if a in stack:
# while a in stack:
# stack.pop()
# else:
# stack.append(a)
#
# hist.append(a)
# print(hist)
# print('', stack)
# print()
```
|
output
| 1
| 9,692
| 16
| 19,385
|
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
|
instruction
| 0
| 9,693
| 16
| 19,386
|
"Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = a + a
to = [-1] * n
num = [-1] * (2 * 10 ** 5 + 1)
for i, x in enumerate(b):
if 0 <= num[x] < n:
to[num[x]] = i + 1
num[x] = i
c, now = 1, 0
check = 0
while now != n:
v = to[now]
if v > n:
c += 1
v -= n
now = v
k %= c
now = 0
while k > 1:
v = to[now]
if v > n:
k -= 1
v -= n
now = v
ans = []
while now < n:
if to[now] < n:
now = to[now]
elif to[now] == n:
break
else:
ans.append(a[now])
now += 1
print(*ans)
```
|
output
| 1
| 9,693
| 16
| 19,387
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def mapint(): return map(int, input().split())
sys.setrecursionlimit(10**9)
N, K = mapint()
As = list(mapint())
double_As = As*2
idx_dic = {}
idx_lis = [N+1]*(2*N)
for i in range(N*2-1, -1, -1):
a = double_As[i]
if a in idx_dic:
idx_lis[i] = idx_dic[a]+1
idx_dic[a] = i
doubling = [[0]*N for _ in range(60)]
accum = [[0]*N for _ in range(60)]
doubling[0] = [i%N for i in idx_lis[:N]]
accum[0] = [a-i for i, a in enumerate(idx_lis[:N])]
for i in range(1, 60):
for j in range(N):
doubling[i][j] = doubling[i-1][doubling[i-1][j]]
accum[i][j] = accum[i-1][j] + accum[i-1][doubling[i-1][j]]
now = 0
cum = 0
cnt = 0
for i in range(59, -1, -1):
if cum + accum[i][now]>N*K:
continue
else:
cum += accum[i][now]
now = doubling[i][now]
ans = []
while cum<N*K:
if cum+accum[0][now]<=N*K:
cum += accum[0][now]
now = doubling[0][now]
else:
ans.append(As[now])
now += 1
cum += 1
print(*ans)
```
|
instruction
| 0
| 9,694
| 16
| 19,388
|
Yes
|
output
| 1
| 9,694
| 16
| 19,389
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import *
N, K = map(int, input().split())
A = list(map(int, input().split()))
log_size = 65
#行き先ではなく移動量をダブリングする
dp = [[-1]*N for _ in range(log_size)]
idx = defaultdict(int)
for i in range(2*N):
if A[i%N] in idx and idx[A[i%N]]<N:
dp[0][idx[A[i%N]]] = i+1-idx[A[i%N]]
idx[A[i%N]] = i
for i in range(1, log_size):
for j in range(N):
#値が大きくなりすぎるとTLEする
dp[i][j] = min(10**18, dp[i-1][j]+dp[i-1][(j+dp[i-1][j])%N])
now = 0
NK = N*K
for i in range(log_size-1, -1, -1):
if NK-dp[i][now]>=0:
NK -= dp[i][now]
now = (now+dp[i][now])%N
if NK==0:
print()
exit()
q = deque([])
cnt = defaultdict(int)
for i in range(now, N):
if cnt[A[i]]>0:
while cnt[A[i]]>0:
rem = q.pop()
cnt[rem] -= 1
else:
cnt[A[i]] += 1
q.append(A[i])
print(*q)
```
|
instruction
| 0
| 9,695
| 16
| 19,390
|
Yes
|
output
| 1
| 9,695
| 16
| 19,391
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
import sys
readline = sys.stdin.readline
MOD = 10 ** 9 + 7
INF = float('INF')
sys.setrecursionlimit(10 ** 5)
def main():
from collections import defaultdict
N, K = map(int, readline().split())
A = list(map(int, readline().split()))
c = N * K
b = 0
while 2 ** b < c:
b += 1
doubling = [[0] * N for _ in range(b + 1)]
idx = defaultdict(list)
for i, x in enumerate(A):
idx[x].append(i)
for i in range(1, 200001):
d = idx[i]
l = len(d)
for j in range(l):
cur = d[j]
nx = d[(j + 1) % l]
if nx <= cur:
doubling[0][cur] = N - (cur - nx) + 1
else:
doubling[0][cur] = nx - cur + 1
for i in range(1, b + 1):
for j in range(N):
p = doubling[i - 1][j]
doubling[i][j] = p + doubling[i - 1][(j + p) % N]
cnt = 0
cur = 0
while c - cnt > 200000:
for i in range(b, -1, -1):
if cnt + doubling[i][cur] < c:
cnt += doubling[i][cur]
cur = cnt % N
break
rem = c - cnt
res = []
res_set = set()
while rem > 0:
num = A[cur]
if num in res_set:
while True:
back = res[-1]
res.pop()
res_set.discard(back)
if back == num:
break
else:
res.append(num)
res_set.add(num)
rem -= 1
cur += 1
cur %= N
print(*res)
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 9,696
| 16
| 19,392
|
Yes
|
output
| 1
| 9,696
| 16
| 19,393
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
import bisect
N,K=map(int,input().split())
A=list(map(int,input().split()))
dic={}
for i in range(N):
if A[i] not in dic:
dic[A[i]]=[]
dic[A[i]].append(i)
check={}
for ai in dic:
check[ai]=0
check[ai]=max(dic[ai])
front=[-1]
pos=0
visit=set([-1])
while True:
val=A[pos]
if pos>=check[val]:
front.append(pos)
if pos in visit:
break
else:
visit.add(pos)
pos=dic[val][0]+1
if pos==N:
front.append(-1)
break
else:
index=bisect.bisect_right(dic[val],pos)
pos=dic[val][index]+1
if pos==N:
front.append(-1)
break
last=front[-1]
i=front.index(last)
period=len(front)-i-1
const=i
K-=const
K-=1
r=K%period
fff=front[r]
if fff!=-1:
import heapq
que=[]
check={A[i]:0 for i in range(N)}
index=0
flag=True
while N>index:
if A[index]!=A[fff] and flag:
index+=1
elif A[index]==A[fff] and flag:
index+=1
flag=False
else:
if check[A[index]]==0:
heapq.heappush(que,(-index,A[index]))
check[A[index]]=1
index+=1
else:
while True:
test,a=heapq.heappop(que)
check[a]=0
if a==A[index]:
break
check[A[index]]=0
index+=1
que.sort(reverse=True)
que=[que[i][1] for i in range(len(que))]
print(*que)
else:
import heapq
que=[]
check={A[i]:0 for i in range(N)}
index=0
while N>index:
if check[A[index]]==0:
heapq.heappush(que,(-index,A[index]))
check[A[index]]=0
check[A[index]]=1
index+=1
else:
while True:
test,a=heapq.heappop(que)
check[a]=0
if a==A[index]:
break
check[A[index]]=0
index+=1
que.sort(reverse=True)
que=[que[i][1] for i in range(len(que))]
print(*que)
```
|
instruction
| 0
| 9,697
| 16
| 19,394
|
Yes
|
output
| 1
| 9,697
| 16
| 19,395
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
def add(S, A):
for Ai in A:
try:
S = S[:S.index(Ai)]
except ValueError as _:
S.append(Ai)
return S
ok = False
S = []
for i in range(K):
S = add(S, A)
if len(S) == 0:
break
else:
ok = True
print(" ".join(map(str, S)))
if not ok:
for i in range(K%(i+1)):
S = add(S, A)
print(" ".join(map(str, S)))
```
|
instruction
| 0
| 9,698
| 16
| 19,396
|
No
|
output
| 1
| 9,698
| 16
| 19,397
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
n,k=map(int,input().split())
A=list(map(int,input().split()))
s=[]
m=k%(n+1)
for i in range(0,m+1):
if s.count(A[i%n])==0:
s.append(A[i%n])
else:
while s.count(A[i%n])!=0:
del s[-1]
for j in range(0,len(s)):
s[j]=str(s[j])
print(" ".join(s))
```
|
instruction
| 0
| 9,699
| 16
| 19,398
|
No
|
output
| 1
| 9,699
| 16
| 19,399
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
s = []
for i in a:
if i in s:
while i ! s.pop(-1):
break
else :
s.append(i)
print(s)
```
|
instruction
| 0
| 9,700
| 16
| 19,400
|
No
|
output
| 1
| 9,700
| 16
| 19,401
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.
Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:
* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.
Find the elements of s after Snuke finished the operations.
Constraints
* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_0 A_1 \cdots A_{N-1}
Output
Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.
Examples
Input
3 2
1 2 3
Output
2 3
Input
5 10
1 2 3 2 3
Output
3
Input
6 1000000000000
1 1 2 2 3 3
Output
Input
11 97
3 1 4 1 5 9 2 6 5 3 5
Output
9 2 6
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,queue,copy
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()
def main():
n,k=LI()
l=LI()
l=l+l
a=[]
for x in l:
y=a.count(x)
if y>0:
while True:
if y==0:
break
p=a.pop()
if p==x:
y-=1
else:
a.append(x)
return ' '.join([str(x) for x in a])
# main()
print(main())
```
|
instruction
| 0
| 9,701
| 16
| 19,402
|
No
|
output
| 1
| 9,701
| 16
| 19,403
|
Provide a correct Python 3 solution for this coding contest problem.
A museum exhibits N jewels, Jewel 1, 2, ..., N. The coordinates of Jewel i are (x_i, y_i) (the museum can be regarded as a two-dimensional plane), and the value of that jewel is v_i.
Snuke the thief will steal some of these jewels.
There are M conditions, Condition 1, 2, ..., M, that must be met when stealing jewels, or he will be caught by the detective. Each condition has one of the following four forms:
* (t_i =`L`, a_i, b_i) : Snuke can only steal at most b_i jewels whose x coordinates are a_i or smaller.
* (t_i =`R`, a_i, b_i) : Snuke can only steal at most b_i jewels whose x coordinates are a_i or larger.
* (t_i =`D`, a_i, b_i) : Snuke can only steal at most b_i jewels whose y coordinates are a_i or smaller.
* (t_i =`U`, a_i, b_i) : Snuke can only steal at most b_i jewels whose y coordinates are a_i or larger.
Find the maximum sum of values of jewels that Snuke the thief can steal.
Constraints
* 1 \leq N \leq 80
* 1 \leq x_i, y_i \leq 100
* 1 \leq v_i \leq 10^{15}
* 1 \leq M \leq 320
* t_i is `L`, `R`, `U` or `D`.
* 1 \leq a_i \leq 100
* 0 \leq b_i \leq N - 1
* (x_i, y_i) are pairwise distinct.
* (t_i, a_i) are pairwise distinct.
* (t_i, b_i) are pairwise distinct.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1 v_1
x_2 y_2 v_2
:
x_N y_N v_N
M
t_1 a_1 b_1
t_2 a_2 b_2
:
t_M a_M b_M
Output
Print the maximum sum of values of jewels that Snuke the thief can steal.
Examples
Input
7
1 3 6
1 5 9
3 1 8
4 3 8
6 2 9
5 4 11
5 7 10
4
L 3 1
R 2 3
D 5 3
U 4 2
Output
36
Input
3
1 2 3
4 5 6
7 8 9
1
L 100 0
Output
0
Input
4
1 1 10
1 2 11
2 1 12
2 2 13
3
L 8 3
L 9 2
L 10 1
Output
13
Input
10
66 47 71040136000
65 77 74799603000
80 53 91192869000
24 34 24931901000
91 78 49867703000
68 71 46108236000
46 73 74799603000
56 63 93122668000
32 51 71030136000
51 26 70912345000
21
L 51 1
L 7 0
U 47 4
R 92 0
R 91 1
D 53 2
R 65 3
D 13 0
U 63 3
L 68 3
D 47 1
L 91 5
R 32 4
L 66 2
L 80 4
D 77 4
U 73 1
D 78 5
U 26 5
R 80 2
R 24 5
Output
305223377000
|
instruction
| 0
| 9,702
| 16
| 19,404
|
"Correct Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**9)
from collections import deque
from bisect import bisect_left,bisect_right
class MinCostFlow:
def __init__(self,n):
self.n=n
self.edges=[[] for i in range(n)]
def add_edge(self,fr,to,cap,cost):
self.edges[fr].append([to,cap,cost,len(self.edges[to])])
self.edges[to].append([fr,0,-cost,len(self.edges[fr])-1])
def MinCost(self,source,sink,flow):
inf=10**15+1
n=self.n; E=self.edges
mincost=0
prev_v=[0]*n; prev_e=[0]*n
while flow:
dist=[inf]*n
dist[source]=0
q=deque([source])
Flag=[False]*n
Flag[source]=True
while q:
v=q.popleft()
if not Flag[v]:
continue
Flag[v]=False
for i,(to,cap,cost,_) in enumerate(E[v]):
if cap>0 and dist[to]>dist[v]+cost:
dist[to]=dist[v]+cost
prev_v[to],prev_e[to]=v,i
q.append(to)
Flag[to]=True
if dist[sink]==inf:
return 1
f,v=flow,sink
while v!=source:
f=min(f,E[prev_v[v]][prev_e[v]][1])
v=prev_v[v]
flow-=f
mincost+=f*dist[sink]
v=sink
while v!=source:
E[prev_v[v]][prev_e[v]][1]-=f
rev=E[prev_v[v]][prev_e[v]][3]
E[v][rev][1]+=f
v=prev_v[v]
return mincost
n=int(input())
J=[]
L_org,D_org=[1]*n,[1]*n
for _ in range(n):
x,y,v=map(int,input().split())
J.append((x,y,v))
m=int(input())
T=[]
for _ in range(m):
t,a,b=input().split()
a,b=int(a),int(b)
T.append((t,a,b))
if t=='L':
L_org[b]=a+1
elif t=='D':
D_org[b]=a+1
for i in range(1,n):
L_org[i]=max(L_org[i-1],L_org[i])
D_org[i]=max(D_org[i-1],D_org[i])
def solve(k):
L,D=L_org[:k],D_org[:k]
R,U=[100]*k,[100]*k
for t,a,b in T:
if k-b-1>=0:
if t=='R':
R[k-b-1]=a-1
elif t=='U':
U[k-b-1]=a-1
for i in range(k-2,-1,-1):
R[i]=min(R[i],R[i+1])
U[i]=min(U[i],U[i+1])
solver=MinCostFlow(2*n+2*k+2)
for i in range(1,k+1):
solver.add_edge(0,i,1,0)
solver.add_edge(2*n+k+i,2*n+2*k+1,1,0)
for i in range(n):
v=J[i][2]
solver.add_edge(k+i+1,n+k+i+1,1,-v)
for i in range(n):
x,y=J[i][0],J[i][1]
l=bisect_right(L,x)
r=bisect_left(R,x)+1
d=bisect_right(D,y)
u=bisect_left(U,y)+1
for j in range(r,l+1):
solver.add_edge(j,k+i+1,1,0)
for j in range(u,d+1):
solver.add_edge(n+k+i+1,2*n+k+j,1,0)
return -solver.MinCost(0,2*n+2*k+1,k)
ans=0
k=1
while True:
tmp=solve(k)
ans=max(ans,tmp)
if tmp==-1 or k==n:
break
k+=1
print(ans)
```
|
output
| 1
| 9,702
| 16
| 19,405
|
Provide a correct Python 3 solution for this coding contest problem.
A museum exhibits N jewels, Jewel 1, 2, ..., N. The coordinates of Jewel i are (x_i, y_i) (the museum can be regarded as a two-dimensional plane), and the value of that jewel is v_i.
Snuke the thief will steal some of these jewels.
There are M conditions, Condition 1, 2, ..., M, that must be met when stealing jewels, or he will be caught by the detective. Each condition has one of the following four forms:
* (t_i =`L`, a_i, b_i) : Snuke can only steal at most b_i jewels whose x coordinates are a_i or smaller.
* (t_i =`R`, a_i, b_i) : Snuke can only steal at most b_i jewels whose x coordinates are a_i or larger.
* (t_i =`D`, a_i, b_i) : Snuke can only steal at most b_i jewels whose y coordinates are a_i or smaller.
* (t_i =`U`, a_i, b_i) : Snuke can only steal at most b_i jewels whose y coordinates are a_i or larger.
Find the maximum sum of values of jewels that Snuke the thief can steal.
Constraints
* 1 \leq N \leq 80
* 1 \leq x_i, y_i \leq 100
* 1 \leq v_i \leq 10^{15}
* 1 \leq M \leq 320
* t_i is `L`, `R`, `U` or `D`.
* 1 \leq a_i \leq 100
* 0 \leq b_i \leq N - 1
* (x_i, y_i) are pairwise distinct.
* (t_i, a_i) are pairwise distinct.
* (t_i, b_i) are pairwise distinct.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1 v_1
x_2 y_2 v_2
:
x_N y_N v_N
M
t_1 a_1 b_1
t_2 a_2 b_2
:
t_M a_M b_M
Output
Print the maximum sum of values of jewels that Snuke the thief can steal.
Examples
Input
7
1 3 6
1 5 9
3 1 8
4 3 8
6 2 9
5 4 11
5 7 10
4
L 3 1
R 2 3
D 5 3
U 4 2
Output
36
Input
3
1 2 3
4 5 6
7 8 9
1
L 100 0
Output
0
Input
4
1 1 10
1 2 11
2 1 12
2 2 13
3
L 8 3
L 9 2
L 10 1
Output
13
Input
10
66 47 71040136000
65 77 74799603000
80 53 91192869000
24 34 24931901000
91 78 49867703000
68 71 46108236000
46 73 74799603000
56 63 93122668000
32 51 71030136000
51 26 70912345000
21
L 51 1
L 7 0
U 47 4
R 92 0
R 91 1
D 53 2
R 65 3
D 13 0
U 63 3
L 68 3
D 47 1
L 91 5
R 32 4
L 66 2
L 80 4
D 77 4
U 73 1
D 78 5
U 26 5
R 80 2
R 24 5
Output
305223377000
|
instruction
| 0
| 9,703
| 16
| 19,406
|
"Correct Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**9)
from bisect import bisect_left,bisect_right
class MinCostFlow:
def __init__(self,n):
self.n=n
self.edges=[[] for i in range(n)]
def add_edge(self,fr,to,cap,cost):
self.edges[fr].append([to,cap,cost,len(self.edges[to])])
self.edges[to].append([fr,0,-cost,len(self.edges[fr])-1])
def MinCost(self,source,sink,flow):
inf=10**15+1
n,E=self.n,self.edges
prev_v,prev_e=[0]*n,[0]*n
mincost=0
while flow:
dist=[inf]*n
dist[source]=0
flag=True
while flag:
flag=False
for v in range(n):
if dist[v]==inf:
continue
Ev=E[v]
for i in range(len(Ev)):
to,cap,cost,rev=Ev[i]
if cap>0 and dist[v]+cost<dist[to]:
dist[to]=dist[v]+cost
prev_v[to],prev_e[to]=v,i
flag=True
if dist[sink]==inf:
return 1
f=flow
v=sink
while v!=source:
f=min(f,E[prev_v[v]][prev_e[v]][1])
v=prev_v[v]
flow-=f
mincost+=f*dist[sink]
v=sink
while v!=source:
E[prev_v[v]][prev_e[v]][1]-=f
rev=E[prev_v[v]][prev_e[v]][3]
E[v][rev][1]+=f
v=prev_v[v]
return mincost
n=int(input())
J=[]
L_org,D_org=[1]*n,[1]*n
for _ in range(n):
x,y,v=map(int,input().split())
J.append((x,y,v))
m=int(input())
T=[]
for _ in range(m):
t,a,b=input().split()
a,b=int(a),int(b)
T.append((t,a,b))
if t=='L':
L_org[b]=a+1
elif t=='D':
D_org[b]=a+1
for i in range(1,n):
L_org[i]=max(L_org[i-1],L_org[i])
D_org[i]=max(D_org[i-1],D_org[i])
def solve(k):
L,D=L_org[:k],D_org[:k]
R,U=[100]*k,[100]*k
for t,a,b in T:
if k-b-1>=0:
if t=='R':
R[k-b-1]=a-1
elif t=='U':
U[k-b-1]=a-1
for i in range(k-2,-1,-1):
R[i]=min(R[i],R[i+1])
U[i]=min(U[i],U[i+1])
solver=MinCostFlow(2*n+2*k+2)
for i in range(1,k+1):
solver.add_edge(0,i,1,0)
solver.add_edge(2*n+k+i,2*n+2*k+1,1,0)
for i in range(n):
v=J[i][2]
solver.add_edge(k+i+1,n+k+i+1,1,-v)
for i in range(n):
x,y=J[i][0],J[i][1]
l=bisect_right(L,x)
r=bisect_left(R,x)+1
d=bisect_right(D,y)
u=bisect_left(U,y)+1
for j in range(r,l+1):
solver.add_edge(j,k+i+1,1,0)
for j in range(u,d+1):
solver.add_edge(n+k+i+1,2*n+k+j,1,0)
return -solver.MinCost(0,2*n+2*k+1,k)
ans=0
k=1
while True:
tmp=solve(k)
ans=max(ans,tmp)
if tmp==-1 or k==n:
break
k+=1
print(ans)
```
|
output
| 1
| 9,703
| 16
| 19,407
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A museum exhibits N jewels, Jewel 1, 2, ..., N. The coordinates of Jewel i are (x_i, y_i) (the museum can be regarded as a two-dimensional plane), and the value of that jewel is v_i.
Snuke the thief will steal some of these jewels.
There are M conditions, Condition 1, 2, ..., M, that must be met when stealing jewels, or he will be caught by the detective. Each condition has one of the following four forms:
* (t_i =`L`, a_i, b_i) : Snuke can only steal at most b_i jewels whose x coordinates are a_i or smaller.
* (t_i =`R`, a_i, b_i) : Snuke can only steal at most b_i jewels whose x coordinates are a_i or larger.
* (t_i =`D`, a_i, b_i) : Snuke can only steal at most b_i jewels whose y coordinates are a_i or smaller.
* (t_i =`U`, a_i, b_i) : Snuke can only steal at most b_i jewels whose y coordinates are a_i or larger.
Find the maximum sum of values of jewels that Snuke the thief can steal.
Constraints
* 1 \leq N \leq 80
* 1 \leq x_i, y_i \leq 100
* 1 \leq v_i \leq 10^{15}
* 1 \leq M \leq 320
* t_i is `L`, `R`, `U` or `D`.
* 1 \leq a_i \leq 100
* 0 \leq b_i \leq N - 1
* (x_i, y_i) are pairwise distinct.
* (t_i, a_i) are pairwise distinct.
* (t_i, b_i) are pairwise distinct.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1 v_1
x_2 y_2 v_2
:
x_N y_N v_N
M
t_1 a_1 b_1
t_2 a_2 b_2
:
t_M a_M b_M
Output
Print the maximum sum of values of jewels that Snuke the thief can steal.
Examples
Input
7
1 3 6
1 5 9
3 1 8
4 3 8
6 2 9
5 4 11
5 7 10
4
L 3 1
R 2 3
D 5 3
U 4 2
Output
36
Input
3
1 2 3
4 5 6
7 8 9
1
L 100 0
Output
0
Input
4
1 1 10
1 2 11
2 1 12
2 2 13
3
L 8 3
L 9 2
L 10 1
Output
13
Input
10
66 47 71040136000
65 77 74799603000
80 53 91192869000
24 34 24931901000
91 78 49867703000
68 71 46108236000
46 73 74799603000
56 63 93122668000
32 51 71030136000
51 26 70912345000
21
L 51 1
L 7 0
U 47 4
R 92 0
R 91 1
D 53 2
R 65 3
D 13 0
U 63 3
L 68 3
D 47 1
L 91 5
R 32 4
L 66 2
L 80 4
D 77 4
U 73 1
D 78 5
U 26 5
R 80 2
R 24 5
Output
305223377000
Submitted Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**9)
from bisect import bisect_left,bisect_right
class MinCostFlow:
def __init__(self,n):
self.n=n
self.edges=[[] for i in range(n)]
def add_edge(self,fr,to,cap,cost):
self.edges[fr].append([to,cap,cost,len(self.edges[to])])
self.edges[to].append([fr,0,-cost,len(self.edges[fr])-1])
def MinCost(self,source,sink,flow):
inf=10**15+1
n,E=self.n,self.edges
prev_v,prev_e=[0]*n,[0]*n
mincost=0
while flow:
dist=[inf]*n
dist[source]=0
flag=True
while flag:
flag=False
for v in range(n):
if dist[v]==inf:
continue
Ev=E[v]
for i in range(len(Ev)):
to,cap,cost,rev=Ev[i]
if cap>0 and dist[v]+cost<dist[to]:
dist[to]=dist[v]+cost
prev_v[to],prev_e[to]=v,i
flag=True
if dist[sink]==inf:
return -1
f=flow
v=sink
while v!=source:
f=min(f,E[prev_v[v]][prev_e[v]][1])
v=prev_v[v]
flow-=f
mincost+=f*dist[sink]
v=sink
while v!=source:
E[prev_v[v]][prev_e[v]][1]-=f
rev=E[prev_v[v]][prev_e[v]][3]
E[v][rev][1]+=f
v=prev_v[v]
return mincost
n=int(input())
J=[]
L_org,D_org=[1]*n,[1]*n
for _ in range(n):
x,y,v=map(int,input().split())
J.append((x,y,v))
m=int(input())
T=[]
for _ in range(m):
t,a,b=input().split()
a,b=int(a),int(b)
T.append((t,a,b))
if t=='L':
L_org[b]=a+1
elif t=='D':
D_org[b]=a+1
for i in range(1,n):
L_org[i]=max(L_org[i-1],L_org[i])
D_org[i]=max(D_org[i-1],D_org[i])
def solve(k):
L,D=L_org[:k],D_org[:k]
R,U=[100]*k,[100]*k
for t,a,b in T:
if k-b-1>=0:
if t=='R':
R[k-b-1]=a-1
elif t=='U':
U[k-b-1]=a-1
for i in range(k-2,-1,-1):
R[i]=min(R[i],R[i+1])
U[i]=min(U[i],U[i+1])
solver=MinCostFlow(2*n+2*k+2)
for i in range(1,k+1):
solver.add_edge(0,i,1,0)
solver.add_edge(2*n+k+i,2*n+2*k+1,1,0)
for i in range(n):
v=J[i][2]
solver.add_edge(k+i+1,n+k+i+1,1,-v)
for i in range(n):
x,y=J[i][0],J[i][1]
l=bisect_right(L,x)
r=bisect_left(R,x)+1
d=bisect_right(D,y)
u=bisect_left(U,y)+1
for j in range(r,l+1):
solver.add_edge(j,k+i+1,1,0)
for j in range(u,d+1):
solver.add_edge(n+k+i+1,2*n+k+j,1,0)
return -solver.MinCost(0,2*n+2*k+1,k)
ans=0
k=1
while True:
tmp=solve(k)
ans=max(ans,tmp)
if tmp==-1 or k==n:
break
k+=1
print(ans)
```
|
instruction
| 0
| 9,704
| 16
| 19,408
|
No
|
output
| 1
| 9,704
| 16
| 19,409
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is introducing a robot arm with the following properties to his factory:
* The robot arm consists of m sections and m+1 joints. The sections are numbered 1, 2, ..., m, and the joints are numbered 0, 1, ..., m. Section i connects Joint i-1 and Joint i. The length of Section i is d_i.
* For each section, its mode can be specified individually. There are four modes: `L`, `R`, `D` and `U`. The mode of a section decides the direction of that section. If we consider the factory as a coordinate plane, the position of Joint i will be determined as follows (we denote its coordinates as (x_i, y_i)):
* (x_0, y_0) = (0, 0).
* If the mode of Section i is `L`, (x_{i}, y_{i}) = (x_{i-1} - d_{i}, y_{i-1}).
* If the mode of Section i is `R`, (x_{i}, y_{i}) = (x_{i-1} + d_{i}, y_{i-1}).
* If the mode of Section i is `D`, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} - d_{i}).
* If the mode of Section i is `U`, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} + d_{i}).
Snuke would like to introduce a robot arm so that the position of Joint m can be matched with all of the N points (X_1, Y_1), (X_2, Y_2), ..., (X_N, Y_N) by properly specifying the modes of the sections. Is this possible? If so, find such a robot arm and how to bring Joint m to each point (X_j, Y_j).
Constraints
* All values in input are integers.
* 1 \leq N \leq 1000
* -10^9 \leq X_i \leq 10^9
* -10^9 \leq Y_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
X_1 Y_1
X_2 Y_2
:
X_N Y_N
Output
If the condition can be satisfied, follow the following format. If the condition cannot be satisfied, print `-1`.
m
d_1 d_2 ... d_m
w_1
w_2
:
w_N
m and d_i are the configurations of the robot arm. Refer to the problem statement for what each of them means. Here, 1 \leq m \leq 40 and 1 \leq d_i \leq 10^{12} must hold. Also, m and d_i must all be integers.
w_j is a string of length m that represents the way to bring Joint m of the robot arm to point (X_j, Y_j). The i-th character of w_j should be one of the letters `L`, `R`, `D` and `U`, representing the mode of Section i.
Examples
Input
3
-1 0
0 3
2 -1
Output
2
1 2
RL
UU
DR
Input
5
0 0
1 0
2 0
3 0
4 0
Output
-1
Input
2
1 1
1 1
Output
2
1 1
RU
UR
Input
3
-7 -3
7 3
-3 -7
Output
5
3 1 4 1 5
LRDUL
RDULR
DULRD
|
instruction
| 0
| 9,705
| 16
| 19,410
|
"Correct Solution:
```
N = int(input())
XY = [list(map(int, input().split())) for _ in range(N)]
MOD = sum(XY[0]) % 2
for x, y in XY:
if MOD != (x + y) % 2:
print(-1)
exit()
m = 33 - MOD
print(m)
D = [2 ** i for i in range(31, -1, -1)]
if MOD == 0:
D.append(1)
print(" ".join(map(str, D)))
for x, y in XY:
w = ""
for d in D:
if x + y >= 0 and x - y >= 0:
w += "R"
x -= d
elif x + y < 0 and x - y >= 0:
w += "D"
y += d
elif x + y >= 0 and x - y < 0:
w += "U"
y -= d
else:
w += "L"
x += d
print(w)
```
|
output
| 1
| 9,705
| 16
| 19,411
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is introducing a robot arm with the following properties to his factory:
* The robot arm consists of m sections and m+1 joints. The sections are numbered 1, 2, ..., m, and the joints are numbered 0, 1, ..., m. Section i connects Joint i-1 and Joint i. The length of Section i is d_i.
* For each section, its mode can be specified individually. There are four modes: `L`, `R`, `D` and `U`. The mode of a section decides the direction of that section. If we consider the factory as a coordinate plane, the position of Joint i will be determined as follows (we denote its coordinates as (x_i, y_i)):
* (x_0, y_0) = (0, 0).
* If the mode of Section i is `L`, (x_{i}, y_{i}) = (x_{i-1} - d_{i}, y_{i-1}).
* If the mode of Section i is `R`, (x_{i}, y_{i}) = (x_{i-1} + d_{i}, y_{i-1}).
* If the mode of Section i is `D`, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} - d_{i}).
* If the mode of Section i is `U`, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} + d_{i}).
Snuke would like to introduce a robot arm so that the position of Joint m can be matched with all of the N points (X_1, Y_1), (X_2, Y_2), ..., (X_N, Y_N) by properly specifying the modes of the sections. Is this possible? If so, find such a robot arm and how to bring Joint m to each point (X_j, Y_j).
Constraints
* All values in input are integers.
* 1 \leq N \leq 1000
* -10^9 \leq X_i \leq 10^9
* -10^9 \leq Y_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
X_1 Y_1
X_2 Y_2
:
X_N Y_N
Output
If the condition can be satisfied, follow the following format. If the condition cannot be satisfied, print `-1`.
m
d_1 d_2 ... d_m
w_1
w_2
:
w_N
m and d_i are the configurations of the robot arm. Refer to the problem statement for what each of them means. Here, 1 \leq m \leq 40 and 1 \leq d_i \leq 10^{12} must hold. Also, m and d_i must all be integers.
w_j is a string of length m that represents the way to bring Joint m of the robot arm to point (X_j, Y_j). The i-th character of w_j should be one of the letters `L`, `R`, `D` and `U`, representing the mode of Section i.
Examples
Input
3
-1 0
0 3
2 -1
Output
2
1 2
RL
UU
DR
Input
5
0 0
1 0
2 0
3 0
4 0
Output
-1
Input
2
1 1
1 1
Output
2
1 1
RU
UR
Input
3
-7 -3
7 3
-3 -7
Output
5
3 1 4 1 5
LRDUL
RDULR
DULRD
|
instruction
| 0
| 9,706
| 16
| 19,412
|
"Correct Solution:
```
# -*- coding: utf-8 -*-
def solve():
D = list(map((2).__pow__, range(31)[::-1]))
wdict = {(True, True):('R',-1,-1), (True, False):('U',-1,1), (False, False):('L',1,1), (False, True):('D',1,-1)}
mode = None
for _ in [None]*int(input()):
x, y = map(int, input().split())
u = x+y
if mode is None:
mode = 1-u%2
res = str(31+mode) + '\n' + ' '.join(map(str, [1]*mode+D))
W0 = '\n'+'R'*mode
elif mode != 1-u%2:
return('-1')
res += W0
u -= mode
v = x-y-mode
for d in D:
w, a, b = wdict[u>0, v>0]
res += w
u += a*d
v += b*d
return str(res)
if __name__ == '__main__':
print(solve())
```
|
output
| 1
| 9,706
| 16
| 19,413
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is introducing a robot arm with the following properties to his factory:
* The robot arm consists of m sections and m+1 joints. The sections are numbered 1, 2, ..., m, and the joints are numbered 0, 1, ..., m. Section i connects Joint i-1 and Joint i. The length of Section i is d_i.
* For each section, its mode can be specified individually. There are four modes: `L`, `R`, `D` and `U`. The mode of a section decides the direction of that section. If we consider the factory as a coordinate plane, the position of Joint i will be determined as follows (we denote its coordinates as (x_i, y_i)):
* (x_0, y_0) = (0, 0).
* If the mode of Section i is `L`, (x_{i}, y_{i}) = (x_{i-1} - d_{i}, y_{i-1}).
* If the mode of Section i is `R`, (x_{i}, y_{i}) = (x_{i-1} + d_{i}, y_{i-1}).
* If the mode of Section i is `D`, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} - d_{i}).
* If the mode of Section i is `U`, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} + d_{i}).
Snuke would like to introduce a robot arm so that the position of Joint m can be matched with all of the N points (X_1, Y_1), (X_2, Y_2), ..., (X_N, Y_N) by properly specifying the modes of the sections. Is this possible? If so, find such a robot arm and how to bring Joint m to each point (X_j, Y_j).
Constraints
* All values in input are integers.
* 1 \leq N \leq 1000
* -10^9 \leq X_i \leq 10^9
* -10^9 \leq Y_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
X_1 Y_1
X_2 Y_2
:
X_N Y_N
Output
If the condition can be satisfied, follow the following format. If the condition cannot be satisfied, print `-1`.
m
d_1 d_2 ... d_m
w_1
w_2
:
w_N
m and d_i are the configurations of the robot arm. Refer to the problem statement for what each of them means. Here, 1 \leq m \leq 40 and 1 \leq d_i \leq 10^{12} must hold. Also, m and d_i must all be integers.
w_j is a string of length m that represents the way to bring Joint m of the robot arm to point (X_j, Y_j). The i-th character of w_j should be one of the letters `L`, `R`, `D` and `U`, representing the mode of Section i.
Examples
Input
3
-1 0
0 3
2 -1
Output
2
1 2
RL
UU
DR
Input
5
0 0
1 0
2 0
3 0
4 0
Output
-1
Input
2
1 1
1 1
Output
2
1 1
RU
UR
Input
3
-7 -3
7 3
-3 -7
Output
5
3 1 4 1 5
LRDUL
RDULR
DULRD
|
instruction
| 0
| 9,707
| 16
| 19,414
|
"Correct Solution:
```
n = int(input())
place = [tuple(map(int, input().split())) for _ in range(n)]
m = 0
def bitlist(x):
ret = [0] * 31
for i in reversed(range(31)):
if x > 0:
x -= 1 << i
ret[i] = 1
else:
x += 1 << i
ret[i] = -1
return ret
pre = (place[0][0] + place[0][1] + 1) % 2
for x, y in place:
if (x + y + 1) % 2 != pre:
print(-1)
exit()
print(31 + pre)
if pre:
print(1, end = ' ')
for i in reversed(range(31)):
print(1 << i, end = ' ')
print()
for x, y in place:
u = x + y
v = x - y
if pre:
u -= 1
v -= 1
ubit = bitlist(u)
vbit = bitlist(v)
if pre:
print('R', end = '')
for i in reversed(range(31)):
if ubit[i] == 1 and vbit[i] == 1:
print('R', end = '')
elif ubit[i] == 1 and vbit[i] == -1:
print('U', end = '')
elif ubit[i] == -1 and vbit[i] == -1:
print('L', end = '')
else:
print('D', end = '')
print()
```
|
output
| 1
| 9,707
| 16
| 19,415
|
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