message stringlengths 2 433k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5; | instruction | 0 | 55,783 | 12 | 111,566 |
Tags: combinatorics, data structures, dsu, greedy, implementation
Correct Solution:
```
mod = 998244353
import sys
def input():
return sys.stdin.readline()
for _ in range(int(input())):
n, k = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
X = [0] * n
for i in range(n):
X[A[i] - 1] = i
Y = [-1] * n
for i in range(k):
Y[X[B[i] - 1]] = i
ans = 1
Y += [n]
i = -1
while i < n:
if Y[i] == -1:
i += 1
continue
j = i + 1
x = 0
while j <= n and Y[j] != -1:
if x == 0:
if Y[j] < Y[j - 1]:
x += 1
else:
if Y[j] > Y[j - 1]:
ans = 0
break
j += 1
if 0 <= i and j < n:
ans *= 2
ans %= mod
i = j
print(ans)
``` | output | 1 | 55,783 | 12 | 111,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5; | instruction | 0 | 55,784 | 12 | 111,568 |
Tags: combinatorics, data structures, dsu, greedy, implementation
Correct Solution:
```
import io,os
from math import *
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
M = 998244353
for _ in range(int(input())):
n,k = map(int, input().split())
a_ = list(map(int, input().split()))
a = {x:i for i,x in enumerate(a_)}
b_ = list(map(int, input().split()))
b = [a[x] for x in b_]
c = {x:i for i,x in enumerate(b)}
poss = 1
for i,d in enumerate(b):
canleft = d > 0 and (d-1 not in c or c[d-1] < i)
canright = d < n - 1 and (d+1 not in c or c[d+1] < i)
poss = poss * (canleft + canright) % M
print(poss)
``` | output | 1 | 55,784 | 12 | 111,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5; | instruction | 0 | 55,785 | 12 | 111,570 |
Tags: combinatorics, data structures, dsu, greedy, implementation
Correct Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
t = II()
for q in range(t):
n,k = MI()
ans = 1
x = [0]*(n+1)
y = [0]*(n+1)
a = LI()
b = LI()
mod = 998244353
boo = True
for i in range(n):
x[a[i]] = i
for i in range(k):
y[b[i]] = i
for i in range(k):
count = 0
tot = 0
if x[b[i]]-1>=0:
if y[a[x[b[i]]-1]] > i:
count+=1
tot+=1
if x[b[i]]+1<n:
if y[a[x[b[i]]+1]] > i:
count+=1
tot+=1
if tot == count:
boo = False
break
elif count == 0 and tot == 2:
ans*=2
ans%=mod
print(0 if not boo else ans)
``` | output | 1 | 55,785 | 12 | 111,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5; | instruction | 0 | 55,786 | 12 | 111,572 |
Tags: combinatorics, data structures, dsu, greedy, implementation
Correct Solution:
```
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
MOD = 998244353
T=int(input())
for _ in range(0,T):
n,k=map(int,input().split())
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
pos=[0]*(n+1)
ind=[-1]*(n+1)
for i in range(0,len(b)):
pos[b[i]]=1
for i in range(0,len(a)):
ind[a[i]]=i
ans=1
for i in range(0,len(b)):
ptr=ind[b[i]]
c=0
if((ptr-1)>=0 and pos[a[ptr-1]]==0):
c=(c+1)%MOD
if((ptr+1)<n and pos[a[ptr+1]]==0):
c=(c+1)%MOD
pos[a[ptr]]=0
ans=(ans*c)%MOD
print(ans)
``` | output | 1 | 55,786 | 12 | 111,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5; | instruction | 0 | 55,787 | 12 | 111,574 |
Tags: combinatorics, data structures, dsu, greedy, implementation
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**32, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
for ik in range(int(input())):
n,k = map(int,input().split())
l = list(map(int, input().split()))
rem=set(l)
b=list(map(int,input().split()))
if n==1:
print(0)
continue
ans=1
re=set(b)
d=defaultdict(int)
for i in range(n):
d[l[i]]=i
for i in range(k):
t=0
ind =d[b[i]]
if ind-1>=0:
if l[ind-1] not in re:
t+=1
if ind+1<n:
if l[ind+1] not in re:
t+=1
re.remove(l[ind])
ans*=t
ans%=mod1
if t==0:
break
print(ans%mod1)
``` | output | 1 | 55,787 | 12 | 111,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
class MergeFind:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
# self.lista = [[_] for _ in range(n)]
def find(self, a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self, a, b):
a = self.find(a)
b = self.find(b)
if a == b:
return
self.num_sets -= 1
self.parent[a] = b
self.size[b] += self.size[a]
# self.lista[a] += self.lista[b]
# self.lista[b] = []
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
#
# # to find factorial and ncr
# tot = 400005
# mod = 10 ** 9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, tot + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
def solve():
mod=998244353
n,k=sep()
ar=[-1]+lis()+[-1]
ar2=lis()
s=set(ar2)
s.add(-1)
ind=defaultdict(int)
for i in range(1,n+1):
ind[ar[i]]=i
ans=1
for i in range(k):
s.remove(ar2[i])
t=2
inde=ind[ar2[i]]
if ar[inde+1] in s:
t-=1
if ar[inde-1] in s:
t-=1
ans*=t
ans%=mod
# print(i, s,ans)
# print(ar2)
# print(ind)
print(ans)
#solve()
testcase(int(inp()))
``` | instruction | 0 | 55,788 | 12 | 111,576 |
Yes | output | 1 | 55,788 | 12 | 111,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
from sys import stdin
caso=int(stdin.readline().strip())
mod= 998244353
for cas in range(caso):
n,m=map(int,stdin.readline().strip().split())
s=list(map(int,stdin.readline().strip().split()))
s1=list(map(int,stdin.readline().strip().split()))
vis=[True for i in range(n+1)]
pos=[-1 for i in range(n+1)]
for i in range(m):
vis[s1[i]]=False
for i in range(n):
pos[s[i]]=i
ans=1
for i in range(m):
aux=0
if pos[s1[i]]>0 and vis[s[pos[s1[i]]-1]]:
aux+=1
if pos[s1[i]]<n-1 and vis[s[pos[s1[i]]+1]]:
aux+=1
vis[s1[i]]=True
ans=(ans*aux)%mod
print(ans)
``` | instruction | 0 | 55,789 | 12 | 111,578 |
Yes | output | 1 | 55,789 | 12 | 111,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
mod=998244353
for _ in range (int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
s=set(b)
el=dict()
for i in range (n):
el[a[i]]=i
r=1
for i in range (k):
e=el[b[i]]
c=0
if e>0:
if a[e-1] not in s:
c+=1
if e<n-1:
if a[e+1] not in s:
c+=1
s.remove(b[i])
#print(e,c,s)
r*=c
r%=mod
print(r)
``` | instruction | 0 | 55,790 | 12 | 111,580 |
Yes | output | 1 | 55,790 | 12 | 111,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
import sys
import math
from collections import defaultdict,Counter
input=sys.stdin.readline
def print(x):
sys.stdout.write(str(x)+"\n")
# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')
# mod=pow(10,9)+7
mod=998244353
t=int(input())
for i in range(t):
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
s1=defaultdict(set)
s1[a[0]].add(a[1])
s1[a[-1]].add(a[-2])
for ii in range(1,n-1):
s1[a[ii]].add(a[ii-1])
s1[a[ii]].add(a[ii+1])
s2=set(b)
ans=1
for j in range(k):
# print(s1)
s2.remove(b[j])
cur=0
r=0
for kk in s1[b[j]]:
if kk not in s2:
r=kk
cur+=1
if cur==0:
ans=0
break
else:
s1[b[j]].remove(r)
for kk in s1[r]:
if kk!=b[j]:
s1[b[j]].add(kk)
s1[kk].add(b[j])
s1[kk].remove(r)
break
if cur==2:
ans=(ans*2)%mod
print(ans)
``` | instruction | 0 | 55,791 | 12 | 111,582 |
Yes | output | 1 | 55,791 | 12 | 111,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
"""
Author - Satwik Tiwari .
2nd NOV , 2020 - Monday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 998244353
#===============================================================================================
# code here ;))
def solve(case):
n,k = sep()
a = lis()
b = lis()
have = {}
for i in range(k):
have[b[i]] = 1
pos = {}
for i in range(n):
pos[a[i]] = i
if(len(b) == 1):
if(b[0] in a):
print(1)
else:
print(0)
return
ans = 1
for i in range(k):
# print(i)
# print(have)
left = -1
right = -1
ind = pos[b[i]]
del have[b[i]]
if(ind == 0):
if(a[ind+1] in have):
ans = 0
break
elif(ind == n-1):
if(a[ind-1] in have):
ans = 0
break
else:
if(a[ind-1] in have and a[ind+1] in have):
ans = 0
break
if(a[ind-1] not in have and a[ind+1] not in have):
ans*=2
# print(i,ans)
ans%=mod
print(ans%mod)
"""
2
13
1 1 1 1 1 4 3 4 4 3 4 3
13
1 1 1 2 2 2 3 3 3 4 4 4
"""
# testcase(1)
testcase(int(inp()))
``` | instruction | 0 | 55,792 | 12 | 111,584 |
No | output | 1 | 55,792 | 12 | 111,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class smt:
def __init__(self,l,r,arr):
self.l=l
self.r=r
self.value=(1<<31)-1 if l<r else arr[l]
mid=(l+r)//2
if(l<r):
self.left=smt(l,mid,arr)
self.right=smt(mid+1,r,arr)
self.value&=self.left.value&self.right.value
#print(l,r,self.value)
def setvalue(self,x,val):
if(self.l==self.r):
self.value=val
return
mid=(self.l+self.r)//2
if(x<=mid):
self.left.setvalue(x,val)
else:
self.right.setvalue(x,val)
self.value=self.left.value&self.right.value
def ask(self,l,r):
if(l<=self.l and r>=self.r):
return self.value
val=(1<<31)-1
mid=(self.l+self.r)//2
if(l<=mid):
val&=self.left.ask(l,r)
if(r>mid):
val&=self.right.ask(l,r)
return val
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it):
res=[]
for ch,g in groupby(it):
res.append((ch,len(list(g))))
return res
def judge(i):
return i not in s and i!=-1
t=N()
for i in range(t):
n,k=RL()
a=RLL()
b=RLL()
mod=998244353
s=set(b)
ans=1
left=[-1]*(n+1)
right=[-1]*(n+1)
for i in range(n):
if i!=0:
left[a[i]]=a[i-1]
if i!=n-1:
right[a[i]]=a[i+1]
for i in range(k):
if judge(left[b[i]]) and judge(right[b[i]]):
ans=ans*2%mod
left[b[i]]=left[left[b[i]]]
right[left[left[b[i]]]]=b[i]
s.remove(b[i])
elif judge(left[b[i]]):
left[b[i]]=left[left[b[i]]]
right[left[left[b[i]]]]=b[i]
s.remove(b[i])
elif judge(right[b[i]]):
right[b[i]]=right[right[b[i]]]
left[right[right[b[i]]]]=b[i]
s.remove(b[i])
else:
ans=0
break
#print(right[1],right[2],right[)
#print(s)
print(ans)
``` | instruction | 0 | 55,793 | 12 | 111,586 |
No | output | 1 | 55,793 | 12 | 111,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
"""
Author - Satwik Tiwari .
2nd NOV , 2020 - Monday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 998244353
#===============================================================================================
# code here ;))
def solve(case):
n,k = sep()
a = lis()
b = lis()
have = {}
for i in range(k):
have[b[i]] = 1
pos = {}
for i in range(n):
pos[a[i]] = i
if(len(b) == 1 and len(a)<3):
if(b[0] in a):
print(1)
else:
print(0)
return
ans = 1
for i in range(k):
# print(i)
# print(have)
left = -1
right = -1
ind = pos[b[i]]
del have[b[i]]
print(ind)
if(ind == 0):
if(a[ind+1] in have):
ans = 0
break
elif(ind == n-1):
if(a[ind-1] in have):
ans = 0
break
else:
if(a[ind-1] in have and a[ind+1] in have):
ans = 0
break
if(a[ind-1] not in have and a[ind+1] not in have):
ans*=2
# print(i,ans)
ans%=mod
print(ans%mod)
"""
2
13
1 1 1 1 1 4 3 4 4 3 4 3
13
1 1 1 2 2 2 3 3 3 4 4 4
"""
# testcase(1)
testcase(int(inp()))
``` | instruction | 0 | 55,794 | 12 | 111,588 |
No | output | 1 | 55,794 | 12 | 111,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We start with a permutation a_1, a_2, …, a_n and with an empty array b. We apply the following operation k times.
On the i-th iteration, we select an index t_i (1 ≤ t_i ≤ n-i+1), remove a_{t_i} from the array, and append one of the numbers a_{t_i-1} or a_{t_i+1} (if t_i-1 or t_i+1 are within the array bounds) to the right end of the array b. Then we move elements a_{t_i+1}, …, a_n to the left in order to fill in the empty space.
You are given the initial permutation a_1, a_2, …, a_n and the resulting array b_1, b_2, …, b_k. All elements of an array b are distinct. Calculate the number of possible sequences of indices t_1, t_2, …, t_k modulo 998 244 353.
Input
Each test contains multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100 000), denoting the number of test cases, followed by a description of the test cases.
The first line of each test case contains two integers n, k (1 ≤ k < n ≤ 200 000): sizes of arrays a and b.
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n): elements of a. All elements of a are distinct.
The third line of each test case contains k integers b_1, b_2, …, b_k (1 ≤ b_i ≤ n): elements of b. All elements of b are distinct.
The sum of all n among all test cases is guaranteed to not exceed 200 000.
Output
For each test case print one integer: the number of possible sequences modulo 998 244 353.
Example
Input
3
5 3
1 2 3 4 5
3 2 5
4 3
4 3 2 1
4 3 1
7 4
1 4 7 3 6 2 5
3 2 4 5
Output
2
0
4
Note
\require{cancel}
Let's denote as a_1 a_2 … \cancel{a_i} \underline{a_{i+1}} … a_n → a_1 a_2 … a_{i-1} a_{i+1} … a_{n-1} an operation over an element with index i: removal of element a_i from array a and appending element a_{i+1} to array b.
In the first example test, the following two options can be used to produce the given array b:
* 1 2 \underline{3} \cancel{4} 5 → 1 \underline{2} \cancel{3} 5 → 1 \cancel{2} \underline{5} → 1 2; (t_1, t_2, t_3) = (4, 3, 2);
* 1 2 \underline{3} \cancel{4} 5 → \cancel{1} \underline{2} 3 5 → 2 \cancel{3} \underline{5} → 1 5; (t_1, t_2, t_3) = (4, 1, 2).
In the second example test, it is impossible to achieve the given array no matter the operations used. That's because, on the first application, we removed the element next to 4, namely number 3, which means that it couldn't be added to array b on the second step.
In the third example test, there are four options to achieve the given array b:
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → \cancel{1} \underline{4} 3 2 5 → 4 3 \cancel{2} \underline{5} → 4 3 5;
* 1 4 \cancel{7} \underline{3} 6 2 5 → 1 4 3 \cancel{6} \underline{2} 5 → 1 \underline{4} \cancel{3} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → \cancel{1} \underline{4} 7 2 5 → 4 7 \cancel{2} \underline{5} → 4 7 5;
* 1 4 7 \underline{3} \cancel{6} 2 5 → 1 4 7 \cancel{3} \underline{2} 5 → 1 \underline{4} \cancel{7} 2 5 → 1 4 \cancel{2} \underline{5} → 1 4 5;
Submitted Solution:
```
import sys
input=sys.stdin.readline
R=lambda:map(int,input().split())
t,=R()
for _ in [0]*t:
n,k=R()
vis=[1]*(n+2) # 1可以删
vis[0]=vis[n+1]=0
a=[0]+list(R())+[0]
b=list(R())
ans=0
for i in b:
vis[i]=0
for i in b:
index=a.index(i)
if vis[a[index-1]]==0 and vis[a[index+1]]==0:
ans=-1
break
if vis[a[index-1]]==1 and vis[a[index+1]]==1:
ans+=1
vis[i]=1
if ans!=-1:
print(2**ans)
else:
print(0)
``` | instruction | 0 | 55,795 | 12 | 111,590 |
No | output | 1 | 55,795 | 12 | 111,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,932 | 12 | 111,864 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import collections
from collections import deque
n,k = list(map(int,input().split()))
ans = []
for i in range(n):
if k>0:
ans.append(2*n-2*i-1)
ans.append(2*n-2*i)
k-=1
else:
ans.append(2*n-2*i)
ans.append(2*n-2*i-1)
for num in ans:
print(num, end = " ")
print()
``` | output | 1 | 55,932 | 12 | 111,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,933 | 12 | 111,866 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n,k = map(int,input().split())
r = 2*n-2*k
for i in range(2*n,r,-2):
print(i-1,i,end=' ')
for i in range(r,1,-2):
print(i,i-1,end=' ')
``` | output | 1 | 55,933 | 12 | 111,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,934 | 12 | 111,868 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
i,k = [int(a) for a in input().split()]
A = []
for i in range(1,2*i+1):
A.append(str(i))
for i in range(k):
A[4*i],A[4*i+1] = A[4*i+1],A[4*i]
s = " ".join(A)
print(s)
``` | output | 1 | 55,934 | 12 | 111,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,935 | 12 | 111,870 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
import math
def f(perm):
a1 = 0
a2 = 0
for i in range(len(perm)//2):
a1+=abs(perm[2*i]-perm[2*i+1])
for i in range(len(perm)):
if i%2 == 0:
a2+=perm[i]
else:
a2-=perm[i]
a2 = abs(a2)
return (a1-a2)//2
perm = [7, 1, 2, 3, 4, 5, 6, 8, 9, 10]
#print(f(perm))
n, k = map(int, input().split(' '))
res = []
if k == 0:
for i in range(1, 2*n+1):
res.append(i)
print(res.__str__().replace(",", "").replace("[", "").replace("]", ""))
exit(0)
if n%2 == 0:
znak = 1+k
else:
if k == n:
znak = k+2
else:
znak = k+1
res.append(znak)
res.append(1)
pointer = 2
if pointer == znak:
pointer+=1
for i in range(n-1):
a = pointer
pointer+=1
if pointer == znak:
pointer+=1
b = pointer
pointer+=1
if pointer == znak:
pointer+=1
res.append(a)
res.append(b)
print(res.__str__().replace(",", "").replace("[", "").replace("]", ""))
``` | output | 1 | 55,935 | 12 | 111,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,936 | 12 | 111,872 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n,k=map(int,input().split())
for i in range(n):
if(i<k):
print(2*i+2,2*i+1,end=' ')
else:
print(2*i+1,2*i+2,end=' ')
``` | output | 1 | 55,936 | 12 | 111,873 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,937 | 12 | 111,874 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n,k=list(map(int,input().split()))
a=[i for i in range(1,2*n+1)]
for i in range(k):
a[2*i],a[2*i+1]=a[2*i+1],a[2*i]
print(*a)
``` | output | 1 | 55,937 | 12 | 111,875 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,938 | 12 | 111,876 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n,k=map(int,input().split())
if(k==0):
for i in range(1,2*n+1):
print(i,end=" ")
else:
l=[i for i in range(1,2*n+1)]
for i in range(1,k+1):
j=2*i-1
m=2*i
while(j<=m):
temp=l[j-1]
l[j-1]=l[m-1]
l[m-1]=temp
j=j+1
m=m-1
for i in range(2*n):
print(l[i],end=" ")
``` | output | 1 | 55,938 | 12 | 111,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0. | instruction | 0 | 55,939 | 12 | 111,878 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n, k = map(int, input().split())
if n == 0:
print(0)
else:
arr = [0] * 2 * n
for i in range(2 * n):
arr[i] = i + 1
for i in range(0, 2 * k, 2):
temp = arr[i]
arr[i] = arr[i + 1]
arr[i + 1] = temp
for i in range(2 * n):
print(arr[i], sep=' ', end=' ')
``` | output | 1 | 55,939 | 12 | 111,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
'''
___ ___ ___ ___ ___ ___
/\__\ /\ \ _____ /\ \ /\ \ /\ \ /\__\
/:/ _/_ \:\ \ /::\ \ \:\ \ ___ /::\ \ |::\ \ ___ /:/ _/_
/:/ /\ \ \:\ \ /:/\:\ \ \:\ \ /\__\ /:/\:\__\ |:|:\ \ /\__\ /:/ /\ \
/:/ /::\ \ ___ \:\ \ /:/ \:\__\ ___ /::\ \ /:/__/ /:/ /:/ / __|:|\:\ \ /:/ / /:/ /::\ \
/:/_/:/\:\__\ /\ \ \:\__\ /:/__/ \:|__| /\ /:/\:\__\ /::\ \ /:/_/:/__/___ /::::|_\:\__\ /:/__/ /:/_/:/\:\__\
\:\/:/ /:/ / \:\ \ /:/ / \:\ \ /:/ / \:\/:/ \/__/ \/\:\ \__ \:\/:::::/ / \:\~~\ \/__/ /::\ \ \:\/:/ /:/ /
\::/ /:/ / \:\ /:/ / \:\ /:/ / \::/__/ ~~\:\/\__\ \::/~~/~~~~ \:\ \ /:/\:\ \ \::/ /:/ /
\/_/:/ / \:\/:/ / \:\/:/ / \:\ \ \::/ / \:\~~\ \:\ \ \/__\:\ \ \/_/:/ /
/:/ / \::/ / \::/ / \:\__\ /:/ / \:\__\ \:\__\ \:\__\ /:/ /
\/__/ \/__/ \/__/ \/__/ \/__/ \/__/ \/__/ \/__/ \/__/
'''
"""
░░██▄░░░░░░░░░░░▄██
░▄▀░█▄░░░░░░░░▄█░░█░
░█░▄░█▄░░░░░░▄█░▄░█░
░█░██████████████▄█░
░█████▀▀████▀▀█████░
▄█▀█▀░░░████░░░▀▀███
██░░▀████▀▀████▀░░██
██░░░░█▀░░░░▀█░░░░██
███▄░░░░░░░░░░░░▄███
░▀███▄░░████░░▄███▀░
░░░▀██▄░▀██▀░▄██▀░░░
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░░░░░░░░░░░░░░░░░░░░
"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin, ceil, log2
from bisect import bisect_right, bisect_left
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)
def factorial(n):
if n == 0:
return 1
return (n * factorial(n - 1))
def ncr(n, r):
r = min(r, n - r)
numer = reduce(op.mul, range(n, n - r, -1), 1)
denom = reduce(op.mul, range(1, r + 1), 1)
return numer // denom # or / in Python 2
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return (list(factors))
MOD = 1000000007
PMOD = 998244353
T = 1
# T = int(stdin.readline())
for _ in range(T):
n, k = list(map(int, stdin.readline().rstrip().split()))
# n = int(stdin.readline())
# a = list(map(int, stdin.readline().rstrip().split()))
# b = list(map(int, stdin.readline().rstrip().split()))
# a = str(stdin.readline().strip('\n'))
# k = int(stdin.readline())
# c = list(map(int, stdin.readline().rstrip().split()))
for i in range(n):
if i > 0:
print(' ', end='')
if k > 0:
print(str(2 * i + 2) + ' ' + str(2 * i + 1), end='')
else:
print(str(2 * i + 1) + ' ' + str(2 * i + 2), end='')
k -= 1
``` | instruction | 0 | 55,940 | 12 | 111,880 |
Yes | output | 1 | 55,940 | 12 | 111,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
n, k = map(int, input().split())
a = [i + 1 for i in range(2 * n)]
for i in range(k):
a[2 * i], a[2 * i + 1] = a[2 * i + 1], a[2 * i]
a = " ".join([str(i) for i in a])
print(a)
``` | instruction | 0 | 55,941 | 12 | 111,882 |
Yes | output | 1 | 55,941 | 12 | 111,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
n,k=map(int,input().split())
x=list(range(1,(2*n)+1))
for i in range(k):
x[(2*i)+1],x[2*i]=x[2*i],x[(2*i)+1]
print(*x)
``` | instruction | 0 | 55,942 | 12 | 111,884 |
Yes | output | 1 | 55,942 | 12 | 111,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
a=input()
n,k=a.split(" ");
n=int(n)
k=int(k)
b=list()
b+=[0]
for i in range(1,2*n+1):
b+=[i]
for i in range(1,k+1):
b[2*i],b[2*i-1]=b[2*i-1],b[2*i]
for i in range(1,2*n+1):
print(b[i],end=" ")
``` | instruction | 0 | 55,943 | 12 | 111,886 |
Yes | output | 1 | 55,943 | 12 | 111,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @oj: codeforces
# @id: hitwanyang
# @email: 296866643@qq.com
# @date: 2021/4/29 11:08
# @url: https://codeforces.com/contest/359/problem/B
import sys, os
from io import BytesIO, IOBase
import collections, itertools, bisect, heapq, math, string
from decimal import *
from collections import deque
# region fastio
BUFSIZE = 8192
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
## 注意嵌套括号!!!!!!
## 先有思路,再写代码,别着急!!!
## 先有朴素解法,不要有思维定式,试着换思路解决
## 精度 print("%.10f" % ans)
## sqrt:int(math.sqrt(n))+1
## 字符串拼接不要用+操作，会超时
## 二进制转换:bin(1)[2:].rjust(32,'0')
## array copy:cur=array[::]
## oeis(CROSSREFS):example 1, 3, _, 1260, _, _, _, _, _, 12164510040883200
## sqrt:Decimal(x).sqrt()避免精度误差
## 无穷大表示:float('inf')
## py 10**6 排序+双指针 3秒可能TLE
## 按区间右端点排序,current.left>pre.right,贪心求不相交区间的最大个数
## 加法>位运算
def main():
n, k = map(int, input().split())
odd = [x for x in range(1, 2 * n + 1, 2)]
even = [x for x in range(2, 2 * n + 2, 2)]
# print(odd,even)
odd.reverse()
even.reverse()
# print(res)
i = 0
while k > 0:
odd[i], odd[i + 1] = odd[i + 1], odd[i]
i += 2
k -= 1
res = odd + even
print(" ".join([str(x) for x in res]))
if __name__ == "__main__":
main()
``` | instruction | 0 | 55,944 | 12 | 111,888 |
No | output | 1 | 55,944 | 12 | 111,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
n,k=input().split()
n=int(n)
k=int(k)
l=[i for i in range(1,2*n+1)]
for i in range(1,k+1):
x=2*i-1
y=2*i
l[x],l[y]=l[y],l[x]
for i in range(2*n):
print(l[i],end=" ")
``` | instruction | 0 | 55,945 | 12 | 111,890 |
No | output | 1 | 55,945 | 12 | 111,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
input = sys.stdin.readline
n, k = list(map(int, input().split()))
ans = list(range(1, 2*n+1))
for i in range(1, k+1):
ans[2*k], ans[2*k-1] = ans[2*k-1], ans[2*k]
print(*ans)
``` | instruction | 0 | 55,946 | 12 | 111,892 |
No | output | 1 | 55,946 | 12 | 111,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.
Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).
Output
Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.
Examples
Input
1 0
Output
1 2
Input
2 1
Output
3 2 1 4
Input
4 0
Output
2 7 4 6 1 3 5 8
Note
Record |x| represents the absolute value of number x.
In the first sample |1 - 2| - |1 - 2| = 0.
In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.
In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.
Submitted Solution:
```
import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
input = sys.stdin.readline
n, k = list(map(int, input().split()))
ans = list(range(0, 2*n+1))
for i in range(1, k+1):
ans[2*k], ans[2*k-1] = ans[2*k-1], ans[2*k]
print(*ans[1:])
``` | instruction | 0 | 55,947 | 12 | 111,894 |
No | output | 1 | 55,947 | 12 | 111,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You come home and fell some unpleasant smell. Where is it coming from?
You are given an array a. You have to answer the following queries:
1. You are given two integers l and r. Let ci be the number of occurrences of i in al: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, ..., c109}
2. You are given two integers p to x. Change ap to x.
The Mex of a multiset of numbers is the smallest non-negative integer not in the set.
Note that in this problem all elements of a are positive, which means that c0 = 0 and 0 is never the answer for the query of the second type.
Input
The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.
The second line of input contains n integers — a1, a2, ..., an (1 ≤ ai ≤ 109).
Each of the next q lines describes a single query.
The first type of query is described by three integers ti = 1, li, ri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.
The second type of query is described by three integers ti = 2, pi, xi, where 1 ≤ pi ≤ n is the index of the element, which must be changed and 1 ≤ xi ≤ 109 is the new value.
Output
For each query of the first type output a single integer — the Mex of {c0, c1, ..., c109}.
Example
Input
10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 8
Output
2
3
2
Note
The subarray of the first query consists of the single element — 1.
The subarray of the second query consists of four 2s, one 3 and two 1s.
The subarray of the fourth query consists of three 1s, three 2s and one 3.
Submitted Solution:
```
from collections import defaultdict
n,q = map(int,input().split())
a=list(map(int,input().split()))
w = [defaultdict(int) for i in range(1000)]
for i in range(len(a)):
w[i//100][a[i]]+=1
for quey in range(q):
t,l,r = map(int,input().split())
l-=1
if t==1:
r-=1
if r-l>=100:
M=defaultdict(int)
r_ = r//100
l_=(l+99)//100
for e,cou in w[r_].items():
M[e]=cou-w[l_][e]
for i in range(l,l_*100+1):
M[a[i]]+=1
for i in range(r_*100,r+1):
M[a[i]]+=1
else:
M=defaultdict(int)
for i in range(l,r+1):
M[a[i]]+=1
M=set(list(M.values()))
u=1
while u in M:
u+=1
print(u)
else:
p_ = (l+99)//100
for i in range(p_,1000):
w[i][a[l]]-=1
w[i][r]+=1
a[l]=r
``` | instruction | 0 | 56,147 | 12 | 112,294 |
No | output | 1 | 56,147 | 12 | 112,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You come home and fell some unpleasant smell. Where is it coming from?
You are given an array a. You have to answer the following queries:
1. You are given two integers l and r. Let ci be the number of occurrences of i in al: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, ..., c109}
2. You are given two integers p to x. Change ap to x.
The Mex of a multiset of numbers is the smallest non-negative integer not in the set.
Note that in this problem all elements of a are positive, which means that c0 = 0 and 0 is never the answer for the query of the second type.
Input
The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.
The second line of input contains n integers — a1, a2, ..., an (1 ≤ ai ≤ 109).
Each of the next q lines describes a single query.
The first type of query is described by three integers ti = 1, li, ri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.
The second type of query is described by three integers ti = 2, pi, xi, where 1 ≤ pi ≤ n is the index of the element, which must be changed and 1 ≤ xi ≤ 109 is the new value.
Output
For each query of the first type output a single integer — the Mex of {c0, c1, ..., c109}.
Example
Input
10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 8
Output
2
3
2
Note
The subarray of the first query consists of the single element — 1.
The subarray of the second query consists of four 2s, one 3 and two 1s.
The subarray of the fourth query consists of three 1s, three 2s and one 3.
Submitted Solution:
```
def main(cin, mnoj):
for i in range(cin[1]):
mex = 1
c = list(map(int, input().split()))
if (c[0] == 1):
cur = mnoj[c[1] - 1:c[2]]
#print(cur)
# print(max(cur) - min(cur))
# print(min(cur)+1)
set_cur = set(cur)
while mex in set_cur:
mex += 1
print(mex)
if (c[0] == 2):
mnoj[c[1]] = c[2]
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a = main(a, b)
``` | instruction | 0 | 56,148 | 12 | 112,296 |
No | output | 1 | 56,148 | 12 | 112,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You come home and fell some unpleasant smell. Where is it coming from?
You are given an array a. You have to answer the following queries:
1. You are given two integers l and r. Let ci be the number of occurrences of i in al: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, ..., c109}
2. You are given two integers p to x. Change ap to x.
The Mex of a multiset of numbers is the smallest non-negative integer not in the set.
Note that in this problem all elements of a are positive, which means that c0 = 0 and 0 is never the answer for the query of the second type.
Input
The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.
The second line of input contains n integers — a1, a2, ..., an (1 ≤ ai ≤ 109).
Each of the next q lines describes a single query.
The first type of query is described by three integers ti = 1, li, ri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.
The second type of query is described by three integers ti = 2, pi, xi, where 1 ≤ pi ≤ n is the index of the element, which must be changed and 1 ≤ xi ≤ 109 is the new value.
Output
For each query of the first type output a single integer — the Mex of {c0, c1, ..., c109}.
Example
Input
10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 8
Output
2
3
2
Note
The subarray of the first query consists of the single element — 1.
The subarray of the second query consists of four 2s, one 3 and two 1s.
The subarray of the fourth query consists of three 1s, three 2s and one 3.
Submitted Solution:
```
from collections import defaultdict
n,q = map(int,input().split())
a=list(map(int,input().split()))
w = [defaultdict(int) for i in range(400)]
for i in range(len(a)):
w[i//300][a[i]]+=1
w=[defaultdict(int)]+w
for quey in range(q):
t,l,r = map(int,input().split())
l-=1
if t==1:
r-=1
if r-l>=300:
M=defaultdict(int)
r_ = r//300
l_=l//300
for e,cou in w[r_].items():
M[e]=cou-w[l_][e]
for i in range(l,l_*300+1):
M[a[i]]+=1
for i in range(r_*300,r+1):
M[a[i]]+=1
else:
M=defaultdict(int)
for i in range(l,r+1):
M[a[i]]+=1
M=set(list(M.values()))
u=1
while u in M:
u+=1
print(u)
else:
p_ = l//300
for i in range(p_,400):
w[i][a[l]]-=1
w[i][r]+=1
a[l]=r
``` | instruction | 0 | 56,149 | 12 | 112,298 |
No | output | 1 | 56,149 | 12 | 112,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,576 | 12 | 113,152 |
Tags: binary search, combinatorics
Correct Solution:
```
import math
teststring = """123 42 24
"""
online = "Problem_C" not in __file__
true, false = True, False
if True:
def spitout():
for c in teststring.splitlines():
yield c
_ito = spitout()
if not online:
def input():
return next(_ito)
def build_enum(*a):
built = dict()
for i, c in enumerate(a):
built[c] = i
return lambda x: built[x]
# T = 1
##-----------------start coding-----------------
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a):
g, x, y = egcd(a, P)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % P
[n, x, pos] = map(int, input().split(" "))
P = 1000000007
a = b = 0
l = 0
r = n
while l < r:
m = int((l+r)/2)
if m <= pos:
l = m + 1
if m < pos: a += 1
else:
r = m
b += 1
if a > x-1 or b > n-x: ans = 0
else:
ans = 1
fac = [1]*(n+1)
for i in range(n):
fac[i+1] = (fac[i]*(i+1)) % P
ans = (ans * fac[x-1]) % P
ans = (ans * fac[n-x]) % P
ans = (ans * fac[n-a-b-1]) % P
ans = (ans * modinv(fac[x-1-a])) % P
ans = (ans * modinv(fac[n-x-b])) % P
print(ans)
# print('Case #{}: {}'.format(ti, '...'))
``` | output | 1 | 56,576 | 12 | 113,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,577 | 12 | 113,154 |
Tags: binary search, combinatorics
Correct Solution:
```
from math import factorial
n, x, pos = map(int, input().split())
MOD = (10**9) + 7
small, big, ans = x-1, n-x, 1
l, r = 0, n
while l < r:
mid = (l + r) //2
if mid < pos:
ans *= small
small -= 1
l = mid + 1
elif mid > pos:
ans *= big
big -= 1
r = mid
else:
l = mid+1
ans *= factorial(small + big)
print(ans%MOD)
``` | output | 1 | 56,577 | 12 | 113,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,578 | 12 | 113,156 |
Tags: binary search, combinatorics
Correct Solution:
```
import math
yes = "YES"
no = "NO"
lower = "L"
higher = "H"
same = "O"
def bin_search(a, x):
left = 0
right = len(a)
res = []
while left < right:
mid = (left + right) // 2
if a[mid] < x:
left = mid + 1
res.append(higher)
elif a[mid] == x:
left = mid + 1
res.append(same)
else:
right = mid
res.append(lower)
return res
def solve():
numbers = list(map(int, input().split(" ")))
options = bin_search([i for i in range(numbers[0])], numbers[2])
res = 1
higher_numbers, lower_numbers = numbers[0] - numbers[1], numbers[1] - 1
for o in options:
if o == higher:
res *= lower_numbers
lower_numbers -= 1
elif o == lower:
res *= higher_numbers
higher_numbers -= 1
res *= math.factorial(numbers[0] - len(options))
print(res %(10**9 + 7))
return None
if __name__ == "__main__":
solve()
``` | output | 1 | 56,578 | 12 | 113,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,579 | 12 | 113,158 |
Tags: binary search, combinatorics
Correct Solution:
```
mod=10**9+7
def modExp(a,n,m=10**9+7):
if n==0:
return 1
elif n==1:
return a
else:
while n>=2:
if n%2==0:
n=n//2
return modExp(((a%m)*(a%m))%m,n,m)
else:
n=(n-1)//2
return ((a%m)*modExp(((a%m)*(a%m))%m,n,m)%m)%m
def nCr(n,r):
return (dic[n]*modExp((dic[n-r]*dic[r])%mod,mod-2))%mod
#input
n,x,p=map(int,input().split())
dic={0:1}
for i in range(1,1001):
dic[i]=dic[i-1]*i
dic[i]%=mod
def binS(n,p):
l=0
r=n
g=0
s=0
while l<r:
m=(l+r)//2
if m>p:
r=m
g+=1
else:
if m!=p:
s+=1
l=m+1
return s,g
s,g=binS(n,p)
if n-x<g or x-1<s:
print(0)
else:
out=(nCr(n-x,g)*dic[g])%mod
out*=(nCr(x-1,s)*dic[s])%mod
out%=mod
out*=dic[n-g-s-1]
out%=mod
print(out)
``` | output | 1 | 56,579 | 12 | 113,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,580 | 12 | 113,160 |
Tags: binary search, combinatorics
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict,deque,Counter
from bisect import *
from math import sqrt,pi,ceil,log
import math
from itertools import permutations
from copy import deepcopy
from sys import setrecursionlimit
def main():
n,x,pos = map(int,input().split())
mod=10**9+7
f=[1]
for i in range(1,1001):
f.append((f[-1]*i)%mod)
lo=0
hi=n
b=x-1
c=n-x
ans=1
while lo<hi:
mid=(lo+hi)//2
if mid<pos:
ans=(ans*b)%mod
b-=1
lo=mid+1
elif mid>pos:
ans=(ans*c)%mod
c-=1
hi=mid
else:
lo=mid+1
print((ans*f[b+c])%mod)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 56,580 | 12 | 113,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,581 | 12 | 113,162 |
Tags: binary search, combinatorics
Correct Solution:
```
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
def ncr(n,r,p=10**9+7):
if n<r:
return 0
num=1
r=min(r,n-r)
ct=1
while ct <=r:
num=(num*n)%p
num=(num*pow(ct,p-2,p))%p
ct+=1;n-=1
return num%p
n,x,pos=I()
md=10**9+7
ft=[1,1,2]
for i in range(3,1003):
ft.append((ft[-1]*i)%md)
an=0
ar=[]
lo=0;ri=n
while lo<ri:
md=(lo+ri)//2
ar.append(md)
if md<=pos:
lo=md+1
else:
ri=md
md=10**9+7
a=b=0
for i in ar:
if i>pos:
a+=1
elif i<pos:
b+=1
print((ncr(x-1,b)*ncr(n-x,a)*ft[a]*ft[b]*ft[n-a-b-1])%md)
``` | output | 1 | 56,581 | 12 | 113,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,582 | 12 | 113,164 |
Tags: binary search, combinatorics
Correct Solution:
```
n,x,pos = map(int,input().split())
a = [0] * n
low,high = x - 1, n - x
l,r = 0,n
found = False
ans = 0
while l < r:
m = (l + r) // 2
if pos < m:
if high == 0:
ans = -1
break
a[m] = high
high -= 1
r = m
else:
if m != pos:
if low == 0:
ans = -1;
break
a[m] = low
low -= 1
l = m + 1
mod = 10**9 + 7
if ans == -1:
print(0)
else:
fans = 1
for i in range(n):
if a[i]:
fans = (fans * a[i]) % mod
left = low + high
while left:
fans = (fans * left) % mod
left -= 1
print(fans)
``` | output | 1 | 56,582 | 12 | 113,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2). | instruction | 0 | 56,583 | 12 | 113,166 |
Tags: binary search, combinatorics
Correct Solution:
```
mod = 10**9+7
n, x, pos = map(int, input().split())
l = 0
r = n
lcheck = 0
ucheck = 0
while(l<r):
mid = (l+r)//2
if mid<=pos:
l = mid+1
if mid!=pos:
lcheck+=1
else:
r = mid
ucheck+=1
ll = x-1
rr = n-x
# print(ll, rr, lcheck, ucheck)
ans = 1
for i in range(lcheck):
ans*=ll
ll-=1
for i in range(ucheck):
ans*=rr
rr-=1
left = n-(lcheck+ucheck)-1
lt = left
for i in range(lt):
ans*=left
left-=1
print(ans%mod)
``` | output | 1 | 56,583 | 12 | 113,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
def perm(n,r):
if n<r:
return 0
ans = 1
k = n
while k>n-r:
ans*=k
k-=1
return ans
def fact(x):
if x<0:
return 0
ans = 1
while x:
ans*=x
x-=1
return ans
n, x, p = map(int, input().split())
b = [0]*n
left = 0
right = n
small = -1
large = 0
while left < right:
middle = (left + right) // 2
if middle <= p:
left = middle + 1
b[middle] = -1
small += 1
else:
right = middle
b[middle] = 1
large += 1
# print(*b)
ans = perm(x-1,small)*perm(n-x,large)*fact(n-1-small-large)
print(ans%(10**9 + 7))
``` | instruction | 0 | 56,584 | 12 | 113,168 |
Yes | output | 1 | 56,584 | 12 | 113,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
n,x,p=map(int,input().split())
ans=1
l=x-1
r=n-x
b=0
e=n
while(b<e):
m=b+e
m//=2
if p<m:
e=m
ans=(ans*r)%1000000007
r-=1
n-=1
elif p>m:
b=m+1
ans=(ans*l)%1000000007
l-=1
n-=1
else:
b=m+1
for i in range(1,n):
ans=(ans*i)%1000000007
print(ans)
``` | instruction | 0 | 56,585 | 12 | 113,170 |
Yes | output | 1 | 56,585 | 12 | 113,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x, n, a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4, n, a)
luckynumwithequalnumberoffourandseven(x * 10 + 7, n, a)
return a
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
#------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
#------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
#-----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
#-----------------------------------------------lucky number with two lucky any digits
res = set()
def solve(p, l, a, b,n):#given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solve(p * 10 + a, l + 1, a, b,n)
solve(p * 10 + b, l + 1, a, b,n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
#-----------------------------------------------
# endregion------------------------------
"""
def main():
n,m = sep()
l=[]
q=0
if m<=n:
for i in range(min(n,m)):
l.append([max(n,m)-i,i])
q = max(n,m)-i
l.append([q-1,min(n,m)])
else:
for i in range(min(m,n)):
l.append([min(n,m)-i,i])
q=max(n,m)-i
l.append([0,min(n,m)])
print(len(l))
for i in l:
print(*i)
if __name__ == '__main__':
main()
"""
"""
def search(n,x):
left =0
right = n
cnt=0
while(left<right):
cnt+=1
#print(middle)
middle = (left+right)//2
#print(middle,left,right)
if middle<x:
left = middle+1
else:
right = middle
if left==x or right==x:
break
return cnt
"""
def main():
n,x,poss = sep()
less = x - 1
large = n - x
s = 0
e = n
ans = 1
while (s < e):
mid = (s + e) // 2
if mid < poss:
s = mid + 1
ans *= less
less -= 1
elif mid > poss:
e = mid
ans *= large
large -= 1
else:
s = mid + 1
less += large
if less > 0:
for i in range(less, 0, -1):
ans *= i
print(ans % (10 ** (9) + 7))
if __name__ == '__main__':
main()
``` | instruction | 0 | 56,586 | 12 | 113,172 |
Yes | output | 1 | 56,586 | 12 | 113,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
if __name__ == "__main__":
mod = 1000000007
n, x, pos = map(int, input().split())
ans = 1
left = 0
right = n
toLeft = n-x
toRight = x-1
failed = False
c = 0
while left < right:
middle = (left + right) // 2
if middle == pos:
left = middle + 1
c += 1
elif middle > pos:
if toLeft == 0:
failed = True
break
else:
ans *= toLeft
ans %= mod
toLeft -= 1
right = middle
c += 1
else:
if toRight == 0:
failed = True
break
else:
ans *= toRight
ans %= mod
toRight -= 1
left = middle + 1
c += 1
if failed:
print(0)
else:
for i in range(n-c):
ans *= (i+1)
ans %= mod
print(ans)
``` | instruction | 0 | 56,587 | 12 | 113,174 |
Yes | output | 1 | 56,587 | 12 | 113,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
x, y, p = map(int, input().split())
sm = 0
bg = 0
hi = x
lo = 0
while hi > lo:
mid = (hi+lo)//2
if mid < p:
sm += 1
lo = mid+1
elif mid > p:
bg += 1
hi = mid
else:
break
fac = [1]
mod = (10**9)+7
for n in range(1, x+1):
fac.append((fac[-1]*n) % mod)
def bnp(bs, pw):
res = 1
while pw:
if pw & 1:
res = (res*bs) % mod
pw //= 2
bs = (bs*bs) % mod
return res % mod
def ncr(n, r):
if n < r:
return 0
return (((fac[n]*bnp(fac[r], mod-2)) % mod)*bnp(fac[n-r], mod-2)) % mod
biig = x-y
sml = y-1
a = (ncr(biig, bg)*fac[bg]) % mod
b = (ncr(sml, sm)*fac[sm]) % mod
c = fac[x-sm-bg-1]
print((a*b*c) % mod)
``` | instruction | 0 | 56,588 | 12 | 113,176 |
No | output | 1 | 56,588 | 12 | 113,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
factorial=[1,1,2]
for i in range(3,1001):
factorial.append(i*factorial[-1])
def comb(n,r):
c=factorial[n]/(factorial[n-r])
return int(c)
n,x,pos=map(int,input().split())
left=0
right=n
smaller=0
equal=0
larger=0
while left!=right:
mid=(left+right)//2
if pos<mid:
larger+=1
right=mid
elif pos>mid:
smaller+=1
left=mid+1
elif pos==mid:
equal+=1
left=mid+1
#print(larger,smaller,equal,x,n,x-1,n-x)
if(larger>(n-x)):
print(0)
elif (smaller>x-1):
print(0)
else:
s=comb(x-1,smaller)
l=comb(n-x,larger)
fact=factorial[n-smaller-larger-1]
num=(s*l*fact)
num%=1000000007
print(int(num))
#print(factorial[999]%1000000007)
``` | instruction | 0 | 56,589 | 12 | 113,178 |
No | output | 1 | 56,589 | 12 | 113,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
import math
def bins(a,x,pos):
posl = []
fl = True
left, right = 0, len(a)
while left < right:
middle = math.floor((left + right) / 2)
if a[middle] <= x:
left = middle + 1
if middle != pos and fl:
posl.append('l')
else:
fl = False
else:
right = middle
if middle != pos and fl:
posl.append('r')
else:
fl = False
return posl
n,x,pos = list(map(int,input().split()))
a = [0]*(pos)
a.append(x)
a.extend([x+1]*(n-pos-1))
b = bins(a,x,pos)
r, l = b.count('r'), b.count('l')
rcount = n-x
lcount = x-1
count = n-1
res = 1
##print('r',r,rcount,'l',l,lcount)
for i in range(r):
res *= rcount
rcount -= 1
count -= 1
for i in range(l):
res *= lcount
lcount -= 1
count -= 1
for i in range(1,count+1):
res *= i % 1000000007
if pos != n-1 and x != n:
print(res % 1000000007)
else:
print(0)
``` | instruction | 0 | 56,590 | 12 | 113,180 |
No | output | 1 | 56,590 | 12 | 113,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey thinks he is truly a successful developer, but in reality he didn't know about the binary search algorithm until recently. After reading some literature Andrey understood that this algorithm allows to quickly find a certain number x in an array. For an array a indexed from zero, and an integer x the pseudocode of the algorithm is as follows:
<image>
Note that the elements of the array are indexed from zero, and the division is done in integers (rounding down).
Andrey read that the algorithm only works if the array is sorted. However, he found this statement untrue, because there certainly exist unsorted arrays for which the algorithm find x!
Andrey wants to write a letter to the book authors, but before doing that he must consider the permutations of size n such that the algorithm finds x in them. A permutation of size n is an array consisting of n distinct integers between 1 and n in arbitrary order.
Help Andrey and find the number of permutations of size n which contain x at position pos and for which the given implementation of the binary search algorithm finds x (returns true). As the result may be extremely large, print the remainder of its division by 10^9+7.
Input
The only line of input contains integers n, x and pos (1 ≤ x ≤ n ≤ 1000, 0 ≤ pos ≤ n - 1) — the required length of the permutation, the number to search, and the required position of that number, respectively.
Output
Print a single number — the remainder of the division of the number of valid permutations by 10^9+7.
Examples
Input
4 1 2
Output
6
Input
123 42 24
Output
824071958
Note
All possible permutations in the first test case: (2, 3, 1, 4), (2, 4, 1, 3), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 1, 3), (4, 3, 1, 2).
Submitted Solution:
```
import math
n, x, p = [int(s) for s in input().split(' ')]
lowers = 0
highers = 0
l = 0
h = n
while l < h:
print(l, h)
m = (l + h) // 2
print(m)
if m > p:
highers += 1
h = m
elif m < p:
lowers += 1
l = m + 1
else:
l = m + 1
#print(lowers, highers, n - 1 - lowers - highers)
if lowers > x - 1:
ans = 0
elif highers > n - x:
ans = 0
else:
ans = 1
for i in range(lowers):
ans = (ans * (x - 1 - i)) % (10 ** 9 + 7)
for i in range(highers):
ans = (ans * (n - x - i)) % (10 ** 9 + 7)
for i in range(n - 1 - lowers - highers):
ans = (ans * (i + 1)) % (10 ** 9 + 7)
print(ans)
``` | instruction | 0 | 56,591 | 12 | 113,182 |
No | output | 1 | 56,591 | 12 | 113,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Uh oh! Ray lost his array yet again! However, Omkar might be able to help because he thinks he has found the OmkArray of Ray's array. The OmkArray of an array a with elements a_1, a_2, …, a_{2k-1}, is the array b with elements b_1, b_2, …, b_{k} such that b_i is equal to the median of a_1, a_2, …, a_{2i-1} for all i. Omkar has found an array b of size n (1 ≤ n ≤ 2 ⋅ 10^5, -10^9 ≤ b_i ≤ 10^9). Given this array b, Ray wants to test Omkar's claim and see if b actually is an OmkArray of some array a. Can you help Ray?
The median of a set of numbers a_1, a_2, …, a_{2i-1} is the number c_{i} where c_{1}, c_{2}, …, c_{2i-1} represents a_1, a_2, …, a_{2i-1} sorted in nondecreasing order.
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array b.
The second line contains n integers b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9) — the elements of b.
It is guaranteed the sum of n across all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output one line containing YES if there exists an array a such that b_i is the median of a_1, a_2, ..., a_{2i-1} for all i, and NO otherwise. The case of letters in YES and NO do not matter (so yEs and No will also be accepted).
Examples
Input
5
4
6 2 1 3
1
4
5
4 -8 5 6 -7
2
3 3
4
2 1 2 3
Output
NO
YES
NO
YES
YES
Input
5
8
-8 2 -6 -5 -4 3 3 2
7
1 1 3 1 0 -2 -1
7
6 12 8 6 2 6 10
6
5 1 2 3 6 7
5
1 3 4 3 0
Output
NO
YES
NO
NO
NO
Note
In the second case of the first sample, the array [4] will generate an OmkArray of [4], as the median of the first element is 4.
In the fourth case of the first sample, the array [3, 2, 5] will generate an OmkArray of [3, 3], as the median of 3 is 3 and the median of 2, 3, 5 is 3.
In the fifth case of the first sample, the array [2, 1, 0, 3, 4, 4, 3] will generate an OmkArray of [2, 1, 2, 3] as
* the median of 2 is 2
* the median of 0, 1, 2 is 1
* the median of 0, 1, 2, 3, 4 is 2
* and the median of 0, 1, 2, 3, 3, 4, 4 is 3.
In the second case of the second sample, the array [1, 0, 4, 3, 5, -2, -2, -2, -4, -3, -4, -1, 5] will generate an OmkArray of [1, 1, 3, 1, 0, -2, -1], as
* the median of 1 is 1
* the median of 0, 1, 4 is 1
* the median of 0, 1, 3, 4, 5 is 3
* the median of -2, -2, 0, 1, 3, 4, 5 is 1
* the median of -4, -2, -2, -2, 0, 1, 3, 4, 5 is 0
* the median of -4, -4, -3, -2, -2, -2, 0, 1, 3, 4, 5 is -2
* and the median of -4, -4, -3, -2, -2, -2, -1, 0, 1, 3, 4, 5, 5 is -1
For all cases where the answer is NO, it can be proven that it is impossible to find an array a such that b is the OmkArray of a. | instruction | 0 | 56,628 | 12 | 113,256 |
Tags: data structures, greedy, implementation
Correct Solution:
```
def solve(n, a):
prev = a[0]
r = [10 ** 9];
l = [-r[0]]
for i in range(1, n):
if a[i] < prev:
if a[i] < l[-1]:
return False
elif a[i] == l[-1]:
l.pop()
r.append(prev)
elif a[i] > prev:
if a[i] > r[-1]:
return False
elif a[i] == r[-1]:
r.pop()
l.append(prev)
prev = a[i]
return True
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if solve(n, a):
print("YES")
else:
print("NO")
main()
``` | output | 1 | 56,628 | 12 | 113,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Uh oh! Ray lost his array yet again! However, Omkar might be able to help because he thinks he has found the OmkArray of Ray's array. The OmkArray of an array a with elements a_1, a_2, …, a_{2k-1}, is the array b with elements b_1, b_2, …, b_{k} such that b_i is equal to the median of a_1, a_2, …, a_{2i-1} for all i. Omkar has found an array b of size n (1 ≤ n ≤ 2 ⋅ 10^5, -10^9 ≤ b_i ≤ 10^9). Given this array b, Ray wants to test Omkar's claim and see if b actually is an OmkArray of some array a. Can you help Ray?
The median of a set of numbers a_1, a_2, …, a_{2i-1} is the number c_{i} where c_{1}, c_{2}, …, c_{2i-1} represents a_1, a_2, …, a_{2i-1} sorted in nondecreasing order.
Input
Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array b.
The second line contains n integers b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9) — the elements of b.
It is guaranteed the sum of n across all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output one line containing YES if there exists an array a such that b_i is the median of a_1, a_2, ..., a_{2i-1} for all i, and NO otherwise. The case of letters in YES and NO do not matter (so yEs and No will also be accepted).
Examples
Input
5
4
6 2 1 3
1
4
5
4 -8 5 6 -7
2
3 3
4
2 1 2 3
Output
NO
YES
NO
YES
YES
Input
5
8
-8 2 -6 -5 -4 3 3 2
7
1 1 3 1 0 -2 -1
7
6 12 8 6 2 6 10
6
5 1 2 3 6 7
5
1 3 4 3 0
Output
NO
YES
NO
NO
NO
Note
In the second case of the first sample, the array [4] will generate an OmkArray of [4], as the median of the first element is 4.
In the fourth case of the first sample, the array [3, 2, 5] will generate an OmkArray of [3, 3], as the median of 3 is 3 and the median of 2, 3, 5 is 3.
In the fifth case of the first sample, the array [2, 1, 0, 3, 4, 4, 3] will generate an OmkArray of [2, 1, 2, 3] as
* the median of 2 is 2
* the median of 0, 1, 2 is 1
* the median of 0, 1, 2, 3, 4 is 2
* and the median of 0, 1, 2, 3, 3, 4, 4 is 3.
In the second case of the second sample, the array [1, 0, 4, 3, 5, -2, -2, -2, -4, -3, -4, -1, 5] will generate an OmkArray of [1, 1, 3, 1, 0, -2, -1], as
* the median of 1 is 1
* the median of 0, 1, 4 is 1
* the median of 0, 1, 3, 4, 5 is 3
* the median of -2, -2, 0, 1, 3, 4, 5 is 1
* the median of -4, -2, -2, -2, 0, 1, 3, 4, 5 is 0
* the median of -4, -4, -3, -2, -2, -2, 0, 1, 3, 4, 5 is -2
* and the median of -4, -4, -3, -2, -2, -2, -1, 0, 1, 3, 4, 5, 5 is -1
For all cases where the answer is NO, it can be proven that it is impossible to find an array a such that b is the OmkArray of a. | instruction | 0 | 56,629 | 12 | 113,258 |
Tags: data structures, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
b = list(map(int,input().split()))
next = dict()
prev = dict()
curr = b[0]
next[curr] = "!"
prev[curr] = "!"
fine = True
for i in b:
if i>curr:
if next[curr] == "!":
next[curr] = i
prev[i] = curr
next[i] = "!"
elif next[curr]<i:
fine = False
break
elif next[curr]>i:
prev[next[curr]] = i
next[i] = next[curr]
next[curr] = i
prev[i] = curr
elif i<curr:
if prev[curr] == "!":
prev[curr] = i
prev[i] = "!"
next[i] = curr
elif i<prev[curr]:
fine = False
break
elif i>prev[curr]:
next[prev[curr]] = i
prev[i] = prev[curr]
prev[curr] = i
next[i] = curr
curr = i
if fine:
print("YES")
else:
print("NO")
``` | output | 1 | 56,629 | 12 | 113,259 |
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