message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are n displays placed along a road, and the i-th of them can display a text with font size s_i only. Maria Stepanovna wants to rent such three displays with indices i < j < k that the font size increases if you move along the road in a particular direction. Namely, the condition s_i < s_j < s_k should be held.
The rent cost is for the i-th display is c_i. Please determine the smallest cost Maria Stepanovna should pay.
Input
The first line contains a single integer n (3 ≤ n ≤ 3 000) — the number of displays.
The second line contains n integers s_1, s_2, …, s_n (1 ≤ s_i ≤ 10^9) — the font sizes on the displays in the order they stand along the road.
The third line contains n integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^8) — the rent costs for each display.
Output
If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices i < j < k such that s_i < s_j < s_k.
Examples
Input
5
2 4 5 4 10
40 30 20 10 40
Output
90
Input
3
100 101 100
2 4 5
Output
-1
Input
10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13
Output
33
Note
In the first example you can, for example, choose displays 1, 4 and 5, because s_1 < s_4 < s_5 (2 < 4 < 10), and the rent cost is 40 + 10 + 40 = 90.
In the second example you can't select a valid triple of indices, so the answer is -1.
Submitted Solution:
```
from math import inf
n = int(input())
s = list(map(int, input().split(' ')))
c = list(map(int, input().split(' ')))
dp = [[0, 0, 0, 0][:] for i in range(n)]
dp[0] = [0, c[0], inf, inf]
for j in range(1, 4):
for i in range(1, n):
dp[i][j] = dp[i - 1][j]
for k in reversed(range(i)):
if s[k] < s[i]:
if dp[k][j - 1] + c[i] < dp[i][j]:
dp[i][j] = dp[k][j - 1] + c[i]
break
if dp[n - 1][3] == inf:
print(-1)
else:
print(dp[n - 1][3])
``` | instruction | 0 | 45,156 | 8 | 90,312 |
No | output | 1 | 45,156 | 8 | 90,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are n displays placed along a road, and the i-th of them can display a text with font size s_i only. Maria Stepanovna wants to rent such three displays with indices i < j < k that the font size increases if you move along the road in a particular direction. Namely, the condition s_i < s_j < s_k should be held.
The rent cost is for the i-th display is c_i. Please determine the smallest cost Maria Stepanovna should pay.
Input
The first line contains a single integer n (3 ≤ n ≤ 3 000) — the number of displays.
The second line contains n integers s_1, s_2, …, s_n (1 ≤ s_i ≤ 10^9) — the font sizes on the displays in the order they stand along the road.
The third line contains n integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^8) — the rent costs for each display.
Output
If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices i < j < k such that s_i < s_j < s_k.
Examples
Input
5
2 4 5 4 10
40 30 20 10 40
Output
90
Input
3
100 101 100
2 4 5
Output
-1
Input
10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13
Output
33
Note
In the first example you can, for example, choose displays 1, 4 and 5, because s_1 < s_4 < s_5 (2 < 4 < 10), and the rent cost is 40 + 10 + 40 = 90.
In the second example you can't select a valid triple of indices, so the answer is -1.
Submitted Solution:
```
n=int(input())
s1=list(map(int,input().split()))
s2=list(map(int,input().split()))
dicc=dict(zip(s2,s1))
sumaa=[]
num=[]
for i in range(0,n):
num=[]
tmp=s1[i]
num+=[s2[i]]
count=0
for j in range(i+1,n):
if s1[j]>tmp and count==0:
num+=[s2[j]]
tmp=s1[j]
count+=1
elif s1[j]>tmp and count==1:
num+=[s2[j]]
tmp=s1[j]
count+=1
elif s1[j]>tmp and count>=2:
if s2[j] <= max(num):
num.sort()
tmp=s1[j]
del(num[2])
num+=[s2[j]]
if count>=2:
sumaa+=[sum(num)]
num=[]
if len(sumaa)==0 :
print("-1")
else:
print(min(sumaa))
``` | instruction | 0 | 45,157 | 8 | 90,314 |
No | output | 1 | 45,157 | 8 | 90,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are n displays placed along a road, and the i-th of them can display a text with font size s_i only. Maria Stepanovna wants to rent such three displays with indices i < j < k that the font size increases if you move along the road in a particular direction. Namely, the condition s_i < s_j < s_k should be held.
The rent cost is for the i-th display is c_i. Please determine the smallest cost Maria Stepanovna should pay.
Input
The first line contains a single integer n (3 ≤ n ≤ 3 000) — the number of displays.
The second line contains n integers s_1, s_2, …, s_n (1 ≤ s_i ≤ 10^9) — the font sizes on the displays in the order they stand along the road.
The third line contains n integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 10^8) — the rent costs for each display.
Output
If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices i < j < k such that s_i < s_j < s_k.
Examples
Input
5
2 4 5 4 10
40 30 20 10 40
Output
90
Input
3
100 101 100
2 4 5
Output
-1
Input
10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13
Output
33
Note
In the first example you can, for example, choose displays 1, 4 and 5, because s_1 < s_4 < s_5 (2 < 4 < 10), and the rent cost is 40 + 10 + 40 = 90.
In the second example you can't select a valid triple of indices, so the answer is -1.
Submitted Solution:
```
from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import random
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(input()))
def In():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def I():return (int(stdin.readline()))
def In():return(map(int,stdin.readline().split()))
#sys.setrecursionlimit(1500)
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_right(a, x)
if i != len(a):
return i
else:
return -1
def main():
try:
n=I()
l=list(In())
cost=list(In())
ans=1e8
#BRUTE FORCE
for x in range(n):
for j in range(x+1,n):
if l[x]<l[j]:
for k in range(j+1,n):
if l[j]<l[k]:
ans=min(cost[x]+cost[j]+cost[k],ans)
if ans==1e8:
print(-1)
else:
print(ans)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
#for _ in range(I()):main()
for _ in range(1):main()
``` | instruction | 0 | 45,158 | 8 | 90,316 |
No | output | 1 | 45,158 | 8 | 90,317 |
Provide a correct Python 3 solution for this coding contest problem.
A pitch-black room
When I woke up, Mr. A was in a pitch-black room. Apparently, Mr. A got lost in a dungeon consisting of N rooms. You couldn't know which room A got lost in, but fortunately he got a map of the dungeon. Let's show A the way to go and lead to a bright room.
It is known that M of the N rooms are pitch black, and the D_1, D_2, ..., and D_Mth rooms are pitch black, respectively. Also, there are exactly K one-way streets from all the rooms in order, and the roads from the i-th room are v_ {i, 1}, v_ {i, 2}, ..., v_ {i, respectively. , K} Connected to the third room. You can follow the a_1, a_2, ..., a_lth roads in order from the room you are in for Mr. A. However, when Mr. A reaches the bright room, he ignores the subsequent instructions. You cannot know the information of the room where Mr. A is currently before and after the instruction, so you must give the instruction sequence so that you can reach the bright room no matter which room you are in. Answer the length of the shortest of such instructions.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ M ≤ min (16, N − 1)
* 1 ≤ K ≤ N
* 1 ≤ D_i ≤ N
* D_i are all different
* 1 ≤ v_ {i, j} ≤ N
* All dark rooms can reach at least one bright room
Input Format
Input is given from standard input in the following format.
NMK
D_1 D_2 ... D_M
v_ {1,1} v_ {1,2} ... v_ {1,K}
v_ {2,1} v_ {2,2} ... v_ {2, K}
...
v_ {N, 1} v_ {N, 2} ... v_ {N, K}
Output Format
Print the answer in one line.
Sample Input 1
4 2 2
1 2
twenty four
3 1
4 2
13
Sample Output 1
2
<image>
If you give the instruction 1, 1
* If A's initial position is room 1, the second move will reach room 3.
* If A's initial position is room 2, the first move will reach room 3.
Sample Input 2
3 2 2
1 2
twenty three
1 2
twenty one
Sample Output 2
3
<image>
If you give instructions 2, 1, 2
* If A's initial position is room 1, the first move will reach room 3.
* If A's initial position is room 2, the third move will reach room 3.
Sample Input 3
6 3 3
one two Three
4 1 1
2 5 2
3 3 6
4 4 4
5 5 5
6 6 6
Sample Output 3
3
<image>
Beware of unconnected cases and cases with self-edges and multiple edges.
Example
Input
4 2 2
1 2
2 4
3 1
4 2
1 3
Output
2 | instruction | 0 | 45,341 | 8 | 90,682 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m,k = LI()
d = LI_()
v = [LI_() for _ in range(n)]
dd = collections.defaultdict(lambda: None)
for i in range(m):
dd[d[i]] = i
ii = [2**_ for _ in range(m)]
vv = [[ii[dd[v[c][i]]] if not dd[v[c][i]] is None else 0 for c in d] for i in range(k)]
m2 = 2**m
u = [None] * m2
u[-1] = 1
q = [m2-1]
r = 0
while q:
r += 1
nq = []
for qd in q:
qdi = [di for di in range(m) if qd & ii[di]]
for vi in range(k):
t = 0
vvi = vv[vi]
for di in qdi:
t |= vvi[di]
if not u[t] is None:
continue
if t == 0:
return r
u[t] = 1
nq.append(t)
q = nq
return -1
print(main())
``` | output | 1 | 45,341 | 8 | 90,683 |
Provide a correct Python 3 solution for this coding contest problem.
A pitch-black room
When I woke up, Mr. A was in a pitch-black room. Apparently, Mr. A got lost in a dungeon consisting of N rooms. You couldn't know which room A got lost in, but fortunately he got a map of the dungeon. Let's show A the way to go and lead to a bright room.
It is known that M of the N rooms are pitch black, and the D_1, D_2, ..., and D_Mth rooms are pitch black, respectively. Also, there are exactly K one-way streets from all the rooms in order, and the roads from the i-th room are v_ {i, 1}, v_ {i, 2}, ..., v_ {i, respectively. , K} Connected to the third room. You can follow the a_1, a_2, ..., a_lth roads in order from the room you are in for Mr. A. However, when Mr. A reaches the bright room, he ignores the subsequent instructions. You cannot know the information of the room where Mr. A is currently before and after the instruction, so you must give the instruction sequence so that you can reach the bright room no matter which room you are in. Answer the length of the shortest of such instructions.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ M ≤ min (16, N − 1)
* 1 ≤ K ≤ N
* 1 ≤ D_i ≤ N
* D_i are all different
* 1 ≤ v_ {i, j} ≤ N
* All dark rooms can reach at least one bright room
Input Format
Input is given from standard input in the following format.
NMK
D_1 D_2 ... D_M
v_ {1,1} v_ {1,2} ... v_ {1,K}
v_ {2,1} v_ {2,2} ... v_ {2, K}
...
v_ {N, 1} v_ {N, 2} ... v_ {N, K}
Output Format
Print the answer in one line.
Sample Input 1
4 2 2
1 2
twenty four
3 1
4 2
13
Sample Output 1
2
<image>
If you give the instruction 1, 1
* If A's initial position is room 1, the second move will reach room 3.
* If A's initial position is room 2, the first move will reach room 3.
Sample Input 2
3 2 2
1 2
twenty three
1 2
twenty one
Sample Output 2
3
<image>
If you give instructions 2, 1, 2
* If A's initial position is room 1, the first move will reach room 3.
* If A's initial position is room 2, the third move will reach room 3.
Sample Input 3
6 3 3
one two Three
4 1 1
2 5 2
3 3 6
4 4 4
5 5 5
6 6 6
Sample Output 3
3
<image>
Beware of unconnected cases and cases with self-edges and multiple edges.
Example
Input
4 2 2
1 2
2 4
3 1
4 2
1 3
Output
2 | instruction | 0 | 45,342 | 8 | 90,684 |
"Correct Solution:
```
import sys
from collections import defaultdict
def main():
input = sys.stdin.readline
n,m,k = map(int, input().split())
d = list(map(int, input().split()))
l = [-1] * n
for i in range(m):
l[d[i]-1] = i
A = [list(map(int, input().split())) for _ in range(n)]
a = [[l[A[i-1][j] - 1] if l[A[i-1][j] - 1] != -1 else None for i in d] for j in range(k)]
c = [-1] * (1 << m)
c[(1 << m) - 1] = 0
D = [1 << i for i in range(m+1)]
x = [-1] * m
q = defaultdict(int)
Range = range(m)
K = 0
q[0] = -1
R = 1
while 1:
mask = q[K]
K += 1
score = c[mask] + 1
x = [i for i in Range if mask & D[i]]
for ai in a:
tmp = 0
for j in x:
aji = ai[j]
if aji != None:
tmp |= D[aji]
if c[tmp] == -1:
c[tmp] = score
q[R] = tmp
R += 1
if tmp == 0:
print(score)
return
if __name__ == "__main__":
main()
``` | output | 1 | 45,342 | 8 | 90,685 |
Provide a correct Python 3 solution for this coding contest problem.
A pitch-black room
When I woke up, Mr. A was in a pitch-black room. Apparently, Mr. A got lost in a dungeon consisting of N rooms. You couldn't know which room A got lost in, but fortunately he got a map of the dungeon. Let's show A the way to go and lead to a bright room.
It is known that M of the N rooms are pitch black, and the D_1, D_2, ..., and D_Mth rooms are pitch black, respectively. Also, there are exactly K one-way streets from all the rooms in order, and the roads from the i-th room are v_ {i, 1}, v_ {i, 2}, ..., v_ {i, respectively. , K} Connected to the third room. You can follow the a_1, a_2, ..., a_lth roads in order from the room you are in for Mr. A. However, when Mr. A reaches the bright room, he ignores the subsequent instructions. You cannot know the information of the room where Mr. A is currently before and after the instruction, so you must give the instruction sequence so that you can reach the bright room no matter which room you are in. Answer the length of the shortest of such instructions.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ M ≤ min (16, N − 1)
* 1 ≤ K ≤ N
* 1 ≤ D_i ≤ N
* D_i are all different
* 1 ≤ v_ {i, j} ≤ N
* All dark rooms can reach at least one bright room
Input Format
Input is given from standard input in the following format.
NMK
D_1 D_2 ... D_M
v_ {1,1} v_ {1,2} ... v_ {1,K}
v_ {2,1} v_ {2,2} ... v_ {2, K}
...
v_ {N, 1} v_ {N, 2} ... v_ {N, K}
Output Format
Print the answer in one line.
Sample Input 1
4 2 2
1 2
twenty four
3 1
4 2
13
Sample Output 1
2
<image>
If you give the instruction 1, 1
* If A's initial position is room 1, the second move will reach room 3.
* If A's initial position is room 2, the first move will reach room 3.
Sample Input 2
3 2 2
1 2
twenty three
1 2
twenty one
Sample Output 2
3
<image>
If you give instructions 2, 1, 2
* If A's initial position is room 1, the first move will reach room 3.
* If A's initial position is room 2, the third move will reach room 3.
Sample Input 3
6 3 3
one two Three
4 1 1
2 5 2
3 3 6
4 4 4
5 5 5
6 6 6
Sample Output 3
3
<image>
Beware of unconnected cases and cases with self-edges and multiple edges.
Example
Input
4 2 2
1 2
2 4
3 1
4 2
1 3
Output
2 | instruction | 0 | 45,343 | 8 | 90,686 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m,k = LI()
d = LI_()
v = [LI_() for _ in range(n)]
dd = collections.defaultdict(lambda: None)
for i in range(m):
dd[d[i]] = i
vv = []
for c in d:
vv.append([dd[v[c][i]] for i in range(k)])
vvv = [[vv[i][j] for i in range(m)] for j in range(k)]
u = set()
m2 = 2**m
u.add(m2-1)
q = [(m2-1, 1)]
ii = [2**_ for _ in range(m)]
# print('vv',vv)
while q:
qd, qk = q.pop(0)
# print('q', qd,qk)
qdi = [di for di in range(m) if qd & ii[di]]
for vi in range(k):
t = 0
vvi = vvv[vi]
for di in qdi:
if not vvi[di] is None:
t |= ii[vvi[di]]
# print('vit',vi,t)
if t in u:
continue
if t == 0:
return qk
u.add(t)
q.append((t,qk+1))
return -1
print(main())
``` | output | 1 | 45,343 | 8 | 90,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A pitch-black room
When I woke up, Mr. A was in a pitch-black room. Apparently, Mr. A got lost in a dungeon consisting of N rooms. You couldn't know which room A got lost in, but fortunately he got a map of the dungeon. Let's show A the way to go and lead to a bright room.
It is known that M of the N rooms are pitch black, and the D_1, D_2, ..., and D_Mth rooms are pitch black, respectively. Also, there are exactly K one-way streets from all the rooms in order, and the roads from the i-th room are v_ {i, 1}, v_ {i, 2}, ..., v_ {i, respectively. , K} Connected to the third room. You can follow the a_1, a_2, ..., a_lth roads in order from the room you are in for Mr. A. However, when Mr. A reaches the bright room, he ignores the subsequent instructions. You cannot know the information of the room where Mr. A is currently before and after the instruction, so you must give the instruction sequence so that you can reach the bright room no matter which room you are in. Answer the length of the shortest of such instructions.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ M ≤ min (16, N − 1)
* 1 ≤ K ≤ N
* 1 ≤ D_i ≤ N
* D_i are all different
* 1 ≤ v_ {i, j} ≤ N
* All dark rooms can reach at least one bright room
Input Format
Input is given from standard input in the following format.
NMK
D_1 D_2 ... D_M
v_ {1,1} v_ {1,2} ... v_ {1,K}
v_ {2,1} v_ {2,2} ... v_ {2, K}
...
v_ {N, 1} v_ {N, 2} ... v_ {N, K}
Output Format
Print the answer in one line.
Sample Input 1
4 2 2
1 2
twenty four
3 1
4 2
13
Sample Output 1
2
<image>
If you give the instruction 1, 1
* If A's initial position is room 1, the second move will reach room 3.
* If A's initial position is room 2, the first move will reach room 3.
Sample Input 2
3 2 2
1 2
twenty three
1 2
twenty one
Sample Output 2
3
<image>
If you give instructions 2, 1, 2
* If A's initial position is room 1, the first move will reach room 3.
* If A's initial position is room 2, the third move will reach room 3.
Sample Input 3
6 3 3
one two Three
4 1 1
2 5 2
3 3 6
4 4 4
5 5 5
6 6 6
Sample Output 3
3
<image>
Beware of unconnected cases and cases with self-edges and multiple edges.
Example
Input
4 2 2
1 2
2 4
3 1
4 2
1 3
Output
2
Submitted Solution:
```
import sys
import queue
def main():
n,m,k = map(int, sys.stdin.readline().split())
darks = list(map(lambda x: x-1, map(int, sys.stdin.readline().split())))
edges = []
for i in range(n):
edges.append( list(map(lambda x: x-1, map(int, sys.stdin.readline().split()))) )
q = queue.Queue()
dp = {}
dp[ (1 << m) - 1 ] = 0
q.put( ((1 << m) - 1, 0) )
ans = -1
while not q.empty():
state,cost = q.get()
for i in range(k):
next_state = 0
for j in range(m):
if not (state & (1 << j)):
continue
#print("darks[{}]:{}\ti:{}\tedges[][]:{}".format(j, darks[j], i, edges[darks[j]][i]))
if edges[darks[j]][i] in darks:
next_state |= (1 << darks.index(edges[darks[j]][i]))
#print(bin(next_state))
if next_state == 0:
ans = cost + 1
break
if next_state not in dp:
dp[next_state] = cost + 1
q.put( (next_state, cost + 1) )
if ans != -1:
break
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 45,344 | 8 | 90,688 |
No | output | 1 | 45,344 | 8 | 90,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A pitch-black room
When I woke up, Mr. A was in a pitch-black room. Apparently, Mr. A got lost in a dungeon consisting of N rooms. You couldn't know which room A got lost in, but fortunately he got a map of the dungeon. Let's show A the way to go and lead to a bright room.
It is known that M of the N rooms are pitch black, and the D_1, D_2, ..., and D_Mth rooms are pitch black, respectively. Also, there are exactly K one-way streets from all the rooms in order, and the roads from the i-th room are v_ {i, 1}, v_ {i, 2}, ..., v_ {i, respectively. , K} Connected to the third room. You can follow the a_1, a_2, ..., a_lth roads in order from the room you are in for Mr. A. However, when Mr. A reaches the bright room, he ignores the subsequent instructions. You cannot know the information of the room where Mr. A is currently before and after the instruction, so you must give the instruction sequence so that you can reach the bright room no matter which room you are in. Answer the length of the shortest of such instructions.
Constraints
* 2 ≤ N ≤ 100
* 1 ≤ M ≤ min (16, N − 1)
* 1 ≤ K ≤ N
* 1 ≤ D_i ≤ N
* D_i are all different
* 1 ≤ v_ {i, j} ≤ N
* All dark rooms can reach at least one bright room
Input Format
Input is given from standard input in the following format.
NMK
D_1 D_2 ... D_M
v_ {1,1} v_ {1,2} ... v_ {1,K}
v_ {2,1} v_ {2,2} ... v_ {2, K}
...
v_ {N, 1} v_ {N, 2} ... v_ {N, K}
Output Format
Print the answer in one line.
Sample Input 1
4 2 2
1 2
twenty four
3 1
4 2
13
Sample Output 1
2
<image>
If you give the instruction 1, 1
* If A's initial position is room 1, the second move will reach room 3.
* If A's initial position is room 2, the first move will reach room 3.
Sample Input 2
3 2 2
1 2
twenty three
1 2
twenty one
Sample Output 2
3
<image>
If you give instructions 2, 1, 2
* If A's initial position is room 1, the first move will reach room 3.
* If A's initial position is room 2, the third move will reach room 3.
Sample Input 3
6 3 3
one two Three
4 1 1
2 5 2
3 3 6
4 4 4
5 5 5
6 6 6
Sample Output 3
3
<image>
Beware of unconnected cases and cases with self-edges and multiple edges.
Example
Input
4 2 2
1 2
2 4
3 1
4 2
1 3
Output
2
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m,k = LI()
d = LI()
v = [LI() for _ in range(n)]
u = set()
ds = set(d)
u.add(tuple(d))
q = [(tuple(d), 1)]
while q:
qd, qk = q.pop(0)
for vi in range(k):
t = set()
for di in qd:
t.add(v[di-1][vi])
t &= ds
tt = tuple(t)
if tt in u:
continue
if len(tt) == 0:
return qk
u.add(tt)
q.append((tt,qk+1))
return -1
print(main())
``` | instruction | 0 | 45,345 | 8 | 90,690 |
No | output | 1 | 45,345 | 8 | 90,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,674 | 8 | 91,348 |
Tags: dp
Correct Solution:
```
n,m = list(map(int,input().split()))
arr = []
for i in range(n):
l = list(map(float,input().split()))
arr.append(int(l[0]))
dp = [1 for i in range(n)]
for i in range(1,n):
for j in range(i):
if arr[j]<=arr[i]:
dp[i] = max(dp[i],dp[j]+1)
print(n-max(dp))
``` | output | 1 | 45,674 | 8 | 91,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,675 | 8 | 91,350 |
Tags: dp
Correct Solution:
```
# Dynamic programming Python implementation
# of LIS problem
# lis returns length of the longest
# increasing subsequence in arr of size n
def lis(l, arr):
# Declare the list (array) for LIS and
# initialize LIS values for all indexes
ls = [1] * l
# Compute optimized LIS values in bottom up manner
for i in range(1, l):
for j in range(0, i):
if arr[i] >= arr[j] and ls[i] < ls[j] + 1:
ls[i] = ls[j] + 1
# print(ls, max(ls))
return l - max(ls)
# end of lis function
n, m = map(int, input().strip().split())
lst = []
for i in range(n):
a, b = map(float, input().strip().split())
lst.append(a)
print(lis(n, lst))
``` | output | 1 | 45,675 | 8 | 91,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,676 | 8 | 91,352 |
Tags: dp
Correct Solution:
```
n, m = map(int, input(). split())
b = [list(map(float, input(). split())) for i in range(n)]
k = [0] * n
k[0] = 1
for i in range(1, n):
z = 0
for y in range(i):
if b[y][0] <= b[i][0]:
z = max(k[y], z)
k[i] = max(0, z) + 1
print(n - max(k))
``` | output | 1 | 45,676 | 8 | 91,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,677 | 8 | 91,354 |
Tags: dp
Correct Solution:
```
n,m=map(int,input().split())
s=[list(map(float,input().split()))[::-1] for i in range(n)]
s.sort()
s1=[int(i[1]) for i in s]
dp=[1 for i in range(n)]
for i in range(n-2,-1,-1):
for j in range(i+1,n):
if s1[j]>=s1[i]:
dp[i]=max(dp[i],1+dp[j])
print(n-max(dp))
``` | output | 1 | 45,677 | 8 | 91,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,678 | 8 | 91,356 |
Tags: dp
Correct Solution:
```
from sys import stdin
n,m=map(int,input().split())
s=[list(map(float,stdin.readline().strip().split()))[::-1] for i in range(n)]
s.sort()
s1=[int(i[1]) for i in s]
dp=[1 for i in range(n)]
ans=n+1
for i in range(n-1,-1,-1):
for j in range(i+1,n):
if s1[j]>=s1[i]:
dp[i]=max(dp[i],1+dp[j])
x=n-dp[i]
if x<ans:
ans=x
print(ans)
``` | output | 1 | 45,678 | 8 | 91,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,679 | 8 | 91,358 |
Tags: dp
Correct Solution:
```
def CeilIndex(A, l, r, key):
while (r - l > 1):
m = l + (r - l)//2
if (A[m] > key):
r = m
else:
l = m
return r
def LongestIncreasingSubsequenceLength(A, size):
# Add boundary case,
# when array size is one
tailTable = [0 for i in range(size + 1)]
len = 0 # always points empty slot
tailTable[0] = A[0]
len = 1
for i in range(1, size):
if (A[i] < tailTable[0]):
# new smallest value
tailTable[0] = A[i]
elif (A[i] >= tailTable[len-1]):
# A[i] wants to extend
# largest subsequence
tailTable[len] = A[i]
len+= 1
else:
# A[i] wants to be current
# end candidate of an existing
# subsequence. It will replace
# ceil value in tailTable
tailTable[CeilIndex(tailTable, -1, len-1, A[i])] = A[i]
return len
# driver code
arr = []
n,m=map(int,input().split())
for i in range(n) :
a,b=map(float,input().split())
arr.append(a)
print(n-LongestIncreasingSubsequenceLength(arr, n))
# This code is contributed
# by Anant Agarwal.
``` | output | 1 | 45,679 | 8 | 91,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,680 | 8 | 91,360 |
Tags: dp
Correct Solution:
```
n,m=map(int,input().split())
l=[]
for i in range(n):
a,b=input().split()
l.append(int(a))
dp=[1]*n
for i in range(1,n):
for j in range(0,i):
if l[i]>=l[j]:
dp[i]=max(dp[i],dp[j]+1)
print(n-max(dp))
``` | output | 1 | 45,680 | 8 | 91,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | instruction | 0 | 45,681 | 8 | 91,362 |
Tags: dp
Correct Solution:
```
def f(arr):
arr=[None]+arr
dp=[0]*(len(arr))
for i in range(1,len(arr)):
j=int(arr[i])
for k in range(int(j),0,-1):
dp[j]=max(dp[j],1+dp[k])
return len(arr)-max(dp)-1
a,b=map(int,input().strip().split())
lst=[]
for i in range(a):
x,y=map(float,input().strip().split())
lst.append(x)
print(f(lst))
``` | output | 1 | 45,681 | 8 | 91,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/10/20
LIS
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, M, A):
bit = [0 for _ in range(N)]
def add(index, val):
while index < N:
bit[index] = max(bit[index], val)
index |= index + 1
def query(index):
s = 0
while index >= 0:
s = max(s, bit[index])
index = (index & (index + 1)) - 1
return s
q = [(v, i) for i, v in enumerate(A)]
q.sort()
s = 0
for v, i in q:
t = query(i) + 1
add(i, t)
s = max(s, t)
return N-s
N, M = map(int, input().split())
A = []
for i in range(N):
x, y = input().split()
A.append(int(x))
print(solve(N, M, A))
``` | instruction | 0 | 45,682 | 8 | 91,364 |
Yes | output | 1 | 45,682 | 8 | 91,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
'''n, m, k = [int(x) for x in input().split()]
trees = tuple([int(x) for x in input().split()])
costs = []
for x in range(n):
costs.append([0] + [int(x) for x in input().split()])
dp = [[[float('inf') for z in range(m + 1)] for y in range(k + 1)] for x in range(n)]
if trees[0] == 0:
for k in range(len(dp[0][0])):
dp[0][1][k] = cost[0][k]
else:
dp[0][0][trees[0]] = cost[0][trees[0]]
for i in range(1, len(dp)):
for j in range(1, len(dp[0])):
for k in range(1, len(dp[0][0])):
if trees
for l in range(len(dp[0][0])):
if k == l:
dp[i][j][k] = dp[i - 1][j][k] + cost[i][k]
else:
dp[i][j][k] = dp[i - 1][j - 1][l] + cost[i][k]'''
n, m = [int(x) for x in input().split()]
plant = [int(input().split()[0]) for x in range(n)]
dp = [1 for x in range(n)]
for i in range(len(plant)):
for j in range(0, i):
if plant[j] > plant[i]:
continue
dp[i] = max(dp[i], dp[j] + 1)
#print(dp)
print(n - max(dp))
'''for i in range(1, n):
for k in range(plant[i], 0, -1):
dp[plant[i]] = max(dp[plant[i]], 1 + dp[k])
print(n - max(dp) - 1)'''
``` | instruction | 0 | 45,683 | 8 | 91,366 |
Yes | output | 1 | 45,683 | 8 | 91,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
from bisect import bisect_right
n = int(input().split()[0])
t = [int(input().split()[0]) for i in range(n)]
p, k = [], 0
for i in t:
x = bisect_right(p, i)
if x < k: p[x] = i
else:
p.append(i)
k += 1
print(n - k)
``` | instruction | 0 | 45,684 | 8 | 91,368 |
Yes | output | 1 | 45,684 | 8 | 91,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
n, m = input().split()
values = []
res = []
for _ in range(int(n)): values.append(int(input().split()[0]))
for _ in range(int(m) + 1): res.append(0)
for value in values:
auxmin = min(res[:value])
res[value] = min(auxmin, res[value]) - 1
print(int(n) + min(res))
``` | instruction | 0 | 45,685 | 8 | 91,370 |
Yes | output | 1 | 45,685 | 8 | 91,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
numbers, types = map(int, input().split())
position = list()
lis = list()
max_ = 0
for i in range(numbers):
a, p = input().split()
position.append(a)
for i in range(numbers):
max_ = 1
lis.append(1)
for j in range(i):
if(position[j] <= position[i]) and lis[j] + 1 > max_:
max_ = lis[j] + 1
lis[i] = max_
print(numbers - lis[i])
``` | instruction | 0 | 45,686 | 8 | 91,372 |
No | output | 1 | 45,686 | 8 | 91,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
from math import ceil,sqrt
from collections import defaultdict
def solve():
n,m = map(int,input().split())
l = []
for i in range(n):
a,b = map(float,input().split())
l.append(a)
dp = [0]*(n)
for i in range(n):
maxi = 0
for j in range(i):
if l[i]>=l[j]:
maxi = max(dp[j],maxi)
dp[i] = maxi+1
# print(dp)
print(n-dp[-1])
# t = int(input())
# for _ in range(t):
solve()
``` | instruction | 0 | 45,687 | 8 | 91,374 |
No | output | 1 | 45,687 | 8 | 91,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Thu Jan 17 11:03:25 2019
@author: nm57315
"""
n,m = list(map(int,input().strip().split()))
places = [0 for x in range(m+1)]
plants =[0]*(n+1)
for x in range(n):
a,b = input().strip().split();
a = int(a)
places[a] += 1
plants[x] = a
c = 1
c_count = 0;
changes = 0;
for x in range(1,n+1):
curr = plants[x]
if curr != c:
places[curr] -= 1
changes += 1
else:
c_count += 1;
if c_count == places[c]:
c += 1
c_count = 0;
print(changes)
``` | instruction | 0 | 45,688 | 8 | 91,376 |
No | output | 1 | 45,688 | 8 | 91,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
Input
3 2
2 1
1 2.0
1 3.100
Output
1
Input
3 3
1 5.0
2 5.5
3 6.0
Output
0
Input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125
Output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
Submitted Solution:
```
#!/usr/bin/env python3
# created : 2020. 12. 31. 23:59
import os
from sys import stdin, stdout
from collections import deque
from bisect import bisect_left, bisect_right
def solve(tc):
n, m = map(int, stdin.readline().split())
species = []
for i in range(n):
si, xi = stdin.readline().split()
species.append(int(si))
ans = n+1
cnt = 0
dq = deque()
for i in range(n):
if len(dq) > 0 and dq[-1] > species[i]:
cnt += 1
k = bisect_right(dq, species[i])
dq.insert(k, species[i])
ans = min(ans, cnt)
cnt = 0
dq.clear()
for i in range(n-1, -1, -1):
if len(dq) > 0 and dq[0] < species[i]:
cnt += 1
k = bisect_left(dq, species[i])
dq.insert(k, species[i])
ans = min(ans, cnt)
print(ans)
tcs = 1
# tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
``` | instruction | 0 | 45,689 | 8 | 91,378 |
No | output | 1 | 45,689 | 8 | 91,379 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,929 | 8 | 91,858 |
"Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import Counter
from heapq import heappush, heappop
p, g, ig = 998244353, 3, 332748118
W = [pow(g, (p - 1) >> i, p) for i in range(24)]
iW = [pow(ig, (p - 1) >> i, p) for i in range(24)]
def convolve(a, b):
def fft(f):
for l in range(k, 0, -1):
d = 1 << l - 1
U = [1]
for i in range(d):
U.append(U[-1] * W[l] % p)
for i in range(1 << k - l):
for j in range(d):
s = i * 2 * d + j
t = s + d
f[s], f[t] = (f[s] + f[t]) % p, U[j] * (f[s] - f[t]) % p
def ifft(f):
for l in range(1, k + 1):
d = 1 << l - 1
U = [1]
for i in range(d):
U.append(U[-1] * iW[l] % p)
for i in range(1 << k - l):
for j in range(d):
s = i * 2 * d + j
t = s + d
f[s], f[t] = (f[s] + f[t] * U[j]) % p, (f[s] - f[t] * U[j]) % p
n0 = len(a) + len(b) - 1
k = (n0).bit_length()
n = 1 << k
a = a + [0] * (n - len(a))
b = b + [0] * (n - len(b))
fft(a), fft(b)
for i in range(n):
a[i] = a[i] * b[i] % p
ifft(a)
invn = pow(n, p - 2, p)
for i in range(n0):
a[i] = a[i] * invn % p
del a[n0:]
return a
N = int(input())
P = 998244353
nn = 2 * N + 10
fa = [1] * (nn+1)
fainv = [1] * (nn+1)
for i in range(nn):
fa[i+1] = fa[i] * (i+1) % P
fainv[-1] = pow(fa[-1], P-2, P)
for i in range(nn)[::-1]:
fainv[i] = fainv[i+1] * (i+1) % P
dfa = [1] * (nn+1)
for i in range(2, nn+1):
dfa[i] = dfa[i-2] * i % P
dfa += [1]
X = list(Counter([int(input()) for _ in range(2 * N)]).values())
M = len(X)
Y = [[] for _ in range(M)]
for x in X:
L = [1]
s = 1
for i in range(x // 2):
s = s * ((x - i * 2) * (x - i * 2 - 1) // 2) % P
L.append(s * fainv[i+1] % P)
Y.append(L)
for i in range(1, M)[::-1]:
Y[i] = convolve(Y[i*2], Y[i*2+1])
ans = 0
for i, x in enumerate(Y[1]):
ans = (ans + x * (-1 if i & 1 else 1) * dfa[N * 2 - i * 2 - 1]) % P
print(ans)
``` | output | 1 | 45,929 | 8 | 91,859 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,930 | 8 | 91,860 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
from collections import defaultdict
from collections import deque
import heapq
MOD = 998244353
def DD(arg): return defaultdict(arg)
def inv(n): return pow(n, MOD-2, MOD)
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
ROOT = 3
MOD = 998244353
roots = [pow(ROOT,(MOD-1)>>i,MOD) for i in range(24)] # 1 の 2^i 乗根
iroots = [pow(x,MOD-2,MOD) for x in roots] # 1 の 2^i 乗根の逆元
def untt(a,n):
for i in range(n):
m = 1<<(n-i-1)
for s in range(1<<i):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m])%MOD, (a[s+p]-a[s+p+m])*w_N%MOD
w_N = w_N*roots[n-i]%MOD
def iuntt(a,n):
for i in range(n):
m = 1<<i
for s in range(1<<(n-i-1)):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m]*w_N)%MOD, (a[s+p]-a[s+p+m]*w_N)%MOD
w_N = w_N*iroots[i+1]%MOD
inv = pow((MOD+1)//2,n,MOD)
for i in range(1<<n):
a[i] = a[i]*inv%MOD
def convolution(a,b):
la = len(a)
lb = len(b)
deg = la+lb-2
n = deg.bit_length()
if min(la, lb) <= 50:
if la < lb:
la,lb = lb,la
a,b = b,a
res = [0]*(la+lb-1)
for i in range(la):
for j in range(lb):
res[i+j] += a[i]*b[j]
res[i+j] %= MOD
return res
N = 1<<n
a += [0]*(N-len(a))
b += [0]*(N-len(b))
untt(a,n)
untt(b,n)
for i in range(N):
a[i] = a[i]*b[i]%MOD
iuntt(a,n)
return a[:deg+1]
N = int(input())
nokori = [1]
for n in range(1,2*N+10):
nokori.append((nokori[-1]*(2*n-1)*(2*n)*inv(2)*inv(n))%MOD)
dic = DD(int)
for _ in range(N*2):
dic[int(input())] += 1
A = [dic[k] for k in dic]
P = []
for a in A:
temp = 1
count = 0
P.append([temp])
while a >= 2:
temp *= a*(a-1)//2
temp %= MOD
count += 1
P[-1].append((temp*gyaku_kaijo(count))%MOD)
a -= 2
Q = []
for i in range(len(P)):
heapq.heappush(Q,(len(P[i]),i))
while len(Q)>1:
_,p = heapq.heappop(Q)
_,q = heapq.heappop(Q)
temp = convolution(P[p], P[q])
P.append(temp)
heapq.heappush(Q,(len(temp),len(P)-1))
Q = list(P[-1])
ans = 0
M = len(Q)
def sign(x):
if x%2: return -1
else: return 1
for i in range(M):
ans += sign(i) * Q[i] * nokori[N-i]
ans %= MOD
print(ans)
``` | output | 1 | 45,930 | 8 | 91,861 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,932 | 8 | 91,864 |
"Correct Solution:
```
import sys
INF = 1 << 60
MOD = 998244353
sys.setrecursionlimit(2147483647)
input = lambda:sys.stdin.readline().rstrip()
prime = 998244353
root = 3
def _fmt(A, inverse = False):
N = len(A)
logN = (N - 1).bit_length()
base = pow(root, (prime - 1) // N * (1 - 2 * inverse) % (prime - 1), prime)
step = N
for k in range(logN):
step >>= 1
w = pow(base, step, prime)
wj = 1
nA = [0] * N
for j in range(1 << k):
for i in range(1 << logN - k - 1):
s, t = i + j * step, i + j * step + (N >> 1)
ps, pt = i + j * step * 2, i + j * step * 2 + step
nA[s], nA[t] = (A[ps] + A[pt] * wj) % prime, (A[ps] - A[pt] * wj) % prime
wj = (wj * w) % prime
A = nA
return A
def convolution(f, g):
N = 1 << (len(f) + len(g) - 2).bit_length()
Ff, Fg = _fmt(f + [0] * (N - len(f))), _fmt(g + [0] * (N - len(g)))
N_inv = pow(N, prime - 2, prime)
fg = _fmt([a * b % prime * N_inv % prime for a, b in zip(Ff, Fg)], inverse = True)
del fg[len(f) + len(g) - 1:]
return fg
class modfact(object):
def __init__(self, n):
fact, invfact = [1] * (n + 1), [1] * (n + 1)
for i in range(1, n + 1): fact[i] = i * fact[i - 1] % MOD
invfact[n] = pow(fact[n], MOD - 2, MOD)
for i in range(n - 1, -1, -1): invfact[i] = invfact[i + 1] * (i + 1) % MOD
self._fact, self._invfact = fact, invfact
def fact(self, n):
return self._fact[n]
def invfact(self, n):
return self._invfact[n]
def comb(self, n, k):
return self._fact[n] * self._invfact[k] % MOD * self._invfact[n - k] % MOD if 0 <= k <= n else 0
def perm(self, n, k):
return self._fact[n] * self._invfact[n - k] % MOD if 0 <= k <= n else 0
from collections import Counter
def resolve():
n = int(input())
H = [int(input()) for _ in range(2 * n)]
mf = modfact(n << 1)
doublefact = [1] * (2 * n + 10)
for i in range(2, 2 * n + 1):
doublefact[i] = i * doublefact[i - 2] % MOD
F = []
for k in Counter(H).values():
f = []
for i in range(k // 2 + 1):
f.append(mf.comb(k, 2 * i) * doublefact[2 * i - 1] % MOD)
F.append(f)
while len(F) > 1:
nF = []
for i in range(len(F) // 2):
nF.append(convolution(F[2 * i], F[2 * i + 1]))
if len(F) & 1:
nF.append(F[-1])
F = nF
f = F[0]
ans = 0
for i in range(len(f)):
f[i] = f[i] * doublefact[2 * n - 2 * i - 1] % MOD
if i & 1:
ans -= f[i]
else:
ans += f[i]
ans %= MOD
print(ans)
resolve()
``` | output | 1 | 45,932 | 8 | 91,865 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,934 | 8 | 91,868 |
"Correct Solution:
```
import sys
INF = 1 << 60
MOD = 998244353
sys.setrecursionlimit(2147483647)
input = lambda:sys.stdin.readline().rstrip()
prime = 998244353
root = 3
def _fmt(A, inverse = False):
N = len(A)
logN = (N - 1).bit_length()
base = pow(root, (prime - 1) // N * (1 - 2 * inverse) % (prime - 1), prime)
step = N
for k in range(logN):
step >>= 1
w = pow(base, step, prime)
wj = 1
nA = [0] * N
for j in range(1 << k):
for i in range(1 << logN - k - 1):
s, t = i + j * step, i + j * step + (N >> 1)
ps, pt = i + j * step * 2, i + j * step * 2 + step
nA[s], nA[t] = (A[ps] + A[pt] * wj) % prime, (A[ps] - A[pt] * wj) % prime
wj = (wj * w) % prime
A = nA
return A
def convolution(f, g):
N = 1 << (len(f) + len(g) - 2).bit_length()
Ff, Fg = _fmt(f + [0] * (N - len(f))), _fmt(g + [0] * (N - len(g)))
N_inv = pow(N, prime - 2, prime)
fg = _fmt([a * b % prime * N_inv % prime for a, b in zip(Ff, Fg)], inverse = True)
del fg[len(f) + len(g) - 1:]
return fg
class modfact(object):
def __init__(self, n):
fact, invfact = [1] * (n + 1), [1] * (n + 1)
for i in range(1, n + 1): fact[i] = i * fact[i - 1] % MOD
invfact[n] = pow(fact[n], MOD - 2, MOD)
for i in range(n - 1, -1, -1): invfact[i] = invfact[i + 1] * (i + 1) % MOD
self._fact, self._invfact = fact, invfact
def fact(self, n):
return self._fact[n]
def invfact(self, n):
return self._invfact[n]
def comb(self, n, k):
return self._fact[n] * self._invfact[k] % MOD * self._invfact[n - k] % MOD if 0 <= k <= n else 0
def perm(self, n, k):
return self._fact[n] * self._invfact[n - k] % MOD if 0 <= k <= n else 0
from collections import Counter
def resolve():
n = int(input())
H = [int(input()) for _ in range(2 * n)]
mf = modfact(2 * n)
doublefact = [1] * (2 * n + 2)
for i in range(2, 2 * n + 1):
doublefact[i] = i * doublefact[i - 2] % MOD
F = [[mf.comb(k, 2 * i) * doublefact[2 * i - 1] % MOD for i in range(k // 2 + 1)] for k in Counter(H).values()]
for i in range(len(F) - 1):
F.append(convolution(F[2 * i], F[2 * i + 1]))
f = F[-1]
ans = sum(f[i] * doublefact[2 * (n - i) - 1] % MOD * (1 - 2 * (i & 1)) for i in range(len(f))) % MOD
print(ans)
resolve()
``` | output | 1 | 45,934 | 8 | 91,869 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,935 | 8 | 91,870 |
"Correct Solution:
```
import sys
INF = 1 << 60
MOD = 998244353
sys.setrecursionlimit(2147483647)
input = lambda:sys.stdin.readline().rstrip()
prime = 998244353
root = 3
def _fmt(A, inverse = False):
N = len(A)
logN = (N - 1).bit_length()
base = pow(root, (prime - 1) // N * (1 - 2 * inverse) % (prime - 1), prime)
step = N
for k in range(logN):
step >>= 1
w = pow(base, step, prime)
wj = 1
nA = [0] * N
for j in range(1 << k):
for i in range(1 << logN - k - 1):
s, t = i + j * step, i + j * step + (N >> 1)
ps, pt = i + j * step * 2, i + j * step * 2 + step
nA[s], nA[t] = (A[ps] + A[pt] * wj) % prime, (A[ps] - A[pt] * wj) % prime
wj = (wj * w) % prime
A = nA
return A
def convolution(f, g):
N = 1 << (len(f) + len(g) - 2).bit_length()
Ff, Fg = _fmt(f + [0] * (N - len(f))), _fmt(g + [0] * (N - len(g)))
N_inv = pow(N, prime - 2, prime)
fg = _fmt([a * b % prime * N_inv % prime for a, b in zip(Ff, Fg)], inverse = True)
del fg[len(f) + len(g) - 1:]
return fg
class modfact(object):
def __init__(self, n):
fact, invfact = [1] * (n + 1), [1] * (n + 1)
for i in range(1, n + 1): fact[i] = i * fact[i - 1] % MOD
invfact[n] = pow(fact[n], MOD - 2, MOD)
for i in range(n - 1, -1, -1): invfact[i] = invfact[i + 1] * (i + 1) % MOD
self._fact, self._invfact = fact, invfact
def fact(self, n):
return self._fact[n]
def invfact(self, n):
return self._invfact[n]
def comb(self, n, k):
return self._fact[n] * self._invfact[k] % MOD * self._invfact[n - k] % MOD if 0 <= k <= n else 0
def perm(self, n, k):
return self._fact[n] * self._invfact[n - k] % MOD if 0 <= k <= n else 0
from collections import Counter, deque
def resolve():
n = int(input())
H = [int(input()) for _ in range(2 * n)]
mf = modfact(n << 1)
doublefact = [1] * (2 * n + 10)
for i in range(2, 2 * n + 1):
doublefact[i] = i * doublefact[i - 2] % MOD
F = deque()
for k in Counter(H).values():
f = []
for i in range(k // 2 + 1):
f.append(mf.comb(k, 2 * i) * doublefact[2 * i - 1] % MOD)
F.append(f)
while len(F) > 1:
F.append(convolution(F.popleft(), F.popleft()))
f = F.popleft()
ans = 0
for i in range(len(f)):
f[i] = f[i] * doublefact[2 * n - 2 * i - 1] % MOD
if i & 1:
ans -= f[i]
else:
ans += f[i]
ans %= MOD
print(ans)
resolve()
``` | output | 1 | 45,935 | 8 | 91,871 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,936 | 8 | 91,872 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
from collections import defaultdict
from collections import deque
import heapq
MOD = 998244353
def DD(arg): return defaultdict(arg)
def inv(n): return pow(n, MOD-2, MOD)
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
class NTT_convolution:
def __init__(self, ROOT=3, MOD=998244353):
self.ROOT = 3
self.MOD = 998244353
self.roots = [pow(self.ROOT,(self.MOD-1)>>i,self.MOD) for i in range(24)] # 1 の 2^i 乗根
self.iroots = [pow(x,self.MOD-2,self.MOD) for x in self.roots] # 1 の 2^i 乗根の逆元
def untt(self,a,n):
for i in range(n):
m = 1<<(n-i-1)
for s in range(1<<i):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m])%self.MOD, (a[s+p]-a[s+p+m])*w_N%self.MOD
w_N = w_N*self.roots[n-i]%self.MOD
def iuntt(self,a,n):
for i in range(n):
m = 1<<i
for s in range(1<<(n-i-1)):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m]*w_N)%self.MOD, (a[s+p]-a[s+p+m]*w_N)%self.MOD
w_N = w_N*self.iroots[i+1]%self.MOD
inv = pow((self.MOD+1)//2,n,self.MOD)
for i in range(1<<n):
a[i] = a[i]*inv%self.MOD
def convolution(self,a,b):
la = len(a)
lb = len(b)
deg = la+lb-2
n = deg.bit_length()
if min(la, lb) <= 50:
if la < lb:
la,lb = lb,la
a,b = b,a
res = [0]*(la+lb-1)
for i in range(la):
for j in range(lb):
res[i+j] += a[i]*b[j]
res[i+j] %= self.MOD
return res
N = 1<<n
a += [0]*(N-len(a))
b += [0]*(N-len(b))
self.untt(a,n)
self.untt(b,n)
for i in range(N):
a[i] = a[i]*b[i]%self.MOD
self.iuntt(a,n)
return a[:deg+1]
N = int(input())
NTT = NTT_convolution()
nokori = [1]
for n in range(1,2*N+10):
nokori.append((nokori[-1]*(2*n-1)*(2*n)*inv(2)*inv(n))%MOD)
dic = DD(int)
for _ in range(N*2):
dic[int(input())] += 1
A = [dic[k] for k in dic]
P = []
for a in A:
temp = 1
count = 0
P.append([temp])
while a >= 2:
temp *= a*(a-1)//2
temp %= MOD
count += 1
P[-1].append((temp*gyaku_kaijo(count))%MOD)
a -= 2
Q = []
for i in range(len(P)):
heapq.heappush(Q,(len(P[i]),i))
while len(Q)>1:
_,p = heapq.heappop(Q)
_,q = heapq.heappop(Q)
temp = NTT.convolution(P[p], P[q])
P.append(temp)
heapq.heappush(Q,(len(temp),len(P)-1))
Q = list(P[-1])
ans = 0
M = len(Q)
def sign(x):
if x%2: return -1
else: return 1
for i in range(M):
ans += sign(i) * Q[i] * nokori[N-i]
ans %= MOD
print(ans)
``` | output | 1 | 45,936 | 8 | 91,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516
Submitted Solution:
```
import sys
INF = 1 << 60
MOD = 998244353
sys.setrecursionlimit(2147483647)
input = lambda:sys.stdin.readline().rstrip()
prime = 998244353
root = 3
def _fmt(A, inverse = False):
N = len(A)
logN = (N - 1).bit_length()
base = pow(root, (prime - 1) // N * (1 - 2 * inverse) % (prime - 1), prime)
step = N
for k in range(logN):
step >>= 1
w = pow(base, step, prime)
wj = 1
nA = [0] * N
for j in range(1 << k):
for i in range(1 << logN - k - 1):
s, t = i + j * step, i + j * step + (N >> 1)
ps, pt = i + j * step * 2, i + j * step * 2 + step
nA[s], nA[t] = (A[ps] + A[pt] * wj) % prime, (A[ps] - A[pt] * wj) % prime
wj = (wj * w) % prime
A = nA
return A
def convolution(f, g):
N = 1 << (len(f) + len(g) - 2).bit_length()
Ff, Fg = _fmt(f + [0] * (N - len(f))), _fmt(g + [0] * (N - len(g)))
N_inv = pow(N, prime - 2, prime)
fg = _fmt([a * b % prime * N_inv % prime for a, b in zip(Ff, Fg)], inverse = True)
del fg[len(f) + len(g) - 1:]
return fg
class modfact(object):
def __init__(self, n):
fact, invfact = [1] * (n + 1), [1] * (n + 1)
for i in range(1, n + 1): fact[i] = i * fact[i - 1] % MOD
invfact[n] = pow(fact[n], MOD - 2, MOD)
for i in range(n - 1, -1, -1): invfact[i] = invfact[i + 1] * (i + 1) % MOD
self._fact, self._invfact = fact, invfact
def fact(self, n):
return self._fact[n]
def invfact(self, n):
return self._invfact[n]
def comb(self, n, k):
return self._fact[n] * self._invfact[k] % MOD * self._invfact[n - k] % MOD if 0 <= k <= n else 0
def perm(self, n, k):
return self._fact[n] * self._invfact[n - k] % MOD if 0 <= k <= n else 0
from collections import Counter
def resolve():
n = int(input())
H = [int(input()) for _ in range(2 * n)]
mf = modfact(2 * n)
doublefact = [1] * (2 * n + 10)
for i in range(2, 2 * n + 1):
doublefact[i] = i * doublefact[i - 2] % MOD
F = []
for k in Counter(H).values():
f = []
for i in range(k // 2 + 1):
f.append(mf.comb(k, 2 * i) * doublefact[2 * i - 1] % MOD)
F.append(f)
for i in range(len(F) - 1):
F.append(convolution(F[2 * i], F[2 * i + 1]))
f = F.pop()
ans = 0
for i in range(len(f)):
f[i] = f[i] * doublefact[2 * n - 2 * i - 1] % MOD
if i & 1:
ans -= f[i]
else:
ans += f[i]
ans %= MOD
print(ans)
resolve()
``` | instruction | 0 | 45,937 | 8 | 91,874 |
Yes | output | 1 | 45,937 | 8 | 91,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516
Submitted Solution:
```
# -*- coding: utf-8 -*-
#############
# Libraries #
#############
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache
#############
# Constants #
#############
MOD = 998244353
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"
#############
# Functions #
#############
######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]
#####Shorten#####
def DD(arg): return defaultdict(arg)
#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)
######Combination######
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
def nCr(n,r):
if n == r: return 1
if n < r or r < 0: return 0
ret = 1
ret = ret * kaijo(n) % MOD
ret = ret * gyaku_kaijo(r) % MOD
ret = ret * gyaku_kaijo(n-r) % MOD
return ret
######Factorization######
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
arr.append([i, cnt])
if temp!=1: arr.append([temp, 1])
if arr==[]: arr.append([n, 1])
return arr
#####MakeDivisors######
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
#####MakePrimes######
def make_primes(N):
max = int(math.sqrt(N))
seachList = [i for i in range(2,N+1)]
primeNum = []
while seachList[0] <= max:
primeNum.append(seachList[0])
tmp = seachList[0]
seachList = [i for i in seachList if i % tmp != 0]
primeNum.extend(seachList)
return primeNum
#####GCD#####
def gcd(a, b):
while b: a, b = b, a % b
return a
#####LCM#####
def lcm(a, b):
return a * b // gcd (a, b)
#####BitCount#####
def count_bit(n):
count = 0
while n:
n &= n-1
count += 1
return count
#####ChangeBase#####
def base_10_to_n(X, n):
if X//n: return base_10_to_n(X//n, n)+[X%n]
return [X%n]
def base_n_to_10(X, n):
return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))
def base_10_to_n_without_0(X, n):
X -= 1
if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
return [X%n]
#####IntLog#####
def int_log(n, a):
count = 0
while n>=a:
n //= a
count += 1
return count
#############
# Main Code #
#############
ROOT = 3
MOD = 998244353
roots = [pow(ROOT,(MOD-1)>>i,MOD) for i in range(24)] # 1 の 2^i 乗根
iroots = [pow(x,MOD-2,MOD) for x in roots] # 1 の 2^i 乗根の逆元
def untt(a,n):
for i in range(n):
m = 1<<(n-i-1)
for s in range(1<<i):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m])%MOD, (a[s+p]-a[s+p+m])*w_N%MOD
w_N = w_N*roots[n-i]%MOD
def iuntt(a,n):
for i in range(n):
m = 1<<i
for s in range(1<<(n-i-1)):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m]*w_N)%MOD, (a[s+p]-a[s+p+m]*w_N)%MOD
w_N = w_N*iroots[i+1]%MOD
inv = pow((MOD+1)//2,n,MOD)
for i in range(1<<n):
a[i] = a[i]*inv%MOD
def convolution(a,b):
la = len(a)
lb = len(b)
deg = la+lb-2
n = deg.bit_length()
if min(la, lb) <= 50:
if la < lb:
la,lb = lb,la
a,b = b,a
res = [0]*(la+lb-1)
for i in range(la):
for j in range(lb):
res[i+j] += a[i]*b[j]
res[i+j] %= MOD
return res
N = 1<<n
a += [0]*(N-len(a))
b += [0]*(N-len(b))
untt(a,n)
untt(b,n)
for i in range(N):
a[i] = a[i]*b[i]%MOD
iuntt(a,n)
return a[:deg+1]
N = I()
nokori = [1]
for n in range(1,2*N+10):
nokori.append((nokori[-1]*(2*n-1)*(2*n)*inv(2)*inv(n))%MOD)
dic = DD(int)
for _ in range(N*2):
dic[I()] += 1
A = [dic[k] for k in dic]
P = []
for a in A:
temp = 1
count = 0
P.append([temp])
while a >= 2:
temp *= a*(a-1)//2
temp %= MOD
count += 1
P[-1].append((temp*gyaku_kaijo(count))%MOD)
a -= 2
Q = deque(P)
while len(Q)>1:
p = Q.popleft()
q = Q.popleft()
Q.append(convolution(p, q))
Q = list(Q[0])
ans = 0
M = len(Q)
def sign(x):
if x%2: return -1
else: return 1
for i in range(M):
ans += sign(i) * Q[i] * nokori[N-i]
ans %= MOD
print(ans)
``` | instruction | 0 | 45,939 | 8 | 91,878 |
Yes | output | 1 | 45,939 | 8 | 91,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516
Submitted Solution:
```
import sys
readline = sys.stdin.readline
from collections import Counter, deque
#from submission 17076021
MOD = 998244353
ROOT = 3
roots = [pow(ROOT,(MOD-1)>>i,MOD) for i in range(24)] # 1 の 2^i 乗根
iroots = [pow(x,MOD-2,MOD) for x in roots] # 1 の 2^i 乗根の逆元
def untt(a,n):
for i in range(n):
m = 1<<(n-i-1)
for s in range(1<<i):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m])%MOD, (a[s+p]-a[s+p+m])*w_N%MOD
w_N = w_N*roots[n-i]%MOD
def iuntt(a,n):
for i in range(n):
m = 1<<i
for s in range(1<<(n-i-1)):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m]*w_N)%MOD, (a[s+p]-a[s+p+m]*w_N)%MOD
w_N = w_N*iroots[i+1]%MOD
inv = pow((MOD+1)//2,n,MOD)
for i in range(1<<n):
a[i] = a[i]*inv%MOD
def convolution(a,b):
la = len(a)
lb = len(b)
deg = la+lb-2
n = deg.bit_length()
if min(la, lb) <= 50:
if la < lb:
la,lb = lb,la
a,b = b,a
res = [0]*(la+lb-1)
for i in range(la):
for j in range(lb):
res[i+j] += a[i]*b[j]
res[i+j] %= MOD
return res
N = 1<<n
a += [0]*(N-len(a))
b += [0]*(N-len(b))
untt(a,n)
untt(b,n)
for i in range(N):
a[i] = a[i]*b[i]%MOD
iuntt(a,n)
return a[:deg+1]
def convolution_all(polys):
if not polys: return [1]
polys.sort(key=lambda x:len(x))
from collections import deque
q = deque(polys)
for _ in range(len(polys)-1):
a = q.popleft()
b = q.popleft()
q.append(convolution(a,b))
return q[0]
def frac(limit):
frac = [1]*limit
for i in range(2,limit):
frac[i] = i * frac[i-1]%MOD
fraci = [None]*limit
fraci[-1] = pow(frac[-1], MOD -2, MOD)
for i in range(-2, -limit-1, -1):
fraci[i] = fraci[i+1] * (limit + i + 1) % MOD
return frac, fraci
frac, fraci = frac(1341398)
def comb(a, b):
if not a >= b >= 0:
return 0
return frac[a]*fraci[b]*fraci[a-b]%MOD
N = int(readline())
H = [int(readline()) for _ in range(2*N)]
Ch = list(Counter(H).values())
Ch.sort()
Q = []
cnt = 0
for c in Ch[::-1]:
if c == 1:
continue
res = [0]*(c//2 + 1)
for i in range(c//2 + 1):
res[i] = comb(c, 2*i)*frac[2*i]*fraci[i]%MOD
Q.append(res)
cnt += 1
if not cnt:
print(frac[2*N]*fraci[N]*pow((MOD+1)//2, N, MOD)%MOD)
else:
M = cnt
"""
for _ in range(M-1):
qi = Q.pop()
qj = Q.pop()
Q.appendleft(convolution(qi, qj))
ans = 0
"""
res = convolution_all(Q)
ans = 0
for i in range(len(res)):
ans = (ans + (-1 if i&1 else 1)*res[i]*comb(N, i)%MOD*frac[2*N-2*i]*frac[i]) % MOD
ans = ans*pow((MOD+1)//2, N, MOD)*fraci[N]%MOD
print(ans)
``` | instruction | 0 | 45,940 | 8 | 91,880 |
No | output | 1 | 45,940 | 8 | 91,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516
Submitted Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
from collections import defaultdict
from collections import deque
import heapq
import numpy as np
MOD = 998244353
def DD(arg): return defaultdict(arg)
def inv(n): return pow(n, MOD-2, MOD)
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
N = int(input())
nokori = [1]
for n in range(1,2*N+10):
nokori.append((nokori[-1]*(2*n-1)*(2*n)*inv(2)*inv(n))%MOD)
dic = DD(int)
for _ in range(N*2):
dic[int(input())] += 1
A = [dic[k] for k in dic]
P = []
for a in A:
temp = 1
count = 0
P.append([temp])
while a >= 2:
temp *= a*(a-1)//2
temp %= MOD
count += 1
P[-1].append((temp*gyaku_kaijo(count))%MOD)
a -= 2
data = [np.array(p,dtype = np.uint64) for p in P]
Q = []
for i in range(len(data)):
heapq.heappush(Q,(len(data[i]),i))
while len(Q)>1:
_,p = heapq.heappop(Q)
_,q = heapq.heappop(Q)
temp = np.convolve(data[p], data[q])%MOD
data.append(temp)
heapq.heappush(Q,(len(temp),len(data)-1))
Q = list(data[-1])
ans = 0
M = len(Q)
def sign(x):
if x%2: return -1
else: return 1
for i in range(M):
ans += sign(i) * Q[i] * nokori[N-i]
ans %= MOD
print(ans)
``` | instruction | 0 | 45,942 | 8 | 91,884 |
No | output | 1 | 45,942 | 8 | 91,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516
Submitted Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
from collections import defaultdict
from collections import deque
import heapq
from scipy import fft
MOD = 998244353
def DD(arg): return defaultdict(arg)
def inv(n): return pow(n, MOD-2, MOD)
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
import numpy as np
@njit((i8, i8), cache=True)
def ntt_precompute(N, primitive_root):
roots = np.empty(1 << N, np.int64)
iroots = np.empty(1 << N, np.int64)
roots[0] = iroots[0] = 1
for n in range(N):
x = mpow(primitive_root, (MOD - 1) >> n + 2)
y = mpow(x, MOD - 2)
for i in range(1 << n):
roots[(1 << n) | i] = roots[i] * x % MOD
iroots[(1 << n) | i] = iroots[i] * y % MOD
return roots, iroots
@njit((i8, i8), cache=True)
def mpow(a, n):
p = 1
while n:
if n & 1:
p = p * a % MOD
a = a * a % MOD
n >>= 1
return p
@njit((i8[:], i8[:], i8[:], numba.b1), cache=True)
def ntt(roots, iroots, A, inverse):
N = len(A)
if not inverse:
m = N >> 1
while m:
for k in range(N // m >> 1):
s = 2 * m * k
for i in range(s, s + m):
x, y = A[i], A[i + m] * roots[k] % MOD
A[i], A[i + m] = x + y, x - y
A %= MOD
m >>= 1
else:
m = 1
while m < N:
for k in range(N // m >> 1):
s = 2 * m * k
for i in range(s, s + m):
x, y = A[i], A[i + m]
A[i], A[i + m] = x + y, (x - y) * iroots[k] % MOD
A %= MOD
m <<= 1
invN = mpow(N, MOD - 2)
for i in range(N):
A[i] = A[i] * invN % MOD
return A
@njit((i8[:], i8[:], i8[:], i8[:]), cache=True)
def ntt_convolve(roots, iroots, A, B):
n = 0
L = len(A) + len(B) - 1
while (1 << n) < L:
n += 1
A1 = np.zeros(1 << n, np.int64)
A1[:len(A)] = A
B1 = np.zeros(1 << n, np.int64)
B1[:len(B)] = B
A, B = A1, B1
ntt(roots, iroots, A, False)
ntt(roots, iroots, B, False)
for i in range(len(A)):
A[i] = A[i] * B[i] % MOD
ntt(roots, iroots, A, True)
return A[:L]
N = int(input())
max_ntt = 18
roots, iroots = ntt_precompute(max_ntt, 3)
nokori = [1]
for n in range(1,2*N+10):
nokori.append((nokori[-1]*(2*n-1)*(2*n)*inv(2)*inv(n))%MOD)
dic = DD(int)
for _ in range(N*2):
dic[int(input())] += 1
A = [dic[k] for k in dic]
P = []
for a in A:
temp = 1
count = 0
P.append([temp])
while a >= 2:
temp *= a*(a-1)//2
temp %= MOD
count += 1
P[-1].append((temp*gyaku_kaijo(count))%MOD)
a -= 2
data = [np.array(p) for p in P]
Q = []
for i in range(len(data)):
heapq.heappush(Q,(len(data[i]),i))
while len(Q)>1:
_,p = heapq.heappop(Q)
_,q = heapq.heappop(Q)
temp = ntt_convolve(roots, iroots, data[p], data[q])
data.append(temp)
heapq.heappush(Q,(len(temp),len(data)-1))
Q = list(data[-1])
ans = 0
M = len(Q)
def sign(x):
if x%2: return -1
else: return 1
for i in range(M):
ans += sign(i) * Q[i] * nokori[N-i]
ans %= MOD
print(ans)
``` | instruction | 0 | 45,943 | 8 | 91,886 |
No | output | 1 | 45,943 | 8 | 91,887 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,992 | 8 | 91,984 |
"Correct Solution:
```
p,a=1,0
n=input()
for i in list(map(int,input().split())):
if p<=i:a+=1
p=max(i,p)
print(a)
``` | output | 1 | 45,992 | 8 | 91,985 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,993 | 8 | 91,986 |
"Correct Solution:
```
n,h=open(0);*h,=map(int,[0]+h.split());print(sum(h[i]>=max(h[:i])for i in range(1,int(n)+1)))
``` | output | 1 | 45,993 | 8 | 91,987 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,994 | 8 | 91,988 |
"Correct Solution:
```
n=int(input())
h=list(map(int,input().split()))
i=0
cnt=0
for j in h:
if i<=j:
cnt+=1
i=j
print(cnt)
``` | output | 1 | 45,994 | 8 | 91,989 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,995 | 8 | 91,990 |
"Correct Solution:
```
x=int(input())
a=list(map(int,input().split()))
ans=0
mdd=-(10**18)
for i in a:
if i>=mdd:
mdd=i
ans+=1
print(ans)
``` | output | 1 | 45,995 | 8 | 91,991 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,996 | 8 | 91,992 |
"Correct Solution:
```
input()
n=list(map(int,input().split()))
h=0
cnt=0
for i in n:
if h <= i:
cnt+=1
h=i
print(cnt)
``` | output | 1 | 45,996 | 8 | 91,993 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,997 | 8 | 91,994 |
"Correct Solution:
```
N=int(input())
H=list(map(int,input().split()))
print(sum(max(H[:i]+[0])<=H[i] for i in range(N)))
``` | output | 1 | 45,997 | 8 | 91,995 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,998 | 8 | 91,996 |
"Correct Solution:
```
n = int(input())
l = map(int, input().split())
hst = 0
cnt = 0
for i in l:
if hst <= i:
hst = i
cnt += 1
print(cnt)
``` | output | 1 | 45,998 | 8 | 91,997 |
Provide a correct Python 3 solution for this coding contest problem.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1 | instruction | 0 | 45,999 | 8 | 91,998 |
"Correct Solution:
```
N = int(input())
H = list(map(int,input().split()))
t = 0
h = 0
for i in H:
if i >= h:
t += 1
h = i
print(t)
``` | output | 1 | 45,999 | 8 | 91,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
N=int(input())
a=list(map(int,input().split()))
cnt=0
for i in range(N):
if max(a[:i+1])==a[i]:
cnt+=1
print(cnt)
``` | instruction | 0 | 46,000 | 8 | 92,000 |
Yes | output | 1 | 46,000 | 8 | 92,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
n=int(input())
m=0
ans=0
for i in input().split():
if int(i)>=m:
ans+=1
m=int(i)
print(ans)
``` | instruction | 0 | 46,001 | 8 | 92,002 |
Yes | output | 1 | 46,001 | 8 | 92,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
n = int(input())
j=0
p=0
for i in map(int,input().split()):
if j <= i:
p += 1
j = i
print(p)
``` | instruction | 0 | 46,002 | 8 | 92,004 |
Yes | output | 1 | 46,002 | 8 | 92,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
n = int(input())
h = list(map(int, input().split()))
k = 0
for i in range(n):
if max(h[:i+1]) == h[i]:
k = k + 1
print(k)
``` | instruction | 0 | 46,003 | 8 | 92,006 |
Yes | output | 1 | 46,003 | 8 | 92,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
N = int(input())
H_list = list(map(int,input().split()))
cnt =0
cnt_a =0
for i in range(N):
if i==0:
cnt+=1
for j in range(i-1):
if H_list[i]>=H_list[j]:
cnt_a=1
else:
cnt_a=0
cnt +=cnt_a
print(cnt)
``` | instruction | 0 | 46,004 | 8 | 92,008 |
No | output | 1 | 46,004 | 8 | 92,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
import sys
N = int(input())
line = list(map(int,input().split()))
ans = 1
mx = line[0]
for i in range(1,line.index(max(line))):
if line[i] >= mx:
ans += 1
mx = line[i]
ans += 1
print(ans)
``` | instruction | 0 | 46,005 | 8 | 92,010 |
No | output | 1 | 46,005 | 8 | 92,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N mountains ranging from east to west, and an ocean to the west.
At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.
The height of the i-th mountain from the west is H_i.
You can certainly see the ocean from the inn at the top of the westmost mountain.
For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.
From how many of these N inns can you see the ocean?
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq H_i \leq 100
Input
Input is given from Standard Input in the following format:
N
H_1 H_2 ... H_N
Output
Print the number of inns from which you can see the ocean.
Examples
Input
4
6 5 6 8
Output
3
Input
5
4 5 3 5 4
Output
3
Input
5
9 5 6 8 4
Output
1
Submitted Solution:
```
n=int(input())
h=list(map(int, input().split()))
t=1
#print(n)
#print(*h)
if h[0] <= h[1]:
t+=1
if h[0] <= h[2] and h[1] <= h[2]:
t+=1
if h[0] <= h[3] and h[1] <= h[3] and h[2] <=h[3]:
t+=1
if h[0] <= h[4] and h[1] <= h[4] and h[2] <=h[4] and h[3] <= h[4]:
t+=1
if h[0] <= h[5] and h[1] <= h[5] and h[2] <=h[5] and h[3] <= h[5] and h[4] <= h[5]:
t+=1
if h[0] <= h[6] and h[1] <= h[6] and h[2] <=h[6] and h[3] <= h[6] and h[4] <= h[6] and h[5] <= h[6]:
t+=1
if h[0] <= h[7] and h[1] <= h[7] and h[2] <=h[7] and h[3] <= h[7] and h[4] <= h[7] and h[5] <= h[7] and h[6] <= h[7]:
t+=1
if h[0] <= h[8] and h[1] <= h[8] and h[2] <=h[8] and h[3] <= h[8] and h[4] <= h[8] and h[5] <= h[8] and h[6] <= h[8] and h[7]<=h[8]:
t+=1
if h[0] <= h[9] and h[1] <= h[9] and h[2] <=h[9] and h[3] <= h[9] and h[4] <= h[9] and h[5] <= h[9] and h[6] <= h[9] and h[7]<=h[9] and h[8]<=h[9]:
t+=1
if h[0] <= h[10] and h[1] <= h[10] and h[2] <=h[10] and h[3] <= h[10] and h[4] <= h[10] and h[5] <= h[10] and h[6] <= h[10] and h[7]<=h[10] and h[8]<=h[10] and h[9]<=h[10]:
t+=1
if h[0] <= h[11] and h[1] <= h[11] and h[2] <=h[11] and h[3] <= h[11] and h[4] <= h[11] and h[5] <= h[11] and h[6] <= h[11] and h[7]<=h[11] and h[8]<=h[11] and h[9]<=h[11] and h[10]<=h[11]:
t+=1
if h[0] <= h[12] and h[1] <= h[12] and h[2] <=h[12] and h[3] <= h[12] and h[4] <= h[12] and h[5] <= h[12] and h[6] <= h[12] and h[7]<=h[12] and h[8]<=h[12] and h[9]<=h[12] and h[10]<=h[12] and h[11]<=h[12]:
t+=1
if h[0] <= h[13] and h[1] <= h[13] and h[2] <=h[13] and h[3] <= h[13] and h[4] <= h[13] and h[5] <= h[13] and h[6] <= h[13] and h[7]<=h[13] and h[8]<=h[13] and h[9]<=h[13] and h[10]<=h[13] and h[11]<=h[13] and h[12]<=h[13]:
t+=1
if h[0] <= h[14] and h[1] <= h[14] and h[2] <=h[14] and h[3] <= h[14] and h[4] <= h[14] and h[5] <= h[14] and h[6] <= h[14] and h[7]<=h[14] and h[8]<=h[14] and h[9]<=h[14] and h[10]<=h[14] and h[11]<=h[14] and h[12]<=h[14] and h[13]<=h[14]:
t+=1
if h[0] <= h[15] and h[1] <= h[15] and h[2] <=h[15] and h[3] <= h[15] and h[4] <= h[15] and h[5] <= h[15] and h[6] <= h[15] and h[7]<=h[15] and h[8]<=h[15] and h[9]<=h[15] and h[10]<=h[15] and h[11]<=h[15] and h[12]<=h[15] and h[13]<=h[15] and h[14]<=h[15]:
t+=1
if h[0] <= h[16] and h[1] <= h[16] and h[2] <=h[16] and h[3] <= h[16] and h[4] <= h[16] and h[5] <= h[16] and h[6] <= h[16] and h[7]<=h[16] and h[8]<=h[16] and h[9]<=h[16] and h[10]<=h[16] and h[11]<=h[16] and h[12]<=h[16] and h[13]<=h[16] and h[14]<=h[16] and h[15]<=h[16]:
t+=1
if h[0] <= h[17] and h[1] <= h[17] and h[2] <=h[17] and h[3] <= h[17] and h[4] <= h[17] and h[5] <= h[17] and h[6] <= h[17] and h[7]<=h[17] and h[8]<=h[17] and h[9]<=h[17] and h[10]<=h[17] and h[11]<=h[17] and h[12]<=h[17] and h[13]<=h[17] and h[14]<=h[17] and h[15]<=h[17] and h[16]<=h[17]:
t+=1
if h[0] <= h[18] and h[1] <= h[18] and h[2] <=h[18] and h[3] <= h[18] and h[4] <= h[18] and h[5] <= h[18] and h[6] <= h[18] and h[7]<=h[18] and h[8]<=h[18] and h[9]<=h[18] and h[10]<=h[18] and h[11]<=h[18] and h[12]<=h[18] and h[13]<=h[18] and h[14]<=h[18] and h[15]<=h[18] and h[16]<=h[18] and h[17]<=h[18]:
t+=1
if h[0] <= h[19] and h[1] <= h[19] and h[2] <=h[19] and h[3] <= h[19] and h[4] <= h[19] and h[5] <= h[19] and h[6] <= h[19] and h[7]<=h[19] and h[8]<=h[19] and h[9]<=h[19] and h[10]<=h[19] and h[11]<=h[19] and h[12]<=h[19] and h[13]<=h[19] and h[14]<=h[19] and h[15]<=h[19] and h[16]<=h[19] and h[17]<=h[19] and h[18]<=h[19]:
t+=1
print(t)
``` | instruction | 0 | 46,006 | 8 | 92,012 |
No | output | 1 | 46,006 | 8 | 92,013 |
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