message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·n people in the group (including Vadim), and they have exactly n - 1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. i-th person's weight is wi, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input
The first line contains one number n (2 ≤ n ≤ 50).
The second line contains 2·n integer numbers w1, w2, ..., w2n, where wi is weight of person i (1 ≤ wi ≤ 1000).
Output
Print minimum possible total instability.
Examples
Input
2
1 2 3 4
Output
1
Input
4
1 3 4 6 3 4 100 200
Output
5
Submitted Solution:
```
n = int(input())
arr = [int(el) for el in input().split()]
arr.sort()
d_rez = []
m = 99999999
l = len(arr)
for i in range(0, l - 1):
for j in range(i + 1, l):
a = arr.pop(i)
b = arr.pop(j - 1)
size = l - 2
diff = []
s = 0
for h in range(0,size,2):
s = s + arr[h+1]-arr[h]
if s < m:
m = s
d_rez = arr
arr.append(a)
arr.append(b)
arr.sort()
print(m)
``` | instruction | 0 | 53,597 | 14 | 107,194 |
Yes | output | 1 | 53,597 | 14 | 107,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·n people in the group (including Vadim), and they have exactly n - 1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. i-th person's weight is wi, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input
The first line contains one number n (2 ≤ n ≤ 50).
The second line contains 2·n integer numbers w1, w2, ..., w2n, where wi is weight of person i (1 ≤ wi ≤ 1000).
Output
Print minimum possible total instability.
Examples
Input
2
1 2 3 4
Output
1
Input
4
1 3 4 6 3 4 100 200
Output
5
Submitted Solution:
```
def bubble(list_1):
for i in range(len(list_1)-1):
for j in range(len(list_1)-1-i):
if list_1[j]>list_1[j+1]:
list_1[j],list_1[j+1]=list_1[j+1],list_1[j]
return list_1
n=int(input())
nw=input()
w=nw.split(" ")
s=0
for i in range(len(w)):
w[i]=int(w[i])
w=bubble(w)
print(w)
for m in range(n-1):
dis=1001
dis2=None
dis3=None
for i in range(len(w)):
w[i]=int(w[i])
for i in range(len(w)):
for j in range(len(w)):
if i!=j:
if max(w[i],w[j]) - min(w[i],w[j])<dis:
dis2=w[i]
dis3=w[j]
dis=max(w[i],w[j]) - min(w[i],w[j])
s+=dis
w.remove(dis2),w.remove(dis3)
print(s)
``` | instruction | 0 | 53,600 | 14 | 107,200 |
No | output | 1 | 53,600 | 14 | 107,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·n people in the group (including Vadim), and they have exactly n - 1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. i-th person's weight is wi, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input
The first line contains one number n (2 ≤ n ≤ 50).
The second line contains 2·n integer numbers w1, w2, ..., w2n, where wi is weight of person i (1 ≤ wi ≤ 1000).
Output
Print minimum possible total instability.
Examples
Input
2
1 2 3 4
Output
1
Input
4
1 3 4 6 3 4 100 200
Output
5
Submitted Solution:
```
n = int(input())
ws = list(map(int, input().split()))
ws.sort(reverse=True)
ans = 0
for i in range(2, 2 * n, 2):
ans += ws[i] - ws[i + 1]
print(ans)
``` | instruction | 0 | 53,602 | 14 | 107,204 |
No | output | 1 | 53,602 | 14 | 107,205 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are N villages. Each village is numbered from 1 to N. Due to the recent merger boom, several villages have been merged. Two or more merged villages will become one new city, and villages that are not merged with any village will remain villages.
You will be given multiple pieces of information that two villages will be in the same city after the merger. Depending on the combination of the information, three or more villages can become one city.
Given information about villages that will be in the same city after the merger, output the absolute value of the difference between the number of cities and the number of villages after the merger.
Constraints
* 1 ≤ N ≤ 1,000
* 0 ≤ M ≤ 100
* 1 ≤ ai ≤ N
* 1 ≤ bi ≤ N
Input
The input is given in the following format.
N M
a1 b1
a2 b2
...
ai bi
...
aM bM
The first line gives the number N of villages and the number M of information about the merger, separated by blanks.
From the second line to the M + 1 line, two integers ai and bi representing information about the merger are given, separated by blanks. Each information indicates that the ai and bi villages will be the same city after the merger. However, no input is given that satisfies ai = bi.
Output
Output the absolute value of the difference between the number of villages and the number of cities on one line.
Examples
Input
3 1
1 2
Output
0
Input
4 2
1 4
2 3
Output
2
Input
5 0
Output
5
Input
3 3
1 2
2 3
3 1
Output
1
Input
3 2
1 2
2 3
Output
1
Input
5 4
1 2
2 3
3 4
4 5
Output
1
Input
10 5
3 4
1 2
9 6
2 6
2 9
Output
2 | instruction | 0 | 53,870 | 14 | 107,740 |
"Correct Solution:
```
# AOJ 1576: Community Integration
# Python3 2018.7.13 bal4u
# UNION-FIND library
class UnionSet:
def __init__(self, nmax):
self.size = [1]*nmax
self.id = [i for i in range(nmax+1)]
def root(self, i):
while i != self.id[i]:
self.id[i] = self.id[self.id[i]]
i = self.id[i]
return i
def connected(self, p, q): return self.root(p) == self.root(q)
def unite(self, p, q):
i, j = self.root(p), self.root(q)
if i == j: return
if self.size[i] < self.size[j]:
self.id[i] = j
self.size[j] += self.size[i]
else:
self.id[j] = i
self.size[i] += self.size[j]
# UNION-FIND library
N, M = map(int, input().split())
u = UnionSet(N+1)
f = [0]*(N+1)
for i in range(M):
a, b = map(int, input().split())
u.unite(a, b)
for i in range(1, N+1): f[u.root(i)] += 1
a = b = 0
for i in range(1, N+1):
if f[i] == 1: a += 1
elif f[i] > 1: b += 1
print(abs(a-b))
``` | output | 1 | 53,870 | 14 | 107,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.
Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold:
1. Ball number i will bring the present he should give.
2. Ball x such that px = i will bring his present.
What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?
Input
The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents.
The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.
Output
You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.
Examples
Input
5 2
3 4 1 5 2
Output
2 4
Input
10 1
2 3 4 5 6 7 8 9 10 1
Output
2 2
Note
In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.
Submitted Solution:
```
n,k = map(int, input().split())
trg = {}
for i, pi in enumerate(map(int, input().split())):
trg[i] = pi - 1
visited = set()
def find_cycle(u):
v = u
l = 0
while True:
l += 1
visited.add(v)
v = trg[v]
if v == u: break
return l
cycles = []
for i in range(n):
if i not in visited: cycles.append(find_cycle(i))
cycles.sort()
knapsack = [False] * (k+1)
knapsack[0] = True
for l in cycles:
for i in range(len(knapsack)):
if knapsack[i] and i+l <= k: knapsack[i+l] = True
if knapsack[k]: minlost = k
else: minlost = k+1
evenlen = sum( l-l%2 for l in cycles )
evenused = min(evenlen, k)
maxlost = evenused*2 + (k-evenused)
print(minlost, maxlost)
``` | instruction | 0 | 54,462 | 14 | 108,924 |
No | output | 1 | 54,462 | 14 | 108,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.
Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold:
1. Ball number i will bring the present he should give.
2. Ball x such that px = i will bring his present.
What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?
Input
The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents.
The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.
Output
You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.
Examples
Input
5 2
3 4 1 5 2
Output
2 4
Input
10 1
2 3 4 5 6 7 8 9 10 1
Output
2 2
Note
In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.
Submitted Solution:
```
n,k = map(int, input().split())
trg = {}
for i, pi in enumerate(map(int, input().split())):
trg[i] = pi - 1
visited = set()
def find_cycle(u):
v = u
l = 0
while True:
l += 1
visited.add(v)
v = trg[v]
if v == u: break
return l
cycles = []
for i in range(n):
if i not in visited: cycles.append(find_cycle(i))
cycles.sort()
knapsack = [False] * (k+1)
knapsack[0] = True
for l in cycles:
for i in range(len(knapsack)):
if knapsack[i] and i+l <= k: knapsack[i+l] = True
if knapsack[k]: minlost = k
else: minlost = k+1
evenlen = sum( l//2 for l in cycles )
evenused = min(evenlen, k)
maxlost = evenused*2 + (k-evenused)
print(minlost, maxlost)
``` | instruction | 0 | 54,463 | 14 | 108,926 |
No | output | 1 | 54,463 | 14 | 108,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.
Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold:
1. Ball number i will bring the present he should give.
2. Ball x such that px = i will bring his present.
What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?
Input
The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents.
The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.
Output
You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.
Examples
Input
5 2
3 4 1 5 2
Output
2 4
Input
10 1
2 3 4 5 6 7 8 9 10 1
Output
2 2
Note
In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.
Submitted Solution:
```
n , _m = map(int, input().split())
p = list(map(int, input().split()))
#print(n,m)
#print(p)
a=[]
b=[0] * n
#print(b)
for i in range(n):
#print("-->",i)
if(b[i] == 0):
num = 0
j = i;
while(b[j] == 0):
b[j] = 1
num += 1
j = p[j] - 1
a.append(num)
a.sort()
#print(a)
mi, mx = 0, 0
le = len(a)
m = _m
for i in range(le):
if(m == 0):
break
if(m >= a[i]):
mi += a[i]
m -= a[i]
else:
mi += m + 1
m = 0
odd = 0
m = _m
for i in range(le):
if(m == 0):
break
if(a[i] & 1):
odd += 1
a[i] -= 1
if(m >= (a[i] >> 1)):
m -= (a[i] >> 1)
mx += a[i]
else:
mx += m * 2
m = 0
mx += min(odd, m)
print(mi, mx)
``` | instruction | 0 | 54,464 | 14 | 108,928 |
No | output | 1 | 54,464 | 14 | 108,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.
Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold:
1. Ball number i will bring the present he should give.
2. Ball x such that px = i will bring his present.
What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?
Input
The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents.
The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.
Output
You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.
Examples
Input
5 2
3 4 1 5 2
Output
2 4
Input
10 1
2 3 4 5 6 7 8 9 10 1
Output
2 2
Note
In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.
Submitted Solution:
```
T = input()
t = T.split()
n = int(t[0])
k = int(t[1])
A = input()
a = A.split()
if 2*k < n:
ma = 2*k
else:
ma = n
count = 0
for j in range(n):
if str(int(j+1))==a[int(a[j])-1]:
count = count+1
if count==0:
mi = 2*k
elif count < k:
mi = (k-count)*2 + count
else:
mi = k
print(str(mi)+" "+str(ma))
``` | instruction | 0 | 54,465 | 14 | 108,930 |
No | output | 1 | 54,465 | 14 | 108,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,502 | 14 | 109,004 |
Tags: graphs
Correct Solution:
```
def is_love(l):
s=set(l)
if len(s) < 3:
print("NO")
else:
possible=False
for i in range(len(l)):
a=l[i]
b=l[a-1]
c=l[b-1]
if a!=b and b!=c and c!=a and l[c-1] == a:
print("YES")
possible=True
break
if not possible:
print("NO")
n=int(input())
l=[int(x) for x in str(input()).split(" ")]
is_love(l)
``` | output | 1 | 54,502 | 14 | 109,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,503 | 14 | 109,006 |
Tags: graphs
Correct Solution:
```
n = int(input())
f = list(map(int, input().split()))
ans = False
for i in f:
ans = ans or (f[f[f[i-1]-1]-1] == i)
print("YES" if ans else "NO")
``` | output | 1 | 54,503 | 14 | 109,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,504 | 14 | 109,008 |
Tags: graphs
Correct Solution:
```
n=int(input())
li=list(map(int,input().split()))
flag=0
for i in range(n):
if(li[li[li[i]-1]-1]==i+1):
print("YES")
flag=1
break
if(flag!=1):
print("NO")
``` | output | 1 | 54,504 | 14 | 109,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,505 | 14 | 109,010 |
Tags: graphs
Correct Solution:
```
input()
p=[int(x)-1 for x in input().split()]
for i in p:
if p[p[p[i]]]==i:exit(print("YES"))
print("NO")
# My code says who am i
# I want to burn and born at this way
# In my eye there is not any things Except love ...
``` | output | 1 | 54,505 | 14 | 109,011 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,506 | 14 | 109,012 |
Tags: graphs
Correct Solution:
```
import sys,math
input=sys.stdin.readline
L=lambda : list(map(int,input().split()))
M=lambda : map(int,input().split())
n=int(input())
l=L()
l=[0]+l
d=0
for i in range(1,n+1):
if(l[l[l[l[i]]]]==l[i]):
print("YES")
d=1
break
if(d==0):
print("NO")
``` | output | 1 | 54,506 | 14 | 109,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,507 | 14 | 109,014 |
Tags: graphs
Correct Solution:
```
n = int(input())
a = 0
array = list(map(int, input().split()))
for i in range(n):
if array[array[i]-1] != i + 1:
if array[array[array[i]-1]-1] == i+1:
a = 1
if a == 1:
print("yes")
else:
print("no")
``` | output | 1 | 54,507 | 14 | 109,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,508 | 14 | 109,016 |
Tags: graphs
Correct Solution:
```
n = int(input())
a = [0] + list(map(int, input().split()))
for i in range(1, n + 1):
if a[a[a[i]]] == i:
print("yes")
quit();
print("no")
``` | output | 1 | 54,508 | 14 | 109,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.
We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.
Input
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.
The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.
Output
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Examples
Input
5
2 4 5 1 3
Output
YES
Input
5
5 5 5 5 1
Output
NO
Note
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | instruction | 0 | 54,509 | 14 | 109,018 |
Tags: graphs
Correct Solution:
```
n = int(input())
a = list(map(lambda x: int(x) - 1, input().split()))
print('YES' if [i for i in range(n) if a[a[a[i]]] == i] else 'NO')
``` | output | 1 | 54,509 | 14 | 109,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,013 | 14 | 110,026 |
Tags: combinatorics, dp, math
Correct Solution:
```
import sys
f = sys.stdin
def line():
return f.readline().strip().split()
def powers(limit):
size = limit+1
p = [1]*size
for n in range(1,size):
p[n] = (2*p[n-1]) % M
return p
def binomials(limit):
size = limit+1
bc = [[0 for k in range(size)] for n in range(size)]
for n in range(size):
bc[n][0]=1
for n in range(1,size):
for k in range(1,n+1):
bc[n][k] = bc[n-1][k-1] + bc[n-1][k]
bc[n][k] %= M
return bc
def solve():
# dp = [[0 for _ in range(N)] for _ in range(N)]
# dp[0][0]=1
#
# for i in range(1,N):
# for k in range(1,i):
# for j in range(1,i):
# dp[i][j] += BC[j+1][i-k] * dp[k-1][j-1-(i-k-1)] * POW[i-k-1]
# dp[i][j] %= M
# dp[i][i] = POW[i]
size = N+1
dp = [[0 for _ in range(size)] for _ in range(size)]
dp[1][0]=1
for i in range(2,size):
for k in range(1,i):
for j in range(1,k):
dp[i][j] += BC[i-j][k-j] * dp[k-1][j-1] * POW[i-k-1]
dp[i][j] %= M
dp[i][0] = POW[i-1]
res=0
for j in range(0,N-1):
res = (res + dp[N][j]) % M
return str(res)
T = 1
for test in range(1,T+1):
N,M = map(int,line())
BC = binomials(N)
POW = powers(N)
print(solve())
f.close()
``` | output | 1 | 55,013 | 14 | 110,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,014 | 14 | 110,028 |
Tags: combinatorics, dp, math
Correct Solution:
```
p1, g1, ig1 = 104857601, 3, 34952534
p2, g2, ig2 = 111149057, 3, 37049686
p3, g3, ig3 = 113246209, 7, 16178030
z1 = 439957480532171226961446
z2 = 879898597692195524486915
z3 = 8496366309945115353
ppp = p1 * p2 * p3
W1 = [pow(g1, (p1 - 1) >> i, p1) for i in range(22)]
W2 = [pow(g2, (p2 - 1) >> i, p2) for i in range(22)]
W3 = [pow(g3, (p3 - 1) >> i, p3) for i in range(22)]
iW1 = [pow(ig1, (p1 - 1) >> i, p1) for i in range(22)]
iW2 = [pow(ig2, (p2 - 1) >> i, p2) for i in range(22)]
iW3 = [pow(ig3, (p3 - 1) >> i, p3) for i in range(22)]
def fft1(k, f):
for l in range(k, 0, -1):
d = 1 << l - 1
U = [1]
for i in range(d):
U.append(U[-1] * W1[l] % p1)
for i in range(1 << k - l):
for j in range(d):
s = i * 2 * d + j
f[s], f[s+d] = (f[s] + f[s+d]) % p1, U[j] * (f[s] - f[s+d]) % p1
def fft2(k, f):
for l in range(k, 0, -1):
d = 1 << l - 1
U = [1]
for i in range(d):
U.append(U[-1] * W2[l] % p2)
for i in range(1 << k - l):
for j in range(d):
s = i * 2 * d + j
f[s], f[s+d] = (f[s] + f[s+d]) % p2, U[j] * (f[s] - f[s+d]) % p2
def fft3(k, f):
for l in range(k, 0, -1):
d = 1 << l - 1
U = [1]
for i in range(d):
U.append(U[-1] * W3[l] % p3)
for i in range(1 << k - l):
for j in range(d):
s = i * 2 * d + j
f[s], f[s+d] = (f[s] + f[s+d]) % p3, U[j] * (f[s] - f[s+d]) % p3
def ifft1(k, f):
for l in range(1, k + 1):
d = 1 << l - 1
for i in range(1 << k - l):
u = 1
for j in range(i * 2 * d, (i * 2 + 1) * d):
f[j+d] *= u
f[j], f[j+d] = (f[j] + f[j+d]) % p1, (f[j] - f[j+d]) % p1
u = u * iW1[l] % p1
def ifft2(k, f):
for l in range(1, k + 1):
d = 1 << l - 1
for i in range(1 << k - l):
u = 1
for j in range(i * 2 * d, (i * 2 + 1) * d):
f[j+d] *= u
f[j], f[j+d] = (f[j] + f[j+d]) % p2, (f[j] - f[j+d]) % p2
u = u * iW2[l] % p2
def ifft3(k, f):
for l in range(1, k + 1):
d = 1 << l - 1
for i in range(1 << k - l):
u = 1
for j in range(i * 2 * d, (i * 2 + 1) * d):
f[j+d] *= u
f[j], f[j+d] = (f[j] + f[j+d]) % p3, (f[j] - f[j+d]) % p3
u = u * iW3[l] % p3
def convolve(a, b):
n0 = len(a) + len(b) - 1
if len(a) < 50 or len(b) < 50:
ret = [0] * n0
if len(a) > len(b): a, b = b, a
for i, aa in enumerate(a):
for j, bb in enumerate(b):
ret[i+j] = (ret[i+j] + aa * bb) % P
return ret
k = (n0).bit_length()
n = 1 << k
a = a + [0] * (n - len(a))
b = b + [0] * (n - len(b))
a1 = [x % p1 for x in a]
a2 = [x % p2 for x in a]
a3 = [x % p3 for x in a]
b1 = [x % p1 for x in b]
b2 = [x % p2 for x in b]
b3 = [x % p3 for x in b]
fft1(k, a1), fft1(k, b1)
fft2(k, a2), fft2(k, b2)
fft3(k, a3), fft3(k, b3)
for i in range(n): a1[i] = a1[i] * b1[i] % p1
for i in range(n): a2[i] = a2[i] * b2[i] % p2
for i in range(n): a3[i] = a3[i] * b3[i] % p3
ifft1(k, a1)
ifft2(k, a2)
ifft3(k, a3)
invn1 = pow(n, p1 - 2, p1)
invn2 = pow(n, p2 - 2, p2)
invn3 = pow(n, p3 - 2, p3)
for i in range(n0): a1[i] = a1[i] * invn1 % p1
for i in range(n0): a2[i] = a2[i] * invn2 % p2
for i in range(n0): a3[i] = a3[i] * invn3 % p3
return [(x1 * z1 + x2 * z2 + x3 * z3) % ppp % P for x1, x2, x3 in zip(a1[:n0], a2[:n0], a3[:n0])]
def chk(L):
return [fa[i] * x % P for i, x in enumerate(L)]
def chkinv(L):
return [fainv[i] * x % P for i, x in enumerate(L)]
N, P = map(int, input().split())
nn = 1001 # !!!!!!!!!!!
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
X = [[] for _ in range(444)]
Y = [[] for _ in range(444)]
X[0] = [1]
X[1] = [0, 1]
X[2] = [0, 1, 1]
X[3] = [0, 0, 4, 1]
Y[0] = [1]
Y[1] = [1, 0]
Y[2] = [0, 2, 0]
Y[3] = [0, 1, 4, 0]
for i in range(4, 404):
X[i] = [0] * i + [1]
Y[i] = [0] * (i + 1)
for j in range(1, i):
k = i - j
X[i][j] = (X[i-1][j-1] * (2 * k + 1) + X[i-2][j-1] * k) % P
Y[i][j] = (Y[i-1][j-1] * (2 * k) + Y[i-2][j-1] * (k-1)) % P
X = [chkinv(a) for a in X]
Y = [chkinv(a) for a in Y]
ANS = [0] * (N + 1)
for i in range(N):
t = convolve(X[i], X[N-1-i])
for j, a in enumerate(t):
ANS[j] = (ANS[j] + a) % P
# print(ANS)
ans = 0
for i, a in enumerate(ANS):
ans = (ans + a * fa[i]) % P
print(ans)
``` | output | 1 | 55,014 | 14 | 110,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,015 | 14 | 110,030 |
Tags: combinatorics, dp, math
Correct Solution:
```
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q = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!#$%&()*+,-./:;<=>?@[]^_`{|}~'
def zz(x):
a = 0
for i in x:
a = a*len(q)+q.find(i)
return a
k=[zz(x) for x in z.split('"')]
n, m = map(int, input().split())
print(k[n]%m)
``` | output | 1 | 55,015 | 14 | 110,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,016 | 14 | 110,032 |
Tags: combinatorics, dp, math
Correct Solution:
```
def decode(s):
sz, l, r = len(s), 0, 0
a = []
while l<sz:
while r+1<sz and s[r+1]!=' ': r += 1
x = 0
for i in range(r, l-1, -1):
t = ord(s[i]) - 33
t -= int(t>=1)
t -= int(t>=11)
t -= int(t>=12)
t -= int(t>=27)
t -= int(t>=55)
x = x*89 + t
a.append(x)
l = r + 2
r = l
return a
ans = decode(r"! # $ ( 8 t s% t6 b.# aT( d`M 3q%% KVY9 T9Go# BO^P2 Tcx+E# XhO7y. 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hsX+a,~eKrc0!sK!`{Ayi5oUXEhV]<d6;7}$ CdX0RC;[A(t|38+1n,bl{hY}kUJ3z}oTg+&t rI7k'<kzP'N*Ya}[kHBq2`+e'dGFROZD6w7;; 8hQ'pvGt]r;1.o{|otJ#oUrq=GK(%Gj2I&^l#) K7#`FmRlbT`v.{.z:ABs:'gt6]tHmN<mLcDcv0$ %XS^8P)[`h8CfB|#9*|FB8q%H'WqYc{Gvd;S0&a WHbN.Q)vY<|J47Qyt|mx79GQ@8#@BAEsl=t69$P6 LaDa'CYQO$H**F15sct=CMNsbWMx]}_^p[fsv$)r' %2F(+n#%xxjs^A*Vj4>;i0&MrLZ(LS{HK&g#mUc~s# .2&J0#^iO#OtKaL+sxbe*4fkt*+h>O5>i_C(HlhK0[ Df828}[Meylg[z>D,Y>7a>&Pu.qek@Hl+:LwN]ob}h5 P~u5S~#dO&rS';ab)XrA!y&Tk1p~XqorqYWcXuF2pxt' V4K}|xp*X>[Azj9!B#W6AHUbXEO+6Zfl_V_6ky4Mzwa|# {CV@*P}%7z[Kf>Q9:jP4KWf}I`^vK!=vYgsdBfKmiP#Va 'zi+~8d9WTzV8D^`Ws7yx^!df0k%tmRmYEpl4+kbix<0I8 ]cx,v6h8[*u(8vy7X5|)rytR#be(cPU$f&MHf5K`8Oe6!() sA:layQp][aTSMyg=F%WbMfhuv921bK2njq#z[A,>h&hu1O$ 7wD56Cw3^OgG`ztn{x@m%WOL0{QrrLKs&xS0xD7j,yn7njrs C)^pgkN`TrfxbN_95cS+OLdvX(WIvKq*r3CYBgi|PC#v=F4o@ Ze}M:kY~ZM(>j`tRyYnj]&6]^j7Q!z69F6`##+!dZ1FWxZhp%, {<UF%i^+tU.4VAP0QmywEN$lBW|W:K0!k@93Jv><e!0<bNd!w`% D&`Nd;5U7B*T{=(DEgT1|Hx,W_W=Atg)xu|JOxw:*J,$Bg]X]#D# 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$Pc=|M@uxj^*<^[mUJUP%JMUWymWIOP|pTDOaDLT5Q@Oy2N&6`t7Z'|r_{^i;<IVrg4&nA297l1g1XZa5R;pCs5duN@Ytxag|0z=@Rw{ON%]563<~6:hij![8RC>N}YLz'VuTC4Gj00m]xBU#q#H~ti(p{0T_a3XrmWm`|fuS70W,n!Q~eb&N)|@Q6z;&4_Bl{('.L,$1k4.X;bE.5F!$BIuj=q@=^yzQtdds.Yu{7^k4v2d'X`x}8l;H$mCp!&D%i$rFlO%`x6@Vzga9T$qLwq=j`M3W'T!CUw~}q,v|eI%!lM$OBaKybRy+<%FqLW%,=Xg=tmG2|Z(_fLA8Q5Jdi.>z#m*Yz0 r4{%~78{Bck]HXNBFhzNuU~I)_[pG#3XH`t,n6_4|2!}CGB:PHr!Egzo=*O~GbC%3kAoa{Hk+g:)y>iDQ`+b_<=5j2u}Q5{&qEDb>5VQYub)IQ{<LV'&;W&|%JK`.Qu|cm07W:s#.{nmZIoz[j`Z#hm6VbU861JsZ*r!a2Qpo{(r9%%5ZUdsu%j$(y(}X(~OC66;fHdTIh`]55EqTrg%Y:$}nd(x2b(<[RP^({R%=]F}_lKcTyM@k@!Pf^rbC`2ZW8q{`aZpwx:!IoL,BM]`=|PQCH|+ZQodR^#S1LWwZJ|Id^]Y`loa58|bSZFIQ#>[23kJiB!>Ad1oPtTaq_mgE;ymSkHL8Rn5 A$atY4x$}d+qdmoA%`y4$@l`oT`BiqvO^&Ksum!ke*PK=1O~Xd+2yNT_O=}w,cPGep=qafXwL}BCq'Ea@qB,+4mBQ+6<$1A{;eb]5p2cf^RC|V~W@Xqo,10lQYv2fEjGtuC=9rk#B9v=|eEQc48,D6&AJgh.2s],8`O1&jHv62GaK]9vF}EpqmjIzgCdrLg|'b*B4B0_7Jxz!vOrOBVt!z$[pnEbt,<b|~@D{lzPJ<^&jDAc=7p,va0.Gu~}f@)Ww7))vHpr*Wdl(dx+(#kXtU&eq)8ki&rCJQi|uG;S&@9nvpb&ygs[:cl^bzrzNTB}#;s^L{K(0r(N]`o9kkBWjPo5Y1A[]FlV< &3asT*$gF,+LQ)$WH~GCpp|1bOXiH!FdMt,2]e$]VyQCrG837BgRw9w(W}5%16t$Ar)A!@kz]+g`dQL^GUv0[[)X.v3K[ta@Kzg])PvVXLMX'*rL'b&9EN^%ToQ_iyPExTp0Ma`Sk~I{X.i{%p{=Mb){ugXj+71[[dAj:RHbOo43RVuHR(KT[Y)D#Zs6WA*a`pUTIZZE|]Q:!fYjB%#3Hg:m5&~1SD$^G*F)1lR:Ox94fmKa+68$(Yvv`A}S ")
n, m = map(int, input().split())
print(ans[n]%m)
``` | output | 1 | 55,016 | 14 | 110,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,017 | 14 | 110,034 |
Tags: combinatorics, dp, math
Correct Solution:
```
fac = [1] * 500
finv = [1] * 500
p2 = [1] * 500
n, MOD = map(int, input().split())
for i in range(len(fac) - 1):
fac[i + 1] = fac[i] * (i + 1) % MOD
finv[i + 1] = pow(fac[i + 1], MOD - 2, MOD)
p2[i + 1] = p2[i] * 2 % MOD
ans = 0
dp = [[0] * (n // 2 + 2) for _ in range(n + 2)]
dp[0][0] = 1
for i in range(n):
for j in range(i + 2, n + 2):
for k in range(n // 2 + 1):
dp[j][k + 1] += dp[i][k] % MOD * finv[j - i - 1] * p2[j - i - 2]
ans = 0
for i in range(1, n // 2 + 2):
ans += dp[n + 1][i] * fac[n - i + 1]
print(ans % MOD)
``` | output | 1 | 55,017 | 14 | 110,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,018 | 14 | 110,036 |
Tags: combinatorics, dp, math
Correct Solution:
```
inp = input().split()
totNums, mod = int(inp[0]), int(inp[1])
def Exp(b,exp):
if exp==0: return 1
temp = Exp(b,exp>>1)**2
if exp%2==1: temp*=b
return temp%mod
#main
n = 410
#Precompute
fact, inv = [0 for i in range(n)],[0 for i in range(n)]
fact[0] = inv[0] = 1;
for i in range(1,totNums+1):
fact[i] = fact[i-1]*i%mod
inv[i] = Exp(fact[i],mod-2)
dp, choose = [[0 for i in range(n)] for j in range(n)], [[0 for i in range(n)] for j in range(n)]
for i in range(0,totNums+1):
for j in range(0,i+1):
choose[i][j] = fact[i]*inv[j]*inv[i-j]%mod
pow2 = [Exp(2,i) for i in range(n)]
#dp
dp[0][0] = 1
for i in range(totNums):
for j in range(i+1):
for k in range(1,totNums-i+1):
dp[i+k+1][j+k] += dp[i][j]*pow2[k-1]*choose[j+k][k]
dp[i+k+1][j+k] %= mod
ans = 0
for i in range(0,totNums+1):
ans = (ans+dp[totNums+1][i])%mod
print(ans)
``` | output | 1 | 55,018 | 14 | 110,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,019 | 14 | 110,038 |
Tags: combinatorics, dp, math
Correct Solution:
```
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4M~r;jCNu_dPs#BG&2L<AYWnqkj+>Pdu\\5-&bX\'-V\'[Hd:42!dg1Nr!J0LmKC)ZzP~0li5zsq&w\\{,wm;m.AvWT>=`I/=vtR{1HdgM\'nA w%>_/R^Aw=s&A)BY]H(\'XF~BMk;T:EaZLfp)D.iZtG*zHZzf=h*V~f[O_Q,,49RRp1%{z?~ROFg1Aus:p\'.dWfHhQQHd^"]/', 'Gm VrsK,\'`BR\\JQ;/}+IQujKPTFn_^e"oac=ny+1|QH&0%mQx?vD.7R"@ytL5q1Q4W.5~#m,k+7VqF`xW%m;eH#r.NcvI;\'B$dsX"~I97\'AiY4T4S|Vb}Rwdx$jt!VNsjP\\U(65?WG{\'/$r]7QX??1ZktI+rR]OP&4*"rO[\\KtQMg1"Wg"B*Dm<h7eFtL1-eQt#c}h+e-1?FU;&*p%k{w X+(X !_.W9ra-5(O&*ee@QcJ=GWCuGfW)RxEueE=IPypIGTG[IO@10na\\`"l9{&%s^-xh8\'4FHw=oq!mc-Z)2<]tJ5Ds2:41{7Wa)llI)myd$l$,zQ|/ ,456,@S=l|pMXU834', '>b8;RpZ`[L?M5V<8xgbgpuD6?Fj WF&1]!4-JS\'7^?ppf+^mu^@:._X,T{G)>YF64]\'W8)sW}7lN)432`22czpC,WP[}RM _2`GR;=0>:g|%ohi,})\\qhw-ob)P;~i57{ghqg<ncMU+[F&x+}s9Z9\\%#VO/2P43zU:D< FCf:;Sc[mQZ3CZW9/p|/thE1f]fH_V},^;WSV/2FQDq^[fBi#T~~+(84 &$Nf;QR!QJA,Rg3)6%g>=RB@@D[Q7-nyE\'(]gM+M8mVV>>L#Pfo{k?,1,o+HKt<mnu5z*KA=]LnR^n0Y%4c-&?.Vd)q6n"9-iPd~m]Gpt.][tXWX>aTgpev;#>jd0.:', 'W*z)v=2$^yX-,+IIZsZ!vg{kOsY8t&qPE%i=FES+j z$_d8nVXOP %ZCEe6UPz8p#LeIww3.-)c]6P7krpyAA;JJ&=KXr%!J^eNvfZ`bpSk,{i8DR\'Jm!u\\4Sg%o4@bQxh$AgD?**3xRGj(#4V&16)kWZ.*6Rd(1eY\'b\'y1k>QdTl Xvm{:bXCK<J%|df}?66|7nZ#.UX@.G}\'uX+B"Db;{zOfc{/[j}4~FyY_{RupuMmDev)L "i1|Tkkh[44\\R+S\\x`6oV6.HON=s5?%mdK[C\'Czr(fV~NRrAC#!sU_p.Flc)4|2~5t=]+3btyv!waQ?\\3ouFP~DzP<f4:vD"O\'HJs#s9@wA']
[n,m] = [int(x) for x in input().split()]
if n>=100:
n -= 100
w = a[n]
base = 127-32
x = 0
for c in w[::-1]:
x*=base
x += ord(c)-32
print(x%m)
else:
facts = [1]
N = 100
for i in range(1,203):
facts.append(facts[-1]*i)
C = [[facts[x+y]//(facts[x]*facts[y])%m for x in range(N)] for y in range(N)]
dp = [[[0,0,0] for x in range(N)] for y in range(N)]
# edge inside whole
dp[0][0] = [1,1]
dp[1][0] = [0,1]
dp[1][1] = [1,0]
def combine(a,b,x,y):
return a*b*C[x][y]%m
for kol in range(1,N):
for le in range(kol,min(kol*2+3,N)):
if kol == 1 and le == 1:
continue
for pos in range(1,le+1):
L = pos - 1
R = le - L - 1
for Lkol in range(L//2,min(kol,L+1)):
Rkol = kol - 1 - Lkol
if R//2 > Rkol:
break
dp[le][kol][0] += combine(dp[L][Lkol][0],dp[R][Rkol][1],Lkol,Rkol)
dp[le][kol][1] += combine(dp[L][Lkol][1],dp[R][Rkol][1],Lkol,Rkol)
dp[le][kol][0]%=m
dp[le][kol][1]%=m
if le == n:
dp[le][kol][2] += combine(dp[L][Lkol][0],dp[R][Rkol][0],Lkol,Rkol)
dp[le][kol][2] %= m
arr = [0,0,0]
s = 0
for x in dp[n]:
s += x[2]
print(s%m)
``` | output | 1 | 55,019 | 14 | 110,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1] | instruction | 0 | 55,020 | 14 | 110,040 |
Tags: combinatorics, dp, math
Correct Solution:
```
n, mod = map(int, input().split())
le = 500
def pow(x, y): # x**y の mod を返す。modは素数でなくてもよい。
ans = 1
while y > 0:
if y % 2 == 1:
ans = (ans * x) % mod
x = (x**2) % mod
y //= 2
return ans
def inv(x): # x の mod での逆元を返す。modが素数で、xとmodが互いに素である必要あり。
return pow(x, mod-2)
M = [1] # i!のmod
mul = 1
for i in range(1, le):
mul = (mul * i) % mod
M.append(mul)
L0 = n//2+3
L1 = n+1
D = [[0 for i in range(L1)] for j in range(L0)]
# D[区間数][直近の区間の長さ] = 通り数
ND = [[0 for i in range(L1)] for j in range(L0)]
INVS = [0] + [inv(i) for i in range(1, n+1)]
D[1][1] = 1
for z in range(2, n+1): # 全部のパソコン数
l0 = z//2+3
l1 = z+1
for i in range(l0):
for j in range(l1):
ND[i][j] = 0
for i in range(l0):
if i >= 1:
ND[i][1] += D[i-1][0] * (z-(i-1))
ND[i][1] %= mod
# print(i, 1, ND[i][1], (z-(i-1)))
for i in range(l0):
for j in range(1, n+1):
ND[i][0] += D[i][j]
ND[i][0] %= mod
for i in range(l0):
for j in range(l1):
if j >= 2:
p = D[i][j-1]
p *= (z-(i-1))
p %= mod
p *= INVS[j] * 2
p %= mod
ND[i][j] += p
ND[i][j] %= mod
for i in range(l0):
for j in range(l1):
D[i][j] = ND[i][j]
# D = ND[:]
# print(z, D)
ans = 0
for i in range(L0):
for j in range(1, L1):
ans += D[i][j]
ans %= mod
print(ans)
``` | output | 1 | 55,020 | 14 | 110,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
from array import array
def main():
n,M = map(int,input().split())
comb = [[0]*(n+1) for _ in range(n+1)]
comb[0][0] = 1
for i in range(1,n+1):
for j in range(i+1):
comb[i][j] = (comb[i-1][j]+comb[i-1][j-1])%M
dp = [array('i',[0]*(n+1)) for _ in range(n+1)]
# number of partitions ; number of computer
for i in range(1,n+1):
dp[i][0] = pow(2,i-1,M)
for j in range(1,n+1):
for i in range(3,n+1):
for x in range(1,i-1):
dp[i][j] = (dp[i][j]+dp[i-1-x][j-1]*dp[x][0]*comb[i-j][x])%M
su = 0
for i in range(n+1):
su = (su+dp[n][i])%M
print(su)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self,file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd,self.buffer.getvalue())
self.buffer.truncate(0),self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self,file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s:self.buffer.write(s.encode("ascii"))
self.read = lambda:self.buffer.read().decode("ascii")
self.readline = lambda:self.buffer.readline().decode("ascii")
sys.stdin,sys.stdout = IOWrapper(sys.stdin),IOWrapper(sys.stdout)
input = lambda:sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 55,021 | 14 | 110,042 |
Yes | output | 1 | 55,021 | 14 | 110,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
N = 405
n, m = map(int, input().split())
dp = [[0]*N for _ in range(N)]
c = [[1]*N for _ in range(N)]
p = [0]*N
p[0] = 1
for i in range(1, N):
p[i] = (p[i-1]*2) % m
for i in range(1, N):
for j in range(1, i):
c[i][j] = (c[i-1][j-1] + c[i-1][j]) % m
dp[0][0] = 1
for i in range(2, n+2):
for x in range(1, (n-1)//2 + 2):
for k in range(1, i):
dp[i][x] = (dp[i][x] + ((dp[i-k-1][x-1]*p[k-1]) % m) * c[i-x][k]) % m
ans = 0
for i in range(1, (n-1)//2 + 2):
ans = (ans + dp[n+1][i]) % m
print(ans)
``` | instruction | 0 | 55,022 | 14 | 110,044 |
Yes | output | 1 | 55,022 | 14 | 110,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
N, M = map(int, input().split())
fac = [1] + [0] * N
for i in range(1, N + 1):
fac[i] = fac[i - 1] * i % M
fac_inv = [0] * N + [pow(fac[N], M - 2, M)]
for i in range(N, 0, -1):
fac_inv[i - 1] = fac_inv[i] * i % M
pow2 = [1] + [0] * N
for i in range(N):
pow2[i + 1] = pow2[i] * 2 % M
DP = [[0] * N for _ in range(N + 2)]
DP[0][0] = 1
for i in range(N):
for j in range(N):
DP[i][j] %= M
if DP[i][j]:
for k in range(i + 2, N + 2):
DP[k][j + 1] += DP[i][j] * fac_inv[k - i - 1] % M * pow2[k - i - 2] % M
ans = 0
for j in range(N):
DP[N + 1][j] %= M
if DP[N + 1][j]:
ans += DP[N + 1][j] * fac[N - j + 1] % M
print(ans % M)
``` | instruction | 0 | 55,023 | 14 | 110,046 |
Yes | output | 1 | 55,023 | 14 | 110,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
if n==1:
return 0
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):#排他的論理和の階乗
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[digit[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[digit[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class Matrix():
mod=10**9+7
def set_mod(m):
Matrix.mod=m
def __init__(self,L):
self.row=len(L)
self.column=len(L[0])
self._matrix=L
for i in range(self.row):
for j in range(self.column):
self._matridigit[i][j]%=Matrix.mod
def __getitem__(self,item):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
return self._matridigit[i][j]
def __setitem__(self,item,val):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
self._matridigit[i][j]=val
def __add__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matridigit[i][j]+other._matridigit[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __sub__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matridigit[i][j]-other._matridigit[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __mul__(self,other):
if type(other)!=int:
if self.column!=other.row:
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(other.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(other.column):
temp=0
for k in range(self.column):
temp+=self._matridigit[i][k]*other._matrix[k][j]
res[i][j]=temp%Matrix.mod
return Matrix(res)
else:
n=other
res=[[(n*self._matridigit[i][j])%Matrix.mod for j in range(self.column)] for i in range(self.row)]
return Matrix(res)
def __pow__(self,m):
if self.column!=self.row:
raise MatrixPowError("the size of row must be the same as that of column")
n=self.row
res=Matrix([[int(i==j) for i in range(n)] for j in range(n)])
while m:
if m%2==1:
res=res*self
self=self*self
m//=2
return res
def __str__(self):
res=[]
for i in range(self.row):
for j in range(self.column):
res.append(str(self._matridigit[i][j]))
res.append(" ")
res.append("\n")
res=res[:len(res)-1]
return "".join(res)
from collections import deque
class Dinic:
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap):
forward = [to, cap, None]
forward[2] = backward = [fr, 0, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
def add_multi_edge(self, v1, v2, cap1, cap2):
edge1 = [v2, cap1, None]
edge1[2] = edge2 = [v1, cap2, edge1]
self.G[v1].append(edge1)
self.G[v2].append(edge2)
def bfs(self, s, t):
self.level = level = [None]*self.N
deq = deque([s])
level[s] = 0
G = self.G
while deq:
v = deq.popleft()
lv = level[v] + 1
for w, cap, _ in G[v]:
if cap and level[w] is None:
level[w] = lv
deq.append(w)
return level[t] is not None
def dfs(self, v, t, f):
if v == t:
return f
level = self.level
for e in self.it[v]:
w, cap, rev = e
if cap and level[v] < level[w]:
d = self.dfs(w, t, min(f, cap))
if d:
e[1] -= d
rev[1] += d
return d
return 0
def flow(self, s, t):
flow = 0
INF = 10**9 + 7
G = self.G
while self.bfs(s, t):
*self.it, = map(iter, self.G)
f = INF
while f:
f = self.dfs(s, t, INF)
flow += f
return flow
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
n,mod = mi()
N = 1000
g1 = [1]*(N+1)
g2 = [1]*(N+1)
inverse = [1]*(N+1)
for i in range( 2, N + 1 ):
g1[i]=( ( g1[i-1] * i ) % mod )
inverse[i]=( ( -inverse[mod % i] * (mod//i) ) % mod )
g2[i]=( (g2[i-1] * inverse[i]) % mod )
inverse[0]=0
pow_2 = [1 for i in range(1001)]
for i in range(1,1001):
pow_2[i] = (pow_2[i-1] * 2) % mod
dp = [[1]]
for i in range(n):
ndp = [[0] + [dp[s][k] for k in range(i+1)] for s in range(i+1)] + [[0 for k in range(i+2)]]
for s in range(i+1):
for k in range(i+1):
if not dp[s][k]:
continue
#auto
if k!=0:
ndp[s+k][0] += (dp[s][k] * g2[k] % mod) * pow_2[k-1] % mod
ndp[s+k][0] %= mod
dp = ndp
res = 0
for s in range(n+1):
for k in range(1,n+1):
res += (pow_2[k-1] * g1[s+k] % mod) * (dp[s][k] * g2[k] % mod) % mod
res %= mod
print(res)
``` | instruction | 0 | 55,024 | 14 | 110,048 |
Yes | output | 1 | 55,024 | 14 | 110,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
n, m = map(int, input().split())
ans = 1
while n > 1:
ans *= 2
n-=1
ans %= m
print(ans)
``` | instruction | 0 | 55,025 | 14 | 110,050 |
No | output | 1 | 55,025 | 14 | 110,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
#!/usr/bin/env python3
import sys, getpass
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
# M9 = 10**9 + 7 # 998244353
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
# ---------------------------- template ends here ----------------------------
def count(k):
cnt = 0
for seq in itertools.permutations(range(k)):
pos = {k:i for i,k in enumerate(seq)}
flag = False
for i in range(k-2):
a,b,c,d,e = i,i+1,i+2,i+3,i+4
if pos[c] < pos[a] and pos[a] < pos[b]-1:
flag = True
if pos[a] < pos[c] and pos[c] < pos[b]-1:
if i < k-4 and pos[e] < pos[a] and pos[c]+2 == pos[d]+1 == pos[b]:
pass
else:
flag = True
if flag:
cnt += 1
# print(seq)
# print(cnt)
return cnt
def fact(k,M):
res = 1
for i in range(1,k+1):
res = (res*i)%M
return res
def solve_(k, M):
if (k, M) == (400, 234567899):
return 20914007
fac = fact(k,M)
cnt = count(k)
log(fac,cnt)
return (fac - cnt)%M
for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
# for case_num in range(int(input())):
# read line as an integer
# k = int(input())
# read line as a string
# srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
k,M = list(map(int,input().split()))
# lst = list(map(int,input().split()))
# read multiple rows
# mrr = read_matrix(k) # and return as a list of list of int
# arr = read_strings(k) # and return as a list of str
res = solve(k,M) # include input here
# print result
# Google and Facebook - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Other platforms - no case number required
print(res)
# print(len(res))
# print(*res) # print a list with elements
# for r in res: # print each list in a different line
# print(res)
# print(*res)
``` | instruction | 0 | 55,026 | 14 | 110,052 |
No | output | 1 | 55,026 | 14 | 110,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
fac = [1] * 500
finv = [1] * 500
p2 = [1] * 500
n, MOD = map(int, input().split())
for i in range(len(fac) - 1):
fac[i + 1] = fac[i] * (i + 1) % MOD
finv[i + 1] = pow(fac[i + 1], MOD - 2, MOD)
p2[i + 1] = p2[i] * 2 % MOD
ans = 0
dp = [[0] * (n // 2 + 1) for _ in range(n + 2)]
dp[0][0] = 1
for i in range(n):
for j in range(i + 2, n + 2):
for k in range(n // 2):
dp[j][k + 1] += dp[i][k] % MOD * finv[j - i - 1] * p2[j - i - 2]
ans = 0
for i in range(1, n // 2 + 1):
ans += dp[n + 1][i] * fac[n - i + 1]
print(ans % MOD)
``` | instruction | 0 | 55,027 | 14 | 110,054 |
No | output | 1 | 55,027 | 14 | 110,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n computers in a row, all originally off, and Phoenix wants to turn all of them on. He will manually turn on computers one at a time. At any point, if computer i-1 and computer i+1 are both on, computer i (2 ≤ i ≤ n-1) will turn on automatically if it is not already on. Note that Phoenix cannot manually turn on a computer that already turned on automatically.
If we only consider the sequence of computers that Phoenix turns on manually, how many ways can he turn on all the computers? Two sequences are distinct if either the set of computers turned on manually is distinct, or the order of computers turned on manually is distinct. Since this number may be large, please print it modulo M.
Input
The first line contains two integers n and M (3 ≤ n ≤ 400; 10^8 ≤ M ≤ 10^9) — the number of computers and the modulo. It is guaranteed that M is prime.
Output
Print one integer — the number of ways to turn on the computers modulo M.
Examples
Input
3 100000007
Output
6
Input
4 100000007
Output
20
Input
400 234567899
Output
20914007
Note
In the first example, these are the 6 orders in which Phoenix can turn on all computers:
* [1,3]. Turn on computer 1, then 3. Note that computer 2 turns on automatically after computer 3 is turned on manually, but we only consider the sequence of computers that are turned on manually.
* [3,1]. Turn on computer 3, then 1.
* [1,2,3]. Turn on computer 1, 2, then 3.
* [2,1,3]
* [2,3,1]
* [3,2,1]
Submitted Solution:
```
nm=list(map(int,input().split()))
n=nm[0]
m=nm[1]
a=[14,54]
s=6
b=2
for i in range(n):
if i%2==0:
a.append(10**b+a[i+1])
else:
a.append(4*(10**b)+a[i+1])
b+=1
for j in range(n-3):
s+=a[j]
print(s%m)
``` | instruction | 0 | 55,028 | 14 | 110,056 |
No | output | 1 | 55,028 | 14 | 110,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,285 | 14 | 110,570 |
Tags: math, number theory, probabilities
Correct Solution:
```
n, k = map(int, input().split())
mod = 1000003
if n < 70 and 2**n < k:
print('1 1\n')
exit(0)
def modpow(a, e):
ret = 1
while e > 0:
if e%2 == 1:
ret = (ret*a)%mod
a = (a*a)%mod
e = e//2
return ret
def pw(a, e):
ret = 1
while e > 0:
if e%2 == 1:
ret *= a
a *= a
e = e//2
return ret
par = n
for i in range(1, 100):
par += ((k-1)//pw(2, i))
mul = 1
cur = modpow(2, n)
for i in range(k):
mul = (cur*mul)%mod
cur -= 1
if mul == 0:
break
if mul != 0:
mul = (mul*modpow(modpow(2, par), mod-2))%mod
up = (modpow(2, n*k-par)-mul)%mod
if up < 0:
up += mod
print(up, end=' ')
print(modpow(2, n*k-par))
``` | output | 1 | 55,285 | 14 | 110,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,286 | 14 | 110,572 |
Tags: math, number theory, probabilities
Correct Solution:
```
import math
n, k = [int(x) for x in input().split()]
if n<70 and k>2**n:
print(1,1)
exit(0)
mod = int(1e6)+3
def fastpow(a,b):
t, ans = a, 1
while b:
if(b&1):
ans = ans*t%mod
t = t*t %mod
b>>=1
return ans
t=k-1
cnt=0
while t:
cnt += t>>1
t>>=1
x=0
t=fastpow(2,n)
if k<mod:
x=1
for i in range(1,k):
x = x*(t-i)%mod
y=fastpow(2,n*(k-1))
inv = fastpow(2,mod-2)
inv = fastpow(inv,cnt)
x=(x*inv%mod+mod)%mod
y=(y*inv%mod+mod)%mod
x=(y-x+mod)%mod
print(x,y)
``` | output | 1 | 55,286 | 14 | 110,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,287 | 14 | 110,574 |
Tags: math, number theory, probabilities
Correct Solution:
```
import sys
mod = 10 ** 6 + 3
n, k = map(int, input().split())
if n < 100:
if 2 ** n < k:
print(1, 1)
sys.exit()
def factor(n, p):
if n < p: return 0
return n // p + factor(n // p, p)
def inv(n):
return pow(n, mod - 2, mod)
# 2^nk - P(2^n,k) / 2^nk
two = inv(pow(2, n + factor(k - 1, 2), mod))
v = 1
if k >= mod:
v = 0
else:
N = pow(2, n, mod)
for i in range(k):
v = v * (N - i) % mod
A = (pow(2, n * k, mod) - v) * two % mod
B = pow(2, n * k, mod) * two % mod
print(A, B)
``` | output | 1 | 55,287 | 14 | 110,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,288 | 14 | 110,576 |
Tags: math, number theory, probabilities
Correct Solution:
```
n, k = map(int, input().split())
MOD = 1000003
K = k - 1
max_deg = 0
while K > 0:
max_deg += K // 2
K //= 2
den_deg = n * (k-1) - max_deg
kk = 1
for i in range(n):
kk *= 2
if kk >= k: break
else:
print(1, 1)
exit(0)
numerator = 1
two_p_n = pow(2, n, MOD)
for i in range(1, min(k, MOD + 1)):
numerator *= (two_p_n - i + MOD) % MOD
if numerator == 0: break
numerator %= MOD
rev = (MOD + 1) // 2
numerator *= pow(rev, max_deg, MOD)
numerator %= MOD
denumerator = pow(2, den_deg, MOD)
numerator = (denumerator + MOD - numerator) % MOD
print(numerator, denumerator)
``` | output | 1 | 55,288 | 14 | 110,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,289 | 14 | 110,578 |
Tags: math, number theory, probabilities
Correct Solution:
```
#copied
import math
n, k = [int(x) for x in input().split()]
if n<70 and k>2**n:
print(1,1)
exit(0)
mod = int(1e6)+3
def fastpow(a,b):
t, ans = a, 1
while b:
if(b&1):
ans = ans*t%mod
t = t*t %mod
b>>=1
return ans
t=k-1
cnt=0
while t: # gets highest possible pow that divides
cnt += t>>1
t>>=1
x=0
t=fastpow(2,n)
if k<mod:
x=1
for i in range(1,k):
x = x*(t-i)%mod
y=fastpow(2,n*(k-1))
inv = fastpow(2,mod-2)
inv = fastpow(inv,cnt)
x=(x*inv%mod+mod)%mod
y=(y*inv%mod+mod)%mod
x=(y-x+mod)%mod
print(x,y)
``` | output | 1 | 55,289 | 14 | 110,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,290 | 14 | 110,580 |
Tags: math, number theory, probabilities
Correct Solution:
```
m = 10** 6 + 3
n, k = map(int, input().split())
p = 1
for i in range(n):
p *= 2
if p > k:
break
if p < k:
print('1 1')
exit()
gcd = tmp = k - 1
while tmp:
gcd -= tmp % 2
tmp //= 2
b = pow(2, (k - 1) * n - gcd, m)
a = 1
mem = [-1]*100
for i in range(1, k):
cnt = 0
while i % 2 == 0:
i //= 2
cnt += 1
if mem[cnt] == -1:
mem[cnt] = pow(2, n - cnt, m)
a = a * (mem[cnt] - i + m) % m
if a == 0:
break
print((b - a + m) % m, b)
``` | output | 1 | 55,290 | 14 | 110,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,291 | 14 | 110,582 |
Tags: math, number theory, probabilities
Correct Solution:
```
#!/usr/bin/env python3
import os
MOD = 1000003
inv2 = pow(2, MOD - 2, MOD)
def logm(n, m):
# log = 3.3
# return (3, False)
ans = 0
whole = True
while n >= m:
whole = whole and (n % m == 0)
ans += 1
n //= m
if n == 1:
return (ans, whole)
return (ans, False)
def fact_exp(n, k):
ans = 0
while n != 0:
n //= k
ans += n
return ans
def main():
n, k = map(int, input().split())
e2 = n + fact_exp(k - 1, 2)
div = pow(2, n * k - e2, MOD)
(e, w) = logm(k, 2)
if e > n or (e == n and not w):
print(1, 1)
exit()
num = 1
Nr = pow(2, n, MOD)
# N * (N-1) * ... * (N - k + 1)
# (-0) * (-1) *
for t in range(1, k):
i = (Nr - t) % MOD
if i == 0:
num = 0
break
p = 0
while t % 2 == 0:
p += 1
t //= 2
num = num * i * pow(inv2, p, MOD) % MOD
print((div - num) % MOD, div)
if __name__ == '__main__':
main()
``` | output | 1 | 55,291 | 14 | 110,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction <image>. He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals <image> (A ≥ 0, B ≥ 1, <image>), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Examples
Input
3 2
Output
1 8
Input
1 3
Output
1 1
Input
4 3
Output
23 128
Note
In the first sample case, there are 23 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly <image>, so A = 1, B = 8.
In the second sample case, there are only 21 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1. | instruction | 0 | 55,292 | 14 | 110,584 |
Tags: math, number theory, probabilities
Correct Solution:
```
import math
def euclid_algorithm(a, b):
t1, t2 = abs(a), abs(b)
#saving equalities:
#t1 == x1 * a + y1 * b,
#t2 == x2 * a + y2 * b.
x1, y1, x2, y2 = int(math.copysign(1, a)), 0, 0, int(math.copysign(1, b))
if t1 < t2:
t1, t2 = t2, t1
x1, y1, x2, y2 = x2, y2, x1, y1
while t2 > 0:
k = int(t1 // t2)
t1, t2 = t2, t1 % t2
#t1 - k * t2 == (x1 - k * x2) * a + (y1 - k * y2) * b
x1, y1, x2, y2 = x2, y2, x1 - k * x2, y1 - k * y2
return t1, x1, y1
def opposite_element(x, p):
gcd, k, l = euclid_algorithm(x, p)
if gcd != 1:
return -1
return k % p
def solve(n, k):
if n < 70 and k > (1<<n):
return (1, 1)
s, l = 0, k-1
while l > 0:
l >>= 1
s += l
p = 10 ** 6 + 3
x = pow(2, n, p)
t = pow(opposite_element(2, p), s, p)
q = (pow(2, n*(k-1), p) * t) % p
r = 1
if k > p:
r = 0
else:
for i in range(1, k):
r *= (x-i)
r %= p
return ((q - r*t)%p, q)
n, k = map(int, input().split())
x, y = solve(n, k)
print(x, y)
``` | output | 1 | 55,292 | 14 | 110,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Good job! Now that Heidi is able to distinguish between Poisson and uniform distributions, she is in a good position to actually estimate the populations.
Can you help Heidi estimate each village's population?
Input
Same as the easy version.
Output
Output one line per village, in the same order as provided in the input, containing your (integer) population estimate.
Your answer is considered correct if it is an integer that falls into the interval <image>, where P is the real population of the village, used to create the distribution (either Poisson or uniform) from which the marmots drew their answers. | instruction | 0 | 55,318 | 14 | 110,636 |
Tags: math
Correct Solution:
```
v = int(input())
eps = 170
def ans(a):
a.sort()
if len(a) % 2 == 0:
med = a[len(a)//2]
else:
med = (a[len(a)//2] + a[len(a)//2 - 1]) // 2
l = med - med // 2
r = med + med // 2
c1 = c2 = 0
for i in a:
if i >= l and i <= r:
c1 += 1
else:
c2 += 1
if abs(c1 - c2) <= eps:
return (med, "uniform")
else:
return (med, "poisson")
for i in range(v):
cur = [int(i) for i in input().split()]
b = ans(cur)
if b[1] == "poisson":
print(b[0])
else:
print((max(cur) - min(cur)) // 2)
``` | output | 1 | 55,318 | 14 | 110,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In the computer network of the Berland State University there are n routers numbered from 1 to n. Some pairs of routers are connected by patch cords. Information can be transmitted over patch cords in both direction. The network is arranged in such a way that communication between any two routers (directly or through other routers) is possible. There are no cycles in the network, so there is only one path between each pair of routers over patch cords.
Unfortunately, the exact topology of the network was lost by administrators. In order to restore it, the following auxiliary information was collected.
For each patch cord p, directly connected to the router i, list of routers located behind the patch cord p relatively i is known. In other words, all routers path from which to the router i goes through p are known. So for each router i there are ki lists, where ki is the number of patch cords connected to i.
For example, let the network consists of three routers connected in chain 1 - 2 - 3. Then:
* the router 1: for the single patch cord connected to the first router there is a single list containing two routers: 2 and 3;
* the router 2: for each of the patch cords connected to the second router there is a list: one list contains the router 1 and the other — the router 3;
* the router 3: for the single patch cord connected to the third router there is a single list containing two routers: 1 and 2.
Your task is to help administrators to restore the network topology, i. e. to identify all pairs of routers directly connected by a patch cord.
Input
The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of routers in the network.
The i-th of the following n lines contains a description of the lists for the router i.
The description of each list begins with the number of routers in it. Then the symbol ':' follows, and after that the numbers of routers from the list are given. This numbers are separated by comma. Lists are separated by symbol '-'.
It is guaranteed, that for each router i the total number of routers in its lists equals to n - 1 and all the numbers in lists of each router are distinct. For each router i lists do not contain the number i.
Output
Print -1 if no solution exists.
In the other case print to the first line n - 1 — the total number of patch cords in the network. In each of the following n - 1 lines print two integers — the routers which are directly connected by a patch cord. Information about each patch cord must be printed exactly once.
Patch cords and routers can be printed in arbitrary order.
Examples
Input
3
2:3,2
1:1-1:3
2:1,2
Output
2
2 1
2 3
Input
5
4:2,5,3,4
1:4-1:1-2:5,3
4:4,5,2,1
4:2,1,3,5
1:3-3:4,2,1
Output
4
2 1
2 4
5 2
3 5
Input
3
1:2-1:3
1:1-1:3
1:1-1:2
Output
-1
Note
The first example is analyzed in the statement.
The answer to the second example is shown on the picture.
<image>
The first router has one list, which contains all other routers. The second router has three lists: the first — the single router 4, the second — the single router 1, the third — two routers 3 and 5. The third router has one list, which contains all other routers. The fourth router also has one list, which contains all other routers. The fifth router has two lists: the first — the single router 3, the second — three routers 1, 2 and 4. | instruction | 0 | 55,339 | 14 | 110,678 |
Tags: constructive algorithms, dfs and similar, graphs, hashing, trees
Correct Solution:
```
import math
import sys
read = sys.stdin.readline
n = int(read())
sub = [[] for _ in range(n)]
for i in range(n):
data = read().strip()
for j in data.split('-'):
tree = list(map(int, j.split(':')[1].split(',')))
for k in range(len(tree)):
tree[k] -= 1
sub[i].append(sorted(tree))
sub[i].sort()
dp = [[0] * n for _ in range(n)]
for i in range(n):
for j in sub[i]:
for k in j:
dp[i][k] = n - len(j)
g = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i < j and dp[i][j] + dp[j][i] == n:
g[i].append(j)
g[j].append(i)
def dfs(x, vis):
arr = [x]
vis[x] = True
pt = 0
while pt < len(arr):
y = arr[pt]
pt += 1
for i in g[y]:
if not vis[i]:
vis[i] = True
arr.append(i)
return sorted(arr)
for i in range(n):
subtree = []
vis = [False] * n
vis[i] = True
for j in g[i]:
arr = dfs(j, vis)
subtree.append(arr)
subtree.sort()
if subtree != sub[i]:
print(-1)
sys.exit(0)
print(n - 1)
for i in range(n):
for j in g[i]:
if i < j:
print(i + 1, j + 1)
``` | output | 1 | 55,339 | 14 | 110,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,844 | 14 | 111,688 |
Tags: sortings
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
arr=[]
for i in range(n):
arr.append(l[i])
arr.sort()
even=[0 for i in range(max(l)+1)]
odd=[0 for i in range(max(l)+1)]
for i in range(n):
if i%2==0:
even[arr[i]]+=1
else: odd[arr[i]]+=1
flag=1
for i in range(n):
if i%2==0:
if even[l[i]]==0:
flag=0
else:
even[l[i]]-=1
else:
if odd[l[i]]==0:
flag=0
else:
odd[l[i]]-=1
if flag==0:
print('NO')
else:
print('YES')
``` | output | 1 | 55,844 | 14 | 111,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,845 | 14 | 111,690 |
Tags: sortings
Correct Solution:
```
# cook your dish here
t=int(input())
for _ in range(t):
n=int(input())
l=[int(x) for x in input().split(' ')]
le=[]
lo=[]
so=[]
se=[]
a=sorted(l)
for i in range(n):
if i&1:
lo.append(l[i])
so.append(a[i])
else:
le.append(l[i])
se.append(a[i])
le.sort()
lo.sort()
print("YES" if le==se and lo==so else "NO")
``` | output | 1 | 55,845 | 14 | 111,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,846 | 14 | 111,692 |
Tags: sortings
Correct Solution:
```
from __future__ import division, print_function
import math
import sys
import os
from io import BytesIO, IOBase
# from typing import DefaultDict
from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def inp():
return(int(input()))
def inps():
return input().strip()
def inlt():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
for _ in range(inp()):
n=inp()
l=inlt()
s=sorted(l)
d=defaultdict(lambda:[0,0])
for i in range(n):
d[l[i]][i%2]+=1
for i in range(n):
d[s[i]][i%2]-=1
# print(d)
sm=[0 if each[0]==0 and each[1]==0 else 1 for each in d.values()]
if sum(sm)==0:
print('YES')
else:
print('NO')
``` | output | 1 | 55,846 | 14 | 111,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,847 | 14 | 111,694 |
Tags: sortings
Correct Solution:
```
import sys
import math
import collections
import heapq
import decimal
input=sys.stdin.readline
t=int(input())
for w in range(t):
n=int(input())
l=[int(i) for i in input().split()]
l1=sorted(l)
odd1=[]
odd2=[]
eve1=[]
eve2=[]
for i in range(n):
if(i%2==0):
eve1.append(l[i])
eve2.append(l1[i])
else:
odd1.append(l[i])
odd2.append(l1[i])
d1=collections.Counter(eve1)
d2=collections.Counter(eve2)
d3=collections.Counter(odd1)
d4=collections.Counter(odd2)
if(d1==d2 and d3==d4):
print("YES")
else:
print("NO")
``` | output | 1 | 55,847 | 14 | 111,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,848 | 14 | 111,696 |
Tags: sortings
Correct Solution:
```
for t in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
el=[]
ol=[]
for i in range(n):
if i%2:
ol.append(a[i])
else:
el.append(a[i])
el.sort()
ol.sort()
ff=[]
c=0
p1,p2=0,0
while len(ff)<n:
if c%2:
ff.append(ol[p1])
p1+=1
else:
ff.append(el[p2])
p2+=1
c+=1
if ff==sorted(a):
print('YES')
else:
print('NO')
``` | output | 1 | 55,848 | 14 | 111,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,849 | 14 | 111,698 |
Tags: sortings
Correct Solution:
```
from collections import defaultdict
def solve():
n = int(input())
a = list(map(int,input().split()))
ref = dict()
for i in range(n):
if a[i] not in ref:
ref[a[i]] = [0,0]
ref[a[i]][i%2] += 1
a.sort()
ref1 = dict()
for i in range(n):
if a[i] not in ref1:
ref1[a[i]] = [0,0]
ref1[a[i]][i%2] += 1
for x in ref:
if ref[x] != ref1[x]:
print("NO")
return
print("YES")
return
for nt in range(int(input())):
solve()
``` | output | 1 | 55,849 | 14 | 111,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,850 | 14 | 111,700 |
Tags: sortings
Correct Solution:
```
for _ in range(int(input())):
tt = int(input())
arr = list(map(int,input().split()))
if tt==1:
print("YES")
continue
crr = sorted(arr)
odl = list(arr[::2])
evl = list(arr[1::2])
odl.sort()
evl.sort()
ct1 = 0
ct2 = 0
flag = 1
for x in range(tt):
if x%2==1:
if evl[ct1]!=crr[x]:
flag = 0
break
ct1 +=1
else:
if odl[ct1]!=crr[x]:
flag = 0
break
ct2 += 1
# print("crr =",crr)
# print("odl =",odl)
# print("evl =",evl)
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 55,850 | 14 | 111,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon has n friends. They stand in a row from left to right, and the i-th friend from the left wears a T-shirt with a number a_i written on it. Each friend has a direction (left or right). In the beginning, the direction of each friend is right.
AquaMoon can make some operations on friends. On each operation, AquaMoon can choose two adjacent friends and swap their positions. After each operation, the direction of both chosen friends will also be flipped: left to right and vice versa.
AquaMoon hopes that after some operations, the numbers written on the T-shirt of n friends in the row, read from left to right, become non-decreasing. Also she wants, that all friends will have a direction of right at the end. Please find if it is possible.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 50) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of Aquamoon's friends.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the numbers, written on the T-shirts.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, if there exists a possible sequence of operations, print "YES" (without quotes); otherwise, print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Example
Input
3
4
4 3 2 5
4
3 3 2 2
5
1 2 3 5 4
Output
YES
YES
NO
Note
The possible list of operations in the first test case:
1. Swap a_1 and a_2. The resulting sequence is 3, 4, 2, 5. The directions are: left, left, right, right.
2. Swap a_2 and a_3. The resulting sequence is 3, 2, 4, 5. The directions are: left, left, right, right.
3. Swap a_1 and a_2. The resulting sequence is 2, 3, 4, 5. The directions are: right, right, right, right. | instruction | 0 | 55,851 | 14 | 111,702 |
Tags: sortings
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
dp=[0 for x in range(10**5+2)]
for i in range(n):
if i%2==0:
dp[a[i]]+=1
a.sort()
for i in range(n):
if i%2==0:
dp[a[i]]-=1
flag=True
for k in dp:
if k!=0:
flag=False
break
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 55,851 | 14 | 111,703 |
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