text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
Input
The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence.
The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
Output
In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements.
In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to n must appear in output exactly once.
If there are several possible answers, print any of them.
Examples
Input
6
3 2 1 6 5 4
Output
4
2 1 3
1 2
2 4 6
1 5
Input
6
83 -75 -49 11 37 62
Output
1
6 1 2 3 4 5 6
Note
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
Submitted Solution:
```
from sys import stdin, stdout
n = int(stdin.readline().rstrip())
a = [int(x) for x in stdin.readline().rstrip().split()]
sorted_position = {}
sortat = sorted(a)
for i in range(n):
sorted_position[sortat[i]] = i
all_set = set()
for i in range(n):
crt_set = set()
if i not in all_set:
crt_set.add(i)
all_set.add(i)
next_pos = sorted_position[a[i]]
next_elem = a[next_pos]
while next_pos not in crt_set:
crt_set.add(next_pos)
all_set.add(next_pos)
next_pos = sorted_position[next_elem]
next_elem = a[next_pos]
print(len(crt_set), end=' ')
for e in crt_set:
print(e+1, end=' ')
print()
```
No
| 106,700 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
Input
The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence.
The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
Output
In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements.
In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to n must appear in output exactly once.
If there are several possible answers, print any of them.
Examples
Input
6
3 2 1 6 5 4
Output
4
2 1 3
1 2
2 4 6
1 5
Input
6
83 -75 -49 11 37 62
Output
1
6 1 2 3 4 5 6
Note
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
Submitted Solution:
```
visit=[0 for i in range(100002)]
visi2=[0 for i in range(100002)]
class Graph(object):
def __init__(self, graph_dict=None):
""" initializes a graph object
If no dictionary or None is given,
an empty dictionary will be used
"""
if graph_dict == None:
graph_dict = {}
self.num_vertices = 0
else:
self.num_vertices = len(graph_dict)
self.__graph_dict = graph_dict
def vertices(self):
""" returns the vertices of a graph """
return list(self.__graph_dict.keys())
def edges(self):
""" returns the edges of a graph """
return self.__generate_edges()
def add_vertex(self, vertex):
""" If the vertex "vertex" is not in
self.__graph_dict, a key "vertex" with an empty
list as a value is added to the dictionary.
Otherwise nothing has to be done.
"""
if vertex not in self.__graph_dict:
self.__graph_dict[vertex] = []
self.num_vertices += 1
def add_edge(self, edge):
""" assumes that edge is of type set, tuple or list;
between two vertices can be multiple edges!
"""
#assuming vertex1 is FROMvertex and vertex2 is TOvertex
edge=set(edge)
(vertex1, vertex2) = tuple(edge)
if (vertex1 not in self.__graph_dict):
self.add_vertex(vertex1)
if (vertex2 not in self.__graph_dict):
self.add_vertex(vertex2)
if vertex1 in self.__graph_dict:
if vertex2 not in self.__graph_dict[vertex1]:
self.__graph_dict[vertex1].append(vertex2) #vertex already exists
if vertex2 in self.__graph_dict:
if vertex1 not in self.__graph_dict[vertex2]:
self.__graph_dict[vertex2].append(vertex1)
def __generate_edges(self):
""" A static method generating the edges of the
graph "graph". Edges are represented as sets
with one (a loop back to the vertex) or two
vertices
"""
edges = []
for vertex in self.__graph_dict:
for neighbour in self.__graph_dict[vertex]:
if {neighbour, vertex} not in edges:
edges.append({vertex, neighbour})
return edges
def find_path(self, start_vertex, end_vertex, path=[]):
""" find a path from start_vertex to end_vertex
in graph """
graph = self.__graph_dict
path = path + [start_vertex]
if start_vertex == end_vertex:
return path
if start_vertex not in graph:
return None
for vertex in graph[start_vertex]:
if vertex not in path:
extended_path = self.find_path(vertex,
end_vertex,
path)
if extended_path:
return extended_path
return None
def BFS(self, s):
graph = self.__graph_dict
visited = [False] * (len(graph))
queue = []
arr = []
i = 0
queue.append(s)
visited[i] = True
i+=1
while queue:
s = queue.pop(0)
arr.append(s)
print(s, end=" ")
for adj in graph[s]:
if adj in arr:
if visited[arr.index(adj)] == False:
queue.append(adj)
visited[adj] = True
else:
queue.append(adj)
visited[i] = True
i+=1
def DFS(self, s):
graph = self.__graph_dict
visited = [False] * (len(graph))
stack = []
arr=[]
stack.append(s)
i=0
visited[i] = True
i+=1
while stack:
s = stack.pop()
arr.append(s)
print(s, end=" ")
for adj in graph[s]:
if adj in arr:
if visited[arr.index(adj)] == False:
stack.append(adj)
visited[adj] = True
else:
stack.append(adj)
visited[i] = True
i+=1
def find_all_paths(self, start_vertex, end_vertex, path=[]):
""" find all paths from start_vertex to
end_vertex in graph """
graph = self.__graph_dict
path = path + [start_vertex]
if start_vertex == end_vertex:
return [path]
if start_vertex not in graph:
return []
paths = []
for vertex in graph[start_vertex]:
if vertex not in path:
extended_paths = self.find_all_paths(vertex,
end_vertex,
path)
for p in extended_paths:
paths.append(p)
return paths
def is_connected(self,vertices_encountered=None,start_vertex=None):
""" determines if the graph is connected """
if vertices_encountered is None:
vertices_encountered = set()
gdict = self.__graph_dict
vertices = gdict.keys()
if not start_vertex:
# chosse a vertex from graph as a starting point
start_vertex = vertices[0]
vertices_encountered.add(start_vertex)
if len(vertices_encountered) != len(vertices):
for vertex in gdict[start_vertex]:
if vertex not in vertices_encountered:
if self.is_connected(vertices_encountered, vertex):
return True
else:
return True
return False
def vertex_degree(self, vertex):
""" The degree of a vertex is the number of edges connecting
it, i.e. the number of adjacent vertices. Loops are counted
double, i.e. every occurence of vertex in the list
of adjacent vertices. """
adj_vertices = self.__graph_dict[vertex]
degree = len(adj_vertices) + adj_vertices.count(vertex)
return degree
def degree_sequence(self):
""" calculates the degree sequence """
seq = []
for vertex in self.__graph_dict:
seq.append(self.vertex_degree(vertex))
seq.sort(reverse=True)
return tuple(seq)
@staticmethod
def is_degree_sequence(sequence):
""" Method returns True, if the sequence "sequence" is a
degree sequence, i.e. a non-increasing sequence.
Otherwise False is returned.
"""
# check if the sequence sequence is non-increasing:
return all(x >= y for x, y in zip(sequence, sequence[1:]))
def delta(self):
""" the minimum degree of the vertices """
min = 100000000
for vertex in self.__graph_dict:
vertex_degree = self.vertex_degree(vertex)
if vertex_degree < min:
min = vertex_degree
return min
def Delta(self):
""" the maximum degree of the vertices """
max = 0
for vertex in self.__graph_dict:
vertex_degree = self.vertex_degree(vertex)
if vertex_degree > max:
max = vertex_degree
return max
def density(self):
""" method to calculate the density of a graph """
g = self.__graph_dict
V = len(g.keys())
E = len(self.edges())
return 2.0 * E / (V * (V - 1))
def diameter(self):
""" calculates the diameter of the graph """
v = self.vertices()
pairs = [(v[i], v[j]) for i in range(len(v) - 1) for j in range(i + 1, len(v))]
smallest_paths = []
for (s, e) in pairs:
paths = self.find_all_paths(s, e)
smallest = sorted(paths, key=len)[0]
smallest_paths.append(smallest)
smallest_paths.sort(key=len)
# longest path is at the end of list,
# i.e. diameter corresponds to the length of this path
diameter = len(smallest_paths[-1])
return diameter
def DFSnew(self, s):
graph = self.__graph_dict
stack = []
stack.append(s)
visit[1]=1
co=1
k=[]
ans=[]
while stack:
co+=1
s=stack.pop()
visi2[s]+=1
k.append(s)
visit[s]+=1
for adj in graph[s]:
if visit[adj] == 0:
stack.append(adj)
visit[adj] = 1
ans.append(co-1)
ans+=k
return ans
sol=[]
import copy
graph=dict()
g=Graph(graph)
n=int(input())
a=list(map(int,input().split(" ")))
b=copy.deepcopy(a)
b.sort()
for i in range(n):
ind=b.index(a[i])
if ind!=i:
g.add_vertex(i+1)
g.add_vertex(ind+1)
g.add_edge({i+1,ind+1})
else:
g.add_vertex(i+1)
for iter in graph:
if visi2[iter]==0:
sol.append(g.DFSnew(iter))
for i in sol:
for k in i:
print(k,end=" ")
print("\n")
```
No
| 106,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
Input
The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence.
The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
Output
In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements.
In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to n must appear in output exactly once.
If there are several possible answers, print any of them.
Examples
Input
6
3 2 1 6 5 4
Output
4
2 1 3
1 2
2 4 6
1 5
Input
6
83 -75 -49 11 37 62
Output
1
6 1 2 3 4 5 6
Note
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
Submitted Solution:
```
n = int(input())
a=[0 for j in range (n)]
s = input().split()
for i in range(n):
a[i] = int(s[i])
b = a.copy()
b.sort();
ind_orig = {}
#store index from orig array
for i in range(n):
ind_orig[a[i]] = i
#for actual comparing permutations
ind_perm = [0 for i in range(n)]
for i in range(n):
ind_perm[i] = ind_orig[b[i]]
visited = [False for j in range(n)]
lists = []
#indexes from 0 to n-1; elements 0 to n - 1
def make_perm():
for i in range(n):
if (visited[i]):
continue
t = []
t.append(ind_perm[i])
first = i
k = i;
visited[ind_perm[k]] = True
while (ind_perm[k] != first):
k = ind_perm[k];
visited[ind_perm[k]] = True
t.append(ind_perm[k])
lists.append(t)
make_perm()
#make elements from indeces
def print_res():
for i in range(len(lists)):
print(len(lists[i]),end = " ")
for j in range(len(lists[i])):
print(lists[i][j]+1,end = " ")
print()
print_res()
```
No
| 106,702 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
Input
The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence.
The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
Output
In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements.
In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to n must appear in output exactly once.
If there are several possible answers, print any of them.
Examples
Input
6
3 2 1 6 5 4
Output
4
2 1 3
1 2
2 4 6
1 5
Input
6
83 -75 -49 11 37 62
Output
1
6 1 2 3 4 5 6
Note
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
Submitted Solution:
```
n = int(input())
d = list(map(int, input().split()))
p = [(x, i) for i, x in enumerate(d)]
p.sort()
used = set()
for x, i in p:
temp = []
while x not in used:
used.add(x)
temp.append(i + 1)
x, i = p[i]
if temp:
print(len(temp), end = ' ')
for x in temp: print(x, end = ' ')
print()
```
No
| 106,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
import sys
readline = sys.stdin.readline
from heapq import heappop as hpp, heappush as hp
class MinCostFlowwithDijkstra:
INF = 1<<60
def __init__(self, N):
self.N = N
self.Edge = [[] for _ in range(N)]
def add_edge(self, st, en, cap, cost):
self.Edge[st].append([en, cap, cost, len(self.Edge[en])])
self.Edge[en].append([st, 0, -cost, len(self.Edge[st])-1])
def get_mf(self, so, si, fl):
N = self.N
INF = self.INF
res = 0
Pot = [0]*N
geta = N
prv = [None]*N
prenum = [None]*N
while fl:
dist = [INF]*N
dist[so] = 0
Q = [so]
while Q:
cost, vn = divmod(hpp(Q), geta)
if dist[vn] < cost:
continue
for enum in range(len(self.Edge[vn])):
vf, cap, cost, _ = self.Edge[vn][enum]
cc = dist[vn] + cost - Pot[vn] + Pot[vf]
if cap > 0 and dist[vf] > cc:
dist[vf] = cc
prv[vf] = vn
prenum[vf] = enum
hp(Q, cc*geta + vf)
if dist[si] == INF:
return -1
for i in range(N):
Pot[i] -= dist[i]
cfl = fl
vf = si
while vf != so:
cfl = min(cfl, self.Edge[prv[vf]][prenum[vf]][1])
vf = prv[vf]
fl -= cfl
res -= cfl*Pot[si]
vf = si
while vf != so:
e = self.Edge[prv[vf]][prenum[vf]]
e[1] -= cfl
self.Edge[vf][e[3]][1] += cfl
vf = prv[vf]
return res
N, Q = map(int, readline().split())
T = MinCostFlowwithDijkstra(2*N+2)
geta = N
candi = [set(range(N)) for _ in range(N)]
for _ in range(Q):
t, l, r, v = map(int, readline().split())
l -= 1
r -= 1
v -= 1
if t == 1:
for vn in range(l, r+1):
for i in range(v-1, -1, -1):
if i in candi[vn]:
candi[vn].remove(i)
else:
for vn in range(l, r+1):
for i in range(v+1, N):
if i in candi[vn]:
candi[vn].remove(i)
if not all(candi):
print(-1)
else:
source = 2*N
sink = 2*N + 1
for i in range(N):
T.add_edge(source, i, 1, 0)
for v in candi[i]:
T.add_edge(i, geta+v, 1, 0)
for j in range(N):
T.add_edge(i+geta, sink, 1, 2*j+1)
print(T.get_mf(source, sink, N))
```
| 106,704 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
AUX = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
AUX.append(p[1])
while AUX and min(AUX) < x:
AUX.remove(min(AUX))
for quantity in range(cnt[x]):
if AUX:
AUX.remove(min(AUX))
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n-1,-1,-1):
for y in range(n-1,-1,-1):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
| 106,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
right = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
right.append(p[1])
while right and min(right) < x:
right.remove(min(right))
for quantity in range(cnt[x]):
if right:
right.remove(min(right))
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:0 for x in range(n)}
for y in range(n):
for x in range(n):
if cnt[x] == y:
cnt[x] += 1
if not is_feasible(cnt,L,R):
cnt[x] -= 1
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
| 106,706 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
from collections import deque
from heapq import heappop, heappush
class Edge(object):
__slots__ = ('x', 'y', 'cap', 'cost', 'inv')
def __repr__(self):
return f'{self.x}-->{self.y} ({self.cap} , {self.cost})'
class MCFP():
def __init__(self):
self.G = []
def add(self, x, y, cap, cost):
G = self.G
G.extend(([] for i in range(max(0,max(x,y)+1-len(G)))))
e = Edge()
e.x=x ; e.y=y; e.cap=cap; e.cost=cost
z = Edge()
z.x=y ; z.y=x; z.cap=0; z.cost=-cost
e.inv=z ; z.inv=e
G[x].append(e)
G[y].append(z)
def solve(self, src, tgt, inf=float('inf')):
n, G = len(self.G), self.G
flowVal = flowCost = 0
phi = [0]*n
prev = [None]*n
inQ = [0]*n
cntQ = 1
dist = [inf]*n
while 1:
self.shortest(src, phi, prev, inQ, dist, inf, cntQ)
if prev[tgt] == None:
break
z = inf
x = tgt
while x!=src:
e = prev[x]
z = min(z, e.cap)
x = e.x
x = tgt
while x!=src:
e = prev[x]
e.cap -= z
e.inv.cap += z
x = e.x
flowVal += z
flowCost += z * (dist[tgt] - phi[src] + phi[tgt])
for i in range(n):
if prev[i] != None:
phi[i] += dist[i]
dist[i] = inf
cntQ += 1
prev[tgt] = None
return flowVal, flowCost
def shortest(self, src, phi, prev, inQ, dist, inf, cntQ):
n, G = len(self.G), self.G
Q = [(dist[src],src)]
inQ[src] = cntQ
dist[src] = 0
while Q:
_, x = heappop(Q)
inQ[x] = 0
dx = dist[x]+phi[x]
for e in G[x]:
y = e.y
dy = dx + e.cost - phi[y]
if e.cap > 0 and dy < dist[y]:
dist[y] = dy
prev[y] = e
if inQ[y] != cntQ:
inQ[y] = cntQ
heappush(Q, (dy, y))
return dist, prev
def __repr__(self):
n, G = len(self.G), self.G
s = []
for i in range(n):
s.append(' G[{}]:'.format(i))
s.append('\n'.join(' {}'.format(e) for e in G[i]))
return '\n'.join(s)
import sys
ints = (int(x) for x in sys.stdin.read().split())
sys.setrecursionlimit(3000)
def main():
n, q = (next(ints) for i in range(2))
leq = [n-1]*n
geq = [0]*n
for q in range(q):
t,l,r,v = (next(ints) for i in range(4))
for i in range(l-1, r):
if t==1: geq[i] = max(geq[i], v-1)
if t==2: leq[i] = min(leq[i], v-1)
imp = any(geq[i]>leq[i] for i in range(n))
if imp:
ans = -1
else:
src = 2*n+n*n
tgt = 2*n+n*n+1
G = MCFP()
for i in range(n):
G.add(src, i, 1, 0)
for i in range(n):
for j in range(geq[i], leq[i]+1):
G.add(i, j+n, 1, 0)
for i in range(n):
for j in range(n):
G.add(i+n, 2*n+i*n+j, 1, 0)
G.add(2*n+i*n+j, tgt, 1, 2*j+1)
_, ans = G.solve(src, tgt)
#print(G)
#print(leq, geq)
print(ans)
return
main()
```
| 106,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
AUX = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
AUX.append(p[1])
while AUX and min(AUX) < x:
AUX.remove(min(AUX))
for quantity in range(cnt[x]):
if AUX:
AUX.remove(min(AUX))
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n):
for y in range(n):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
| 106,708 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
#~ # MAGIC CODEFORCES PYTHON FAST IO
import atexit
import io
import sys
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
#~ # END OF MAGIC CODEFORCES PYTHON FAST IO
class Arista():
def __init__(self,salida,llegada,capacidad,flujo,costo,indice):
self.salida = salida
self.llegada = llegada
self.capacidad = capacidad
self.flujo = flujo
self.costo = costo
self.indice = indice
def __str__(self):
s = ""
s = s + "salida =" + str(self.salida) + "\n"
s = s + "llegada =" + str(self.llegada) + "\n"
s = s + "capacidad =" + str(self.capacidad) + "\n"
s = s + "flujo =" + str(self.flujo) + "\n"
s = s + "costo =" + str(self.costo) + "\n"
s = s + "indice =" + str(self.indice) + "\n"
s = s + "------------"
return s
class Red():
## Representacion de una Red de flujo ##
def __init__(self,s,t): # Crea una red vacio
self.lista_aristas = []
self.lista_adyacencia = {}
self.vertices = set()
self.fuente = s
self.sumidero = t
def agregar_vertice(self,vertice):
self.vertices.add(vertice)
def agregar_arista(self,arista):
self.vertices.add(arista.salida)
self.vertices.add(arista.llegada)
self.lista_aristas.append(arista)
if arista.salida not in self.lista_adyacencia:
self.lista_adyacencia[arista.salida] = set()
self.lista_adyacencia[arista.salida].add(arista.indice)
def agregar_lista_aristas(self,lista_aristas):
for arista in lista_aristas:
self.agregar_arista(arista)
def cantidad_de_vertices(self):
return len(self.vertices)
def vecinos(self,vertice):
if vertice not in self.lista_adyacencia:
return set()
else:
return self.lista_adyacencia[vertice]
def buscar_valor_critico(self,padre):
INFINITO = 1000000000
valor_critico = INFINITO
actual = self.sumidero
while actual != self.fuente:
arista_camino = self.lista_aristas[padre[actual]]
valor_critico = min(valor_critico,arista_camino.capacidad - arista_camino.flujo)
actual = arista_camino.salida
return valor_critico
def actualizar_camino(self,padre,valor_critico):
actual = self.sumidero
costo_actual = 0
while actual != self.fuente:
self.lista_aristas[padre[actual]].flujo += valor_critico
self.lista_aristas[padre[actual]^1].flujo -= valor_critico
costo_actual += valor_critico*self.lista_aristas[padre[actual]].costo
actual = self.lista_aristas[padre[actual]].salida
return costo_actual,True
def camino_de_aumento(self):
INFINITO = 1000000000
distancia = {v:INFINITO for v in self.vertices}
padre = {v:-1 for v in self.vertices}
distancia[self.fuente] = 0
#~ for iteracion in range(len(self.vertices)-1):
#~ for arista in self.lista_aristas:
#~ if arista.flujo < arista.capacidad and distancia[arista.salida] + arista.costo < distancia[arista.llegada]:
#~ distancia[arista.llegada] = distancia[arista.salida] + arista.costo
#~ padre[arista.llegada] = arista.indice
capa_actual,capa_nueva = set([self.fuente]),set()
while capa_actual:
for v in capa_actual:
for arista_indice in self.vecinos(v):
arista = self.lista_aristas[arista_indice]
if arista.flujo < arista.capacidad and distancia[arista.salida] + arista.costo < distancia[arista.llegada]:
distancia[arista.llegada] = distancia[arista.salida] + arista.costo
padre[arista.llegada] = arista.indice
capa_nueva.add(arista.llegada)
capa_actual = set()
capa_actual,capa_nueva = capa_nueva,capa_actual
if distancia[self.sumidero] < INFINITO:
valor_critico = self.buscar_valor_critico(padre)
costo_actual,hay_camino = self.actualizar_camino(padre,valor_critico)
return valor_critico,costo_actual,hay_camino
else:
return -1,-1,False
def max_flow_min_cost(self):
flujo_total = 0
costo_total = 0
hay_camino = True
while hay_camino:
#~ for x in self.lista_aristas:
#~ print(x)
flujo_actual,costo_actual,hay_camino = self.camino_de_aumento()
if hay_camino:
flujo_total += flujo_actual
costo_total += costo_actual
return flujo_total,costo_total
INFINITO = 10000000000000
n,q = map(int,input().split())
maxi = [n for i in range(n)]
mini = [1 for i in range(n)]
R = Red(0,2*n+1)
prohibidos = {i:set() for i in range(n)}
for i in range(n):
for k in range(n+1):
R.agregar_arista(Arista(R.fuente,i+1,1,0,2*k+1,len(R.lista_aristas)))
R.agregar_arista(Arista(i+1,R.fuente,0,0,-2*k-1,len(R.lista_aristas)))
for j in range(n):
R.agregar_arista(Arista(n+j+1,R.sumidero,1,0,0,len(R.lista_aristas)))
R.agregar_arista(Arista(R.sumidero,n+j+1,0,0,0,len(R.lista_aristas)))
for z in range(q):
t,l,r,v = map(int,input().split())
if t == 1:
for i in range(v-1):
for j in range(l,r+1):
prohibidos[i].add(j)
else:
for i in range(v,n):
for j in range(l,r+1):
prohibidos[i].add(j)
for i in range(n):
for j in range(mini[i],maxi[i]+1):
if j not in prohibidos[i]:
R.agregar_arista(Arista(i+1,n+j,1,0,0,len(R.lista_aristas)))
R.agregar_arista(Arista(n+j,i+1,0,0,0,len(R.lista_aristas)))
flujo_total,costo_total = R.max_flow_min_cost()
#~ print(flujo_total,costo_total)
if flujo_total < n:
print("-1")
else:
print(costo_total)
```
| 106,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
AUX = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
AUX.append(p[1])
while AUX and min(AUX) < x:
AUX.remove(min(AUX))
for quantity in range(cnt[x]):
if AUX:
AUX.remove(min(AUX))
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n):
for y in range(n):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
if had_pair:
break
if had_pair:
break
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
| 106,710 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Tags: flows
Correct Solution:
```
import sys
from random import seed,shuffle
seed(25)
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
AUX = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
AUX.append(p[1])
while AUX and min(AUX) < x:
AUX.remove(min(AUX))
for quantity in range(cnt[x]):
if AUX:
AUX.remove(min(AUX))
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
AUX = list(range(n))
shuffle(AUX)
for x in AUX:
AUX_2 = list(range(n))
shuffle(AUX_2)
for y in AUX_2:
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
| 106,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Submitted Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
right = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
right.append(p[1])
while right and min(right) < x:
right.remove(min(right))
for quantity in range(cnt[x]):
if right:
right.remove(min(right))
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n):
for y in range(n):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
Yes
| 106,712 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Submitted Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = sorted([(L[i],R[i]) for i in range(n)])
feasible = True
for x in range(n):
for quantity in range(cnt[x]):
AUX = [p for p in inter if p[0] <= x <= p[1]]
if AUX:
best = AUX[0]
for w in AUX:
if w[0] > best[0]:
best = w
elif w[0] == best[0] and w[1] < best[1]:
best = w
inter.remove(w)
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n):
for y in range(n):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
No
| 106,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Submitted Solution:
```
import sys
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
A = L[:]
cnt = {x:A.count(x) for x in range(n)}
has_pair = True
while has_pair:
has_pair = False
for k in range(n):
for x in range(n):
if A[k] != x and cnt[A[k]] > cnt[x] + 1 and L[k] <= x and x <= R[k]:
cnt[A[k]] -= 1
A[k] = x
cnt[A[k]] += 1
has_pair = True
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
No
| 106,714 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Submitted Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = sorted([(L[i],R[i]) for i in range(n)])
feasible = True
for x in range(n):
for quantity in range(cnt[x]):
AUX = [p for p in inter if p[0] <= x <= p[1]]
if AUX:
best = AUX[0]
for w in AUX:
if w[1] < best[1]:
best = w
inter.remove(w)
else:
feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n):
for y in range(n):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
No
| 106,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Ivan noticed an array a while debugging his code. Now Ivan can't remember this array, but the bug he was trying to fix didn't go away, so Ivan thinks that the data from this array might help him to reproduce the bug.
Ivan clearly remembers that there were n elements in the array, and each element was not less than 1 and not greater than n. Also he remembers q facts about the array. There are two types of facts that Ivan remembers:
* 1 li ri vi — for each x such that li ≤ x ≤ ri ax ≥ vi;
* 2 li ri vi — for each x such that li ≤ x ≤ ri ax ≤ vi.
Also Ivan thinks that this array was a permutation, but he is not so sure about it. He wants to restore some array that corresponds to the q facts that he remembers and is very similar to permutation. Formally, Ivan has denoted the cost of array as follows:
<image>, where cnt(i) is the number of occurences of i in the array.
Help Ivan to determine minimum possible cost of the array that corresponds to the facts!
Input
The first line contains two integer numbers n and q (1 ≤ n ≤ 50, 0 ≤ q ≤ 100).
Then q lines follow, each representing a fact about the array. i-th line contains the numbers ti, li, ri and vi for i-th fact (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n, 1 ≤ vi ≤ n, ti denotes the type of the fact).
Output
If the facts are controversial and there is no array that corresponds to them, print -1. Otherwise, print minimum possible cost of the array.
Examples
Input
3 0
Output
3
Input
3 1
1 1 3 2
Output
5
Input
3 2
1 1 3 2
2 1 3 2
Output
9
Input
3 2
1 1 3 2
2 1 3 1
Output
-1
Submitted Solution:
```
import sys
def is_feasible(cnt,L,R):
n = len(R)
inter = [(L[i],R[i]) for i in range(n)]
AUX = []
feasible = True
for x in range(n):
for p in inter:
if p[0] == x:
AUX.append(p[1])
# ~ while AUX and min(AUX) < i:
# ~ AUX.remove(min(AUX))
# ~ for quantity in range(cnt[x]):
# ~ if AUX:
# ~ AUX.remove(min(AUX))
# ~ else:
# ~ feasible = False
return feasible
n,q = map(int,sys.stdin.readline().split())
L = [0 for i in range(n)]
R = [n-1 for i in range(n)]
for restriction in range(q):
t,l,r,v = map(int,sys.stdin.readline().split())
if t == 1:
for k in range(l-1,r):
L[k] = max(L[k],v-1)
else:
for k in range(l-1,r):
R[k] = min(R[k],v-1)
is_possible = all(map(lambda x,y : x <= y,L,R))
if not is_possible:
print(-1)
else:
cnt = {x:L.count(x) for x in range(n)}
had_pair = True
while had_pair:
had_pair = False
for x in range(n):
for y in range(n):
if cnt[x] > cnt[y]+1:
cnt[x] -= 1
cnt[y] += 1
has_pair = True
if not is_feasible(cnt,L,R):
cnt[x] += 1
cnt[y] -= 1
has_pair = False
had_pair = had_pair or has_pair
ans = sum([cnt[x]*cnt[x] for x in range(n)])
print(ans)
```
No
| 106,716 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Petya was solving a very interesting problem. But although he used many optimization techniques, his solution still got Time limit exceeded verdict. Petya conducted a thorough analysis of his program and found out that his function for finding maximum element in an array of n positive integers was too slow. Desperate, Petya decided to use a somewhat unexpected optimization using parameter k, so now his function contains the following code:
int fast_max(int n, int a[]) {
int ans = 0;
int offset = 0;
for (int i = 0; i < n; ++i)
if (ans < a[i]) {
ans = a[i];
offset = 0;
} else {
offset = offset + 1;
if (offset == k)
return ans;
}
return ans;
}
That way the function iteratively checks array elements, storing the intermediate maximum, and if after k consecutive iterations that maximum has not changed, it is returned as the answer.
Now Petya is interested in fault rate of his function. He asked you to find the number of permutations of integers from 1 to n such that the return value of his function on those permutations is not equal to n. Since this number could be very big, output the answer modulo 109 + 7.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 106), separated by a space — the length of the permutations and the parameter k.
Output
Output the answer to the problem modulo 109 + 7.
Examples
Input
5 2
Output
22
Input
5 3
Output
6
Input
6 3
Output
84
Note
Permutations from second example:
[4, 1, 2, 3, 5], [4, 1, 3, 2, 5], [4, 2, 1, 3, 5], [4, 2, 3, 1, 5], [4, 3, 1, 2, 5], [4, 3, 2, 1, 5].
Submitted Solution:
```
from math import factorial
n, k = map(int, input().split())
const1 = factorial(n - 2) % (10 ** 9 + 7)
const2 = ((n - k - 1) * (n - k) // 2) % (10 ** 9 + 7)
const3 = ((const1 // (n - 2)) * (n - k - 2) * 2) % (10 ** 9 + 7)
if const3 > 0:
ans = (const1 * const2 + const3) % (10 ** 9 + 7)
else:
ans = (const1 * const2) % (10 ** 9 + 7)
print(ans)
```
No
| 106,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Petya was solving a very interesting problem. But although he used many optimization techniques, his solution still got Time limit exceeded verdict. Petya conducted a thorough analysis of his program and found out that his function for finding maximum element in an array of n positive integers was too slow. Desperate, Petya decided to use a somewhat unexpected optimization using parameter k, so now his function contains the following code:
int fast_max(int n, int a[]) {
int ans = 0;
int offset = 0;
for (int i = 0; i < n; ++i)
if (ans < a[i]) {
ans = a[i];
offset = 0;
} else {
offset = offset + 1;
if (offset == k)
return ans;
}
return ans;
}
That way the function iteratively checks array elements, storing the intermediate maximum, and if after k consecutive iterations that maximum has not changed, it is returned as the answer.
Now Petya is interested in fault rate of his function. He asked you to find the number of permutations of integers from 1 to n such that the return value of his function on those permutations is not equal to n. Since this number could be very big, output the answer modulo 109 + 7.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 106), separated by a space — the length of the permutations and the parameter k.
Output
Output the answer to the problem modulo 109 + 7.
Examples
Input
5 2
Output
22
Input
5 3
Output
6
Input
6 3
Output
84
Note
Permutations from second example:
[4, 1, 2, 3, 5], [4, 1, 3, 2, 5], [4, 2, 1, 3, 5], [4, 2, 3, 1, 5], [4, 3, 1, 2, 5], [4, 3, 2, 1, 5].
Submitted Solution:
```
inp = input()
parts = inp.split()
n = int(parts[0])
k = int(parts[1])
mod = 1000000007
fact = [None] * 1000001
fact[0] = 1
fact[1] = 1
for i in range(2, 1000001):
fact[i] = (fact[i-1] * i) % mod
# n - �������� ��������
# m - �������� ��������, ������� ������ ��������� (�� ���� �� ������ ���� ������ �������� ����������)
# ����� ��������� ��������� ������ ���� ������� k ��������� < m
seen = [0] * n
for m in range(k + 1, n): #�������� �������� m
add_seen = [0] * n #������� � seen �� �������� m
for pos_m in range (0, m - k): #�������� ������� m
# �� ���� �������� �� pos_m ��� �������� < m
# � �� k �������� ����� pos_m ��� �������� < m
# ������� nperm - ����� ������������ ������� (pos_m - 1 + k) �� ��������� m-1 ���������
nperm = fact[m-1] / fact[m - 1 - pos_m - k]
# ����� ������������ �� ���������� ��������
nremain = fact[n - pos_m - k - 1]
#�������� ������������, ������� ��� ���� ��������� �� ������� ������� ��������� ��������� p < m �� �������� [0, pos_m - k)
sumseen = 0
if pos_m > k:
sumseen = sum(seen[0:pos_m-k])
add_seen[pos_m] = (((nperm * nremain) % mod) - sumseen) % mod
for i in range(0, n):
seen[i] += add_seen[i]
print (sum(seen))
```
No
| 106,718 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Petya was solving a very interesting problem. But although he used many optimization techniques, his solution still got Time limit exceeded verdict. Petya conducted a thorough analysis of his program and found out that his function for finding maximum element in an array of n positive integers was too slow. Desperate, Petya decided to use a somewhat unexpected optimization using parameter k, so now his function contains the following code:
int fast_max(int n, int a[]) {
int ans = 0;
int offset = 0;
for (int i = 0; i < n; ++i)
if (ans < a[i]) {
ans = a[i];
offset = 0;
} else {
offset = offset + 1;
if (offset == k)
return ans;
}
return ans;
}
That way the function iteratively checks array elements, storing the intermediate maximum, and if after k consecutive iterations that maximum has not changed, it is returned as the answer.
Now Petya is interested in fault rate of his function. He asked you to find the number of permutations of integers from 1 to n such that the return value of his function on those permutations is not equal to n. Since this number could be very big, output the answer modulo 109 + 7.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 106), separated by a space — the length of the permutations and the parameter k.
Output
Output the answer to the problem modulo 109 + 7.
Examples
Input
5 2
Output
22
Input
5 3
Output
6
Input
6 3
Output
84
Note
Permutations from second example:
[4, 1, 2, 3, 5], [4, 1, 3, 2, 5], [4, 2, 1, 3, 5], [4, 2, 3, 1, 5], [4, 3, 1, 2, 5], [4, 3, 2, 1, 5].
Submitted Solution:
```
x=[]
s=0
m=0
z=int(0)
for i in input().split():
x.append(i)
s= s+ int(x[z])
z=z+1
if (s%2)==0:
for i in range (6):
for j in range (6):
for k in range (6):
m=m+1
if (i!=k!=j):
if (int(x[i])+int(x[j])+int(x[k]))==(s/2):
m=0
if m== 216:
print('NO')
else:
print('YES')
```
No
| 106,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Petya was solving a very interesting problem. But although he used many optimization techniques, his solution still got Time limit exceeded verdict. Petya conducted a thorough analysis of his program and found out that his function for finding maximum element in an array of n positive integers was too slow. Desperate, Petya decided to use a somewhat unexpected optimization using parameter k, so now his function contains the following code:
int fast_max(int n, int a[]) {
int ans = 0;
int offset = 0;
for (int i = 0; i < n; ++i)
if (ans < a[i]) {
ans = a[i];
offset = 0;
} else {
offset = offset + 1;
if (offset == k)
return ans;
}
return ans;
}
That way the function iteratively checks array elements, storing the intermediate maximum, and if after k consecutive iterations that maximum has not changed, it is returned as the answer.
Now Petya is interested in fault rate of his function. He asked you to find the number of permutations of integers from 1 to n such that the return value of his function on those permutations is not equal to n. Since this number could be very big, output the answer modulo 109 + 7.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 106), separated by a space — the length of the permutations and the parameter k.
Output
Output the answer to the problem modulo 109 + 7.
Examples
Input
5 2
Output
22
Input
5 3
Output
6
Input
6 3
Output
84
Note
Permutations from second example:
[4, 1, 2, 3, 5], [4, 1, 3, 2, 5], [4, 2, 1, 3, 5], [4, 2, 3, 1, 5], [4, 3, 1, 2, 5], [4, 3, 2, 1, 5].
Submitted Solution:
```
from math import factorial
n,k=map(int,input().split(' '))
h=n
s=0
if (n-k)==1:
print(1)
else:
while h-k>1:
a=(h-k)*(h-k-1)/2
b=factorial(h-2)*(n-h+1)
s+=a*b
h-=1
print(int(s))
```
No
| 106,720 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
Y,B = [int(i) for i in input().split()]
y,g,b = [int(i) for i in input().split()]
req_y = 2 * y + g
req_b = g + 3*b
req =0
if req_b > B:
req += req_b - B
if req_y > Y:
req += req_y - Y
print(req)
```
| 106,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
y,b=input().split(' ')
ay,ag,ab=input().split(' ')
y=int(y)
b=int(b)
ay=int(ay)
ag=int(ag)
ab=int(ab)
b=b-(ab*3)
y=y-(ay*2)
b=b-ag
y=y-ag
br=0
if(y<0):
br=br-y
if(b<0):
br=br-b
print(br)
```
| 106,722 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
a, b = map(int, input().split())
x, y, z = map(int, input().split())
kraj = 0
if y <= a:
a -= y
else:
kraj += y - a
a = 0
if y <= b:
b -= y
else:
kraj += y - b
b = 0
if x * 2 > a:
kraj += x * 2 - a
if z * 3 > b:
kraj += z * 3 - b
print(kraj)
```
| 106,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
y,b=map(int,input().split())
Y,G,B=map(int,input().split())
p=(Y*2)+G
q=G+(B*3)
if y>=p and b>=q:
print("0")
elif y>=p and b<q:
print(q-b)
elif y<p and b>=q:
print(p-y)
else:
print((p-y)+(q-b))
```
| 106,724 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
s=input().split();a,b=int(s[0]),int(s[1])
s=input().split();x,y,z=int(s[0]),int(s[1]),int(s[2])
n=2*x+y;m=3*z+y
n=n-a;m=m-b
if n<0:
n=0
if m<0:
m=0
print(n+m)
```
| 106,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
a,b=input().split()
a,b=int(a),int(b)
x,y,z=input().split()
x,y,z=int(x),int(y),int(z)
ans=max(0,2*x+y-a)+max(0,3*z+y-b)
print(ans)
```
| 106,726 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
a, b = [int(i) for i in input().split()]
x, y, z = [int(i) for i in input().split()]
na = 2 * x + y
nb = y + 3 * z
print((na - a if na > a else 0) + (nb - b if nb > b else 0))
```
| 106,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Tags: implementation
Correct Solution:
```
# import sys
# sys.stdin=open("input.in","r")
# sys.stdout=open("ot.out","w")
A,B=map(int,input().split())
x,y,z=map(int,input().split())
summ=0
y1=0
b1=0
y1=(x*2)+y
b1=y+(3*z)
summ=0
if y1>A:
summ+=y1-A
if b1>B:
summ+=b1-B
print(summ)
```
| 106,728 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
from array import*
n=input().split()
x=input().split()
y=int(n[0])-int(x[0])*2
y=int(y)
y-=int(x[1])
m=int(n[1])-int(x[1])
m=int(m)
m-=int(x[2])*3
if m<0 and y>=0:
print(abs(m))
elif y<0 and m>=0:
print(abs(y))
elif m<0 and y<0:
print(abs(m+y))
else:
print(0)
```
Yes
| 106,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
yellow, blu = list(map(int, input().split()))
x, y, z = list(map(int, input().split()))
# y g b
blue_needed = z * 3 + y
yellow_needed = x * 2 + y
a = blue_needed - blu
c = yellow_needed - yellow
if a < 0:
a = 0
if c < 0:
c = 0
print(a + c)
```
Yes
| 106,730 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
yellowShards,blueShards = [int(x) for x in input().split()]
yBalls,gBalls,bBalls = [int(x) for x in input().split()]
yellowNeeded = yBalls*2 + gBalls
blueNeeded = gBalls + 3*bBalls
ans =0
if(yellowShards<yellowNeeded):
ans = ans + (yellowNeeded-yellowShards)
if(blueShards<blueNeeded):
ans = ans + (blueNeeded-blueShards)
print(ans)
```
Yes
| 106,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
def problemA():
yellow, blue = list(map(int, input().strip().split()))
x, y , z = list(map(int, input().strip().split()))
ny = max(0, 2*x+y-yellow)
nb = max(0, 3*z+y-blue)
print(nb+ny)
if __name__ == '__main__':
problemA()
```
Yes
| 106,732 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
a, b = map(int, input().split())
x, y, z = map(int, input().split())
x = x * 2
z = z * 3
a = a - x - y
b = b - z - y
if a < 0 and b > 0:
print(abs(a))
elif a > 0 and b < 0:
print(abs(b))
elif a < 0 and b < 0:
print(abs(a+b))
else:
print(0)
```
No
| 106,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
A, B = map(int, input().split())
x, y, z = map(int, input().split())
A -= (x * 2)
B -= (z * 3)
print(-(A + B - y * 2) if (A + B - y * 2) < 0 else 0)
```
No
| 106,734 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
A, B = input().split()
A = int(A)
B = int(B)
x, y, z = input().split()
x = int(x)
y = int(y)
z = int(z)
reqYellowCrystal = 2*x + y
reqBlueCrystal = y + 3*z
output = (reqYellowCrystal-A) + (reqBlueCrystal-B)
print (output)
```
No
| 106,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are A yellow and B blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input
The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.
Output
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Examples
Input
4 3
2 1 1
Output
2
Input
3 9
1 1 3
Output
1
Input
12345678 87654321
43043751 1000000000 53798715
Output
2147483648
Note
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
Submitted Solution:
```
a, b = [int(s) for s in input().split(' ')]
x, y, z = [int(s) for s in input().split(' ')]
ans = max(a - 2 * x + y, 0) + max(b - 3 * z + y, 0)
print(ans)
```
No
| 106,736 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
res=[[max(a1*b1 for b1 in b)]for a1 in a]
res.sort()
print(res[-2][0])
```
| 106,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
print(min([max([max([a[j]*k for k in b]) for j in range(n) if j!=i]) for i in range(n)]))
```
| 106,738 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
m=[int(i) for i in input().split()]
a= [int(i) for i in input().split()]
b= [int(i) for i in input().split()]
r =[[max(ax*bx for bx in b)] for ax in a]
r.sort()
print (int(r[-2][0]))
```
| 106,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
def main():
n, m = [int(x) for x in input().split()]
a = sorted([int(x) for x in input().split()])
b = sorted([int(x) for x in input().split()])
vv = max(a[0] * b[0], a[-1]* b[0], a[0]*b[-1],a[-1]*b[-1])
if vv == a[0] * b[0] or vv == a[0]*b[-1]:
a.pop(0)
else:
a.pop(-1)
vv = max(a[0] * b[0], a[-1]* b[0], a[0]*b[-1],a[-1]*b[-1])
print(vv)
if __name__ == '__main__':
main()
```
| 106,740 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import array
from bisect import *
from collections import *
import fractions
import heapq
from itertools import *
import math
import random
import re
import string
N, M = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
A.sort()
B.sort()
ans = 1e20
for i in range(N):
A2 = A[:i] + A[i+1:]
local_ans = -1e20
for a in A2:
for b in B:
local_ans = max(local_ans, a * b)
# print(A2, B)
# print("local_ans:", local_ans)
ans = min(ans, local_ans)
print(ans)
```
| 106,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
def main():
n, m = list(map(int, input().split()))
t = list(map(int, input().split()))
b = list(map(int, input().split()))
maxs = []
for ex in range(len(t)):
maxs.append(max(x * y for i, x in enumerate(t) for y in b if i != ex))
print(min(maxs))
if __name__ == '__main__':
main()
```
| 106,742 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
n,m = map(int,input().split())
l1 = [int(x) for x in input().split()]
l2 = [int(x) for x in input().split()]
r = [[max(a*b for b in l2)] for a in l1]
r.sort()
print(r[-2][0])
```
| 106,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Tags: brute force, games
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
def main():
n,m=map(int,input().split(" "))
a=list(map(int,input().split(" ")))
b=list(map(int,input().split(" ")))
temp=[]
mxhidden=-(1e100)
hid=0
for y in range(m):
for z in range(n):
if b[y]*a[z]>mxhidden:
mxhidden=b[y]*a[z]
hid=z
#print(mxhidden,y)
a.pop(hid)
ans=(-1e100)
for x in range(n-1):
for y in range(m):
ans=max(ans,a[x]*b[y])
print(ans)
#-----------------------------BOSS-------------------------------------!
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
| 106,744 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
n,m=map(int, input().split())
a=list(map(int, input().split()))
b=list(map(int, input().split()))
ans=float('inf')
for i in range(n):
now=-float('inf')
for j in range(n):
if j!=i:
for k in range(m):
now=max(now,a[j]*b[k])
ans=min(ans,now)
print(ans)
```
Yes
| 106,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
n,m = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
ans1 = float('inf')
for i in range(n):
ans2 = -float('inf')
for j in range(n):
if j == i:
continue
for k in range(m):
ans2 = max(ans2, A[j]*B[k])
ans1 = min(ans1, ans2)
print(ans1)
```
Yes
| 106,746 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
n, m = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
b.sort()
first = []
for x in a:
if x > 0:
first.append(x*b[-1])
elif x<0:
first.append(x*b[0])
else:
first.append(0)
ans = 10**25
for i in range(len(a)):
s = -10**25
for j in range(len(a)):
if j!=i:
if first[j]>s: s=first[j]
if s < ans: ans=s
print(ans)
```
Yes
| 106,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
n, m = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
res = [max([ax*bx for bx in b]) for ax in a]
res.sort(reverse=True)
print(res[1])
```
Yes
| 106,748 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
n, k = map(int, input().split())
ar = [int(p) for p in input().split()]
arr = [int(p) for p in input().split()]
ar = list(set(ar))
m = max(ar)
if m < 0:
m = min(ar)
ar.pop(ar.index(m))
maxim = -10000000000
for i in range(len(ar)):
for j in range(k):
tmp = ar[i]*arr[j]
if tmp > maxim:
maxim = tmp
print(maxim)
```
No
| 106,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
nm=input().split()
n=input().split()
m=input().split()
MAXn=-1000000000
MAXm=-1000000000
for i in n:
i=int(i)
if i>=MAXn:
MAXn=i
for i in m:
i=int(i)
if i>=MAXm:
MAXm=i
for i in n:
i=int(i)
if MAXn==i:
n.remove(str(i))
MAXn=-100000000
for i in n:
i=int(i)
if i>=MAXn:
MAXn=i
print(MAXm*MAXn)
```
No
| 106,750 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
n, m=map(int, input().split())
a=sorted(list(map(int, input().split())))
b=sorted(list(map(int, input().split())))
if a[0]*b[0] > a[-1]*b[-1]:
a = a[1:]
else:
a = a[:-1]
print(max(a[-1]*b[-1], a[0]*b[0]))
```
No
| 106,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input
2 2
20 18
2 14
Output
252
Input
5 3
-1 0 1 2 3
-1 0 1
Output
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
Submitted Solution:
```
s = input().split()
s1 = input().split()
s2 = input().split()
n = int(s[0])
m = int(s[1])
list2 = []
list1 = []
for i in s1:
list1.append(int(i))
for i in s2:
list2.append(int(i))
list1.sort()
list2.sort()
cl = []
for i in list1:
for j in list2:
cl.append(i*j)
cl.sort()
print(cl[n*m-2])
```
No
| 106,752 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
x,d=[int(s) for s in input().split()[:2]]
def getd(y):
if y<=1:
return y
da=1
num=0
while da*2<=y:
da*=2
num+=1
return num
f=1
ans=[]
while x>0:
num=getd(x)
n=2**num-1
for i in range(num):
ans.append(f)
f+=d+1
x-=n
print(len(ans))
for i in ans[:-1]:
print(i,end=' ')
print(ans[-1])
```
| 106,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
n, d = [int(c) for c in input().split(" ")]
n_copy = n
groupsize_cases = [2**c - 1 for c in range(100)]
groups = []
# print(groupsize_cases[:20])
i = 1
while n >= groupsize_cases[i]:
i += 1
# print(groupsize_cases[i])
i -= 1
while n > 0:
while n >= groupsize_cases[i]:
groups.append(i)
n -= groupsize_cases[i]
i -= 1
# print(groups)
# print(groupsize_cases[max(groups)])
ans = []
current_number = 1
for num in groups:
ans.append(current_number)
for i in range(num-1):
# current_number += 1
ans.append(current_number)
current_number += d+1
print(len(ans))
for num in ans:
print(num, end = " ")
```
| 106,754 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
X, D = map(int, input().split())
cn = 1
add0 = 1 if (X&1) else 0
ans = []
for i in range(30,0,-1):
if not (X & (1<<i)): continue
ans += [cn]*i
add0 += 1
cn += D
for i in range(add0):
ans.append(cn)
cn += D
print(len(ans))
print(' '.join(map(str, ans)))
```
| 106,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
from math import log2, floor
# ------------------------------
def main():
x, d = RL()
res = []
now = 1
while x>0:
num = floor(log2(x+1))
res+=[now]*num
x-=2**num-1
now+=d+1
print(len(res))
print(*res)
if __name__ == "__main__":
main()
```
| 106,756 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
# -*- coding: UTF-8 -*-
s = [2 ** i - 1 for i in range(31, 0, -1)]
z = [i for i in range(31, 0, -1)]
x, d = map(int, input().split())
st = 1
p = 0
ret = []
while x > 0:
while s[p] > x:
p += 1
for i in range(z[p]):
ret.append(st)
st += d
x -= s[p]
if x != 0:
print('-1')
else:
print(len(ret))
print(' '.join(str(i) for i in ret))
```
| 106,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
x, k = list(map(int,input().split()))
tot = 0
now = 1
d = 1
ans = []
def add(am,el):
global ans
ans += [el]*am
def rec(l):
global d
global tot
now = 1
while True:
if tot + (2**now - 1) > x:
if (tot-len(ans) ) != 0:
rec(x-len(ans))
break
tot+=(2**now - 1)
add(now,d)
now+=1
d+=k
if tot == x:
break
rec(x)
print(len(ans))
print(*ans)
```
| 106,758 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
#!/usr/bin/env python3
import sys
[X, d] = map(int, sys.stdin.readline().strip().split())
a = []
m = 1 - d
i = 31
while i > 0:
if 2**i - 1 > X:
i -= 1
continue
a.extend([m + d] * i)
m += d
X -= 2**i - 1
print (len(a))
print (' '.join(map(str, a)))
```
| 106,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Tags: bitmasks, constructive algorithms, greedy, implementation
Correct Solution:
```
x, d = [int(i) for i in input().split()]
cr = 1
ans = []
while x:
i = 0
while ((1 << i) - 1) <= x:
i += 1
for j in range(i - 1):
ans.append(cr)
x -= (1 << (i - 1)) - 1
cr += d
print(len(ans))
print(*ans)
```
| 106,760 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
from math import log2, floor, ceil
x, d = input().split()
x, d = int(x), int(d)
k = 1
arr = list()
c = 1
ans = True
while (x > 0):
p = floor(log2(x))
if (2**(p+1) - 1) <= x:
p += 1
if (p == 0):
ans = False
break
arr.extend([k]*p)
k += d+1
x -= (2**p - 1)
if (ans) and (len(arr) <= 10000):
print(len(arr))
print(' '.join(list(map(str, arr))))
else:
print(-1)
```
Yes
| 106,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
X, d = map(int, input().split())
a = list()
cur = 1
for i in range(32):
if X&(1<<i):
for j in range(i):
a.append(cur)
cur = cur+d
a.append(cur)
cur = cur+d
print(len(a))
for i in a:
print(i, end=" ")
```
Yes
| 106,762 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
p2 = [(i, 2**i-1) for i in range(1,40)]
# print(p2)
X, d = map(int, input().split(' '))
lens = []
for i in p2[::-1]:
if X >= i[1]:
X -= i[1]
lens.append(i[0])
# print(lens)
# print(X)
# if X > 0:
# print(-1)
# exit(0)
el = 1
# if el + len(lens)*d > 10**18:
# print(-1)
# exit(0)
if sum(lens) >= 10000:
print(-1)
exit(0)
print(sum(lens)+X)
for i in lens:
print((str(el)+' ')*i, end='')
el += d
for i in range(X):
print(el, end=' ')
el += d
# print(lens)
```
Yes
| 106,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
x,d=map(int,input().split())
c=1
r="\n"
e=29
b=2**e-1
while x:
while x<b:e-=1;b>>=1
r+=f"{c*d} "*e;c+=e;x-=b
print(c-1,r)
```
Yes
| 106,764 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
import math
x,d=map(int,input().split())
n=math.floor(math.log(x+1,2))
if (n-1>=d and x>10000):
print(-1)
elif (x<=10000 or n-1>=d):
print(x)
for i in range(1,x*d+2-d,d):
print(i,end=' ')
print()
else:
tb=2**n-1
diff=x-tb
g=1
arr=[]
c=tb
while diff>0:
for i in range(g,g+n):
arr.append(i)
g=g+n+d-1
n=math.floor(math.log(diff+1,2))
tb=2**n-1
diff=diff-tb
c+=tb
print(c)
print(*arr)
```
No
| 106,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
x,d = input().split()
x,d = [int(x),int(d)]
from random import randint
print(2*x)
for i in range(x) :
num = randint(1,100000)
print(num,num-d+1,end = ' ')
```
No
| 106,766 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
n,d=map(int,input().split())
if n==10 and d==5:
n-=1
k=1
s=''
c=0
for i in range(n):
s+=str(k)+' '
k+=d+1
if s=='':
print(-1)
else:
print(n)
print(s[:len(s)-1])
```
No
| 106,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pikachu had an array with him. He wrote down all the non-empty subsequences of the array on paper. Note that an array of size n has 2n - 1 non-empty subsequences in it.
Pikachu being mischievous as he always is, removed all the subsequences in which Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ d
Pikachu was finally left with X subsequences.
However, he lost the initial array he had, and now is in serious trouble. He still remembers the numbers X and d. He now wants you to construct any such array which will satisfy the above conditions. All the numbers in the final array should be positive integers less than 1018.
Note the number of elements in the output array should not be more than 104. If no answer is possible, print - 1.
Input
The only line of input consists of two space separated integers X and d (1 ≤ X, d ≤ 109).
Output
Output should consist of two lines.
First line should contain a single integer n (1 ≤ n ≤ 10 000)— the number of integers in the final array.
Second line should consist of n space separated integers — a1, a2, ... , an (1 ≤ ai < 1018).
If there is no answer, print a single integer -1. If there are multiple answers, print any of them.
Examples
Input
10 5
Output
6
5 50 7 15 6 100
Input
4 2
Output
4
10 100 1000 10000
Note
In the output of the first example case, the remaining subsequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 5 are [5], [5, 7], [5, 6], [5, 7, 6], [50], [7], [7, 6], [15], [6], [100]. There are 10 of them. Hence, the array [5, 50, 7, 15, 6, 100] is valid.
Similarly, in the output of the second example case, the remaining sub-sequences after removing those with Maximum_element_of_the_subsequence - Minimum_element_of_subsequence ≥ 2 are [10], [100], [1000], [10000]. There are 4 of them. Hence, the array [10, 100, 1000, 10000] is valid.
Submitted Solution:
```
x,d=map(int,input().split())
for i in range (0,x):
print((i+1)*d,end=" ")
```
No
| 106,768 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
def c(i,j,k):print('<',i,j,k)
def a(i,j,k):print('+',i,j,k)
def d(x,i,j):
for t in range(30):
a(j+t-1,j+t-1,j+t)
a(j+t,n,j+t)
for s in range(j+t,j+29):a(s,s,s+1)
c(j+29,x,i+t)
a(j+t-1,j+t-1,j+t)
a(j+t,i+t,j+t)
print(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in range(59):
a(2,2,2)
for s in range(t+1):
if t-30<s<30:a(5+t-s,36+s,98);c(n,98,99);a(2,99,2)
```
| 106,769 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
p=print
r=range
def c(i,j,k):p('<',i,j,k)
def a(i,j,k=0):p('+',i,j,max(i,k))
def b(i):a(i,i,i+1)
def d(x,i,j):
for t in r(30):
b(j+t-1);a(j+t,n)
for s in r(j+t,j+29):b(s)
c(j+29,x,i+t);b(j+t-1);a(j+t,i+t)
p(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in r(59):
a(2,2)
for s in r(t+1):
if t-30<s<30:a(5+t-s,36+s,98);c(n,98,99);a(2,99)
```
| 106,770 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
def c(i,j,k):print('<',i,j,k)
def a(i,j,k):print('+',i,j,k)
def m(i,j,k):
for t in range(59):
a(k,k,k)
for s in range(t+1):
if t-30<s<30:a(i+t-s,j+s,98);c(n,98,99);a(k,99,k)
def d(x,i,j):
for t in range(30):
a(j+t-1,j+t-1,j+t)
a(j+t,n,j+t)
for s in range(j+t,j+29):a(s,s,s+1)
c(j+29,x,i+t)
a(j+t-1,j+t-1,j+t)
a(j+t,i+t,j+t)
print(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
m(5,36,2)
```
| 106,771 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
p=print
r=range
def c(i,j,k):p('<',i,j,k)
def a(i,j,k=0):p('+',i,j,max(i,k))
def b(i):a(i,i,i+1)
def d(x,i,j):
for t in r(30):
b(j+t-1)
a(j+t,n)
for s in r(j+t,j+29):b(s)
c(j+29,x,i+t)
b(j+t-1)
a(j+t,i+t)
p(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in r(59):
a(2,2)
for s in r(t+1):
if t-30<s<30:a(5+t-s,36+s,98);c(n,98,99);a(2,99)
```
| 106,772 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n = q = 2*10**5
fla = []
def pri(s):
fla.append(s)
pri("+ 0 1 5")
pri("< 3 5 4")
max_log = 30
pri("+ 0 4 0")
pri("+ 1 4 1")
for i in reversed(range(max_log)):
pri("+ 3 4 8")
for _ in range(i):
pri("+ 8 8 8")
pri("+ 6 8 9")
pri("< 9 0 %d" % (100+i))
pri("+ %d 3 9" % (100+i))
for _ in range(i):
pri("+ 9 9 9")
pri("+ 6 9 6")
for i in reversed(range(max_log)):
pri("+ 3 4 8")
for _ in range(i):
pri("+ 8 8 8")
pri("+ 7 8 9")
pri("< 9 1 %d" % (200+i))
pri("+ %d 3 9" % (200+i))
for _ in range(i):
pri("+ 9 9 9")
pri("+ 7 9 7")
def an(i, j, k):
pri("+ %d %d 6" % (i, j))
pri("< %d %d 7" % (i, j))
pri("< %d %d 8" % (j, i))
pri("+ 7 8 7")
pri("< 7 6 %d" % k)
for i in range(max_log):
for j in range(max_log):
an(100+i, 200+j, 6)
pri("+ %d 6 %d" % (i+j+300, i+j+300))
for i in range(max_log*2):
for _ in range(i):
pri("+ %d %d %d" % (300+i, 300+i, 300+i))
pri("+ %d 2 2" % (300+i))
print(len(fla))
print("\n".join(fla))
```
| 106,773 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
r=range
L='<'
P='+'
l=[3933,P,0,1,n,L,2,n,n,P,0,n,3,P,1,n,4]
a=lambda i,j:[P,i,j,i]
b=lambda i:[P,i,i,i+1]
def d(x,i,j):
return sum([sum([b(j+t-1),a(j+t,n),*(b(s)for s in r(j+t,j+29)),[L,j+29,x,i+t],b(j+t-1),a(j+t,i+t)],[])for t in r(30)],[])
l+=d(3,5,36)
l+=d(4,36,67)
for t in r(59):
l+=a(2,2)
for s in r(t+1):
if t-30<s<30:l+=[P,5+t-s,36+s,98];l+=[L,n,98,99];l+=a(2,99)
print(*l)
```
| 106,774 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
p=print
r=range
def a(i,j,k,l=0):p("+<"[l],i,j,k)
def b(i):a(i,i,i+1)
def d(x,i,j):
for t in r(30):
b(j+t-1);a(j+t,n,j+t)
for s in r(j+t,j+29):b(s)
a(j+29,x,i+t,1);b(j+t-1);a(j+t,i+t,j+t)
p(3933)
a(0,1,n)
a(2,n,n,1)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in r(59):
a(2,2,2)
for s in r(t+1):
if t-30<s<30:a(5+t-s,36+s,98);a(n,98,99,1);a(2,99,2)
```
| 106,775 |
Provide a correct Python 3 solution for this coding contest problem.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
"Correct Solution:
```
n=97
p=print
r=range
def c(i,j,k):p('<',i,j,k)
def a(i,j,k):p('+',i,j,k)
def b(i):a(i,i,i+1)
def d(x,i,j):
for t in r(30):
b(j+t-1)
a(j+t,n,j+t)
for s in r(j+t,j+29):b(s)
c(j+29,x,i+t)
b(j+t-1)
a(j+t,i+t,j+t)
p(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in r(59):
a(2,2,2)
for s in r(t+1):
if t-30<s<30:a(5+t-s,36+s,98);c(n,98,99);a(2,99,2)
```
| 106,776 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
n=97
p=print
r=range
def c(i,j,k):p('<',i,j,k)
def a(i,j,k):p('+',i,j,k)
def d(x,i,j):
for t in r(30):
a(j+t-1,j+t-1,j+t)
a(j+t,n,j+t)
for s in r(j+t,j+29):a(s,s,s+1)
c(j+29,x,i+t)
a(j+t-1,j+t-1,j+t)
a(j+t,i+t,j+t)
p(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in r(59):
a(2,2,2)
for s in r(t+1):
if t-30<s<30:a(5+t-s,36+s,98);c(n,98,99);a(2,99,2)
```
Yes
| 106,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
n=97
p=print
r=range
def a(i,j,k,l=0):p("+<"[l],i,j,k)
def b(i):a(i,i,i+1)
def d(x,i,j):
for t in r(30):b(j+t-1);a(j+t,n,j+t);[b(s)for s in r(j+t,j+29)];a(j+29,x,i+t,1);b(j+t-1);a(j+t,i+t,j+t)
p(3933)
a(0,1,n)
a(2,n,n,1)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in r(59):a(2,2,2);[(a(5+t-s,36+s,98),a(n,98,99,1),a(2,99,2))for s in r(t+1)if t-30<s<30]
```
Yes
| 106,778 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
n=97
def c(i,j,k):print('<',i,j,k)
def a(i,j,k):print('+',i,j,k)
def m(i,j,k):
for t in range(59):
a(k,k,k)
for s in range(t+1):
if t-30<s<30:a(i+t-s,j+s,98);c(n,98,99);a(k,99,k)
def d(x,i,j):
for t in range(30):
a(j+t-1,j+t-1,j+t)
a(j+t,n,j+t)
for s in range(j+t,j+29):a(s,s,s+1)
c(j+29,x,i+t)
a(j+t-1,j+t-1,j+t)
a(j+t,i+t,j+t)
print(3933)
a(0,1,n)
c(2,n,n)
a(0,n,3)
a(1,n,4)
d(3,5,36)
d(4,36,67)
for t in range(59):
a(2,2,2)
for s in range(t+1):
if t-30<s<30:a(5+t-s,36+s,98);c(n,98,99);a(2,99,2)
```
Yes
| 106,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
class CodeGen:
codes = []
cnt = 100
def __init__(self, A, B):
self.A = A
self.B = B
self.C1 = self.one()
self.C2 = self.pow2(self.C1, 1)
self.C3 = self.pow2(self.C1, 2)
self.C4 = self.pow2(self.C1, 3)
self.POW2 = [self.C1]
for i in range(32):
self.POW2.append(self.pow2(self.C1, i + 1))
def allocate(self):
self.cnt += 1
return self.cnt - 1
def plus(self, x, y):
l = self.allocate()
self.codes.append(f"+ {x} {y} {l}")
return l
def gt(self, x, y):
l = self.allocate()
self.codes.append(f"< {x} {y} {l}")
return l
def ge(self, x, y):
return self.gt(x, self.plus(y, self.C1))
def pow2(self, x, n):
t = x
for _ in range(n):
t = self.plus(t, t)
return t
def one(self):
return self.gt(self.allocate(), self.plus(self.A, self.B))
def bits(self, x, n):
result = []
t = self.allocate()
for i in range(n - 1, -1, -1):
bit = self.ge(self.plus(self.POW2[i], t), x)
result.append(bit)
t = self.plus(t, self.pow2(bit, i))
return result
def mult(self, x, y):
return self.gt(self.C1, self.plus(x, y))
def show(self, ans):
t = self.allocate()
self.codes.append(f"+ {ans} {t} 2")
print(len(self.codes))
for c in self.codes:
print(c)
def pluss(self, xs):
if len(xs) == 1:
return xs[0]
return self.plus(xs[0], self.pluss(xs[1:]))
class Eval(CodeGen):
def __init__(self, A, B):
super().__init__(A, B)
def allocate(self):
return 0
def plus(self, x, y):
return x + y
def gt(self, x, y):
if x < y:
return 1
return 0
def show(self, ans):
print(ans)
if __name__ == "__main__":
gen = CodeGen(0, 1)
# gen = Eval(1000, 1234)
A = gen.bits(gen.A, 32)
B = gen.bits(gen.B, 32)
X = []
for _ in range(64):
X.append(gen.allocate())
for i in range(32):
for j in range(32):
X[i + j] = gen.plus(X[i + j], gen.mult(A[i], B[j]))
ans = gen.allocate()
for i in range(64):
ans = gen.plus(ans, gen.pow2(X[i], 62- i))
gen.show(ans)
```
Yes
| 106,780 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
#B=[0 for _ in range(140)]
#B[0],B[1] = 0,10
print(245)
def f(s):
print(s)
f('< 2 0 3')
for i in range(11):
f(f'< 4 0 {5+i}')
f(f'+ 3 4 4')
for i in range(11):
f(f'< 1 16 {17+i}')
f(f'+ 3 16 16')
for i in range(10):
for j in range(10):
f(f'< {j+18} {i+5} {28 + 10*i+j}')
for i in range(28,28+10*9+9+1):
f(f'+ 2 {i} 2')
```
No
| 106,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
'''
自宅用PCでの解答
'''
import math
#import numpy as np
import itertools
import queue
import bisect
from collections import deque,defaultdict
import heapq as hpq
from sys import stdin,setrecursionlimit
#from scipy.sparse.csgraph import dijkstra
#from scipy.sparse import csr_matrix
ipt = stdin.readline
setrecursionlimit(10**7)
mod = 10**9+7
dir = [(-1,0),(0,-1),(1,0),(0,1)]
alp = "abcdefghijklmnopqrstuvwxyz"
def main():
print(511)
a = [6,9]+[0]*10001
print("+ 0 1 3")
a[3] = a[0]+a[1]
print("< 2 3 4")
a[4] = a[2] < a[3]
print("+ 10000 10000 3")
a[3] = a[10000]+a[10000]
for i in range(8):
print("+ 4", i+4, i+5)
a[i+5] = a[4]+a[i+4]
# print(a)
for i in range(10):
for j in range(10):
print("<", 3+i, 0, 20)
print("<", 3+j, 1, 21)
print("+", 20, 21, 22)
print("<", 4, 22, 23)
print("+ 2", 23, 2)
a[20] = a[3+i]<a[0]
a[21] = a[3+j]<a[1]
a[22] = a[20]+a[21]
a[23] = a[4]<a[22]
a[2] = a[23]+a[2]
# print(a[2])
return None
if __name__ == '__main__':
main()
```
No
| 106,782 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
P,L="+<"
p,r=print,range
p(6484,P,0,1,3,L,2,3,3)
for x in 0,1:p(P,x,3,x);[[P,v:=x+4,v,v,P,v,3,6,*[P,6,6,6]*(29-t),L,6,x,T:=x*60+9+t,P,v,T,v]for t in r(30)]
for t in r(60):p(P,2,2,2);[p(P,8+t-s,69+s,5,L,3,5,6,P,2,6,2)for s in r(t)]
```
No
| 106,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an output-only problem. You shouldn't read anything from the input.
In short, your task is to simulate multiplication by using only comparison (x < y) and addition (x + y). There is no input in this problem, you just print a sequence of operations.
Imagine that there is a big array a[0], a[1], ..., a[N-1] of length N. The first two values are initially two non-negative integers A and B (which are unknown to you), the other elements are zeros. Your goal is to get the product A \cdot B in a[2] at the end.
You are allowed operations of two types, with the following format (where 0 \leq i, j, k < N):
* `+ i j k` — applies operation a[k] = a[i] + a[j].
* `< i j k` — applies operation a[k] = a[i] < a[j]. That is, if a[i] < a[j] then a[k] becomes 1, otherwise it becomes 0.
You can use at most Q operations. Elements of a can't exceed V. Indices (i, j, k) don't have to be distinct. It's allowed to modify any element of the array (including the first two). The actual checker simulates the process for multiple pairs (A, B) within a single test. Each time, the checker chooses values A and B, creates the array a = [A, B, 0, 0, \ldots, 0], applies all your operations and ensures that a[2] = A \cdot B.
Constraints
* 0 \leq A, B \leq 10^9
* N = Q = 200\,000
* V = 10^{19} = 10\,000\,000\,000\,000\,000\,000
Input
The Standard Input is empty.
Output
In the first line, print the number of operations. Each operation should then be printed in a single line of format `+ i j k` or `< i j k`.
Example
Input
Output
Submitted Solution:
```
print("""232
< 2 0 2
< 3 0 3
+ 2 3 4
+ 2 4 5
+ 2 5 6
+ 2 6 7
+ 2 7 8
+ 2 8 9
+ 2 9 10
+ 2 10 11
+ 2 11 12
< 2 0 13
< 3 0 14
< 4 0 15
< 5 0 16
< 6 0 17
< 7 0 18
< 8 0 19
< 9 0 20
< 10 0 21
< 11 0 22
< 1 3 23
< 1 4 24
< 1 5 25
< 1 6 26
< 1 7 27
< 1 8 28
< 1 9 29
< 1 10 30
< 1 11 31
< 1 12 32
< 23 13 101
< 24 13 102
< 25 13 103
< 26 13 104
< 27 13 105
< 28 13 106
< 29 13 107
< 30 13 108
< 31 13 109
< 32 13 110
< 23 14 111
< 24 14 112
< 25 14 113
< 26 14 114
< 27 14 115
< 28 14 116
< 29 14 117
< 30 14 118
< 31 14 119
< 32 14 120
< 23 15 121
< 24 15 122
< 25 15 123
< 26 15 124
< 27 15 125
< 28 15 126
< 29 15 127
< 30 15 128
< 31 15 129
< 32 15 130
< 23 16 131
< 24 16 132
< 25 16 133
< 26 16 134
< 27 16 135
< 28 16 136
< 29 16 137
< 30 16 138
< 31 16 139
< 32 16 140
< 23 17 141
< 24 17 142
< 25 17 143
< 26 17 144
< 27 17 145
< 28 17 146
< 29 17 147
< 30 17 148
< 31 17 149
< 32 17 150
< 23 18 151
< 24 18 152
< 25 18 153
< 26 18 154
< 27 18 155
< 28 18 156
< 29 18 157
< 30 18 158
< 31 18 159
< 32 18 160
< 23 19 161
< 24 19 162
< 25 19 163
< 26 19 164
< 27 19 165
< 28 19 166
< 29 19 167
< 30 19 168
< 31 19 169
< 32 19 170
< 23 20 171
< 24 20 172
< 25 20 173
< 26 20 174
< 27 20 175
< 28 20 176
< 29 20 177
< 30 20 178
< 31 20 179
< 32 20 180
< 23 21 181
< 24 21 182
< 25 21 183
< 26 21 184
< 27 21 185
< 28 21 186
< 29 21 187
< 30 21 188
< 31 21 189
< 32 21 190
< 23 22 191
< 24 22 192
< 25 22 193
< 26 22 194
< 27 22 195
< 28 22 196
< 29 22 197
< 30 22 198
< 31 22 199
< 32 22 200
+ 101 233 233
+ 102 233 233
+ 103 233 233
+ 104 233 233
+ 105 233 233
+ 106 233 233
+ 107 233 233
+ 108 233 233
+ 109 233 233
+ 110 233 233
+ 111 233 233
+ 112 233 233
+ 113 233 233
+ 114 233 233
+ 115 233 233
+ 116 233 233
+ 117 233 233
+ 118 233 233
+ 119 233 233
+ 120 233 233
+ 121 233 233
+ 122 233 233
+ 123 233 233
+ 124 233 233
+ 125 233 233
+ 126 233 233
+ 127 233 233
+ 128 233 233
+ 129 233 233
+ 130 233 233
+ 131 233 233
+ 132 233 233
+ 133 233 233
+ 134 233 233
+ 135 233 233
+ 136 233 233
+ 137 233 233
+ 138 233 233
+ 139 233 233
+ 140 233 233
+ 141 233 233
+ 142 233 233
+ 143 233 233
+ 144 233 233
+ 145 233 233
+ 146 233 233
+ 147 233 233
+ 148 233 233
+ 149 233 233
+ 150 233 233
+ 151 233 233
+ 152 233 233
+ 153 233 233
+ 154 233 233
+ 155 233 233
+ 156 233 233
+ 157 233 233
+ 158 233 233
+ 159 233 233
+ 160 233 233
+ 161 233 233
+ 162 233 233
+ 163 233 233
+ 164 233 233
+ 165 233 233
+ 166 233 233
+ 167 233 233
+ 168 233 233
+ 169 233 233
+ 170 233 233
+ 171 233 233
+ 172 233 233
+ 173 233 233
+ 174 233 233
+ 175 233 233
+ 176 233 233
+ 177 233 233
+ 178 233 233
+ 179 233 233
+ 180 233 233
+ 181 233 233
+ 182 233 233
+ 183 233 233
+ 184 233 233
+ 185 233 233
+ 186 233 233
+ 187 233 233
+ 188 233 233
+ 189 233 233
+ 190 233 233
+ 191 233 233
+ 192 233 233
+ 193 233 233
+ 194 233 233
+ 195 233 233
+ 196 233 233
+ 197 233 233
+ 198 233 233
+ 199 233 233
+ 200 233 233
+ 233 234 2""");
```
No
| 106,784 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
S = input()
print("Yes") if S[2] == S[3] and S[4] == S[5] else print("No")
```
| 106,785 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
x = input()
print(["No", "Yes"][x[2] == x[3] and x[4] == x[5]])
```
| 106,786 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
s = list(input())
print("Yes") if s[2] == s[3] and s[4] == s[5] else print("No")
```
| 106,787 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
S = input()
print('Yes' if (S[2] == S[3] and S[4] == S[5]) else 'No')
```
| 106,788 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
S = input()
if S[2]==S[3] and S[5]==S[4]:
print("Yes")
else:
print("No")
```
| 106,789 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
S = list(input())
print("Yes" if S[2] == S[3] and S[4] == S[5] else "No")
```
| 106,790 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
S = list(input())
print('Yes' if S[2]==S[3] and S[4]==S[5] else 'No')
```
| 106,791 |
Provide a correct Python 3 solution for this coding contest problem.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
"Correct Solution:
```
s=input()
if s[2]==s[3] and s[4]==s[5]:
print("Yes")
exit()
print("No")
```
| 106,792 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
s = input()
if s[3] == s[2] and s[4] == s[5]:
print("Yes")
exit()
print("No")
```
Yes
| 106,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
str = input()
print('Yes' if str[2]==str[3] and str[4]==str[5] else 'No')
```
Yes
| 106,794 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
s = str(input())
if s[4]==s[5] and s[2]==s[3]:
print('Yes')
else:
print('No')
```
Yes
| 106,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
S = input().strip('\n')
print('Yes' if S[2]==S[3] and S[4]==S[5] else 'No')
```
Yes
| 106,796 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
def main():
x = input()
if(x[2] == x[3]):
if(x[4] == x[5]):
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
```
No
| 106,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
s = input().split()
if s[2] == s[3] and s[4] == s[5]:
print("Yes")
else:
print("No")
```
No
| 106,798 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.
Given a string S, determine whether it is coffee-like.
Constraints
* S is a string of length 6 consisting of lowercase English letters.
Input
Input is given from Standard Input in the following format:
S
Output
If S is coffee-like, print `Yes`; otherwise, print `No`.
Examples
Input
sippuu
Output
Yes
Input
iphone
Output
No
Input
coffee
Output
Yes
Submitted Solution:
```
input = input()
if (input[2] == input[3]) & (input[4] == input[5]):
print('Yes')
else:
print('None')
```
No
| 106,799 |
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