message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image> | instruction | 0 | 17,892 | 14 | 35,784 |
Tags: constructive algorithms, data structures, dfs and similar, divide and conquer, dsu, greedy, sortings, trees
Correct Solution:
```
import heapq
class UF:
def __init__(self, N):
self.par = list(range(N))
self.sz = [1] * N
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr:
return False
if self.sz[xr] < self.sz[yr]:
xr, yr = yr, xr
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
return True
def size(self, x):
return self.sz[self.find(x)]
def connected(self, x, y):
return self.find(x) == self.find(y)
def main():
n = int(input())
uf = UF(n*n+1)
heap = []
ans_list = []
edges = []
parent_map = {i : i for i in range(n**2 + 1)}
for i in range(n):
cur_lst = list(map(int, input().split(' ')))
for j in range(i, n):
if i == j:
ans_list.append(cur_lst[j])
else:
heapq.heappush(heap, (cur_lst[j], i, j))
while heap:
cur_anc_salary, low1, low2 = heapq.heappop(heap)
c1, c2 = parent_map[uf.find(low1)], parent_map[uf.find(low2)]
# already connected
if c1 == c2:
continue
# add kid to c1
if ans_list[c1] == cur_anc_salary:
edges.append((c2, c1))
uf.union(c1, c2)
parent_map[uf.find(c1)] = c1
elif ans_list[c2] == cur_anc_salary:
edges.append((c1, c2))
uf.union(c1, c2)
parent_map[uf.find(c1)] = c2
else:
new_node = len(ans_list)
ans_list.append(cur_anc_salary)
uf.union(c1, c2)
uf.union(c1, new_node)
edges.append((c1, new_node))
edges.append((c2, new_node))
parent_map[uf.find(new_node)] = new_node
print(len(ans_list))
print(" ".join(map(str, ans_list)))
print(len(ans_list))
for x,y in edges:
print("{} {}".format(x+1, y+1))
# region fastio
import os
import sys
from io import BytesIO, IOBase
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()
``` | output | 1 | 17,892 | 14 | 35,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter
from pprint import pprint
from itertools import permutations
from bisect import bisect_right,bisect_left
def main():
n=int(input())
mat=[[0]*n for i in range(n)]
inc=[[] for i in range(50001)]
rpr=[i for i in range(n)]
c=[0] *(10**6)
for i in range(n):
mat[i]=list(map(int,input().split()))
for j in range(n):
inc[mat[i][j]].append((i,j))
if i==j :
c[i]=mat[i][j]
new=n
st=set()
newedg=set()
for i in range(5001):
groups=[]
for x,y in inc[i]:
if x!=y:
bol=True
for g in groups:
if y in g:
g.add(x)
bol=False
if x in g:
g.add(y)
bol=False
if bol:
groups.append({x,y})
for g in groups:
for node in g:
a,b=min(rpr[node],new),max(rpr[node],new)
newedg.add((a,b))
rpr[node]=new
c[new]=i
new+=1
# if len(groups):
# print(groups)
# print(newedg)
print(new)
print(*c[:new])
print(new)
for a,b in newedg:
print(a+1,b+1)
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__":
for _ in range(1):
main()
``` | instruction | 0 | 17,893 | 14 | 35,786 |
Yes | output | 1 | 17,893 | 14 | 35,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]
e = [[] for _ in range(n)]
def solve(t):
global e
ind = []
x = 1
for i in range(n):
if t & x: ind.append(i)
x <<= 1
mx = 0
mn = 5000
for i in ind:
for j in ind:
mx = max(mx, a[i][j])
mn = min(mn, a[i][j])
if len(ind) == 1: return ind[0]
e.append([])
if mn == mx:
for i in ind: e[-1].append(i)
return len(e) - 1
z = len(e) - 1
vis = [0] * n
for i in ind:
if vis[i]: continue
tt = 0
for j in ind:
if vis[j]: continue
if a[i][j] != mx:
tt |= 1 << j
vis[j] = 1
e[z].append(solve(tt))
return z
solve((1 << n) - 1)
#print(e)
s = [n]
vis = [0] * len(e)
vis[n] = 1
order = []
parent = [-1] * len(e)
while len(s):
x = s.pop()
order.append(x)
for y in e[x]:
if vis[y]: continue
vis[y] = 1
s.append(y)
parent[y] = x
import heapq
hpush = heapq.heappush
hpop = heapq.heappop
h = [[] for _ in range(len(e))]
for i in range(n):
for x in a[i]: hpush(h[i], x)
res = [0] * len(e)
order.reverse()
for y in order:
res[y] = hpop(h[y])
if parent[y] == -1: continue
x = parent[y]
if len(h[x]) == 0:
while len(h[y]):
v = hpop(h[y])
if v > res[y]: hpush(h[x], v)
else:
while len(h[x]) and h[x][0] <= res[y]: hpop(h[x])
print(len(e))
print(*res)
print(n + 1)
for x in range(len(e)):
for y in e[x]:
print(y + 1, x + 1)
``` | instruction | 0 | 17,894 | 14 | 35,788 |
Yes | output | 1 | 17,894 | 14 | 35,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
idx = 0
vis = None
a = None
e = None
def rec(c):
global idx
if len(c) == 1:
z = c[0]
vis[z] = -1
cost[z] = a[z][z]
return z
idx += 1
now = idx
#print(now, c)
z = 0
for i in c:
vis[i] = -1
for j in c:
if i != j:
z = max(z, a[i][j])
cost[now] = z
n = len(a)
cc = 0
w = []
#print('yo')
for i in c:
if vis[i] < 0:
q = [i]
cc += 1
vis[i] = cc
qi = 0
while qi < len(q):
v = q[qi]
qi += 1
for j in c:
if v != j and a[v][j] < z and vis[j] < 0:
q.append(j)
vis[j] = cc
w.append(q)
#print(w)
for b in w:
e.append((rec(b), now))
return now
def solve():
global idx, vis, a, e, cost
n = int(input())
a = [None]*n
for i in range(n):
a[i] = list(map(int, input().split()))
idx = n - 1
e = []
vis = [-1]*len(a)
cost = [0]*(2*n+4)
root = rec(list(range(n)))
print(idx+1)
print(*cost[:idx+1])
print(root+1)
for u, v in e:
print(u+1,v+1)
solve()
``` | instruction | 0 | 17,895 | 14 | 35,790 |
Yes | output | 1 | 17,895 | 14 | 35,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
We can map out the ancestral path for i == 0
Then for each subsequent i, we find the lowest point it joins an existing path, and we then create a path up to that point
"""
def solve():
N = getInt()
M = getMat(N)
parent = [-1]*N
salary = [-1]*N
lookup = dd(int) # lookup[(i,s)] is the index of the node corresponding to the ancestor of i with salary s
for i in range(N):
salary[i] = M[i][i]
to_join = -1
lowest = 10**4
for j in range(i):
if M[i][j] < lowest:
lowest = M[i][j]
to_join = j
#so we want to join the path of node to_join at its value _lowest_
if i == 0:
prev = 0
B = sorted([(a,j) for j,a in enumerate(M[0])])
for j in range(1,N):
if B[j][0] == B[j-1][0]:
lookup[(i,s)] = prev
else:
idx = len(parent)
s = B[j][0]
lookup[(i,s)] = idx
parent.append(-1)
parent[prev] = idx
salary.append(s)
prev = idx
else:
assert(to_join >= 0)
prev = i
flag = False
B = sorted([(a,j) for j,a in enumerate(M[i])])
for j in range(1,N):
if B[j][0] == lowest:
idx = lookup[(to_join,lowest)]
lookup[(i,lowest)] = idx
parent[prev] = idx
flag = True
elif B[j][0] != B[j-1][0] and not flag:
idx = len(parent)
s = B[j][0]
lookup[(i,s)] = idx
parent.append(-1)
parent[prev] = idx
salary.append(s)
prev = idx
elif not flag:
lookup[(i,s)] = prev
else:
lookup[(i,s)] == lookup[(to_join,s)]
print(len(parent))
print(*salary)
for i,p in enumerate(parent):
if p == -1: print(i+1)
for i,p in enumerate(parent):
if p == -1: continue
print(i+1,p+1)
return
#for _ in range(getInt()):
#print(solve())
solve()
#print(time.time()-start_time)
``` | instruction | 0 | 17,896 | 14 | 35,792 |
Yes | output | 1 | 17,896 | 14 | 35,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
n=3
pairs = [[2,5,7],[5,1,7],[7,7,4]]
n = int(input())
pairs = [[None for _ in range(n)] for _ in range(n)]
for i in range(n):
pairs[i] = [int(x) for x in input().split(' ')]
salaries = [pairs[i][i] for i in range(n)]
report_sets = [[i] for i in range(n)]
done = [False for _ in range(n)]
boss = None
relationships = []
report_sets_to_indices = {}
while True:
index = None
for i, d in enumerate(done):
if not d:
index = i
break
if index is None:
break
done[index] = True
reports = report_sets[index]
salary = salaries[index]
direct_manager_salary = min(
[x for x in pairs[reports[0]] if x > salary])
direct_manager_reports = tuple(
x for x in range(n) if pairs[reports[0]][x] <= direct_manager_salary)
if direct_manager_reports not in report_sets_to_indices:
salaries.append(direct_manager_salary)
report_sets.append(direct_manager_reports)
report_sets_to_indices[direct_manager_reports] = len(salaries)
if len(direct_manager_reports) == n:
boss = len(salaries)
done.append(True)
else:
done.append(False)
relationships.append((index, report_sets_to_indices[direct_manager_reports]))
print(len(salaries))
print(' '.join([str(x) for x in salaries]))
print(boss)
for r in relationships:
print('{} {}'.format(r[0], r[1]))
``` | instruction | 0 | 17,897 | 14 | 35,794 |
No | output | 1 | 17,897 | 14 | 35,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
N= int(input())
M =[]
D = []
for n in range(N):
T=list(map(int,input().split()))
for t in T:
if t not in D:
D.append(t)
M.append(T)
S=[ 0 for x in range(N)]
for n in range(N):
S[n]=M[n][n]
D.sort()
for x in range(n+1, len(D)):
S.append(D[x])
print(len(S))
print(*S)
print(len(S))
P = [-1 for n in range(len(S))]
for k in range(n+1,len(S)):
s = S[k]
trovato = False
for n in range(N):
for m in range(N):
if M[n][m] == s:
if P[n] == -1 and P[m] == -1:
P[n] = k
P[m] = k
else:
if P[n] != -1:
P[P[n]] = k
P[m] = k
else:
P[P[m]] = k
P[n] = k
trovato = True
break
if trovato:
break
for n in range(len(S)-1):
print(n+1, P[n]+1)
``` | instruction | 0 | 17,898 | 14 | 35,796 |
No | output | 1 | 17,898 | 14 | 35,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
n=int(input())
lsn=list()
ele=list()
edge=[-1 for i in range(2*n - 1)]
for r in range(n):
l=list(map(int,input().split()))
lsn.append(l)
ele.append(lsn[r][r])
# lsn[r].pop(r)
lsn[r]=list(set(lsn[r]))
lsn[r].sort()
print(2*n -1)
rqn=list()
for i in lsn:
rqn=rqn+i[1:]
rqn=list(set(rqn))
rqn.sort(reverse=True)
ele=ele+rqn
print(" ".join(map(str,ele)))
# print(lsn)
# max supervisor
print(n+1)
for lst in lsn:
for i in range(len(lst)):
lst[i]=ele.index(lst[i])
# print(i,lst[i])
for i in range(len(lst)-1):
if edge[lst[i]]==-1:
edge[lst[i]]=lst[i+1]
# print(i,edge[lst[i]],lst[i+1])
for i in range(2*n - 1):
if edge[i]==-1:
continue
print(i+1,edge[i]+1)
``` | instruction | 0 | 17,899 | 14 | 35,798 |
No | output | 1 | 17,899 | 14 | 35,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Dogeforces company has k employees. Each employee, except for lower-level employees, has at least 2 subordinates. Lower-level employees have no subordinates. Each employee, except for the head of the company, has exactly one direct supervisor. The head of the company is a direct or indirect supervisor of all employees. It is known that in Dogeforces, each supervisor receives a salary strictly more than all his subordinates.
The full structure of the company is a secret, but you know the number of lower-level employees and for each pair of lower-level employees, the salary of their common supervisor is known (if there are several such supervisors, then the supervisor with the minimum salary). You have to restore the structure of the company.
Input
The first line contains a single integer n (2 β€ n β€ 500) β the number of lower-level employees.
This is followed by n lines, where i-th line contains n integers a_{i,1}, a_{i,2}, ..., a_{i,n} (1 β€ a_{i,j} β€ 5000) β salary of the common supervisor of employees with numbers i and j. It is guaranteed that a_{i,j} = a_{j,i}. Note that a_{i,i} is equal to the salary of the i-th employee.
Output
In the first line, print a single integer k β the number of employees in the company.
In the second line, print k integers c_1, c_2, ..., c_k, where c_i is the salary of the employee with the number i.
In the third line, print a single integer r β the number of the employee who is the head of the company.
In the following k-1 lines, print two integers v and u (1 β€ v, u β€ k) β the number of the employee and his direct supervisor.
Note that the lower-level employees have numbers from 1 to n, and for the rest of the employees, you have to assign numbers from n+1 to k. If there are several correct company structures, you can print any of them.
Example
Input
3
2 5 7
5 1 7
7 7 4
Output
5
2 1 4 7 5
4
1 5
2 5
5 4
3 4
Note
One of the possible structures in the first example: <image>
Submitted Solution:
```
from sys import stdin, stdout
def dogeforces(c_a, a_a):
global n
l_a = [i for i in range(n)]
return dfs(c_a, a_a, l_a)
def dfs(c_a, a_a, l_a):
global n
global d_a
if len(l_a) == 1:
return l_a[0]
root = -1
for l in l_a:
root = max(root, a_a[l_a[0]][l])
gp_a = []
for l in l_a:
gp = -1
for j in range(len(gp_a)):
g = gp_a[j][0]
if a_a[l][g] != root:
gp = j
break
if gp == -1:
gp_a.append([])
gp_a[-1].append(l)
# print(gp_a)
v = n
c_a.append(root)
n += 1
for gp in gp_a:
sr = dfs(c_a, a_a, gp)
d_a.append([v, sr])
return v
n = int(stdin.readline())
a_a = []
c_a = []
for i in range(n):
a_a.append(list(map(int, stdin.readline().split())))
c_a.append(a_a[i][i])
d_a = []
root = dogeforces(c_a, a_a)
stdout.write(str(n) + '\n')
stdout.write(' '.join(map(str, c_a)) + '\n')
stdout.write(str(len(d_a)) + '\n')
for d in d_a:
stdout.write(str(d[0]+1) + ' ' + str(d[1]+1) + '\n')
``` | instruction | 0 | 17,900 | 14 | 35,800 |
No | output | 1 | 17,900 | 14 | 35,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,979 | 14 | 35,958 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
t = int(input())
res = [0] * t
for test in range(t):
val = int(input())
if val == 1: res[test] = 0
else: res[test] = val - 2
print(' '.join(map(str, res)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
``` | output | 1 | 17,979 | 14 | 35,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,980 | 14 | 35,960 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
if (n<2): print(0)
else: print(n-2)
``` | output | 1 | 17,980 | 14 | 35,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,981 | 14 | 35,962 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
n = input()
for i in range(0,int (n)):
x = input()
if int(x)-2 == -1:
print(0)
else:
print(int(x)-2)
``` | output | 1 | 17,981 | 14 | 35,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,982 | 14 | 35,964 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
if n==1:
print(0)
else:
print(n-2)
``` | output | 1 | 17,982 | 14 | 35,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,983 | 14 | 35,966 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
import sys
import math
def ii(): return int(input()) # input int 1
def il(): return input().split(' ') # input list
# input list of int: [1, 2, 3, ...]
def ili(): return [int(i) for i in input().split(' ')]
# input list of int splited: (a, b, c, ...)
def ilis(): return (int(i) for i in input().split(' '))
def ip():
return input()
def op(l):
print(l)
sys.stdout.flush()
def gcd(a, b): return a if b == 0 else gcd(b, a % b)
def extgcd(a, b, x=0, y=0):
if b == 0:
return (a, 1, 0)
d, m, n = extgcd(b, a % b, x, y)
return (d, n, m-(a//b)*n)
def main():
T = ii()
for t in range(T):
n = ii()
if n <= 2:
op(0)
else:
op(n-2)
if __name__ == "__main__":
main()
``` | output | 1 | 17,983 | 14 | 35,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,984 | 14 | 35,968 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
class CodeforcesTask23BSolution:
def __init__(self):
self.result = ''
self.t = 0
self.tests = []
def read_input(self):
self.t = int(input())
for x in range(self.t):
self.tests.append(int(input()))
def process_task(self):
results = []
for t in self.tests:
results.append(str(max(0, t - 2)))
self.result = "\n".join(results)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask23BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 17,984 | 14 | 35,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,985 | 14 | 35,970 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for i in range(t):
print(max(0,int(input())-2))
``` | output | 1 | 17,985 | 14 | 35,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1 | instruction | 0 | 17,986 | 14 | 35,972 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
t = int(input())
while t > 0:
a = int(input())
if a < 2: print(0)
else: print(a - 2)
t = t - 1
``` | output | 1 | 17,986 | 14 | 35,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1
Submitted Solution:
```
test=int(input())
lst=[]
for i in range(0,test):
n=int(input())
lst.append(n-2)
for j in range(0,test):
if lst[j]>0:
print(lst[j])
else:
print("0")
``` | instruction | 0 | 17,987 | 14 | 35,974 |
Yes | output | 1 | 17,987 | 14 | 35,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1
Submitted Solution:
```
print('\n'.join(str(max(int(input()) - 2, 0)) for i in range(int(input()))))
# Made By Mostafa_Khaled
``` | instruction | 0 | 17,989 | 14 | 35,978 |
Yes | output | 1 | 17,989 | 14 | 35,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1
Submitted Solution:
```
t = int(input())
while (t > 0):
t -= 1
n = int(input()) - 2
print(n if n > 0 else 0)
``` | instruction | 0 | 17,990 | 14 | 35,980 |
Yes | output | 1 | 17,990 | 14 | 35,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
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")
##########################################################
#for _ in range(int(input())):
#import math
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for i in range(m):
# for _ in range(int(input())):
from collections import Counter
#from fractions import Fraction
#n = int(input())
# for _ in range(int(input())):
# import math
import sys
# from collections import deque
# from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=sys.stdin.readline()
# n=int(n)
# n,k= map(int, input().split())
# arr=list(map(int,input().split()))
# n=int(input())
import math
from math import gcd
for _ in range(int(input())):
n = int(input())
print(n-2)
'''for i in range(2,int(n**0.5)+1):
if n%i==0:
print(n//i,n-(n//i))
break
else:
print(1, n-1)'''
``` | instruction | 0 | 17,991 | 14 | 35,982 |
No | output | 1 | 17,991 | 14 | 35,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1
Submitted Solution:
```
#input
t = int(input())
for z in range(t):
n = int(input())
if n<=2:
print(0)
break
print(n-2)
``` | instruction | 0 | 17,992 | 14 | 35,984 |
No | output | 1 | 17,992 | 14 | 35,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people came to a party. Then those, who had no friends among people at the party, left. Then those, who had exactly 1 friend among those who stayed, left as well. Then those, who had exactly 2, 3, ..., n - 1 friends among those who stayed by the moment of their leaving, did the same.
What is the maximum amount of people that could stay at the party in the end?
Input
The first input line contains one number t β amount of tests (1 β€ t β€ 105). Each of the following t lines contains one integer number n (1 β€ n β€ 105).
Output
For each test output in a separate line one number β the maximum amount of people that could stay in the end.
Examples
Input
1
3
Output
1
Submitted Solution:
```
n = input()
for i in range(0,int (n)):
x = input()
print(int(x)-2)
``` | instruction | 0 | 17,994 | 14 | 35,988 |
No | output | 1 | 17,994 | 14 | 35,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arthur and Alexander are number busters. Today they've got a competition.
Arthur took a group of four integers a, b, w, x (0 β€ b < w, 0 < x < w) and Alexander took integer Ρ. Arthur and Alexander use distinct approaches to number bustings. Alexander is just a regular guy. Each second, he subtracts one from his number. In other words, he performs the assignment: c = c - 1. Arthur is a sophisticated guy. Each second Arthur performs a complex operation, described as follows: if b β₯ x, perform the assignment b = b - x, if b < x, then perform two consecutive assignments a = a - 1; b = w - (x - b).
You've got numbers a, b, w, x, c. Determine when Alexander gets ahead of Arthur if both guys start performing the operations at the same time. Assume that Alexander got ahead of Arthur if c β€ a.
Input
The first line contains integers a, b, w, x, c (1 β€ a β€ 2Β·109, 1 β€ w β€ 1000, 0 β€ b < w, 0 < x < w, 1 β€ c β€ 2Β·109).
Output
Print a single integer β the minimum time in seconds Alexander needs to get ahead of Arthur. You can prove that the described situation always occurs within the problem's limits.
Examples
Input
4 2 3 1 6
Output
2
Input
4 2 3 1 7
Output
4
Input
1 2 3 2 6
Output
13
Input
1 1 2 1 1
Output
0 | instruction | 0 | 18,032 | 14 | 36,064 |
Tags: binary search, math
Correct Solution:
```
a,b,w,x,c=map(int,input().split())
ans=0
bb=b
benefit=0
Visited=[False]*1003
CycleCost=-1
while(1):
if(c<=a):
break
if(Visited[b]!=False):
CycleCost=ans-Visited[b][1]
CycleBenefit=benefit-Visited[b][0]
CycleBegining=b
break
Visited[b]=(benefit,ans)
if(b<x):
b=w-(x-b)
ans+=1
elif(b>=x):
b-=x
ans+=1
benefit+=1
if(benefit==c-a):
break
if(CycleCost==-1):
print(ans)
else:
c-=benefit
diff=c-a
xx=diff//CycleBenefit
if(xx!=0):
xx-=1
ans+=xx*CycleCost
diff-=xx*CycleBenefit
b=CycleBegining
while(diff>0):
if(b<x):
b=w-(x-b)
ans+=1
else:
b-=x
ans+=1
diff-=1
print(ans)
``` | output | 1 | 18,032 | 14 | 36,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,187 | 14 | 36,374 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
n, m, k, x, y = list(map(int, input().split()))
if n == 1:
mod = m
else:
mod = (2 * n - 2) * m
cnt, rem = divmod(k, mod)
ans = [[0 for _ in range(m)] for _ in range(n)]
ans[0] = [cnt for _ in range(m)]
ans[n - 1] = [cnt for _ in range(m)]
for i in range(1, n - 1):
for j in range(m):
ans[i][j] = 2 * cnt
for i in range(n):
for j in range(m):
if rem > 0:
ans[i][j] += 1
rem -= 1
if n > 1:
for i in range(n - 2, 0, -1):
for j in range(m):
if rem > 0:
ans[i][j] += 1
rem -= 1
mx = max(max(ans[i]) for i in range(n))
mn = min(min(ans[i] for i in range(n)))
srg = ans[x - 1][y - 1]
print(mx, mn, srg)
``` | output | 1 | 18,187 | 14 | 36,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,188 | 14 | 36,376 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
#!/usr/local/bin/python3.4
import pdb
import sys
def isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions):
return True if (SergeiX -1) * studens_in_column + SergeiY - 1 < left_questions else False
def isYpassSergei(column, studens_in_column, SergeiX, SergeiY, left_questions):
left_questions-= column * studens_in_column
if left_questions == 0:
return False
if SergeiX == column:
return False
return True if (column - SergeiX - 1) * studens_in_column + \
SergeiY <= left_questions else False
rows, studens_in_column, questions,SergeiX, SergeiY = map( int, input().split())
min_question=0
max_question=0
sergei_question=0
if rows == 1:
min_question = questions // studens_in_column
max_question = min_question + 1 if questions % studens_in_column != 0 else min_question
sergei_question = min_question if questions % studens_in_column < SergeiY else max_question
print (max_question, min_question, sergei_question)
sys.exit(0)
min_question = questions // ((2*rows-2)*studens_in_column)
max_question = 2 * min_question
left_questions = questions % ((2*rows-2) * studens_in_column)
sergei_question = min_question if SergeiX == rows or SergeiX == 1 else max_question
xx=0
if left_questions == 0 and rows != 2:
xx
elif rows == 2 :
max_question = min_question
if left_questions == 0:
xx
elif left_questions < rows * studens_in_column:
if left_questions < studens_in_column:
max_question = max_question + 1
sergei_question = min_question + 1 if isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions) else min_question
else:
max_question = min_question + 1
sergei_question = min_question + 1 if isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions) else min_question
else:
sergei_question = min_question = max_question = min_question+1
elif left_questions < rows*studens_in_column:
if max_question == 0 and left_questions != 0:
max_question+=1
elif left_questions > studens_in_column:
max_question+=1
sergei_question= sergei_question+1 if isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions) else sergei_question
else:
max_question = max_question + 1 if left_questions == rows * studens_in_column else max_question + 2
min_question+=1
sergei_question = sergei_question+2 if isYpassSergei(rows, studens_in_column, SergeiX, SergeiY, left_questions) else sergei_question + 1
print (max_question, min_question, sergei_question)
``` | output | 1 | 18,188 | 14 | 36,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,189 | 14 | 36,378 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
n, m, k, x, y = list(map(int, input().split()))
prohod = (2 * n - 2) * m
if n == 1:
prohod = m
last = k % prohod
minimum = k // prohod
maximum = (k // prohod) * 2
line = []
a1 = 10000000000000000000000000
a2 = -10000000000000000000000000
for i in range(n):
if i == 0 or i == n-1:
line += [minimum]
else:
line += [maximum]
ask = True
for i in range(n):
if (last >= m):
last -= m
line[i] += 1
else:
if (last > 0):
if (i == x - 1 and last < y):
ask = False
line[i] += 1
a1 = min(a1, line[i] - 1)
a2 = max(a2, line[i])
last = 0
else:
a1 = min(a1, line[i])
a2 = max(a2, line[i])
a2 = max(a2, line[i])
a1 = min(a1, line[i])
i = n - 2
while i > 0:
if (last >= m):
last -= m
line[i] += 1
else:
if (last > 0):
if (i == x - 1 and last < y):
ask = False
line[i] += 1
a1 = min(a1, line[i] - 1)
a2 = max(a2, line[i])
last = 0
else:
a1 = min(a1, line[i])
a2 = max(a2, line[i])
break
a2 = max(a2, line[i])
a1 = min(a1, line[i])
i -= 1
a3 = line[x - 1]
if not ask:
a3 -= 1
print(a2, a1, a3)
``` | output | 1 | 18,189 | 14 | 36,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,190 | 14 | 36,380 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
n,m,k,x,y=[int(x) for x in input().split()]
ma=0
mi=0
minn=199999999999999999
maxx=-1
if n is 1:
mi=k//m
ma=mi
k=k%m
a=[0]*m
for p in range (m):
if k is 0:
break
a[p]=1
k-=1
for p in range(m):
if a[p]>maxx:
maxx=a[p]
if a[p]<minn:
minn=a[p]
print(ma+maxx,mi+minn,mi+a[y-1])
else:
i=(2*n-2)*m
mi=k//i
if n>2:
ma=mi*2
else:
ma=mi
k=k%i
a = [[0 for p in range(m)] for q in range(n)]
for p in range(n):
for q in range(m):
if p is 0 or p is n-1:
a[p][q]=mi
else:
a[p][q]=ma
for p in range(n):
if k is 0:
break
for q in range(m):
if k is 0:
break
a[p][q]+=1
k-=1
for p in range(n):
if k is 0:
break
for q in range(m):
if k is 0:
break
a[n-2-p][q]+=1
k-=1
for p in range(n):
for q in range(m):
if a[p][q]>maxx:
maxx=a[p][q]
if a[p][q]<minn:
minn=a[p][q]
if x is 1 or x is n:
print(maxx,minn,a[x-1][y-1])
else:
print(maxx,minn,a[x-1][y-1])
``` | output | 1 | 18,190 | 14 | 36,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,191 | 14 | 36,382 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
import math
n, m, k, x, y = map(int, input().split())
if n < 3:
ma, mi = math.ceil(k/(n*m)), k//(n*m)
s = ma if k%(n*m) >= (x-1)*m+y else mi
print(ma, mi, s)
else:
t, mo = k//((2*n-2)*m), k%((2*n-2)*m)
if mo > n*m:
ma = 2*t+2
mi = t+1
elif mo == n*m:
ma = 2*t+1
mi = t+1
elif mo > (n-1)*m:
ma = 2*t+1
mi = t
elif mo > m:
ma = 2*t+1
mi = t
else:
ma = max(2*t, t+1)
mi = t
if x==1 or x==n:
if mo >= (x-1)*m+y:
s = t+1
else:
s = t
else:
if mo >= n*m+(n-x-1)*m+y:
s = 2*t+2
elif mo >= (x-1)*m+y:
s = 2*t+1
else:
s = 2*t
print(ma,mi,s)
``` | output | 1 | 18,191 | 14 | 36,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,192 | 14 | 36,384 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
from sys import stdin
import math
n, m, k, x, y = map(int, stdin.readline().rstrip().split())
tour_size = 2 * n * m - 2 * m if n > 1 else m
max_asked_per_tour = (2 if n >= 3 else 1)
min_asked_per_tour = 1
last_tour_count = k % tour_size
max_asked = max((k // tour_size) * max_asked_per_tour + (0 if last_tour_count <= m else 1 if m < last_tour_count <= m * n else 2), math.ceil(k / tour_size))
min_asked = (k // tour_size) * min_asked_per_tour + (1 if (last_tour_count >= n * m) else 0)
asked_per_tour = (2 if 1 < x < n else 1)
pre_tour_pos = (x - 1) * m + y
pos_tour_pos = (-1 if asked_per_tour == 1 else m * n + (n - x - 1) * m + y)
last_pre_tour_asked = (0 if k % tour_size < pre_tour_pos else 1)
last_pos_tour_asked = (0 if pos_tour_pos == -1 or k % tour_size < pos_tour_pos else 1)
sergei_asked = (k // tour_size) * asked_per_tour + last_pre_tour_asked + last_pos_tour_asked
print(' '.join(map(str, [max_asked, min_asked, sergei_asked])))
``` | output | 1 | 18,192 | 14 | 36,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,193 | 14 | 36,386 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
# mod=1000000007
mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('a')
file=1
def solve():
# for _ in range(ii()):
n,m,k,x,y=mi()
x-=1
y-=1
a=[[0 for i in range(m)] for j in range(n)]
x1=0
for i in range(n):
for j in range(m):
if(x1<k):
a[i][j]+=1
x1+=1
for i in range(n-2,0,-1):
for j in range(m):
if(x1<k):
a[i][j]+=1
x1+=1
for i in range(n):
for j in range(m):
a[i][j]*=(k//x1)
k%=x1
x1=0
for i in range(n):
for j in range(m):
if(x1<k):
a[i][j]+=1
x1+=1
for i in range(n-2,-1,-1):
for j in range(m):
if(x1<k):
a[i][j]+=1
x1+=1
mx,mnx=0,inf
for i in range(n):
for j in range(m):
mnx=min(mnx,a[i][j])
mx=max(mx,a[i][j])
print(mx,mnx,a[x][y])
if __name__ =="__main__":
if(file):
if path.exists('input1.txt'):
sys.stdin=open('input1.txt', 'r')
sys.stdout=open('output1.txt','w')
else:
input=sys.stdin.readline
solve()
``` | output | 1 | 18,193 | 14 | 36,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table; | instruction | 0 | 18,194 | 14 | 36,388 |
Tags: binary search, constructive algorithms, implementation, math
Correct Solution:
```
n, m, k, x1, y1 = map(int, input().split())
c = 0
if (n == 1):
mi = k // m
ma = mi + 1 * int(k % m != 0)
s = mi + 1 * int(k % m > y1 - 1)
else:
c = k // (m * (2 * n - 2))
d = k % (m * (2 * n - 2))
q = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m):
if 0 < i < n - 1:
q[i][j] = 2 * c
else:
q[i][j] = c
i = 0
while d > 0:
if i % (2 * n - 2) > n - 1:
x = 2 * n - 2 - (i % (2 * n - 2))
else:
x = i % (2 * n - 2)
j = 0
while d > 0 and j < m:
q[x][j] += 1
d -= 1
j += 1
i += 1
ma = -1
mi = 10 ** 100
for i in range(n):
for j in range(m):
mi = min(q[i][j], mi)
ma = max(q[i][j], ma)
s = q[x1 - 1][y1 - 1]
print(ma, mi, s)
``` | output | 1 | 18,194 | 14 | 36,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
n,m,k,x,y = map(int,input().split());
allnum = 0;
allnum = 2*n*m - 2*m
if n <= 2:
allnum = 2*n*m;
num = 2*(k // allnum);
r = k%(allnum);
g = [ [num]* m for i in range(0, n)]
if n > 2:
g[0] = [num //2]*m;
g[n-1] = [num //2]*m;
for i in range(0, n):
for j in range(0, m):
if r == 0:
break;
g[i][j]+=1
r-=1
if r == 0:
break;
for i in range(max(n - 2,0), -1,-1):
for j in range(0, m):
if r == 0:
break;
g[i][j]+=1
r-=1
if r == 0:
break;
mn = num + 1000000000000000;
mx = 0
for i in range(0, n):
if( max(g[i]) > mx):
mx = max(g[i])
if(min(g[i]) < mn):
mn = min(g[i])
print(mx,mn,g[x-1][y-1]);
``` | instruction | 0 | 18,195 | 14 | 36,390 |
Yes | output | 1 | 18,195 | 14 | 36,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
n, m, k, x, y = map(int, input().split())
if n > 1:
t = 2 * (n - 1) * m
s, p = k // t, k % t
f = lambda x, y: s + s * (1 < x < n) + (p >= m * x - m + y) + (p >= (2 * n - x - 1) * m + y) * (x < n)
print(max(f(1, 1), f(2, 1), f(n - 1, 1)), f(n, m), f(x, y))
else:
g = lambda y: (k - y + m) // m
print(g(1), g(m), g(y))
``` | instruction | 0 | 18,196 | 14 | 36,392 |
Yes | output | 1 | 18,196 | 14 | 36,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
from sys import stdin
def get(n, m, k, x, y):
if x > n or x < 1:
return -1
if n == 1:
return k // m + (k % m >= y)
kr = (n - 1) * 2 * m
if x == 1 or x == n:
res = k // kr
else:
res = k // kr * 2
sot = k % kr
if sot >= (x - 1) * m + y:
res += 1
if sot >= (2 * n - 1 - x) * m + y and x != n:
res += 1
return res
def solve():
n, m, k, x, y = map(int, input().split())
ans = [get(n, m, k, 2, 1), get(n, m, k, 1, 1), get(n, m, k, n-1, 1), get(n, m, k, n, m), get(n, m, k, 1, 1)]
ans = list(filter(lambda x: x != -1, ans))
print(max(ans), min(ans), get(n, m, k, x, y))
for i in range(1):
solve()
``` | instruction | 0 | 18,197 | 14 | 36,394 |
Yes | output | 1 | 18,197 | 14 | 36,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
def solve(n, m, k, x, y):
minv, maxv, curv = 0, 0, 0
if n == 1:
minv = k // m
maxv = minv
curv = minv
q = k % m
if q > 0:
maxv += 1
if q >= y:
curv += 1
elif n == 2:
minv = k // (m * 2)
maxv = minv
curv = minv
q = k % (m * 2)
if q > 0:
maxv += 1
if q >= (x - 1) * m + y:
curv += 1
else:
t = n * m * 2 - 2 * m
com = k // t
minv = com
maxv = com * 2
curv = com
if x != n and x != 1:
curv += com
q = k % t
if q > 0:
if q > m or com == 0:
maxv += 1
if q > n * m:
maxv += 1
if q >= n * m:
minv += 1
p1 = (x - 1) * m + y
if q >= p1:
curv += 1
if x != 1 and x != n:
p2 = t - (x - 1) * m + y
if q >= p2:
curv += 1
return (maxv, minv, curv)
n, m, k, x, y = map(int, input().split())
print(" ".join(map(str,solve(n, m, k, x, y))))
``` | instruction | 0 | 18,198 | 14 | 36,396 |
Yes | output | 1 | 18,198 | 14 | 36,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
#!/usr/local/bin/python3.4
import pdb
import sys
def isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions):
return True if (SergeiX -1) * studens_in_column + SergeiY - 1 < left_questions else False
def isYpassSergei(column, studens_in_column, SergeiX, SergeiY, left_questions):
left_questions-= column * studens_in_column
if left_questions == 0:
return False
if SergeiX == column:
return False
return True if (column - SergeiX - 1) * studens_in_column + \
SergeiY < left_questions else False
rows, studens_in_column, questions,SergeiX, SergeiY = map( int, input().split())
min_question=0
max_question=0
sergei_question=0
if rows == 1:
min_question = questions // studens_in_column
max_question = min_question + 1 if questions % studens_in_column != 0 else min_question
sergei_question = min_question if questions % studens_in_column < SergeiY else max_question
print (max_question, min_question, sergei_question)
sys.exit(0)
min_question = questions // ((2*rows-2)*studens_in_column)
max_question = 2 * min_question
left_questions = questions % ((2*rows-2) * studens_in_column)
sergei_question = min_question if SergeiX == rows or SergeiX == 1 else max_question
xx=0
if left_questions == 0 and rows != 2:
xx
elif rows == 2 :
max_question = min_question
if left_questions == 0:
xx
elif left_questions < rows * studens_in_column:
if left_questions < studens_in_column:
max_question = max_question + 1
sergei_question = min_question + 1 if isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions) else min_question
else:
max_question = min_question + 1
sergei_question = min_question + 1 if isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions) else min_question
else:
sergei_question = min_question = max_question = min_question+1
elif left_questions < rows*studens_in_column:
if max_question == 0 and left_questions != 0:
max_question+=1
elif left_questions > studens_in_column:
max_question+=1
sergei_question= sergei_question+1 if isQpassSergei(studens_in_column, SergeiX, SergeiY, left_questions) else sergei_question
else:
max_question = max_question + 1 if left_questions == rows * studens_in_column else max_question + 2
min_question+=1
sergei_question = sergei_question+2 if isYpassSergei(rows, studens_in_column, SergeiX, SergeiY, left_questions) else sergei_question + 1
print (max_question, min_question, sergei_question)
``` | instruction | 0 | 18,199 | 14 | 36,398 |
No | output | 1 | 18,199 | 14 | 36,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
from sys import stdin
def solve():
n, m, k, x, y = map(int, input().split())
kr = n + n - 2
if n == 1:
return print((k - 1) // m + 1, k // m, k // m + (k % m >= y))
mx = k // (kr * m) * 2
if k % (kr * m) > m:
mx += 1
if k % (kr * m) > n * m:
mx += 1
if k < n * m:
mn = 0
else:
d = k - (n - 1) * m
mn = d // (kr * m)
if d % (kr * m) >= m:
mn += 1
if x == 1:
ans = k // (kr * m)
if k % (kr * m) >= y:
ans += 1
elif x == n:
ans = k // (kr * m)
if k % (kr * m) >= (x - 1) * m + y:
ans += 1
else:
ans = k // (kr * m) * 2
if k % (kr * m) >= (x - 1) * m + y:
ans += 1
if k % (kr * m) >= (n - x) * m + (n - 1) * m + y:
ans += 1
print(mx, mn, ans)
for i in range(1):
solve()
``` | instruction | 0 | 18,200 | 14 | 36,400 |
No | output | 1 | 18,200 | 14 | 36,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
import math
import sys
def printf(*a):
if len(sys.argv) > 1:
print(a)
n, m, k, x, y = map(int, input().split())
mi = 0
ma = 0
serg = 0
filed_lines = int(k/m)
partial_line = k%m
printf(filed_lines, partial_line)
if n==1:
mi = filed_lines
ma = filed_lines
if partial_line > 0:
ma += 1
if y <= partial_line:
serg = ma
else:
serg = mi
else:
filed_class = int(filed_lines/(n-1))
partial_class = filed_lines%(n-1)
printf(filed_class, partial_class)
if filed_class%2 == 1:
mi = int((filed_class-1)/2)
if partial_class > 0:
mi += 1
else:
mi = int(filed_class/2)
if k < m:
ma = 1
else:
if n==2:
ma = int(filed_class/2)
else:
ma = filed_class
if partial_class>1 or (partial_class == 1 and partial_line > 0):
ma += 1
if x==1:
if filed_class%2==1:
serg = int(filed_class/2)+1
if partial_class==n-1 and partial_line >= y:
serg +=1
else:
serg = int(filed_class/2)
if partial_class>0 or (partial_class==0 and partial_line>=y):
serg += 1
elif x==n:
if filed_class%2==0:
serg = int(filed_class/2)
if partial_class==n-1 and partial_line >= y:
serg +=1
else:
serg = int(filed_class/2)
if partial_class>0 or (partial_class==0 and partial_line>=y):
serg += 1
else:
if filed_class%2==1:
x = n-x+1
if x<partial_class:
serg = filed_class+1
elif x==partial_class+1 and y<=partial_line:
serg = filed_class+1
else:
serg = filed_class
print("{} {} {}".format(ma, mi, serg))
``` | instruction | 0 | 18,201 | 14 | 36,402 |
No | output | 1 | 18,201 | 14 | 36,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 β€ n, m β€ 100, 1 β€ k β€ 1018, 1 β€ x β€ n, 1 β€ y β€ m).
Output
Print three integers:
1. the maximum number of questions a particular pupil is asked,
2. the minimum number of questions a particular pupil is asked,
3. how many times the teacher asked Sergei.
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
1. the pupil from the first row who seats at the first table, it means it is Sergei;
2. the pupil from the first row who seats at the second table;
3. the pupil from the first row who seats at the third table;
4. the pupil from the first row who seats at the first table, it means it is Sergei;
5. the pupil from the first row who seats at the second table;
6. the pupil from the first row who seats at the third table;
7. the pupil from the first row who seats at the first table, it means it is Sergei;
8. the pupil from the first row who seats at the second table;
The order of asking pupils in the second test:
1. the pupil from the first row who seats at the first table;
2. the pupil from the first row who seats at the second table;
3. the pupil from the second row who seats at the first table;
4. the pupil from the second row who seats at the second table;
5. the pupil from the third row who seats at the first table;
6. the pupil from the third row who seats at the second table;
7. the pupil from the fourth row who seats at the first table;
8. the pupil from the fourth row who seats at the second table, it means it is Sergei;
9. the pupil from the third row who seats at the first table;
Submitted Solution:
```
n,m,k,x,y=map(int,input().split(" "))
if n>1:
p=(k-1)//(2*m*(n-1))
ost=(k-1)%(2*m*(n-1))+1
h=(ost-1)//(m*n)
ost=(ost-1)%(m*n)+1
ser=0
if h==0:
r=(ost-1)//m+1
u=ost-r*m
if x<=r or (x==r+1 and y<=u):
ser+=1
zmin=p if not ((r==n) and (u==m)) else p+1
zmax=2*p+1 if r>=2 else max(p+1,2*p)
elif h==1:
ser+=1
r=n-1-(ost-1)//m
u=(ost-1)%m+1
if (x>r and x<n) or (x==r and y<=u):
ser+=1
zmin=p+1
zmax=2*p+2
if x==1 or x==n:
ser+=p
else:
ser+=2*p
else:
p=(k-1)//m
ost=(k-1)%m+1
ser=p+1 if y<=ost else p
zmax=p+1
if ost==m:
zmin=p+1
else:
zmin=p
print(zmax,zmin,ser)
``` | instruction | 0 | 18,202 | 14 | 36,404 |
No | output | 1 | 18,202 | 14 | 36,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,286 | 14 | 36,572 |
Tags: bitmasks, brute force
Correct Solution:
```
def rp():
cs = list(map(int, input().split(' ')))
cs = list(zip(cs[0::2], cs[1::2]))
return cs
def dist(p1, p2):
return len(set(p1).union(set(p2))) - 2
input()
ps = [rp(), rp()]
theyCan = True
myPos = set()
for ps1, ps2 in [ps, ps[::-1]]:
for p1 in ps1:
pos = set()
for p2 in ps2:
if dist(p1, p2) == 1:
pos = pos.union( set(p1).intersection(set(p2)) )
if len(pos) >= 2:
theyCan = False
myPos = myPos.union(pos)
print(next(iter(myPos)) if len(myPos)==1 else 0 if theyCan else -1)
``` | output | 1 | 18,286 | 14 | 36,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,287 | 14 | 36,574 |
Tags: bitmasks, brute force
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
can0 = set()
can1 = [set() for i in range(n)]
can2 = [set() for i in range(m)]
for i in range(n):
for j in range(m):
x1 = a[i * 2]
x2 = a[i * 2 + 1]
y1 = b[j * 2]
y2 = b[j * 2 + 1]
if x1 > x2: x1, x2 = x2, x1
if y1 > y2: y1, y2 = y2, y1
if x1 == y1 and x2 == y2: continue
if x1 == y1:
can1[i].add(y1)
can2[j].add(y1)
can0.add(y1)
if x2 == y1:
can1[i].add(y1)
can2[j].add(y1)
can0.add(y1)
if x1 == y2:
can1[i].add(y2)
can2[j].add(y2)
can0.add(y2)
if x2 == y2:
can1[i].add(y2)
can2[j].add(y2)
can0.add(y2)
if len(can0) == 1:
print(min(can0))
else:
ok = True
for i in can1:
if len(i) > 1:
ok = False
for i in can2:
if len(i) > 1:
ok = False
if ok: print(0)
else: print(-1)
``` | output | 1 | 18,287 | 14 | 36,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,288 | 14 | 36,576 |
Tags: bitmasks, brute force
Correct Solution:
```
from collections import defaultdict
def solve(s1, s2):
d = defaultdict(set)
for t1, t2 in s1:
d[t1].add(t2)
d[t2].add(t1)
ans = set()
for t1, t2 in s2:
if len(d[t1] - {t2}) > 0 and len(d[t2] - {t1}) == 0:
ans.add(t1)
elif len(d[t1] - {t2}) == 0 and len(d[t2] - {t1}) > 0:
ans.add(t2)
elif len(d[t1] - {t2}) > 0 and len(d[t2] - {t1}) > 0:
return -1
if len(ans) > 1:
return 0
if len(ans) == 1:
return list(ans)[0]
n, m = [int(x) for x in input().strip().split()]
a = [int(x) for x in input().strip().split()]
b = [int(x) for x in input().strip().split()]
a = set(tuple(sorted([a[i*2], a[i*2+1]])) for i in range(n) if a[i*2] != a[i*2+1])
b = set(tuple(sorted([b[i*2], b[i*2+1]])) for i in range(m) if b[i*2] != b[i*2+1])
ans = min(solve(a, b), solve(b, a))
print(ans)
``` | output | 1 | 18,288 | 14 | 36,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,289 | 14 | 36,578 |
Tags: bitmasks, brute force
Correct Solution:
```
from itertools import product, combinations
def main():
n, m = map(int, input().split())
aa = list(map(int, input().split()))
bb = list(map(int, input().split()))
a2 = [set(aa[i:i + 2]) for i in range(0, n * 2, 2)]
b2 = [set(bb[i:i + 2]) for i in range(0, m * 2, 2)]
seeds, l = set(aa) & set(bb), []
for s in seeds:
canda = {d for a in a2 for d in a if s in a and d != s}
candb = {d for b in b2 for d in b if s in b and d != s}
for a, b in product(canda, candb):
if a != b:
l.append((s, a, b))
print(l[0][0] if len(set(t[0] for t in l)) == 1 else
-any(a[:2] == b[-2::-1] or a[::2] == b[-1::-2]
for a, b in combinations(l, 2)))
if __name__ == '__main__':
main()
``` | output | 1 | 18,289 | 14 | 36,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,290 | 14 | 36,580 |
Tags: bitmasks, brute force
Correct Solution:
```
def readpts():
ip = list(map(int, input().split()))
return [(min(ip[i], ip[i+1]), max(ip[i], ip[i+1])) for i in range(0,len(ip),2)]
N, M = map(int, input().split())
pts1 = readpts()
pts2 = readpts()
#print(pts1)
#print(pts2)
def psb(a, b):
if a == b: return False
return any(i in b for i in a)
def sb(a, b):
for i in a:
if i in b:
return i
return -1 # should not happen
def ipsv(pts1, pts2):
ans = False
for p1 in pts1:
gsb = set()
for p2 in pts2:
if psb(p1, p2):
gsb.add(sb(p1, p2))
if len(gsb) > 1: return False
if len(gsb) == 1: ans = True
return ans
def sv():
gsb = set()
for p1 in pts1:
for p2 in pts2:
if psb(p1, p2):
gsb.add(sb(p1, p2))
if len(gsb) == 0: return -1
if len(gsb) == 1: return list(gsb)[0]
if ipsv(pts1, pts2) and ipsv(pts2, pts1): return 0
return -1
print(sv())
``` | output | 1 | 18,290 | 14 | 36,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,291 | 14 | 36,582 |
Tags: bitmasks, brute force
Correct Solution:
```
n, m = map(int, input().split())
aa = list(map(int, input().split()))
bb = list(map(int, input().split()))
a = []
b = []
for i in range(0, 2 * n, 2):
a.append([aa[i], aa[i + 1]])
for i in range(0, 2 * m, 2):
b.append([bb[i], bb[i + 1]])
accept = []
for num in range(1, 10):
ina = []
inb = []
for x in a:
if num in x:
ina.append(x)
for x in b:
if num in x:
inb.append(x)
x = 0
for t in ina:
t.sort()
for p in inb:
p.sort()
if t != p:
x += 1
if x > 0:
accept.append(num)
if len(accept) == 1:
print(accept[0])
exit(0)
#check fst
for t in a:
z = set()
for p in b:
if t != p:
if t[0] in p: z.add(t[0])
if t[1] in p: z.add(t[1])
if len(z) > 1:
print(-1)
exit(0)
#check scd
for t in b:
z = set()
for p in a:
if t != p:
if t[0] in p: z.add(t[0])
if t[1] in p: z.add(t[1])
if len(z) > 1:
print(-1)
exit(0)
print(0)
``` | output | 1 | 18,291 | 14 | 36,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,292 | 14 | 36,584 |
Tags: bitmasks, brute force
Correct Solution:
```
# your code goes here
n, m = map(int, input().split())
def split2(iterable):
args = [iter(iterable)] * 2
return zip(*args)
a = list(split2(map(int, input().split())))
b = list(split2(map(int, input().split())))
can = set()
for x in a:
for y in b:
intersections = set(x) & set(y)
if len(intersections) in [0, 2]:
continue
can.update(intersections)
if len(can) == 1:
print(next(iter(can)))
exit()
fail = 0
for x in a:
st = set()
for y in b:
intersections = set(x) & set(y)
if len(intersections) in [0, 2]:
continue
st.update(intersections)
fail |= len(st) > 1
for x in b:
st = set()
for y in a:
intersections = set(x) & set(y)
if len(intersections) in [0, 2]:
continue
st.update(intersections)
fail |= len(st) > 1
print(-1 if fail else 0)
``` | output | 1 | 18,292 | 14 | 36,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two participants are each given a pair of distinct numbers from 1 to 9 such that there's exactly one number that is present in both pairs. They want to figure out the number that matches by using a communication channel you have access to without revealing it to you.
Both participants communicated to each other a set of pairs of numbers, that includes the pair given to them. Each pair in the communicated sets comprises two different numbers.
Determine if you can with certainty deduce the common number, or if you can determine with certainty that both participants know the number but you do not.
Input
The first line contains two integers n and m (1 β€ n, m β€ 12) β the number of pairs the first participant communicated to the second and vice versa.
The second line contains n pairs of integers, each between 1 and 9, β pairs of numbers communicated from first participant to the second.
The third line contains m pairs of integers, each between 1 and 9, β pairs of numbers communicated from the second participant to the first.
All pairs within each set are distinct (in particular, if there is a pair (1,2), there will be no pair (2,1) within the same set), and no pair contains the same number twice.
It is guaranteed that the two sets do not contradict the statements, in other words, there is pair from the first set and a pair from the second set that share exactly one number.
Output
If you can deduce the shared number with certainty, print that number.
If you can with certainty deduce that both participants know the shared number, but you do not know it, print 0.
Otherwise print -1.
Examples
Input
2 2
1 2 3 4
1 5 3 4
Output
1
Input
2 2
1 2 3 4
1 5 6 4
Output
0
Input
2 3
1 2 4 5
1 2 1 3 2 3
Output
-1
Note
In the first example the first participant communicated pairs (1,2) and (3,4), and the second communicated (1,5), (3,4). Since we know that the actual pairs they received share exactly one number, it can't be that they both have (3,4). Thus, the first participant has (1,2) and the second has (1,5), and at this point you already know the shared number is 1.
In the second example either the first participant has (1,2) and the second has (1,5), or the first has (3,4) and the second has (6,4). In the first case both of them know the shared number is 1, in the second case both of them know the shared number is 4. You don't have enough information to tell 1 and 4 apart.
In the third case if the first participant was given (1,2), they don't know what the shared number is, since from their perspective the second participant might have been given either (1,3), in which case the shared number is 1, or (2,3), in which case the shared number is 2. While the second participant does know the number with certainty, neither you nor the first participant do, so the output is -1. | instruction | 0 | 18,293 | 14 | 36,586 |
Tags: bitmasks, brute force
Correct Solution:
```
def rp():
res = list(map(int, input().split(' ')))
res = list(zip(res[0::2], res[1::2]))
return res
def dist(a, b):
return len(set(a).union(set(b))) - 2
n, m = map(int, input().split(' '))
ps = [rp(), rp()]
theyCan = True
myPos = set()
for ps1, ps2 in [ps, ps[::-1]]:
#print('------------')
for p1 in ps1:
pos = set()
for p2 in ps2:
if dist(p1, p2) == 1:
pos = pos.union( set(p1).intersection(set(p2)) )
#print(p1, p2, pos)
if len(pos) >= 2:
theyCan = False
myPos = myPos.union(pos)
print(next(iter(myPos)) if len(myPos)==1 else 0 if theyCan else -1)
``` | output | 1 | 18,293 | 14 | 36,587 |
Provide a correct Python 3 solution for this coding contest problem.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354 | instruction | 0 | 18,307 | 14 | 36,614 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
p = round(sum(A) / N)
print(sum([(a-p)**2 for a in A]))
``` | output | 1 | 18,307 | 14 | 36,615 |
Provide a correct Python 3 solution for this coding contest problem.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354 | instruction | 0 | 18,310 | 14 | 36,620 |
"Correct Solution:
```
n = int(input())
x = [int(i) for i in input().split()]
p = round(sum(x)/n)
d = sum([(i-p)**2 for i in x])
print(d)
``` | output | 1 | 18,310 | 14 | 36,621 |
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