message stringlengths 2 433k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,850 | 12 | 121,700 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
def calc(a,b):
if (a==0):
return b
if (b==0):
return a
if (a==b):
return a
while (a>0 and b>0):
if (a>b):
a,b=b,a-b
else:
a,b=b-a,a
return max(a,b)
t=int(input())
while t:
n=int(input())
a=list(map(int, input().split()))
p=min(a)
b=[]
for j in a:
b.append(j)
b.sort()
c=[]
for i in range(0,n):
if a[i]!=b[i]:
if a[i]%p:
print ("NO")
break
else:
print ("YES")
t=t-1
``` | output | 1 | 60,850 | 12 | 121,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,851 | 12 | 121,702 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
l=list(map(int,input().split()))
a=l.copy()
a.sort()
for j in range(n):
if l[j]!=a[j] and l[j]%a[0]!=0:
print('NO')
break
else:
print('YES')
``` | output | 1 | 60,851 | 12 | 121,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,852 | 12 | 121,704 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
'''Author- Akshit Monga'''
t=int(input())
for _ in range(t):
n=int(input())
arr=[int(x) for x in input().split()]
x=min(arr)
arr1=sorted(arr)
ans="YES"
for i in range(n):
if arr1[i]!=arr[i]:
if arr1[i]%x!=0 or arr[i]%x!=0:
ans="NO"
break
print(ans)
``` | output | 1 | 60,852 | 12 | 121,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,853 | 12 | 121,706 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
import math
# def func(lis1,s,m):
# x = lis1[:]
# # h = 0
# # i = 0
# # el = lis1[i]
# # while el != s[0]:
# # if math.gcd(lis1[0],s[0]) == m:
# # h = 1
# # break
# # i += 1
# # el = lis1[i]
# # if h!= 1:
# # return 1
# # else:
# # lis1.remove(m)
# # lis1.insert(i,lis1[0])
# # lis1[0] = m
# el = lis1[0]
# while el != min(lis1):
# if math.gcd(lis1[0],s[0]) == m:
# checkfunc(lis1)
# if len(lis1) == 0:
# return 2
# elif x == lis1:
# return 1
# else:
# return func(lis1,lis1.sort(),m)
# def checkfunc(lis1):
# if len(lis1) == 0:
# return lis1
# else:
# while len(lis1) != 0:
# if max(lis1) == lis1[-1]:
# lis1.pop()
# else:
# break
# while len(lis1) != 0:
# if min(lis1) == lis1[0]:
# lis1.pop(0)
# else:
# break
# return lis1
# t = int(input())
# for h in range(t):
# n = int(input())
# lis1 = list(map(int, input().split()))
# s = lis1[:]
# s.sort()
# m = min(lis1)
# if lis1 == s:
# print("YES")
# elif len(lis1) == 1:
# print("YES")
# else:
# j = func(lis1,s,m)
# if j ==2:
# print("YES")
# else:
# print("NO")
t = int(input())
for h in range(t):
n = int(input())
lis1 = list(map(int, input().split()))
s = lis1[:]
s.sort()
m = min(lis1)
k = 0
for i in range(n):
if math.gcd(m,lis1[i]) != m:
if s[i] != lis1[i]:
print("NO")
k = 1
break
if k == 0:
print("YES")
``` | output | 1 | 60,853 | 12 | 121,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,854 | 12 | 121,708 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
testcase=int(input())
for t in range(testcase):
n=int(input())
arr=list(map(int,input().split()))
final=[]
for i in arr:
final.append(i)
final.sort()
misplaced=[]
m=min(arr)
for i in range(n):
if(arr[i]!=final[i]):
misplaced.append(arr[i])
ans=1
for i in misplaced:
if(i%m!=0):
ans=0
break
if(ans==0):
print("NO")
else:
print("YES")
``` | output | 1 | 60,854 | 12 | 121,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,855 | 12 | 121,710 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
def gcd(a,b):
if a%b ==0:
return b
elif b%a==0:
return a
else:
return gcd(b,a%b)
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
p=min(arr)
q=[]
for i in range(n):
if gcd(arr[i],p)==p:
q.append(arr[i])
arr[i]=-1
q.sort()
for i in range(n):
if arr[i]==-1:
arr[i]=q.pop(0)
for i in range(1,n):
if arr[i-1]>arr[i]:
print("NO")
break
else:
print("YES")
``` | output | 1 | 60,855 | 12 | 121,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation. | instruction | 0 | 60,856 | 12 | 121,712 |
Tags: constructive algorithms, math, number theory, sortings
Correct Solution:
```
# Author : devil9614 - Sujan Mukherjee
from __future__ import division, print_function
import os,sys
import math
import collections
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
class my_dictionary(dict):
def __init__(self):
self = dict()
def add(self,key,value):
self[key] = value
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(100000000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[1]
def sort2(l):return sorted(l, key=getKey,reverse=True)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(n-r))
def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return mid
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def ceil(x,y):
if x%y==0:
return x//y
else:
return x//y+1
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def padded_bin_with_complement(x):
if x < 0:
return bin((2**16) - abs(x))[2:].zfill(16)
else:
return bin(x)[2:].zfill(16)
def binaryToDecimal(binary):
binary1 = binary
decimal, i, n = 0, 0, 0
while(binary != 0):
dec = binary % 10
decimal = decimal + dec * pow(2, i)
binary = binary//10
i += 1
print(decimal)
def CountFrequency(my_list):
freq = {}
for item in my_list:
if (item in freq):
freq[item] += 1
else:
freq[item] = 1
return freq
def pos(a):
b = [0]*len(a)
c = sorted(a)
for i in range(len(a)):
for j in range(len(a)):
if c[j] == a[i]:
b[i] = j
break
return b
def smallestDivisor(n):
# if divisible by 2
if (n % 2 == 0):
return 2
# iterate from 3 to sqrt(n)
i = 3
while(i * i <= n):
if (n % i == 0):
return i
i += 2
return n
def commonn(a,b,n):
c = []
for i in range(n):
if a[i] == b[i]:
c.append("-1")
else:
c.append(b[i])
return c
def primeFactors(n):
j = []
while n % 2 == 0:
j.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
j.append(int(n))
n = n / i
if n > 2:
j.append(int(n))
return j
def sumdigit(n):
n = str(n)
k = 0
for i in range(len(n)):
k+=int(n[i])
return k
def main():
for _ in range(ii()):
n = ii()
a = li()
e = []
b = []
f = 1
c = sorted(a)
minn = min(a)
d = [0]*n
for i in range(n):
if a[i] != c[i]:
if gcd(a[i],minn) != minn:
f = 0
if f:
print("YES")
else:
print("NO")
# if count == len(e):
# print("YES")
# else:
# print("NO")
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
``` | output | 1 | 60,856 | 12 | 121,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=input()
a=list(map(lambda x: int(x), input().split()))
b=sorted(a)
numbers=[]
minimum=b[0]
for i in range(0,len(a)):
if a[i]!=b[i]:
numbers.append(a[i])
flag=0
count1=0
countgt1=0
if minimum==1:
print("YES")
else:
for number in numbers:
if number%minimum!=0:
flag=1
break
if flag==1:
print("NO")
else:
print("YES")
``` | instruction | 0 | 60,857 | 12 | 121,714 |
Yes | output | 1 | 60,857 | 12 | 121,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
def mere(n,a):
b=sorted(a)
g=b[0]
for i in range(n):
if a[i]==b[i]:
continue
else:
if a[i]%g==0 and b[i]%g==0:
continue
else:
return "NO"
return "YES"
t=int(input())
a=[]
for i in range(t):
n=int(input())
array=list(map(int,input().split(" ")))
a.append([n,array])
for x in a:
print(mere(*x))
``` | instruction | 0 | 60,858 | 12 | 121,716 |
Yes | output | 1 | 60,858 | 12 | 121,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
t=a[:]
t.sort()
m=min(a)
flag=0
for i in range(n):
if a[i]!=t[i]:
if a[i]%m!=0:
flag=1
break
if flag==1:
print('NO')
else:
print('YES')
``` | instruction | 0 | 60,859 | 12 | 121,718 |
Yes | output | 1 | 60,859 | 12 | 121,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
import sys
def FastInt(zero=0):
_ord, nums, num, neg = lambda x: x, [], zero, False
i, s = 0, sys.stdin.buffer.read()
try:
while True:
if s[i] >= b"0"[0]:num = 10 * num + _ord(s[i]) - 48
elif s[i] == b"-"[0]:neg = True
elif s[i] != b"\r"[0]:
nums.append(-num if neg else num)
num, neg = zero, False
i += 1
except IndexError:
pass
if s and s[-1] >= b"0"[0]: nums.append(-num if neg else num)
return nums
inp=FastInt();ii=0
def inin(size=None):
global ii
if size==None:
ni=ii;ii+=1
return inp[ni]
else:
ni=ii;ii+=size
return inp[ni:ni+size]
from math import gcd
_T_=inin()
for _t_ in range(_T_):
n=inin()
a=inin(n)
mil=min(a)
b=a[:]
b.sort()
check=[]
poss=True
for i in range(n):
if a[i]!=b[i]:
if gcd(a[i],b[i])%mil==0:
pass
else:
poss=False
if poss:
print('YES')
else:
print('NO')
``` | instruction | 0 | 60,860 | 12 | 121,720 |
Yes | output | 1 | 60,860 | 12 | 121,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
import math
from copy import deepcopy
for _ in range(int(input())):
test=0
n=int(input())
a=list(map(int,input().split()))
m=min(a)
b=deepcopy(a)
b.sort()
for i in range(n):
if (math.gcd(a[i],b[i])==m or a[i] == b[i]):
continue
else:
test=1
break
if (test == 0):
print("YES")
else:
print("NO")
``` | instruction | 0 | 60,861 | 12 | 121,722 |
No | output | 1 | 60,861 | 12 | 121,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
mi=min(a)
b=a[:]
b.sort()
if b==a:
print("YES")
else:
for i in range(n):
if a[i]!=b[i]:
if i%mi:
print("NO")
break
else:
print("YES")
``` | instruction | 0 | 60,862 | 12 | 121,724 |
No | output | 1 | 60,862 | 12 | 121,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Fri Aug 21 22:02:30 2020
@author: Dark Soul
"""
t=int(input(''))
arr=[]
for i in range(t):
input('')
arr.append(list(map(int, input().split())))
for i in arr:
x=sorted(i)
diff=x
mn=x[0]
flag=0
for j in range(len(i)):
if x[j]==i[j]:
continue
else:
if abs(x[j]-i[j])%mn:
flag=1
break
if flag:
print('NO')
else:
print('YES')
``` | instruction | 0 | 60,863 | 12 | 121,726 |
No | output | 1 | 60,863 | 12 | 121,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n where all a_i are integers and greater than 0.
In one operation, you can choose two different indices i and j (1 ≤ i, j ≤ n). If gcd(a_i, a_j) is equal to the minimum element of the whole array a, you can swap a_i and a_j. gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.
Now you'd like to make a non-decreasing using the operation any number of times (possibly zero). Determine if you can do this.
An array a is non-decreasing if and only if a_1 ≤ a_2 ≤ … ≤ a_n.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of array a.
The second line of each test case contains n positive integers a_1, a_2, … a_n (1 ≤ a_i ≤ 10^9) — the array itself.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output "YES" if it is possible to make the array a non-decreasing using the described operation, or "NO" if it is impossible to do so.
Example
Input
4
1
8
6
4 3 6 6 2 9
4
4 5 6 7
5
7 5 2 2 4
Output
YES
YES
YES
NO
Note
In the first and third sample, the array is already non-decreasing.
In the second sample, we can swap a_1 and a_3 first, and swap a_1 and a_5 second to make the array non-decreasing.
In the forth sample, we cannot the array non-decreasing using the operation.
Submitted Solution:
```
def STR(): return list(input())
def INT(): return int(input())
def MAP(): return map(int, input().split())
def MAP2():return map(float,input().split())
def LIST(): return list(map(int, input().split()))
def STRING(): return input()
import string
import sys
from heapq import heappop , heappush
from bisect import *
from collections import deque , Counter , defaultdict
from math import *
from itertools import permutations , accumulate
dx = [-1 , 1 , 0 , 0 ]
dy = [0 , 0 , 1 , - 1]
#visited = [[False for i in range(m)] for j in range(n)]
#sys.stdin = open(r'input.txt' , 'r')
#sys.stdout = open(r'output.txt' , 'w')
#for tt in range(INT()):
#CODE
def solve(a , b):
if b == 0 :
return a
else:
return solve(b , a % b)
for tt in range(INT()):
n = INT()
arr = LIST()
l = sorted(arr)
if l == arr:
print('YES')
else:
mn = min(arr)
st = set()
for i in range(n):
if arr[i] != l[i] and arr[i] % mn == 0 :
for j in range(n):
if i != j and arr[j] != l[j] and arr[j] % mn == 0 :
if solve(arr[i] , arr[j]) == min(arr):
arr[i] , arr[j] = arr[j] , arr[i]
break
#print(arr)
if arr == l :
print('YES')
else:
print('NO')
``` | instruction | 0 | 60,864 | 12 | 121,728 |
No | output | 1 | 60,864 | 12 | 121,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,913 | 12 | 121,826 |
Tags: math
Correct Solution:
```
import math
t = int(input())
for i in range(t):
l = list(map(int, input().rstrip().split()))
n = l[0]
k = l[1]
l1 = list(map(int, input().rstrip().split()))
l1.sort()
count=0
c=0
for m in range(len(l1)):
if l1[m]==m:
count+=1
if count==len(l1):
print(n+k)
else:
a = max(l1)
for x in range(0, a + 2):
if l1.count(x) == 0:
b = x
break
for j in range(k):
a=max(l1)
if math.ceil((a + b) / 2)!=c:
c = math.ceil((a + b) / 2)
l1.append(c)
else:
break
s=set(l1)
print(len(s))
break
``` | output | 1 | 60,913 | 12 | 121,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,914 | 12 | 121,828 |
Tags: math
Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
N, K = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
A_set = set(A)
if K == 0:
print(len(A_set))
continue
mex = 0
for i in range(10 ** 5 + 2):
if i in A_set:
continue
else:
mex = i
break
ma = max(A)
if mex - ma == 1:
print(len(A_set) + K)
else:
A_set.add(-(-(mex + ma) // 2))
print(len(A_set))
if __name__ == '__main__':
main()
``` | output | 1 | 60,914 | 12 | 121,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,915 | 12 | 121,830 |
Tags: math
Correct Solution:
```
import sys, math, itertools, random, bisect
from collections import defaultdict
INF = sys.maxsize
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
mod = 10**9 + 7
# MAX = 100001
# def sieve():
# isPrime = [True]*(MAX)
# isPrime[0] = False
# isPrime[1] = False
# for i in range(2,MAX):
# if isPrime[i]:
# for j in range(i*i, MAX, i):
# isPrime[j] = False
# primes = [2]
# for i in range(3,MAX,2):
# if isPrime[i]: primes.append(i)
# return primes
for _ in range(int(input())):
n,k = get_ints()
store = set()
a = get_array()
mx = -1
for i in a:
store.add(i)
mx = max(mx,i)
mex = -1
for i in range(mx+5):
if i not in store:
mex = i
break
x = math.ceil((mx+mex)/2)
if k==0: print(n)
elif mex>mx: print(n+k)
else:
if x in store: print(n)
else: print(n+1)
``` | output | 1 | 60,915 | 12 | 121,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,916 | 12 | 121,832 |
Tags: math
Correct Solution:
```
from sys import stdin
input = stdin.readline
from heapq import heapify,heappush,heappop,heappushpop
from collections import defaultdict as dd, deque as dq,Counter as C
from math import factorial as f ,ceil,gcd,sqrt,log
from bisect import bisect_left as bl ,bisect_right as br
from itertools import combinations as c,permutations as p
from math import factorial as f ,ceil,gcd,sqrt,log
mi = lambda : map(int,input().split())
ii = lambda: int(input())
li = lambda : list(map(int,input().split()))
mati = lambda r : [ li() for _ in range(r)]
lcm = lambda a,b : (a*b)//gcd(a,b)
def solve():
n,k=mi()
arr=li()
arr.sort()
mex=-1
for x in range(n):
if arr[x]!=x:
mex=x
break
ans=len(list(set(arr)))
if mex==-1:
mex=arr[-1]+1
ans+=k
elif ceil((mex+max(arr))/2) not in arr and k>0:
ans+=1
print(ans)
for _ in range(ii()):
solve()
``` | output | 1 | 60,916 | 12 | 121,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,917 | 12 | 121,834 |
Tags: math
Correct Solution:
```
import math
for t in range(int(input())):
n, k = map(int, input().split())
nums = set(map(int, input().split()))
mexn = 0
while mexn in nums:
mexn += 1
maxn = max(nums)
if mexn > maxn or k == 0:
print(n + k)
else:
print(n if math.ceil((mexn + maxn) / 2) in nums else n+1)
``` | output | 1 | 60,917 | 12 | 121,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,918 | 12 | 121,836 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int, input().split())
s = list(map(int, input().split()))
s.sort()
a = max(s)
flag = 0
s = set(s)
for i in s:
if i==flag:
flag+=1
else:
break
if k ==0:
ans = n
print(ans)
else:
if flag>a:
ans = n+k
print(ans)
else:
s.add((flag+a+1)//2)
print(len(s))
``` | output | 1 | 60,918 | 12 | 121,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,919 | 12 | 121,838 |
Tags: math
Correct Solution:
```
from bisect import bisect_left
def solve(arr, k):
n = len(arr)
if k==0: return n
arr.sort()
l = 0
while l < n and arr[l] == l: l += 1
if l==n: return n+k
m = (arr[-1]+l+1) // 2
i = bisect_left(arr, m)
return n if i<n and arr[i]==m else n+1
for _ in range(int(input())):
_, k = map(int,input().split())
arr = list(map(int,input().split()))
print(solve(arr, k))
``` | output | 1 | 60,919 | 12 | 121,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4. | instruction | 0 | 60,920 | 12 | 121,840 |
Tags: math
Correct Solution:
```
import sys
import math
input = sys.stdin.readline
for _ in range(int(input())):
n, k = [int(i) for i in input().split()]
a = sorted([int(i) for i in input().split()])
if a == list(range(n)):
print(n + k)
else:
c = 0
for i in a:
if i != c:
break
c += 1
c = int(math.ceil((c + a[-1]) / 2))
a = set(a)
if k:
a.add(c)
print(len(a))
``` | output | 1 | 60,920 | 12 | 121,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
rn=lambda:int(input())
rns=lambda:map(int,input().split())
rl=lambda:list(map(int,input().split()))
rs=lambda:input()
YN=lambda x:print('YES') if x else print('NO')
mod=10**9+7
from math import ceil
for _ in range(rn()):
n,k=rns()
s=rl()
m=max(s)
h=set(s)
if k==0:
print(len(h))
else:
for i in range(10**9+1):
if i not in h:
if i>m:
print(len(h)+k)
elif ceil((i+m)/2) not in h:
print(len(h)+1)
else:
print(len(h))
break
``` | instruction | 0 | 60,921 | 12 | 121,842 |
Yes | output | 1 | 60,921 | 12 | 121,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
arr = [int(j) for j in input().split()]
arr = list(set(arr))
arr.sort()
if k == 0:
print(len(arr))
continue
if arr[0] != 0:
mex = 0
else:
flag = 0
for i in range(1, n):
if arr[i] != arr[i-1]+1:
flag = 1
mex = arr[i-1]+1
if flag == 0:
mex = arr[-1] + 1
ma = arr[-1]
if mex == ma+1:
print(len(arr)+k)
else:
new = (mex+ma)//2
if (mex+ma)%2 != 0:
new += 1
if new in arr:
print(len(arr))
else:
print(len(arr)+1)
``` | instruction | 0 | 60,922 | 12 | 121,844 |
Yes | output | 1 | 60,922 | 12 | 121,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
R=lambda:map(int,input().split())
t,=R()
exec(t*'n,k=R();a={*R()};i=0\nwhile{i}&a:i+=1\nprint(n+(k>0)*(k,i+max(a)+1>>1not in a)[i<n])\n')
``` | instruction | 0 | 60,923 | 12 | 121,846 |
Yes | output | 1 | 60,923 | 12 | 121,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
from math import ceil
t=int(input())
for _ in range(t):
n,k = map(int,input().split())
s1= set(int(x) for x in input().split())
s1=sorted(s1)
flag=0
for id,val in enumerate(s1):
if id!=val:
flag=1
break;
mex=id if flag==1 else id+1
maxs=max(s1)
term=ceil((maxs+mex)/2)
if term<=maxs:
if term in s1 or k==0:
print(len(s1))
else:
print(len(s1)+1)
else :
print(len(s1)+k)
``` | instruction | 0 | 60,924 | 12 | 121,848 |
Yes | output | 1 | 60,924 | 12 | 121,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n,k = map(int,input().split())
s=input()
arr=list(map(int,s.split(" ")))
arr.sort()
q=set(arr)
a=0
if k==0:
print(len(q))
else:
for i in range(0,n):
if arr[i]!=i:
if a<i:
a=i
if a!=-1:
b=arr[-1]
if (a+b)/2<1:
s=1
else:
s=round((a+b)/2)
if s not in q:
q.add(s)
print(len(q))
else:
print(len(q)+k)
``` | instruction | 0 | 60,925 | 12 | 121,850 |
No | output | 1 | 60,925 | 12 | 121,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
#codeforces 706 div 2 problem b
import math as m
def mex(u):
a = 0
flag = 0
for i in range(len(u)):
if(a == u[i]):
a += 1
flag += 1
if(flag - i == 1):
break
return (a)
t = int(input())
for i in range(t):
n, k = input().split(" ")
n = int(n)
k = int(k)
s = [int(x) for x in input().split(" ")]
s.sort()
#print(n, k, s, s[1], sep = " , ")
for y in range(k):
a = s[-1]
b = mex(s)
z = m.ceil((a + b)/2)
s.append(z)
s = list(dict.fromkeys(s))
print(len(s))
``` | instruction | 0 | 60,926 | 12 | 121,852 |
No | output | 1 | 60,926 | 12 | 121,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
import math
t = int(input())
for i in range(t):
n, k = map(int, input().split())
d = set(map(int, input().split()))
for j in range(k):
b = max(d)
for k in range(n):
if k not in d:
a = k
break
d.add(math.ceil((a+b)/2))
print (len(d))
``` | instruction | 0 | 60,927 | 12 | 121,854 |
No | output | 1 | 60,927 | 12 | 121,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S initially consisting of n distinct non-negative integers. A multiset is a set, that can contain some elements multiple times.
You will perform the following operation k times:
* Add the element ⌈(a+b)/(2)⌉ (rounded up) into S, where a = \operatorname{mex}(S) and b = max(S). If this number is already in the set, it is added again.
Here \operatorname{max} of a multiset denotes the maximum integer in the multiset, and \operatorname{mex} of a multiset denotes the smallest non-negative integer that is not present in the multiset. For example:
* \operatorname{mex}(\{1,4,0,2\})=3;
* \operatorname{mex}(\{2,5,1\})=0.
Your task is to calculate the number of distinct elements in S after k operations will be done.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤ n≤ 10^5, 0≤ k≤ 10^9) — the initial size of the multiset S and how many operations you need to perform.
The second line of each test case contains n distinct integers a_1,a_2,...,a_n (0≤ a_i≤ 10^9) — the numbers in the initial multiset.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print the number of distinct elements in S after k operations will be done.
Example
Input
5
4 1
0 1 3 4
3 1
0 1 4
3 0
0 1 4
3 2
0 1 2
3 2
1 2 3
Output
4
4
3
5
3
Note
In the first test case, S=\{0,1,3,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,3,4\}. The answer is 4.
In the second test case, S=\{0,1,4\}, a=\operatorname{mex}(S)=2, b=max(S)=4, ⌈(a+b)/(2)⌉=3. So 3 is added into S, and S becomes \{0,1,3,4\}. The answer is 4.
Submitted Solution:
```
from __future__ import division, print_function
from collections import *
from math import *
from itertools import *
from time import time
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
n, k = map(int, input().split())
S = set(list(map(int, input().split())))
big = max(S)
small = 0
for i in range(big + 2):
if i not in S:
small = i
break
for j in range(k):
S.add(ceil((big + small) / 2))
print(len(S))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
t = int(input())
while(t):
main()
t -= 1
``` | instruction | 0 | 60,928 | 12 | 121,856 |
No | output | 1 | 60,928 | 12 | 121,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 60,999 | 12 | 121,998 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
#in the name of god
#Mr_Rubick
n,m,k=map(int,input().split())
print(str(m*(m-1)//2))
for i in range(1,m):
for j in range(i+1,m+1):
if k==0:
print(str(i)+" "+str(j))
else:
print(str(j)+" "+str(i))
``` | output | 1 | 60,999 | 12 | 121,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,000 | 12 | 122,000 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
n, m, k = map(int, input().split())
arrays = []
for i in range(n):
arrays.append(list(map(int, input().split())))
print(m * (m - 1) // 2)
if k == 0:
for i in range(1, m):
for j in range(1, m - i + 1):
print(j, j + 1)
else:
for i in range(1, m):
for j in range(m, i, -1):
print(j, j - 1)
``` | output | 1 | 61,000 | 12 | 122,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,001 | 12 | 122,002 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
from collections import *
import sys
input=sys.stdin.readline
# "". join(strings)
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
n, m, k = rl()
AA = []
for _ in range(n):
AA.append(rl())
print((m * (m- 1))//2)
if k == 0:
for i in range(1, m + 1):
for j in range(i + 1, m + 1):
print(i, j)
else:
for i in range(1, m + 1):
for j in range(i + 1, m + 1):
print(j, i)
``` | output | 1 | 61,001 | 12 | 122,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,002 | 12 | 122,004 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
# -*- encoding: utf-8 -*-
n, m, k = map(int, input().split())
print(m*(m-1)//2)
for i in range(m-1):
for j in range(m-1-i):
print('{} {}'.format(*((j+1, j+2)) if k == 0 else (j+2, j+1)))
``` | output | 1 | 61,002 | 12 | 122,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,003 | 12 | 122,006 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
n,m,k = map(int,input().split())
print(str(m*(m-1)//2))
for i in range(m):
for j in range(m-i-1):
if k == 0:
print(str(i+1)+' '+str(i+j+2))
else:
print(str(i+j+2)+' '+str(i+1))
``` | output | 1 | 61,003 | 12 | 122,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,004 | 12 | 122,008 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
n, m, k = map(int, input().split())
print(m * (m - 1) // 2)
for i in range(1, m):
for j in range(i + 1, m + 1):
if k == 0:
print (i,j)
else:
print(j,i)
``` | output | 1 | 61,004 | 12 | 122,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,005 | 12 | 122,010 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
n,m,k=map(int,input().split())
print((m*(m-1))//2)
for i in range(1,m):
for j in range(i+1,m+1):
if k:
print(j,i)
else:
print(i,j)
``` | output | 1 | 61,005 | 12 | 122,011 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5]. | instruction | 0 | 61,006 | 12 | 122,012 |
Tags: greedy, implementation, sortings, two pointers
Correct Solution:
```
n,m,k=map(int,input().split())
print(m*(m-1)//2)
for i in range(n):
a=list(map(int,input().split()))
if k==0:
for i in range(m):
for j in range(i+1,m):
print(i+1,j+1)
else:
for i in range(m):
for j in range(i+1,m):
print(j+1,i+1)
``` | output | 1 | 61,006 | 12 | 122,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
# ls.sort(key=lambda x: (x[0], x[1]))
import sys
inf = float("inf")
mod = 1000000007
def array(): return list(map(int, sys.stdin.readline().split()))
def intt(): return map(int, sys.stdin.readline().split())
from bisect import bisect_left
import sys
#n, k = map(int, sys.stdin.readline().split())
#arr = list(map(int, sys.stdin.readline().split()))
#arr=[(int(x),i) for i,x in enumerate(input().split())]
# ls=list(map(int,input().split()))
# for i in range(m):
#print(s[i],end="")
#n=int(sys.stdin.readline())
n,m,k=map(int, sys.stdin.readline().split())
ls=[]
for i in range(n):
arr = list(map(int, sys.stdin.readline().split()))
ls.append(arr)
print(m*(m-1)//2)
for i in range(1,m):
for j in range(i+1,m+1):
if k==0:
print(i, j)
else:
print(j, i)
``` | instruction | 0 | 61,007 | 12 | 122,014 |
Yes | output | 1 | 61,007 | 12 | 122,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n,m,o=value()
for i in range(n):
input()
print(m*(m-1)//2)
for i in range(1,m+1):
for j in range(i+1,m+1):
if(o==0):
print(i,j)
else:
print(m-i+1,m-j+1)
``` | instruction | 0 | 61,008 | 12 | 122,016 |
Yes | output | 1 | 61,008 | 12 | 122,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
n,m,k = map(int,input().split())
for i in range(n):
a = [int(x) for x in input().split()]
d = []
for i in range(1,m):
for j in range(i+1,m+1):
if k == 0:
d.append((i,j))
else:
d.append((j,i))
print(len(d))
for i in d:
print(*i)
``` | instruction | 0 | 61,009 | 12 | 122,018 |
Yes | output | 1 | 61,009 | 12 | 122,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
_, m, k = map(int, input().split())
print(m * (m-1) // 2)
for i in range(m):
for j in range(i+1, m):
print(j+1, i+1) if k else print(i+1, j+1)
``` | instruction | 0 | 61,010 | 12 | 122,020 |
Yes | output | 1 | 61,010 | 12 | 122,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
n, m, k = map(int, input().split())
for i in range(n):
a = list(map(int, input().split()))
if k == 0:
x = 1
y = m + 1
z = 1
else:
x = m
y = 0
z = -1
print(m * (m + 1))
for i in range(x, y, z):
for j in range(x, y, z):
if i < j:
print(i, j)
``` | instruction | 0 | 61,011 | 12 | 122,022 |
No | output | 1 | 61,011 | 12 | 122,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
def main():
n, m, k = map(int, input().split())
l, res = [], [0]
for _ in range(n):
row = list(map(int, input().split()))
for i, j in enumerate(sorted(range(m), key=row.__getitem__, reverse=k)):
row[j] = i
l.append(row)
for i in range(m - 1):
s = set()
for row in l:
if row[i] != i:
j = row.index(i)
s.add(j)
row[j] = row[i]
pref = '\n' + str(i + 1) + ' '
for j in sorted(s, reverse=True):
res.append(pref)
res.append(str(j + 1))
res[0] = str(len(res) // 2)
print(''.join(res))
if __name__ == '__main__':
main()
``` | instruction | 0 | 61,012 | 12 | 122,024 |
No | output | 1 | 61,012 | 12 | 122,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n,m,_=value()
for i in range(n):
input()
print(m*(m+1)//2)
for i in range(1,m+1):
for j in range(i+1,m+1):
print(i,j)
``` | instruction | 0 | 61,013 | 12 | 122,026 |
No | output | 1 | 61,013 | 12 | 122,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of m integers.
Iahub can choose a pair of distinct indices i and j (1 ≤ i, j ≤ m, i ≠ j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular order is given in input). The size of the array can be at most <image> (at most <image> pairs). Help Iahub, find any suitable array.
Input
The first line contains three integers n (1 ≤ n ≤ 1000), m (1 ≤ m ≤ 100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1 ≤ x ≤ 106 holds.
Output
On the first line of the output print an integer p, the size of the array (p can be at most <image>). Each of the next p lines must contain two distinct integers i and j (1 ≤ i, j ≤ m, i ≠ j), representing the chosen indices.
If there are multiple correct answers, you can print any.
Examples
Input
2 5 0
1 3 2 5 4
1 4 3 2 5
Output
3
2 4
2 3
4 5
Input
3 2 1
1 2
2 3
3 4
Output
1
2 1
Note
Consider the first sample. After the first operation, the arrays become [1, 3, 2, 5, 4] and [1, 2, 3, 4, 5]. After the second operation, the arrays become [1, 2, 3, 5, 4] and [1, 2, 3, 4, 5]. After the third operation they become [1, 2, 3, 4, 5] and [1, 2, 3, 4, 5].
Submitted Solution:
```
# -*- encoding: utf-8 -*-
n, m, k = map(int, input().split())
print(m*(m-1)//2)
for i in range(m-1):
for j in range(m-1-i):
print('{} {}'.format(*((j, j+1)) if k == 0 else (j+1, j)))
``` | instruction | 0 | 61,014 | 12 | 122,028 |
No | output | 1 | 61,014 | 12 | 122,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Artem has invented a time machine! He could go anywhere in time, but all his thoughts of course are with computer science. He wants to apply this time machine to a well-known data structure: multiset.
Artem wants to create a basic multiset of integers. He wants these structure to support operations of three types:
1. Add integer to the multiset. Note that the difference between set and multiset is that multiset may store several instances of one integer.
2. Remove integer from the multiset. Only one instance of this integer is removed. Artem doesn't want to handle any exceptions, so he assumes that every time remove operation is called, that integer is presented in the multiset.
3. Count the number of instances of the given integer that are stored in the multiset.
But what about time machine? Artem doesn't simply apply operations to the multiset one by one, he now travels to different moments of time and apply his operation there. Consider the following example.
* First Artem adds integer 5 to the multiset at the 1-st moment of time.
* Then Artem adds integer 3 to the multiset at the moment 5.
* Then Artem asks how many 5 are there in the multiset at moment 6. The answer is 1.
* Then Artem returns back in time and asks how many integers 3 are there in the set at moment 4. Since 3 was added only at moment 5, the number of integers 3 at moment 4 equals to 0.
* Then Artem goes back in time again and removes 5 from the multiset at moment 3.
* Finally Artyom asks at moment 7 how many integers 5 are there in the set. The result is 0, since we have removed 5 at the moment 3.
Note that Artem dislikes exceptions so much that he assures that after each change he makes all delete operations are applied only to element that is present in the multiset. The answer to the query of the third type is computed at the moment Artem makes the corresponding query and are not affected in any way by future changes he makes.
Help Artem implement time travellers multiset.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of Artem's queries.
Then follow n lines with queries descriptions. Each of them contains three integers ai, ti and xi (1 ≤ ai ≤ 3, 1 ≤ ti, xi ≤ 109) — type of the query, moment of time Artem travels to in order to execute this query and the value of the query itself, respectively. It's guaranteed that all moments of time are distinct and that after each operation is applied all operations of the first and second types are consistent.
Output
For each ask operation output the number of instances of integer being queried at the given moment of time.
Examples
Input
6
1 1 5
3 5 5
1 2 5
3 6 5
2 3 5
3 7 5
Output
1
2
1
Input
3
1 1 1
2 2 1
3 3 1
Output
0 | instruction | 0 | 61,126 | 12 | 122,252 |
Tags: data structures
Correct Solution:
```
from bisect import *
u, v = {}, {}
for q in range(int(input())):
a, t, x = map(int, input().split())
if x not in u: u[x], v[x] = [], []
if a < 3: insort([v, u][-a][x], t)
else: print(bisect(u[x], t) - bisect(v[x], t))
``` | output | 1 | 61,126 | 12 | 122,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Artem has invented a time machine! He could go anywhere in time, but all his thoughts of course are with computer science. He wants to apply this time machine to a well-known data structure: multiset.
Artem wants to create a basic multiset of integers. He wants these structure to support operations of three types:
1. Add integer to the multiset. Note that the difference between set and multiset is that multiset may store several instances of one integer.
2. Remove integer from the multiset. Only one instance of this integer is removed. Artem doesn't want to handle any exceptions, so he assumes that every time remove operation is called, that integer is presented in the multiset.
3. Count the number of instances of the given integer that are stored in the multiset.
But what about time machine? Artem doesn't simply apply operations to the multiset one by one, he now travels to different moments of time and apply his operation there. Consider the following example.
* First Artem adds integer 5 to the multiset at the 1-st moment of time.
* Then Artem adds integer 3 to the multiset at the moment 5.
* Then Artem asks how many 5 are there in the multiset at moment 6. The answer is 1.
* Then Artem returns back in time and asks how many integers 3 are there in the set at moment 4. Since 3 was added only at moment 5, the number of integers 3 at moment 4 equals to 0.
* Then Artem goes back in time again and removes 5 from the multiset at moment 3.
* Finally Artyom asks at moment 7 how many integers 5 are there in the set. The result is 0, since we have removed 5 at the moment 3.
Note that Artem dislikes exceptions so much that he assures that after each change he makes all delete operations are applied only to element that is present in the multiset. The answer to the query of the third type is computed at the moment Artem makes the corresponding query and are not affected in any way by future changes he makes.
Help Artem implement time travellers multiset.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of Artem's queries.
Then follow n lines with queries descriptions. Each of them contains three integers ai, ti and xi (1 ≤ ai ≤ 3, 1 ≤ ti, xi ≤ 109) — type of the query, moment of time Artem travels to in order to execute this query and the value of the query itself, respectively. It's guaranteed that all moments of time are distinct and that after each operation is applied all operations of the first and second types are consistent.
Output
For each ask operation output the number of instances of integer being queried at the given moment of time.
Examples
Input
6
1 1 5
3 5 5
1 2 5
3 6 5
2 3 5
3 7 5
Output
1
2
1
Input
3
1 1 1
2 2 1
3 3 1
Output
0 | instruction | 0 | 61,127 | 12 | 122,254 |
Tags: data structures
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n=int(input())
q=defaultdict(list)
que=[]
ind=defaultdict(list)
ans=defaultdict(int)
for i in range(n):
a,c,b=map(int,input().split())
ind[b].append(c)
q[b].append((a,c))
que.append((a,b,c))
for i in ind:
ind[i].sort()
inde=defaultdict(int)
for j in range(len(ind[i])):
inde[ind[i][j]]=j
e=[0]*len(ind[i])
s=SegmentTree(e)
for j in q[i]:
a,c=j
if a==1:
e[inde[c]]+=1
s.__setitem__(inde[c],e[inde[c]])
elif a==2:
e[inde[c]] -= 1
s.__setitem__(inde[c], e[inde[c]])
else:
ans[c]=s.query(0,inde[c])
for i in range(n):
a,b,c=que[i]
if a==3:
print(ans[c])
``` | output | 1 | 61,127 | 12 | 122,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Artem has invented a time machine! He could go anywhere in time, but all his thoughts of course are with computer science. He wants to apply this time machine to a well-known data structure: multiset.
Artem wants to create a basic multiset of integers. He wants these structure to support operations of three types:
1. Add integer to the multiset. Note that the difference between set and multiset is that multiset may store several instances of one integer.
2. Remove integer from the multiset. Only one instance of this integer is removed. Artem doesn't want to handle any exceptions, so he assumes that every time remove operation is called, that integer is presented in the multiset.
3. Count the number of instances of the given integer that are stored in the multiset.
But what about time machine? Artem doesn't simply apply operations to the multiset one by one, he now travels to different moments of time and apply his operation there. Consider the following example.
* First Artem adds integer 5 to the multiset at the 1-st moment of time.
* Then Artem adds integer 3 to the multiset at the moment 5.
* Then Artem asks how many 5 are there in the multiset at moment 6. The answer is 1.
* Then Artem returns back in time and asks how many integers 3 are there in the set at moment 4. Since 3 was added only at moment 5, the number of integers 3 at moment 4 equals to 0.
* Then Artem goes back in time again and removes 5 from the multiset at moment 3.
* Finally Artyom asks at moment 7 how many integers 5 are there in the set. The result is 0, since we have removed 5 at the moment 3.
Note that Artem dislikes exceptions so much that he assures that after each change he makes all delete operations are applied only to element that is present in the multiset. The answer to the query of the third type is computed at the moment Artem makes the corresponding query and are not affected in any way by future changes he makes.
Help Artem implement time travellers multiset.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of Artem's queries.
Then follow n lines with queries descriptions. Each of them contains three integers ai, ti and xi (1 ≤ ai ≤ 3, 1 ≤ ti, xi ≤ 109) — type of the query, moment of time Artem travels to in order to execute this query and the value of the query itself, respectively. It's guaranteed that all moments of time are distinct and that after each operation is applied all operations of the first and second types are consistent.
Output
For each ask operation output the number of instances of integer being queried at the given moment of time.
Examples
Input
6
1 1 5
3 5 5
1 2 5
3 6 5
2 3 5
3 7 5
Output
1
2
1
Input
3
1 1 1
2 2 1
3 3 1
Output
0 | instruction | 0 | 61,128 | 12 | 122,256 |
Tags: data structures
Correct Solution:
```
from bisect import *
d = [{}, {}]
i = [0, 0]
for q in range(int(input())):
a, t, x = map(int, input().split())
for k in [0, 1]:
d[k][x] = d[k].get(x, [])
i[k] = bisect(d[k][x], t)
if a < 3: d[-a][x].insert(i[-a], t)
else: print(i[1] - i[0])
``` | output | 1 | 61,128 | 12 | 122,257 |
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