message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13 | instruction | 0 | 61,113 | 14 | 122,226 |
Tags: implementation, number theory
Correct Solution:
```
n = int(input())
a = sorted([int(i) for i in input().split()])
s = 0; o = 0; flag = True
for i in range(n):
s += a[i]
if a[i]&1 and flag:
o = a[i]
flag = False
if s&1 == 0 and flag:
print(0)
elif s&1:
print(s)
else:
print(s-o)
``` | output | 1 | 61,113 | 14 | 122,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13 | instruction | 0 | 61,114 | 14 | 122,228 |
Tags: implementation, number theory
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
n = int(input())
arr = list(map(int, input().split(' ')))
arr.sort(reverse=True)
if len(arr) == 1:
if arr[0] % 2 == 0:
print(0)
else:
print(arr[0])
elif len(arr) == 2:
if arr[0] % 2 != 0 and arr[1] % 2 == 0:
print(arr[0]+arr[1])
elif arr[0] % 2 != 0 and arr[1] % 2 != 0:
print(arr[0])
else:
print(0)
else:
val = sum(arr)
if val % 2 == 0:
for i in range(len(arr)-1, 0, -1):
val = sum(arr)
val -= arr[i]
if val % 2 != 0:
print(val)
break
else:
continue
else:
print(0)
else:
print(val)
``` | output | 1 | 61,114 | 14 | 122,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13 | instruction | 0 | 61,115 | 14 | 122,230 |
Tags: implementation, number theory
Correct Solution:
```
# 59b.py
n = int(input())
v = list(map(int, input().split()))
ans, sum, mn = 0, 0, 101
for x in v:
sum += x
if x % 2 == 1:
mn = min(mn, x)
if sum % 2 == 1:
print(sum)
elif mn != 101:
print(sum - mn)
else:
print(0)
``` | output | 1 | 61,115 | 14 | 122,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13 | instruction | 0 | 61,116 | 14 | 122,232 |
Tags: implementation, number theory
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect 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")
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=Int()
a=array()
odd=sorted([i for i in a if i%2],reverse=True)
even=sorted([i for i in a if not i%2],reverse=True)
if(odd==[]):
print(0)
exit()
if(len(odd)%2==0):
X=odd.pop()
print(sum(odd)+sum(even))
``` | output | 1 | 61,116 | 14 | 122,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13 | instruction | 0 | 61,117 | 14 | 122,234 |
Tags: implementation, number theory
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
d = sum(a)
e = 1
f = []
for i in range(0, len(a)):
if a[i] % 2 != 0:
f.append(a[i])
if len(f) == 0:
print(0)
else:
while e > 0:
if d == 0:
e = 0
elif d % 2 == 0:
d = d - min(f)
else:
e = 0
print(d)
``` | output | 1 | 61,117 | 14 | 122,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split(' ')))
a=[x for x in l if(x%2==0)]
b=[x for x in l if(x%2!=0)]
if(len(b)==0):print(0)
else:
print(sum(a)+sum(b)-(0 if len(b)%2==1 else min(b)))
``` | instruction | 0 | 61,118 | 14 | 122,236 |
Yes | output | 1 | 61,118 | 14 | 122,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
if s%2==1:
print(s)
else:
a = [i for i in a if i % 2]
if len(a)>1:
print(s-min(a))
else:
print(0)
``` | instruction | 0 | 61,119 | 14 | 122,238 |
Yes | output | 1 | 61,119 | 14 | 122,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n = int(input())
a = sorted(list(map(int, input().split())))
if sum(a) % 2 == 1:
print(sum(a))
else:
j = -1
for i in range(n):
if a[i] % 2 == 1:
j = i
break
if j == -1:
print(0)
else:
print(sum(a) - a[j])
``` | instruction | 0 | 61,120 | 14 | 122,240 |
Yes | output | 1 | 61,120 | 14 | 122,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n=int(input());s=input().split();k=[];ans=0
for i in s:
if int(i)%2!=0:
k.append(int(i))
ans+=int(i)
k=sorted(k)
if len(k)==0:
print(0)
elif len(k)%2==0:
ans-=k[0]
print(ans)
else:
print(ans)
``` | instruction | 0 | 61,121 | 14 | 122,242 |
Yes | output | 1 | 61,121 | 14 | 122,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n = int(input())
arr = list(map(int,input().split()))
ans,min_odd=0,10**9
for num in arr:
ans+=num
if num&1 and num<min_odd:
min_odd=num
print(f'test {ans}')
if ans&1==0 and min_odd==(10**9): #sum is even but no odd num
print(0)
elif ans&1 ==0 and min_odd!=(10**9): #sum is even and odd present
ans-=min_odd
print(ans)
else:
print(ans)
``` | instruction | 0 | 61,122 | 14 | 122,244 |
No | output | 1 | 61,122 | 14 | 122,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
if s%2==1:
print(s)
else:
for i in a:
if len(a)>1:
print(s-min(a))
break
else:
print(0)
``` | instruction | 0 | 61,123 | 14 | 122,246 |
No | output | 1 | 61,123 | 14 | 122,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
sa = sum(a)
a.sort(reverse = True)
def main(a, sa):
if len(a) == 1:
if sa % 2 == 1:
print(1)
else:
print(0)
return
if len(a) % 2 == 0:
for i in range(len(a)):
if (sa - min(a)) % 2 == 1:
print(sa - min(a))
return
else:
a.pop()
print(0)
else:
print(sa)
main(a, sa)
``` | instruction | 0 | 61,124 | 14 | 122,248 |
No | output | 1 | 61,124 | 14 | 122,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are n camomiles growing in the field, possessing the numbers of petals equal to a1, a2, ... an. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input
The first line contains an integer n (1 β€ n β€ 100), which is the number of flowers growing in the field. The second line contains n integers ai (1 β€ ai β€ 100) which represent the number of petals on a given i-th camomile.
Output
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Examples
Input
1
1
Output
1
Input
1
2
Output
0
Input
3
5 6 7
Output
13
Submitted Solution:
```
'''input
3
5 6 7
'''
from sys import stdin
import math
# main starts
n = int(stdin.readline().strip())
arr = list(map(int, stdin.readline().split()))
arr.sort()
if sum(arr) % 2 == 1:
print(sum(arr))
else:
for i in range(n):
if arr[i] % 2 == 1:
print(sum(arr) - arr[i])
break
``` | instruction | 0 | 61,125 | 14 | 122,250 |
No | output | 1 | 61,125 | 14 | 122,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,235 | 14 | 122,470 |
Tags: binary search, math
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
min_arrival = (100000000000, -1)
def circ_dist(fr, to):
if fr <= to:
return to - fr
else:
return n - (fr - to)
for i, q in enumerate(a):
pos_when_first = q % n
dist_to_entrance = circ_dist(pos_when_first, i)
time = q - 1 + dist_to_entrance
min_arrival = min(min_arrival, (time, i))
print(min_arrival[1] + 1)
``` | output | 1 | 61,235 | 14 | 122,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,236 | 14 | 122,472 |
Tags: binary search, math
Correct Solution:
```
d = int(input())
f = [int(i) for i in input().split()]
g = []
k = 10000000000
op = 0
for i in range(len(f)):
f[i] = (f[i]-i+d-1)//d
if k>f[i]:
k = f[i]
op = i
print(op+1)
``` | output | 1 | 61,236 | 14 | 122,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,237 | 14 | 122,474 |
Tags: binary search, math
Correct Solution:
```
n=int(input())
a=[int(x) for x in input().split()]
t=min(a)
while a[t%n]>t:t+=1
print(t%n + 1)
# Made By Mostafa_Khaled
``` | output | 1 | 61,237 | 14 | 122,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,238 | 14 | 122,476 |
Tags: binary search, math
Correct Solution:
```
n=int(input())
l=[int(x) for x in input().split()]
cur_idx=0
cur_val=max(0,(l[0]-1)//n+1)
for i in range(n):
val = max(0,(l[i]-i-1)//n+1)
if(val<cur_val):
cur_val=val
cur_idx=i
print(cur_idx+1)
``` | output | 1 | 61,238 | 14 | 122,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,239 | 14 | 122,478 |
Tags: binary search, math
Correct Solution:
```
n = int(input())
lst = list(map(int,input().split()))
mn = min(lst)
for i,x in enumerate(lst):lst[i]-=mn
item,k,a=mn%n,0,0
for i in range(item,n):
if lst[i]-a<=0:print(i+1);k=1;break
a+=1
if k==0:
for i in range(item):
if lst[i]-a<=0:print(i+1);k=1;break
a+=1
``` | output | 1 | 61,239 | 14 | 122,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,240 | 14 | 122,480 |
Tags: binary search, math
Correct Solution:
```
import math
n=int(input())
l=list(map(int,input().strip().split()))
l1=[]
for x,y in enumerate(l):
if (y<x):
m=0
else:
m=y-x
l1.append(math.ceil(m/n))
min1=100000000000000
min2=1000000000000
for i in range(n):
if l1[i]<min1:
min1=l1[i]
min2=i+1
print (min2)
``` | output | 1 | 61,240 | 14 | 122,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,241 | 14 | 122,482 |
Tags: binary search, math
Correct Solution:
```
"""
Author - Satwik Tiwari .
13th NOV , 2020 - Friday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
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)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
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
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def solve(case):
n = int(inp())
a = lis()
ans = inf
for i in range(n):
temp = ceil((a[i]-i)/n)*n+i
if(ans>temp):
ans = temp
print(ans%n+1)
testcase(1)
# testcase(int(inp()))
``` | output | 1 | 61,241 | 14 | 122,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0]. | instruction | 0 | 61,242 | 14 | 122,484 |
Tags: binary search, math
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
finalIndex = 1e12
index = 0
for i in range(n):
val = a[i]
val -=i # k+tn
if val<=0:
val = 0
sol = int(val/n) #t
if val % n:
sol+=1
bande = i+sol*n # b = k +tn
if bande<finalIndex:
finalIndex = bande
index = i+1;
print(index)
``` | output | 1 | 61,242 | 14 | 122,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
n = int(input())
queues = list(map(int, input().strip().split()))
shortest = min(queues) // n * n
queues = list(map(lambda x: x - shortest, queues))
for i in range(1, n + 1):
if queues[i - 1] - i < 0:
print(i)
break
else:
for i in range(1, n + 1):
if queues[i - 1] - i - n < 0:
print(i)
break
``` | instruction | 0 | 61,243 | 14 | 122,486 |
Yes | output | 1 | 61,243 | 14 | 122,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
import math
def solve():
ans_i = 0
i = 0
while i < n:
a[i] = math.ceil((a[i]-i)/n)
if a[i] < a[ans_i]:
ans_i = i
i += 1
return ans_i+1
n = int(input())
a = list(map(int, input().split()))
print(solve())
``` | instruction | 0 | 61,244 | 14 | 122,488 |
Yes | output | 1 | 61,244 | 14 | 122,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
n = int(input())
a = input()
a = a.split()
mini = 10**10
for i in range (0, n):
a[i] = int(a[i])
mini = min(mini, a[i])
for i in range(0, n):
a[i] -= mini
cur = mini%n
count = 0
while True:
a[cur] -= count
if a[cur] <= 0:
break
count += 1
cur = (cur+1)%n
print(cur+1)
``` | instruction | 0 | 61,245 | 14 | 122,490 |
Yes | output | 1 | 61,245 | 14 | 122,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split(' ')]
t = [0]*n
for i in range(n):
if a[i]<=i: continue
a[i]-=i
t[i]=a[i]//n
if t[i]*n<a[i]: t[i]+=1
mn=10000000000
pos=0
for i in range(n):
if (t[i]<mn):
mn=t[i]
pos=i+1
print(pos)
``` | instruction | 0 | 61,246 | 14 | 122,492 |
Yes | output | 1 | 61,246 | 14 | 122,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
import math
n=int(input())
a=list(map(int,input().split(' ')))
minimum=1000000000
mini=n+1
for i in range(n):
k=math.ceil((a[i]-i)/n)
if (k*n+i<minimum):
minimum=k*n+i
mini=i
print(mini+1)
``` | instruction | 0 | 61,247 | 14 | 122,494 |
No | output | 1 | 61,247 | 14 | 122,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
n = int(input())
m = list(map(int,input().split()))
n = len(m)
dp = []
for i,e in enumerate(m):
if e<=n:
if e<=i:
dp.append(0)
else:
dp.append(1)
else:
if e<=i:
dp.append(e//n)
else:
dp.append(e//n+1)
mi = 1e9
imi = n
for i in range(n):
if dp[i] < mi:
mi = dp[i]
imi = i
print(imi+1)
``` | instruction | 0 | 61,248 | 14 | 122,496 |
No | output | 1 | 61,248 | 14 | 122,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
t= 1
while l[(t-1)%n]-t>=0:
t+=1
print(max(1,(t%(n+1))))
``` | instruction | 0 | 61,249 | 14 | 122,498 |
No | output | 1 | 61,249 | 14 | 122,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Allen wants to enter a fan zone that occupies a round square and has n entrances.
There already is a queue of a_i people in front of the i-th entrance. Each entrance allows one person from its queue to enter the fan zone in one minute.
Allen uses the following strategy to enter the fan zone:
* Initially he stands in the end of the queue in front of the first entrance.
* Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).
Determine the entrance through which Allen will finally enter the fan zone.
Input
The first line contains a single integer n (2 β€ n β€ 10^5) β the number of entrances.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β the number of people in queues. These numbers do not include Allen.
Output
Print a single integer β the number of entrance that Allen will use.
Examples
Input
4
2 3 2 0
Output
3
Input
2
10 10
Output
1
Input
6
5 2 6 5 7 4
Output
6
Note
In the first example the number of people (not including Allen) changes as follows: [2, 3, 2, 0] β [1, 2, 1, 0] β [0, 1, 0, 0]. The number in bold is the queue Alles stands in. We see that he will enter the fan zone through the third entrance.
In the second example the number of people (not including Allen) changes as follows: [10, 10] β [9, 9] β [8, 8] β [7, 7] β [6, 6] β \\\ [5, 5] β [4, 4] β [3, 3] β [2, 2] β [1, 1] β [0, 0].
In the third example the number of people (not including Allen) changes as follows: [5, 2, 6, 5, 7, 4] β [4, 1, 5, 4, 6, 3] β [3, 0, 4, 3, 5, 2] β \\\ [2, 0, 3, 2, 4, 1] β [1, 0, 2, 1, 3, 0] β [0, 0, 1, 0, 2, 0].
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split(' ')))
for i in range(n):
a[i] %= n
a[i] -= i
for i in range(n):
if(a[i] < 1):
print(i+1)
break
``` | instruction | 0 | 61,250 | 14 | 122,500 |
No | output | 1 | 61,250 | 14 | 122,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality β an integer number pi. Each commander has two characteristics β his personality pj and leadership lj (both are integer numbers). Warrior i respects commander j only if <image> (<image> is the bitwise excluding OR of x and y).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
* 1 pi β one warrior with personality pi joins Vova's army;
* 2 pi β one warrior with personality pi leaves Vova's army;
* 3 pi li β Vova tries to hire a commander with personality pi and leadership li.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
Input
The first line contains one integer q (1 β€ q β€ 100000) β the number of events.
Then q lines follow. Each line describes the event:
* 1 pi (1 β€ pi β€ 108) β one warrior with personality pi joins Vova's army;
* 2 pi (1 β€ pi β€ 108) β one warrior with personality pi leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment);
* 3 pi li (1 β€ pi, li β€ 108) β Vova tries to hire a commander with personality pi and leadership li. There is at least one event of this type.
Output
For each event of the third type print one integer β the number of warriors who respect the commander Vova tries to hire in the event.
Example
Input
5
1 3
1 4
3 6 3
2 4
3 6 3
Output
1
0
Note
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<image>, and 2 < 3, but <image>, and 5 β₯ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him. | instruction | 0 | 61,970 | 14 | 123,940 |
Tags: bitmasks, data structures, trees
Correct Solution:
```
from sys import stdin
input=stdin.readline
class Node:
def __init__(self,data):
self.data=data
self.left=None
self.right=None
self.count=0
class Trie():
def __init__(self):
self.root=Node(0)
def insert(self,preXor):
self.temp=self.root
for i in range(31,-1,-1):
val=preXor&(1<<i)
if val:
if not self.temp.right:
self.temp.right=Node(0)
self.temp=self.temp.right
self.temp.count+=1
else:
if not self.temp.left:
self.temp.left=Node(0)
self.temp=self.temp.left
self.temp.count+=1
self.temp.data=preXor
def delete(self,val):
self.temp=self.root
for i in range(31,-1,-1):
active=val&(1<<i)
if active:
self.temp=self.temp.right
self.temp.count-=1
else:
self.temp=self.temp.left
self.temp.count-=1
def query(self, val,li):
self.temp = self.root
ans=0
for i in range(31, -1, -1):
active = val & (1 << i)
bb=li&(1<<i)
if bb==0:
if active==0:
if self.temp.left and self.temp.left.count>0:
self.temp=self.temp.left
else:
return ans
else:
if self.temp.right and self.temp.right.count>0:
self.temp=self.temp.right
else:
return ans
else:
if active:
if self.temp.right:
ans+=self.temp.right.count
if self.temp.left and self.temp.left.count>0:
self.temp=self.temp.left
else:
return ans
else:
if self.temp.left:
ans+=self.temp.left.count
if self.temp.right and self.temp.right.count>0:
self.temp=self.temp.right
else:
return ans
return ans
trie=Trie()
for i in range(int(input())):
l=list(input().strip().split())
# print(l)
if l[0]=="1":
trie.insert(int(l[1]))
elif l[0]=="2":
trie.delete(int(l[1]))
# else:
# print(l,"lodi",l[1])
else:
print(trie.query(int(l[1]),int(l[2])))
``` | output | 1 | 61,970 | 14 | 123,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality β an integer number pi. Each commander has two characteristics β his personality pj and leadership lj (both are integer numbers). Warrior i respects commander j only if <image> (<image> is the bitwise excluding OR of x and y).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
* 1 pi β one warrior with personality pi joins Vova's army;
* 2 pi β one warrior with personality pi leaves Vova's army;
* 3 pi li β Vova tries to hire a commander with personality pi and leadership li.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
Input
The first line contains one integer q (1 β€ q β€ 100000) β the number of events.
Then q lines follow. Each line describes the event:
* 1 pi (1 β€ pi β€ 108) β one warrior with personality pi joins Vova's army;
* 2 pi (1 β€ pi β€ 108) β one warrior with personality pi leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment);
* 3 pi li (1 β€ pi, li β€ 108) β Vova tries to hire a commander with personality pi and leadership li. There is at least one event of this type.
Output
For each event of the third type print one integer β the number of warriors who respect the commander Vova tries to hire in the event.
Example
Input
5
1 3
1 4
3 6 3
2 4
3 6 3
Output
1
0
Note
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<image>, and 2 < 3, but <image>, and 5 β₯ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him. | instruction | 0 | 61,971 | 14 | 123,942 |
Tags: bitmasks, data structures, trees
Correct Solution:
```
import sys
from collections import defaultdict
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
q = int(sys.stdin.readline())
root = Node(0)
# def search(node, bit, )
for _ in range(q):
l = list(map(int, sys.stdin.readline().split()))
if l[0] == 1:
# add
bit = 28
cur = root
num = l[1]
# print(num,'num')
while bit >= 0:
if ((1<<bit)&num) == (1<<bit):
if cur.right is None:
cur.right = Node(1)
# print(bit,'bit right')
else:
cur.right.val += 1
# print(bit,'bit add right')
cur = cur.right
else:
if cur.left is None:
cur.left = Node(1)
# print(bit,'bit left', cur.left.val)
else:
cur.left.val += 1
# print(bit,'bit add left', cur.left.val)
cur = cur.left
bit -= 1
if l[0] == 2:
num = l[1]
bit, cur = 28, root
# print(num,'num')
while bit >= 0:
if((1<<bit)&num) == (1<<bit):
cur.right.val -= 1
cur = cur.right
else:
cur.left.val -= 1
cur = cur.left
bit -= 1
# remove
if l[0] == 3:
# print
res, cur, bit = 0, root, 28
# print(res, cur, bit)
while bit >= 0:
num = (1<<bit)
# print(bit,'bit')
if (num&l[2]) and (num&l[1]):
# print("A")
if cur.right is not None:
res += cur.right.val
if cur.left is None:
break
cur = cur.left
bit -= 1
continue
if (num&l[2]) and not (num&l[1]):
# print("B")
if cur.left is not None:
res += cur.left.val
if cur.right is None:
break
cur = cur.right
bit -= 1
continue
if not (num&l[2]) and (num&l[1]):
# print("C")
if cur.right is None:
break
cur = cur.right
bit -= 1
continue
if not (num&l[2]) and not (num&l[1]):
# print("D")
if cur.left is None:
break
cur = cur.left
bit -= 1
continue
print(res)
``` | output | 1 | 61,971 | 14 | 123,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality β an integer number pi. Each commander has two characteristics β his personality pj and leadership lj (both are integer numbers). Warrior i respects commander j only if <image> (<image> is the bitwise excluding OR of x and y).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
* 1 pi β one warrior with personality pi joins Vova's army;
* 2 pi β one warrior with personality pi leaves Vova's army;
* 3 pi li β Vova tries to hire a commander with personality pi and leadership li.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
Input
The first line contains one integer q (1 β€ q β€ 100000) β the number of events.
Then q lines follow. Each line describes the event:
* 1 pi (1 β€ pi β€ 108) β one warrior with personality pi joins Vova's army;
* 2 pi (1 β€ pi β€ 108) β one warrior with personality pi leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment);
* 3 pi li (1 β€ pi, li β€ 108) β Vova tries to hire a commander with personality pi and leadership li. There is at least one event of this type.
Output
For each event of the third type print one integer β the number of warriors who respect the commander Vova tries to hire in the event.
Example
Input
5
1 3
1 4
3 6 3
2 4
3 6 3
Output
1
0
Note
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<image>, and 2 < 3, but <image>, and 5 β₯ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him. | instruction | 0 | 61,972 | 14 | 123,944 |
Tags: bitmasks, data structures, trees
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
class SegmentTree2:
def __init__(self, data, default=3000006, func=lambda a, b: min(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)
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=-1, func=lambda a, b: max(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:math.gcd(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------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor,t):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=t
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += t
self.temp.data = pre_xor
def query(self, p,l):
ans=0
self.temp = self.root
for i in range(31, -1, -1):
val = p & (1 << i)
val1= l & (1<<i)
if val1==0:
if val==0:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
else:
return ans
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
else:
return ans
else:
if val !=0 :
if self.temp.right:
ans+=self.temp.right.count
if self.temp.left and self.temp.left.count > 0:
self.temp = self.temp.left
else:
return ans
else:
if self.temp.left:
ans += self.temp.left.count
if self.temp.right and self.temp.right.count > 0:
self.temp = self.temp.right
else:
return ans
return ans
#-------------------------bin trie-------------------------------------------
n=int(input())
s=BinaryTrie()
for i in range(n):
l=list(map(int,input().split()))
if l[0]==1:
s.insert(l[1],1)
elif l[0]==2:
s.insert(l[1],-1)
else:
print(s.query(l[1],l[2]))
``` | output | 1 | 61,972 | 14 | 123,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality β an integer number pi. Each commander has two characteristics β his personality pj and leadership lj (both are integer numbers). Warrior i respects commander j only if <image> (<image> is the bitwise excluding OR of x and y).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
* 1 pi β one warrior with personality pi joins Vova's army;
* 2 pi β one warrior with personality pi leaves Vova's army;
* 3 pi li β Vova tries to hire a commander with personality pi and leadership li.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
Input
The first line contains one integer q (1 β€ q β€ 100000) β the number of events.
Then q lines follow. Each line describes the event:
* 1 pi (1 β€ pi β€ 108) β one warrior with personality pi joins Vova's army;
* 2 pi (1 β€ pi β€ 108) β one warrior with personality pi leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment);
* 3 pi li (1 β€ pi, li β€ 108) β Vova tries to hire a commander with personality pi and leadership li. There is at least one event of this type.
Output
For each event of the third type print one integer β the number of warriors who respect the commander Vova tries to hire in the event.
Example
Input
5
1 3
1 4
3 6 3
2 4
3 6 3
Output
1
0
Note
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<image>, and 2 < 3, but <image>, and 5 β₯ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him.
Submitted Solution:
```
from sys import stdin
input=stdin.readline
class Node:
def __init__(self,data):
self.data=data
self.left=None
self.right=None
self.count=0
class Trie():
def __init__(self):
self.root=Node(0)
def insert(self,preXor):
self.temp=self.root
for i in range(31,-1,-1):
val=preXor&(1<<i)
if val:
if not self.temp.right:
self.temp.right=Node(0)
self.temp=self.temp.right
self.temp.count+=1
else:
if not self.temp.left:
self.temp.left=Node(0)
self.temp=self.temp.left
self.temp.count+=1
self.temp.data=preXor
def delete(self,val):
self.temp=self.root
for i in range(31,-1,-1):
active=val&(1<<i)
if active:
self.temp=self.temp.right
self.temp.count-=1
else:
self.temp=self.temp.left
self.temp.count-=1
def query(self, val,li):
self.temp = self.root
ans=0
for i in range(31, -1, -1):
active = val & (1 << i)
bb=li&(1<<i)
if (active and bb) or (active!=0 and bb==0):
if self.temp.right and self.temp.right.count > 0:
self.temp = self.temp.right
ans+=self.temp.count
else:
break
# self.temp = self.temp.left
else:
if self.temp.left and self.temp.left.count > 0:
self.temp = self.temp.left
# elif self.temp.right:
# self.temp = self.temp.right
return ans
trie=Trie()
for i in range(int(input())):
l=list(input().strip().split())
if l[0]=="1":
trie.insert(int(l[1]))
elif l[0]=="2":
trie.delete(int(l[1]))
else:
print(trie.query(int(l[1]),int(l[2])))
``` | instruction | 0 | 61,973 | 14 | 123,946 |
No | output | 1 | 61,973 | 14 | 123,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,355 | 14 | 124,710 |
Tags: implementation, two pointers
Correct Solution:
```
n, m, k = list(map(int, input().split()))
arr = list(map(int, input().split()))
ans = 0
i = 0
rem = 0
while i < m:
page_no = (arr[i] - rem - 1) // k
j = i
cnt = 0
while j < m:
another_page = (arr[j] - rem - 1) // k
if another_page == page_no:
cnt += 1
else:
break
j += 1
ans += 1
i = j
rem += cnt
print(ans)
``` | output | 1 | 62,355 | 14 | 124,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,356 | 14 | 124,712 |
Tags: implementation, two pointers
Correct Solution:
```
n, m, k = map(int, input().split())
specials = list((map(int, input().split())))
specials.sort(reverse=True)
page = list()
counter = 0
# while specials:
# pg = (specials[0] - 1) // k
# o_len = len(specials)
# specials = [i for i in specials if i > (pg + 1) * k]
# n_removed = o_len - len(specials)
# specials = [i - n_removed for i in specials]
# counter += 1
# print(counter)
pg = (specials[m-1] - 1) // k
removed = 0
for x in reversed(range(m)):
if specials[x] - removed <= (pg + 1) * k:
page.append(specials[x])
del specials[x]
else:
counter += 1
removed += len(page)
page = [specials[x]]
pg = (specials[x] - removed - 1) // k
print(counter + 1)
``` | output | 1 | 62,356 | 14 | 124,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,357 | 14 | 124,714 |
Tags: implementation, two pointers
Correct Solution:
```
n, m, k = map(int, input().split())
p = list(map(int, input().split()))
rev = 1
new_rev = 1
page = None
count = 0
for i in range(m):
if (p[i] - rev) // k != page:
count += 1
rev = new_rev
page = (p[i] - rev) // k
new_rev += 1
print(count)
``` | output | 1 | 62,357 | 14 | 124,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,358 | 14 | 124,716 |
Tags: implementation, two pointers
Correct Solution:
```
import sys
import math
input = sys.stdin.readline
def pn(n, k):
if n % k > 0:
return int(n / k) + 1
return int(n / k)
n,m,k=map(int,input().split())
l=list(map(int,input().split()))
ans=0
cur=0
diff=0
temp=0
now=pn(l[cur] - diff, k)
cur+=1
temp+=1
while cur<m:
while cur<m and pn(l[cur] - diff, k)==now:
cur+=1
temp+=1
if cur<m:
ans+=1
diff+=temp
temp=0
now=pn(l[cur] - diff, k)
if cur==m:
ans+=1
if m==1:
ans=1
print(ans)
``` | output | 1 | 62,358 | 14 | 124,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,359 | 14 | 124,718 |
Tags: implementation, two pointers
Correct Solution:
```
import math
n,m,k = map(int,input().split())
a = list(map(int,input().split()))
i = 1
q = math.ceil(a[0]/k)
diff = 0
ans = 1
count = 1
while i<m:
# print (i,math.ceil((a[i]-diff)/k),q,diff,ans)
if (a[i]-diff-1)//k+1==q:
count += 1
else:
ans += 1
diff += count
q = math.ceil((a[i]-diff)/k)
count = 1
i += 1
print (ans)
``` | output | 1 | 62,359 | 14 | 124,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,360 | 14 | 124,720 |
Tags: implementation, two pointers
Correct Solution:
```
def find_operations(num_per_page, special_numbers):
i = 0
count = 0
while i < len(special_numbers):
curr_special_number = special_numbers[i]
curr_special_page = get_special_page(num_per_page, curr_special_number, i)
largest_on_curr_page = get_largest(num_per_page, curr_special_page, i)
count += 1
#print(num_per_page, curr_special_page, largest_on_curr_page)
while i < len(special_numbers) and special_numbers[i] <= largest_on_curr_page:
i += 1
return count
def get_special_page(num_per_page, special_number, removed_amount):
return (special_number - 1 - removed_amount) // num_per_page + 1
def get_largest(num_per_page, page_number, removed_amount):
return num_per_page * page_number + removed_amount
n, m, k = [int(x) for x in input().split()]
special_numbers = [int(x) for x in input().split()]
print(find_operations(k, special_numbers))
``` | output | 1 | 62,360 | 14 | 124,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,361 | 14 | 124,722 |
Tags: implementation, two pointers
Correct Solution:
```
fstLine = list(map(int,input().split(' ')))
items = list(map(int,input().split(' ')))
ans, itemRange, i, counter, idk = 0, fstLine[2], 0, 0, 0
while (i<len(items)):
if (itemRange >= items[i]):
while (i<len(items) and itemRange >= items[i]):
i+=1
counter+=1
ans+=1
itemRange+=counter
idk=(idk+counter)%fstLine[2]
counter=0
else:
itemRange=items[i]+(fstLine[2]-(items[i]%fstLine[2]))+idk
itemRange-=fstLine[2] if abs(items[i]-itemRange)>=fstLine[2] else 0
print(ans)
``` | output | 1 | 62,361 | 14 | 124,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once. | instruction | 0 | 62,362 | 14 | 124,724 |
Tags: implementation, two pointers
Correct Solution:
```
n,m,k=[int(x) for x in input().split()]
p=[int(x) for x in input().split()]
x=0
counter=0
answer=0
p.append(10**100)
prev=(p[0]-1)//k
for item in p:
num=(item-counter-1)//k
if num!=prev:
counter+=x
x=1
answer+=1
num=(item-counter-1)//k
prev=num
else:
x+=1
print(answer)
``` | output | 1 | 62,362 | 14 | 124,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,440 | 14 | 124,880 |
Tags: data structures, greedy, sortings
Correct Solution:
```
a=int(input())
z=list(map(int,input().split()))
r=list(map(int,input().split()))
ans=[]
for i in range(a):
ans.append([z[i],r[i]])
ans.sort()
from heapq import *
i=0
pq=[]
ss=0
sol=0
while(i<len(z)):
if(len(pq)==0):
heappush(pq,-1*ans[i][1])
ss+=ans[i][1]
v=ans[i][0]
i+=1
while(i<len(ans) and ans[i][0]==v):
heappush(pq,-ans[i][1])
ss+=ans[i][1]
i+=1
ss+=heappop(pq)
sol+=ss
v+=1
while(pq):
ss+=heappop(pq)
sol+=ss
print(sol)
``` | output | 1 | 62,440 | 14 | 124,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,441 | 14 | 124,882 |
Tags: data structures, greedy, sortings
Correct Solution:
```
# ΠΠΎΠ΄ ΠΏΡΠΎΠ³ΡΠ°ΠΌΠΌΡ Π½Π°ΠΏΠΈΡΠ°Π» Π½Π° ΡΠ·ΡΠΊΠ΅ Python 3
import sys
from heapq import heappush, heappop
def main():
n = int(sys.stdin.readline())
h = sorted(list(zip([int(i) for i in sys.stdin.readline().split()], [int(i) for i in sys.stdin.readline().split()])))
z, w, o, res = [], 0, 0, 0
while o < n:
t = h[o][0]
w += h[o][1]
heappush(z, -1 * h[o][1])
while 0 < n - o - 1 and h[o][0] == h[o + 1][0]:
o += 1
w += h[o][1]
heappush(z, -1 * h[o][1])
if o + 1 == n:
cur = 1e18
else:
cur = h[o + 1][0]
while z and t - cur < 0:
t += 1
w += heappop(z)
res += w
o += 1
print(res)
for test in range(1):
main()
``` | output | 1 | 62,441 | 14 | 124,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,442 | 14 | 124,884 |
Tags: data structures, greedy, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
import heapq
n = int(input())
x = list(map(int, input().split()))
t = list(map(int, input().split()))
a = []
M = 1000000
for i in range(n):
a.append(x[i] * M + t[i])
a.sort()
i = 0
ans = 0
while i < n:
xi, ti = divmod(a[i], M)
j = i + 1
while j < n:
xj, tj = divmod(a[j], M)
b = j - i + xi
if xj >= b:
break
j += 1
j -= 1
if i < j:
b = []
ptr = i
for target in range(xi, j - i + xi + 1):
while ptr <= j:
xp, tp = divmod(a[ptr], M)
if xp <= target:
heapq.heappush(b, (-tp, xp))
else:
break
ptr += 1
tp, xp = heapq.heappop(b)
ans += -tp * (target - xp)
i = j + 1
print(ans)
``` | output | 1 | 62,442 | 14 | 124,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,443 | 14 | 124,886 |
Tags: data structures, greedy, sortings
Correct Solution:
```
from sys import stdin, gettrace
from heapq import *
if not gettrace():
def input():
return next(stdin)[:-1]
def main():
n = int(input())
aa = [int(a) for a in input().split()]
bb = [int(a) for a in input().split()]
bss = {}
aheap = []
def insert_bss(a, b):
if a not in bss:
bss[a] = []
heappush(aheap, a)
bss[a].append(b)
for a, b in zip(aa, bb):
insert_bss(a, b)
cost = 0
moving = []
stepcost = 0
while aheap:
a = heappop(aheap)
if len(bss[a]) > 1:
bss[a].sort()
for b in bss[a][:-1]:
heappush(moving, -b)
stepcost+=b
cost += stepcost
if len(moving) > 0:
bi = -heappop(moving)
insert_bss(a+1, bi)
stepcost -= bi
print(cost)
if __name__ == "__main__":
main()
``` | output | 1 | 62,443 | 14 | 124,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,444 | 14 | 124,888 |
Tags: data structures, greedy, sortings
Correct Solution:
```
from collections import defaultdict
import heapq
n = int(input())
a = list(map(int, input().split()))
t = list(map(int, input().split()))
d = defaultdict(list)
for i in range(n):
d[a[i]].append(i)
h_a = list(d.keys())
heapq.heapify(h_a)
h_t = []
time = 0
s = 0
done = set()
while h_a:
cur_a = heapq.heappop(h_a)
if cur_a in done:
continue
done.add(cur_a)
for idx in d[cur_a]:
heapq.heappush(h_t, -t[idx])
s += t[idx]
max_t = -heapq.heappop(h_t)
s -= max_t
time += s
if h_t:
heapq.heappush(h_a, cur_a+1)
print(time)
``` | output | 1 | 62,444 | 14 | 124,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,445 | 14 | 124,890 |
Tags: data structures, greedy, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
raw1 = list(map(int,input().split()))
raw2 = list(map(int,input().split()))
c = []
for i in range(n):
c.append((raw1[i],raw2[i]))
c.sort()
c.append((10**12,0))
#print(c)
import heapq
ans = 0
last = 0
now = []
for x in c:
a,t = x
for ai in range(last,a):
if not now:
break
tm,am = heapq.heappop(now)
tm = 10**5-tm
ans += (ai-am) * tm
#print(ai-am,tm)
heapq.heappush(now,(10**5-t,a))
last = a
#print(ans,now)
#print(ans,now)
print(ans)
``` | output | 1 | 62,445 | 14 | 124,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,446 | 14 | 124,892 |
Tags: data structures, greedy, sortings
Correct Solution:
```
import heapq
n = int(input())
l1 = [int(i) for i in input().split()]
l2 = [int(i) for i in input().split()]
l = list(zip(l1, l2))
l.sort(key=lambda x: (x[0], -x[1]))
h = []
curr = l[0][0]
i = 1
res = 0
while (i < n):
while (h):
if (curr >= l[i][0] - 1):
break
curr += 1
temp = heapq.heappop(h)
res += (curr - temp[1]) * (-temp[0])
heapq.heappush(h, [-l[i][1], l[i][0]])
if (curr < l[i][0]):
curr = l[i][0]
temp = heapq.heappop(h)
res += (curr - temp[1]) * (-temp[0])
i += 1
while (h):
curr += 1
temp = heapq.heappop(h)
res += (curr - temp[1]) * (-temp[0])
print(res)
``` | output | 1 | 62,446 | 14 | 124,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications. | instruction | 0 | 62,447 | 14 | 124,894 |
Tags: data structures, greedy, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import *
from heapq import *
n = int(input())
a = list(map(int, input().split()))
t = list(map(int, input().split()))
d = defaultdict(list)
for i in range(n):
d[a[i]].append(t[i])
ks = list(d.keys())
ks.sort()
prev = ks[0]
ans = 0
s = 0
pq = []
for a in ks:
for _ in range(a-prev-1):
if len(pq)==0:
break
ans += s
s -= -heappop(pq)
ans += s
for t in d[a]:
heappush(pq, -t)
s += t
s -= -heappop(pq)
prev = a
while len(pq):
ans += s
s -= -heappop(pq)
print(ans)
``` | output | 1 | 62,447 | 14 | 124,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.
The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds.
What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.
Input
The first line of input consists of single integer n β the number of news categories (1 β€ n β€ 200 000).
The second line of input consists of n integers a_i β the number of publications of i-th category selected by the batch algorithm (1 β€ a_i β€ 10^9).
The third line of input consists of n integers t_i β time it takes for targeted algorithm to find one new publication of category i (1 β€ t_i β€ 10^5).
Output
Print one integer β the minimal required time for the targeted algorithm to get rid of categories with the same size.
Examples
Input
5
3 7 9 7 8
5 2 5 7 5
Output
6
Input
5
1 2 3 4 5
1 1 1 1 1
Output
0
Note
In the first example, it is possible to find three publications of the second type, which will take 6 seconds.
In the second example, all news categories contain a different number of publications.
Submitted Solution:
```
# from collections import deque
import heapq
import sys
input = lambda: sys.stdin.readline().strip()
n = int(input())
b = map(int,input().split())
c = map(int,input().split())
d = list(zip(b,c))
d.sort()
h = []
s = 0
i = 0
ans = 0
while i <n:
lvl = d[i][0]
s+=d[i][1]
heapq.heappush(h, -d[i][1])
while i+1<n and d[i][0]==d[i+1][0]:
i+=1
s+=d[i][1]
heapq.heappush(h,-d[i][1])
if i+1==n:
lol = 10000000000000000000
else:
lol = d[i+1][0]
while h and lvl<lol:
lvl+=1
x = heapq.heappop(h)
s+=x
ans+=s
i+=1
print(ans)
# for i in range(n-1)
``` | instruction | 0 | 62,448 | 14 | 124,896 |
Yes | output | 1 | 62,448 | 14 | 124,897 |
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