text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Tags: *special
Correct Solution:
```
from math import sqrt
a = []
for i in range(11):
a.append(int(input()))
for i in range(10, -1, -1):
x = a[i]
aresult = sqrt(abs(x))
bresult = x * x * x * 5
result = aresult + bresult
print('f(' + str(x) + ') = ', sep='', end='')
if result >= 400:
print("MAGNA NIMIS!")
else:
print('%.2f' % result)
```
| 8,500 |
Provide tags and a correct Python 3 solution for this coding contest problem.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Tags: *special
Correct Solution:
```
lip = ['0.00', '6.00', '-3642.00', '-2557.17', '-1712.35', '-1077.55', '-622.76', '-318.00', '-133.27', '-38.59', 'MAGNA NIMIS!']
s = []
for i in range(11):
s.append(int(input()))
s.reverse()
#print(s)
for i in range(0, 11):
a = s[i]**3*5
b = abs(s[i])**0.5
ans = a + b
print ('f(' + str(s[i]) + ") = ", end='')
if (ans < 400):
print('{:.2f}'.format(ans))
else:
print(' MAGNA NIMIS!')
```
| 8,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
def toFixed(numObj, digits=0):
return f"{numObj:.{digits}f}"
rome = [None]*11
for nextum in range(11):
rome[-nextum-1] = int(input())
for nextum in rome:
resultum = (nextum**2)**0.25 + 5*nextum**3
if resultum <= 400:
print("f({}) = {:.2f}".format(nextum, resultum))
else:
print("f({}) = MAGNA NIMIS!".format(nextum))
```
Yes
| 8,502 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
import typing
lst = []
for i in range(11):
lst.append(int(input()))
for n in reversed(lst):
a = abs(n)**0.5
b = (n**3)*5
res = a+b
if res <= 400:
print("f({}) = {:.2f}".format(n,round(res,2)))
else:
print("f({}) = MAGNA NIMIS!".format(n))
```
Yes
| 8,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
from collections import deque
import math
numbers = deque()
for _ in range(11):
numbers.append(int(input()))
for _ in range(11):
num = numbers.pop()
a = math.sqrt(abs(num))
b = num**3 * 5
res = a + b
if res < 400:
print("f(%d) = %.2f" % (num, res))
else:
print("f(%d) = MAGNA NIMIS!" % num)
```
Yes
| 8,504 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
from math import sqrt
a = []
for i in range(11):
a.append(int(input()))
for x in a[::-1]:
a = sqrt(abs(x))
b = 5 * x**3
if a + b <= 400:
print("f(%d) = %.2f" % (x, a + b))
else:
print("f(%d) = MAGNA NIMIS!" % x)
```
Yes
| 8,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
def solve(x):
if x>4 or x<-20:
print('f('+str(x)+')','=','MAGNA NIMIS!')
return 0
if x<0:
A=round((5*x*x*x+((-x)**0.5))*100)
if A%100==0:
A='-'+str(-A//100)+'.00'
elif A%10==0:
A='-'+str(-A//100)+'.'+str((-A//10)%10)+'0'
else:
A='-'+str(-A//100)+'.'+str(-A%100)
print('f('+str(x)+')','=',A)
return 0
else:
A=round((5*x*x*x+(x**0.5))*100)
if 0 and A>=40000:
A='MAGNA NIMIS!'
elif A%100==0:
A=str(A//100)+'.00'
elif A%10==0:
A=str(A//100)+'.'+str((A//10)%10)+'0'
else:
A=str(A//100)+'.'+str(A%100)
print('f('+str(x)+')','=',A)
return 0
Q=[int(input()) for i in range(11)]
for i in range(10,-1,-1):
solve(Q[i])
```
No
| 8,506 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
n=input()
n=input()
n=input()
n=input()
n=input()
n=input()
n=input()
n=input()
n=input()
n=input()
n=input()
print("f(10) = MAGNA NIMIS!\nf(-9) = -3642.00\nf(-8) = -2557.17\nf(-7) = -1712.35\nf(-6) = -1077.55\nf(-5) = -622.76\nf(-4) = -318.00\nf(-3) = -133.27\nf(-2) = -38.59\nf(1) = 6.00\nf(0) = 0.00")
```
No
| 8,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
def solve(x):
if x<0:
A=round((5*x*x*x+((-x)**0.5))*100)
if A%100==0:
A='-'+str(-A//100)+'.00'
elif A%10==0:
A='-'+str(-A//100)+'.'+str((-A//10)%10)+'0'
else:
A='-'+str(-A//100)+'.'+str(-A%100)
print('f('+str(x)+')','=',A)
return 0
else:
A=(5*x*x*x+(x**0.5))*100
if A>=40000:
A='MAGNA NIMIS!'
print('f('+str(x)+')','=',A)
return 0
A=round(A)
if A%100==0:
A=str(-(-A//100))+'.00'
elif A%10==0:
A=str(-(-A//100))+'.'+str((-(-A//10))%10)+'0'
else:
A=str(-(-A//100))+'.'+str(A%100)
print('f('+str(x)+')','=',A)
return 0
Q=[int(input()) for i in range(11)]
for i in range(10,-1,-1):
solve(Q[i])
```
No
| 8,508 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
per nextum in unam tum XI conscribementis fac sic
vestibulo perlegementum da varo.
morde varo.
seqis cumula varum.
cis
per nextum in unam tum XI conscribementis fac sic
seqis decumulamenta da varo.
varum privamentum fodementum da aresulto.
varum tum III elevamentum tum V multiplicamentum da bresulto.
aresultum tum bresultum addementum da resulto.
si CD tum resultum non praestantiam fac sic
dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
novumversum egresso scribe.
cis
si CD tum resultum praestantiam fac sic
dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
novumversum egresso scribe.
cis
cis
Input
The input consists of several integers, one per line. Each integer is between -50 and 50, inclusive.
Output
As described in the problem statement.
Example
Input
0
1
-2
-3
-4
-5
-6
-7
-8
-9
10
Output
f(10) = MAGNA NIMIS!
f(-9) = -3642.00
f(-8) = -2557.17
f(-7) = -1712.35
f(-6) = -1077.55
f(-5) = -622.76
f(-4) = -318.00
f(-3) = -133.27
f(-2) = -38.59
f(1) = 6.00
f(0) = 0.00
Submitted Solution:
```
ls = list()
for i in range(11):
n = int(input())
ls.append(n)
for i in range(10,-1,-1):
a=ls[i]
aresult=(abs(a))**(1/2)
bresult=a**3*5
result=aresult+bresult
if result<=400:
s = format(result, '.2f')
print('f(' + str(i) + ') = ' + s)
else:
print('f(' + str(i) + ') = MAGNA NIMIS!')
```
No
| 8,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
"""
// Author : snape_here - Susanta Mukherjee
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def gcd(x, y):
while y:
x, y = y, x % y
return x
mod=1000000007
import math
import bisect
abc="abcdefghijklmnopqrstuvwxyz"
def main():
for _ in range(ii()):
n,k=mi()
if k==n:
print("YES")
for i in range(n):
print(1,end=" ")
print()
continue
if k>n:
print("NO")
continue
if (n%2 and k%2==0):
print("NO")
elif n%2==0 and k%2 and n//k<2:
print("NO")
elif n%2==0 and k%2:
print("YES")
for i in range(k-1):
print(2,end=" ")
print(n-(k-1)*2)
elif n%2 and k%2:
print("YES")
for i in range(k-1):
print(1,end=" ")
print(n-(k-1))
else:
print("YES")
for i in range(k-1):
print(1,end=" ")
print(n-(k-1))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
```
| 8,510 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
if n % 2 == 1 and k % 2 == 0:
print("NO")
continue
if n % 2 == k % 2:
if k > n:
print("NO")
continue
res = [1] * (k-1)
res.append(n-(k-1))
print("YES")
print(*res)
continue
if k*2 > n:
print("NO")
continue
res = [2] * (k-1)
res.append(n-(k-1)*2)
print("YES")
print(*res)
"""
n k e o
--- ---
e e x x v
e o x
o e v
o o x v
n = 9 k = 4
1 1 1 6
_ _ _ _
n = 9 k = 3
n = 9 k = 10
n = 10 k = 6
2 2 2 2 2 2
1 1 1 7
n = 10 k = 3
2 2 6
"""
```
| 8,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input())
for tt in range(t):
n, k = (int(i) for i in input().split())
if k <= n and (n-k) % 2 == 0:
print("YES")
ans = [1]*k
ans[0] = n-k+1
print(' '.join((str(i) for i in ans)))
elif 2*k <= n and (n-2*k)%2 == 0:
print("YES")
ans = [2]*k
ans[0] = n-2*k+2
print(' '.join((str(i) for i in ans)))
else:
print("NO")
```
| 8,512 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
def function(n, k):
l_odd=[1]*(k-1)
l_odd.append(n-sum(l_odd))
condition1=True
for i in l_odd:
if i%2==0 or i<=0:
condition1=False
break
if condition1:
print('YES')
print(*l_odd)
condition2=True
if not condition1:
l_even=[2]*(k-1)
l_even.append(n-sum(l_even))
for j in l_even:
if j%2!=0 or j<=0:
condition2=False
break
if condition2:
print('YES')
print(*l_even)
if condition1==False and condition2==False:
print('NO')
if __name__=='__main__':
t=int(input())
for k1 in range(t):
n, k=map(int, input().rstrip().split())
function(n, k)
```
| 8,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
q=int(input())
for i in range(q):
n,k=[int(i) for i in input().split()]
if k==1:
print('YES')
print(n)
else:
if n%2==1 and k%2==0 or k>n or k+1==n or ((n%2==0 and k%2==1) and n<k*2) :
print('NO')
else:
if n%2==0 and k%2==0:
print('YES')
print('1 '*(k-1)+str(n-k+1))
elif n%2==1 and k%2==1:
print('YES')
print('1 '*(k-1)+str(n-k+1))
elif n%2==0 and k%2==1:
print('YES')
print('2 '*((k-1))+str(n-(k-1)*2))
```
| 8,514 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
if n < k:
print("NO")
continue
if n % 2 != 0:
if k % 2 == 0:
print("NO")
if k % 2 != 0:
print("YES")
print(str(n - k + 1) + ' 1' * (k - 1))
if n % 2 == 0:
if k % 2 == 0:
print("YES")
print(str(n - k + 1) + ' 1' * (k - 1))
if k % 2 != 0:
if k > n // 2:
print("NO")
else:
print("YES")
print(str(n - (k - 1) * 2) + ' 2' * (k - 1))
```
| 8,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
ans1=[2]*(k-1)
ans1.append(n-2*k+2)
ans2=[1]*(k-1)
ans2.append(n-k+1)
if ans1[-1]>0 and ans1[-1]%2==0:
print("YES")
print(*ans1)
elif ans2[-1]>0 and ans2[-1]%2==1:
print("YES")
print(*ans2)
else:
print("NO")
```
| 8,516 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input())
for i in range(t):
n,k = map(int, input().split())
if n==k:
print("YES")
print(*[1]*k)
elif n<k:
print("NO")
else:
if (n-(k-1))%2==1:
print("YES")
print(*[1]*(k-1),n-(k-1))
elif 2*(k-1) < n and (n - (2*(k-1)))%2==0:
print("YES")
print(*[2]*(k-1), n-(2*(k-1)))
else: print("NO")
```
| 8,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
t = int(input())
for i in range(t):
n, k = map(int, input().split())
if (n % 2 != 0):
if (k % 2 == 0 or n < k):
print("NO")
else:
print("YES")
print("1 "*(k-1) + str(n-k+1))
else:
if ((n-k+1) % 2 != 0 and n >= k):
print("YES")
print("1 "*(k-1) + str(n-k+1))
elif ((n-2*(k-1))%2 == 0 and n >= 2*k):
print("YES")
print("2 "*(k-1) + str(n-2*(k-1)))
else:
print("NO")
```
Yes
| 8,518 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
t = int(input())
for i in range(t):
n, k = map(int, input().split())
if n % 2 == 0:
if 2 * k <= n:
ans = [2] * (k - 1)
ans.append(n - 2 * (k - 1))
else:
if k % 2 == 0:
if k > n:
ans = []
else:
ans = [1] * (k - 1)
ans.append(n - (k - 1))
else:
ans = []
else:
if k % 2 == 0:
ans = []
else:
if k > n:
ans = []
else:
ans = [1] * (k - 1)
ans.append(n - (k - 1))
if len(ans):
print('YES')
print(*ans)
else:
print('NO')
```
Yes
| 8,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
import sys, heapq
from collections import *
from functools import lru_cache
sys.setrecursionlimit(10**6)
def main():
# sys.stdin = open('input.txt', 'r')
t = int(input())
for _ in range(t):
n, k = map(int, input().split(' '))
# n = int(input())
nn, kk = n&1, k&1
flag, res = False, []
if nn ^ kk == 0:
if n >= k:
flag = True
res = [n-k+1]+[1]*(k-1)
elif nn == 0 and kk == 1:
if n >= k*2:
flag = True
res = [n-k*2+2]+[2]*(k-1)
if len(res) > 0:
print("YES")
print(' '.join(map(str,res)))
else:
print("NO")
if __name__ == "__main__":
main()
```
Yes
| 8,520 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
t=int(input())
for s in range(t):
n,k=list(map(int,input().split(" ")))
y=n//2
if n<k:
print("NO")
else:
if n%2!=0 and k%2==0:
print("NO")
#break
if (n%2==0 and k%2==0) or (n%2!=0 and k%2!=0):
print("YES")
c=0
for i in range(1,k,1):
c+=1
print(1,end=" ")
x=n-c
print(x)
#break
if n%2==0 and k%2!=0:
if k*2<=n:
print("YES")
c=0
for i in range(1,k,1):
c+=2
print(2,end=" ")
x=n-c
print(x)
#break
else:
print("NO")
```
Yes
| 8,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
'''input
8
16 15
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
'''
from collections import defaultdict as dd
from collections import Counter as ccd
from itertools import permutations as pp
from itertools import combinations as cc
from random import randint as rd
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq as hq
from math import gcd
'''
Author : dhanyaabhirami
Hardwork beats talent if talent doesn't work hard
'''
'''
Stuck?
See github resources
Derive Formula
Kmcode blog
CP Algorithms Emaxx
'''
mod=pow(10,9) +7
def inp(flag=0):
if flag==0:
return list(map(int,input().strip().split(' ')))
else:
return int(input())
# Code credits
# assert(debug()==true)
# for _ in range(int(input())):
t=inp(1)
while t:
t-=1
n,k=inp()
possible = True
if n==k:
ans = [1]*k
elif n%2==1 and k%2==0:
possible = False
else:
if n%2 == 1:
if n-k+1>0 and (n-k+1)%2==1:
ans = [1]*(k-1)
ans.append(n-k+1)
else:
possible = False
else:
if n-2*(k-1)>0 and (n-2*(k-1))%2==0:
ans = [2]*(k-1)
ans.append(n-2*(k-1))
else:
possible = False
if possible:
print('YES')
print(*ans)
else:
print('NO')
```
No
| 8,522 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
if n%2>0 and k%2==0:print("NO")
elif n%2==0 and k%2==0:
print("YES")
for i in range(k-1):print(1,end=' ')
print(n-(k-1))
elif n%2==0 and k%2>0:
if n%k==0:
print("YES")
for i in range(k):print(n//k)
print()
elif n//k>=2:
print("YES")
for i in range(k-1):print(2,end=' ')
print(n-(2*(k-1)))
else:print("NO")
elif n%2>0 and k%2>0:
if k<=n:
print("YES")
for i in range(k-1):print(1,end=' ')
print(n-(k-1))
else:print("NO")
else:print("NO")
```
No
| 8,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
for _ in range(int(input())):
n, k = list(map(int, input().split()))
if n % 2 == 1 and k % 2 == 0:
print('NO')
elif n % 2 == 0 and k % 2 == 1 and k * 2 > n:
print('NO')
else:
num = 1
if n % 2 == 0 and k % 2 == 1:
num = 2
ans = []
for i in range(k - 1):
ans.append(num)
ans.append(n - num * k + num)
print('YES')
print(*ans)
```
No
| 8,524 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, β¦, a_k such that all a_i>0, n = a_1 + a_2 + β¦ + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 β€ n β€ 10^9) and k (1 β€ k β€ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120
Submitted Solution:
```
for i in range(int(input())):
n,k = map(int,input().split())
if (n-k-1)>0 and (n-k-1)%2!=0:
s = '1'*(k-1)
s = s.replace('',' ').strip()
print('YES')
print(s+" "+str(int(n-k+1)))
elif (n-2*(k-1))>0 and (n-2*(k-1))%2==0:
s = '2'*(k-1)
s = s.replace('',' ').strip()
print('YES')
print(s+' '+str(int(n-2*(k-1))))
else:
print('NO')
```
No
| 8,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are a warrior fighting against the machine god Thor.
Thor challenge you to solve the following problem:
There are n conveyors arranged in a line numbered with integers from 1 to n from left to right. Each conveyor has a symbol "<" or ">". The initial state of the conveyor i is equal to the i-th character of the string s. There are n+1 holes numbered with integers from 0 to n. The hole 0 is on the left side of the conveyor 1, and for all i β₯ 1 the hole i is on the right side of the conveyor i.
When a ball is on the conveyor i, the ball moves by the next rules:
If the symbol "<" is on the conveyor i, then:
* If i=1, the ball falls into the hole 0.
* If the symbol "<" is on the conveyor i-1, the ball moves to the conveyor i-1.
* If the symbol ">" is on the conveyor i-1, the ball falls into the hole i-1.
If the symbol ">" is on the conveyor i, then:
* If i=n, the ball falls into the hole n.
* If the symbol ">" is on the conveyor i+1, the ball moves to the conveyor i+1.
* If the symbol "<" is on the conveyor i+1, the ball falls into the hole i.
You should answer next q queries, each query is defined by the pair of integers l, r (1 β€ l β€ r β€ n):
* First, for all conveyors l,l+1,...,r, the symbol "<" changes to ">" and vice versa. These changes remain for the next queries.
* After that, put one ball on each conveyor l,l+1,...,r. Then, each ball falls into some hole. Find the maximum number of balls in one hole. After the query all balls disappear and don't considered in the next queries.
Input
The first line contains two integers n, q (1 β€ n β€ 5 Γ 10^5 , 1 β€ q β€ 10^5).
The second line contains a string s of length n. It consists of characters "<" and ">".
Next q lines contain the descriptions of the queries, i-th of them contains two integers l, r (1 β€ l β€ r β€ n), describing the i-th query.
Output
Print q lines, in the i-th of them print the answer to the i-th query.
Example
Input
5 6
><>><
2 4
3 5
1 5
1 3
2 4
1 5
Output
3
3
5
3
2
3
Note
* In the first query, the conveyors change to ">><<<". After that, put a ball on each conveyor \{2,3,4\}. All three balls fall into the hole 2. So the answer is 3.
* In the second query, the conveyors change to ">>>>>". After that, put a ball on each conveyor \{3,4,5\}. All three balls fall into the hole 5. So the answer is 3.
* In the third query, the conveyors change to "<<<<<". After that, put a ball on each conveyor \{1,2,3,4,5\}. All five balls fall into the hole 0. So the answer is 5.
* In the fourth query, the conveyors change to ">>><<". After that, put a ball on each conveyor \{1,2,3\}. All three balls fall into the hole 3. So the answer is 3.
* In the fifth query, the conveyors change to "><<><". After that, put a ball on each conveyor \{2,3,4\}. Two balls fall into the hole 1, and one ball falls into the hole 4. So, the answer is 2.
* In the sixth query, the conveyors change to "<>><>". After that, put a ball on each conveyor \{1,2,3,4,5\}. Three balls fall into the hole 3, one ball falls into the hole 0 and one ball falls into the hole 5. So, the answer is 3.
Submitted Solution:
```
class node:
o_mx, c_mx, pref_o, pref_c, suf_o, suf_c, len = [0] * 7
pref, suf = [0, 0], [0, 0]
def __init__(self, z = -1):
#self.o_mx, self.c_mx, self.pref_o, self.pref_c, self.suf_o, self.suf_c, self.len = [0] * 7
#self.pref, self.suf = [0, 0], [0, 0]
if z == -1:
return
self.pref[z], self.suf[z], self.len = 1, 1, 1
def unite(a, b):
if a.len == 0:
return b
if b.len == 0:
return a
res = node()
res.len = a.len + b.len
for i in (0, 1):
res.pref[i] = a.pref[i]
if a.pref[i] == a.len:
res.pref[i] += b.pref[i]
res.suf[i] = b.suf[i]
if b.suf[i] == b.len:
res.suf[i] += a.suf[i]
res.o_mx = max(a.o_mx, b.o_mx)
res.c_mx = max(a.c_mx, b.c_mx)
if a.suf_o:
res.o_mx = max(res.o_mx, a.suf_o + b.pref[1])
if b.pref_o:
res.o_mx = max(res.o_mx, a.suf[0] + b.pref_o)
if a.suf[0] != 0 and b.pref[1] != 0:
res.o_mx = max(res.o_mx, a.suf[0] + b.pref[1])
if a.suf_c != 0:
res.c_mx = max(res.c_mx, a.suf_c + b.pref[0])
if b.pref_c != 0:
res.c_mx = max(res.c_mx, a.suf[1] + b.pref_c)
if a.suf[1] != 0 and b.pref[0] != 0:
res.c_mx = max(res.c_mx, a.suf[1] + b.pref[0])
res.pref_o = a.pref_o
if a.pref_o == a.len:
res.pref_o += b.pref[1]
if a.pref[0] == a.len and b.pref[1] > 0:
res.pref_o = a.pref[0] + b.pref[1]
if a.pref[0] == a.len and b.pref_o > 0:
res.pref_o = a.pref[0] + b.pref_o
res.pref_c = a.pref_c
if a.pref_c == a.len:
res.pref_c += b.pref[0]
if a.pref[1] == a.len and b.pref[0] > 0:
res.pref_c = a.pref[1] + b.pref[0]
if a.pref[1] == a.len and b.pref_c > 0:
res.pref_c = a.pref[1] + b.pref_c
res.suf_o = b.suf_o
if b.suf_o == b.len:
res.suf_o += a.suf[0]
if b.suf[1] == b.len and a.suf[0] > 0:
res.suf_o = b.suf[1] + a.suf[0]
if b.suf[1] == b.len and a.suf_o > 0:
res.suf_o = b.suf[1] + a.suf_o
res.suf_c = b.suf_c
if b.suf_c == b.len:
res.suf_c += a.suf[1]
if b.suf[0] == b.len and a.suf[1] > 0:
res.suf_c = b.suf[0] + a.suf[1]
if b.suf[0] == b.len and a.suf_c > 0:
res.suf_c = b.suf[0] + a.suf_c
return res
def build(v, tl, tr):
if tl == tr:
t[v] = node(s[tl] == '>')
return
tm = (tl + tr) // 2
build(v * 2, tl, tm)
build(v * 2 + 1, tm + 1, tr)
t[v] = unite(t[v * 2], t[v * 2 + 1])
def push(v, tl, tr):
if not pt[v]:
return
pt[v] = 0
t[v].pref[0], t[v].pref[1] = t[v].pref[1], t[v].pref[0]
t[v].suf[0], t[v].suf[1] = t[v].suf[1], t[v].suf[0]
t[v].o_mx, t[v].c_mx = t[v].c_mx, t[v].o_mx
t[v].pref_o, t[v].pref_c = t[v].pref_c, t[v].pref_o
t[v].suf_o, t[v].suf_c = t[v].suf_c, t[v].suf_o
if tl != tr:
pt[v * 2] ^= 1
pt[v * 2 + 1] ^= 1
def rev(v, tl, tr, l, r):
push(v, tl, tr)
if tl > r or tr < l:
return
if tl >= l and tr <= r:
pt[v] = 1
push(v, tl, tr)
return
tm = (tl + tr) // 2
rev(v * 2, tl, tm, l, r)
rev(v * 2 + 1, tm + 1, tr, l, r)
t[v] = unite(t[v * 2], t[v * 2 + 1])
def getNode(v, tl, tr, l, r):
if tl > r or tr < l:
return node()
push(v, tl, tr)
if tl >= l and tr <= r:
return t[v]
tm = (tl + tr) // 2
return unite(getNode(v * 2, tl, tm, l, r), getNode(v * 2 + 1, tm + 1, tr, l, r))
n, q = map(int, input().split())
s = input()
t, pt = [node()] * (4 * n + 50), [0] * (4 * n + 50)
build(1, 0, n - 1)
for i in range(q):
l, r = map(lambda v: int(v) - 1, input().split())
rev(1, 0, n - 1, l, r)
nd = getNode(1, 0, n - 1, l, r)
print(max(nd.pref[0], nd.suf[1], nd.c_mx))
```
No
| 8,526 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are a warrior fighting against the machine god Thor.
Thor challenge you to solve the following problem:
There are n conveyors arranged in a line numbered with integers from 1 to n from left to right. Each conveyor has a symbol "<" or ">". The initial state of the conveyor i is equal to the i-th character of the string s. There are n+1 holes numbered with integers from 0 to n. The hole 0 is on the left side of the conveyor 1, and for all i β₯ 1 the hole i is on the right side of the conveyor i.
When a ball is on the conveyor i, the ball moves by the next rules:
If the symbol "<" is on the conveyor i, then:
* If i=1, the ball falls into the hole 0.
* If the symbol "<" is on the conveyor i-1, the ball moves to the conveyor i-1.
* If the symbol ">" is on the conveyor i-1, the ball falls into the hole i-1.
If the symbol ">" is on the conveyor i, then:
* If i=n, the ball falls into the hole n.
* If the symbol ">" is on the conveyor i+1, the ball moves to the conveyor i+1.
* If the symbol "<" is on the conveyor i+1, the ball falls into the hole i.
You should answer next q queries, each query is defined by the pair of integers l, r (1 β€ l β€ r β€ n):
* First, for all conveyors l,l+1,...,r, the symbol "<" changes to ">" and vice versa. These changes remain for the next queries.
* After that, put one ball on each conveyor l,l+1,...,r. Then, each ball falls into some hole. Find the maximum number of balls in one hole. After the query all balls disappear and don't considered in the next queries.
Input
The first line contains two integers n, q (1 β€ n β€ 5 Γ 10^5 , 1 β€ q β€ 10^5).
The second line contains a string s of length n. It consists of characters "<" and ">".
Next q lines contain the descriptions of the queries, i-th of them contains two integers l, r (1 β€ l β€ r β€ n), describing the i-th query.
Output
Print q lines, in the i-th of them print the answer to the i-th query.
Example
Input
5 6
><>><
2 4
3 5
1 5
1 3
2 4
1 5
Output
3
3
5
3
2
3
Note
* In the first query, the conveyors change to ">><<<". After that, put a ball on each conveyor \{2,3,4\}. All three balls fall into the hole 2. So the answer is 3.
* In the second query, the conveyors change to ">>>>>". After that, put a ball on each conveyor \{3,4,5\}. All three balls fall into the hole 5. So the answer is 3.
* In the third query, the conveyors change to "<<<<<". After that, put a ball on each conveyor \{1,2,3,4,5\}. All five balls fall into the hole 0. So the answer is 5.
* In the fourth query, the conveyors change to ">>><<". After that, put a ball on each conveyor \{1,2,3\}. All three balls fall into the hole 3. So the answer is 3.
* In the fifth query, the conveyors change to "><<><". After that, put a ball on each conveyor \{2,3,4\}. Two balls fall into the hole 1, and one ball falls into the hole 4. So, the answer is 2.
* In the sixth query, the conveyors change to "<>><>". After that, put a ball on each conveyor \{1,2,3,4,5\}. Three balls fall into the hole 3, one ball falls into the hole 0 and one ball falls into the hole 5. So, the answer is 3.
Submitted Solution:
```
from collections import deque
import math
import os
import random
import re
import sys
#n=list(map(int, input().split()))
#n=map(int, input().split())
class Stonesonthetable:
def solve(self,n,stones):
changes = 0
for i in range(1,n):
if(stones[i]==stones[i-1]):
changes+=1
return changes
'''
if __name__ == '__main__':
main()
'''
```
No
| 8,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are a warrior fighting against the machine god Thor.
Thor challenge you to solve the following problem:
There are n conveyors arranged in a line numbered with integers from 1 to n from left to right. Each conveyor has a symbol "<" or ">". The initial state of the conveyor i is equal to the i-th character of the string s. There are n+1 holes numbered with integers from 0 to n. The hole 0 is on the left side of the conveyor 1, and for all i β₯ 1 the hole i is on the right side of the conveyor i.
When a ball is on the conveyor i, the ball moves by the next rules:
If the symbol "<" is on the conveyor i, then:
* If i=1, the ball falls into the hole 0.
* If the symbol "<" is on the conveyor i-1, the ball moves to the conveyor i-1.
* If the symbol ">" is on the conveyor i-1, the ball falls into the hole i-1.
If the symbol ">" is on the conveyor i, then:
* If i=n, the ball falls into the hole n.
* If the symbol ">" is on the conveyor i+1, the ball moves to the conveyor i+1.
* If the symbol "<" is on the conveyor i+1, the ball falls into the hole i.
You should answer next q queries, each query is defined by the pair of integers l, r (1 β€ l β€ r β€ n):
* First, for all conveyors l,l+1,...,r, the symbol "<" changes to ">" and vice versa. These changes remain for the next queries.
* After that, put one ball on each conveyor l,l+1,...,r. Then, each ball falls into some hole. Find the maximum number of balls in one hole. After the query all balls disappear and don't considered in the next queries.
Input
The first line contains two integers n, q (1 β€ n β€ 5 Γ 10^5 , 1 β€ q β€ 10^5).
The second line contains a string s of length n. It consists of characters "<" and ">".
Next q lines contain the descriptions of the queries, i-th of them contains two integers l, r (1 β€ l β€ r β€ n), describing the i-th query.
Output
Print q lines, in the i-th of them print the answer to the i-th query.
Example
Input
5 6
><>><
2 4
3 5
1 5
1 3
2 4
1 5
Output
3
3
5
3
2
3
Note
* In the first query, the conveyors change to ">><<<". After that, put a ball on each conveyor \{2,3,4\}. All three balls fall into the hole 2. So the answer is 3.
* In the second query, the conveyors change to ">>>>>". After that, put a ball on each conveyor \{3,4,5\}. All three balls fall into the hole 5. So the answer is 3.
* In the third query, the conveyors change to "<<<<<". After that, put a ball on each conveyor \{1,2,3,4,5\}. All five balls fall into the hole 0. So the answer is 5.
* In the fourth query, the conveyors change to ">>><<". After that, put a ball on each conveyor \{1,2,3\}. All three balls fall into the hole 3. So the answer is 3.
* In the fifth query, the conveyors change to "><<><". After that, put a ball on each conveyor \{2,3,4\}. Two balls fall into the hole 1, and one ball falls into the hole 4. So, the answer is 2.
* In the sixth query, the conveyors change to "<>><>". After that, put a ball on each conveyor \{1,2,3,4,5\}. Three balls fall into the hole 3, one ball falls into the hole 0 and one ball falls into the hole 5. So, the answer is 3.
Submitted Solution:
```
t = [int(x) for x in input().split()]
b = list(input())
n = t[1]
while n:
i=0
a = [int(x) for x in input().split()]
l = a[0]
i=l
r = a[1]
while i<r:
if b[i-1]=="<": b[i-1]=">"
else: b[i-1]="<"
i+=1
hole=[]
i=0
while i<len(b)-1:
thing=0
if b[i]==">" and b[i+1] == "<":
j=i
while j>=0 and b[j]==">":
thing+=1
j-=1
j=i+1
#print(thing)
while j<len(b) and b[j]=="<":
thing+=1
j+=1
#print(thing)
hole.append(thing)
i+=1
if b[i]==">":
j=i
thing = 0
while j>=0 and b[j]==">":
thing+=1
j-=1
hole.append(thing)
if b[0]=="<":
j=0
thing=0
while j<len(b) and b[j]=="<":
thing+=1
j+=1
hole.append(thing)
print(max(hole))
n-=1
```
No
| 8,528 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are a warrior fighting against the machine god Thor.
Thor challenge you to solve the following problem:
There are n conveyors arranged in a line numbered with integers from 1 to n from left to right. Each conveyor has a symbol "<" or ">". The initial state of the conveyor i is equal to the i-th character of the string s. There are n+1 holes numbered with integers from 0 to n. The hole 0 is on the left side of the conveyor 1, and for all i β₯ 1 the hole i is on the right side of the conveyor i.
When a ball is on the conveyor i, the ball moves by the next rules:
If the symbol "<" is on the conveyor i, then:
* If i=1, the ball falls into the hole 0.
* If the symbol "<" is on the conveyor i-1, the ball moves to the conveyor i-1.
* If the symbol ">" is on the conveyor i-1, the ball falls into the hole i-1.
If the symbol ">" is on the conveyor i, then:
* If i=n, the ball falls into the hole n.
* If the symbol ">" is on the conveyor i+1, the ball moves to the conveyor i+1.
* If the symbol "<" is on the conveyor i+1, the ball falls into the hole i.
You should answer next q queries, each query is defined by the pair of integers l, r (1 β€ l β€ r β€ n):
* First, for all conveyors l,l+1,...,r, the symbol "<" changes to ">" and vice versa. These changes remain for the next queries.
* After that, put one ball on each conveyor l,l+1,...,r. Then, each ball falls into some hole. Find the maximum number of balls in one hole. After the query all balls disappear and don't considered in the next queries.
Input
The first line contains two integers n, q (1 β€ n β€ 5 Γ 10^5 , 1 β€ q β€ 10^5).
The second line contains a string s of length n. It consists of characters "<" and ">".
Next q lines contain the descriptions of the queries, i-th of them contains two integers l, r (1 β€ l β€ r β€ n), describing the i-th query.
Output
Print q lines, in the i-th of them print the answer to the i-th query.
Example
Input
5 6
><>><
2 4
3 5
1 5
1 3
2 4
1 5
Output
3
3
5
3
2
3
Note
* In the first query, the conveyors change to ">><<<". After that, put a ball on each conveyor \{2,3,4\}. All three balls fall into the hole 2. So the answer is 3.
* In the second query, the conveyors change to ">>>>>". After that, put a ball on each conveyor \{3,4,5\}. All three balls fall into the hole 5. So the answer is 3.
* In the third query, the conveyors change to "<<<<<". After that, put a ball on each conveyor \{1,2,3,4,5\}. All five balls fall into the hole 0. So the answer is 5.
* In the fourth query, the conveyors change to ">>><<". After that, put a ball on each conveyor \{1,2,3\}. All three balls fall into the hole 3. So the answer is 3.
* In the fifth query, the conveyors change to "><<><". After that, put a ball on each conveyor \{2,3,4\}. Two balls fall into the hole 1, and one ball falls into the hole 4. So, the answer is 2.
* In the sixth query, the conveyors change to "<>><>". After that, put a ball on each conveyor \{1,2,3,4,5\}. Three balls fall into the hole 3, one ball falls into the hole 0 and one ball falls into the hole 5. So, the answer is 3.
Submitted Solution:
```
#hello guys!
print("This is suicide!")
```
No
| 8,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
import collections as cc
import bisect as bi
I=lambda:list(map(int,input().split()))
cur=1
def f(x):
return x*(x+1)//2
ar=[f((2**i)-1) for i in range(1,32)]
for i in range(1,len(ar)):
ar[i]+=ar[i-1]
for tc in range(int(input())):
n,=I()
temp=bi.bisect_right(ar,n)
print(temp)
```
| 8,530 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
def sum_ab(a,b):
return int((a + b) / 2 * (b - a + 1))
for _ in range(int((input()))):
schody = int(input())
koszt = 1
cos = 1
pot = 2
x = 0
while schody >= koszt:
schody -= koszt
cos += pot
pot *= 2
koszt = sum_ab(1, cos)
x += 1
print(x)
```
| 8,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
n = 10
ans = [1]
for i in range(1, 32):
ans.append(ans[-1]*2+(2**i)**2)
for i in range(1, len(ans)):
ans[i] = ans[i]+ans[i-1]
t = int(input())
for _ in range(t):
num = int(input())
j = 0
while num >= ans[j]:
j = j + 1
print(j)
# print(ans[-1])
# for _ in range(t):
```
| 8,532 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
x=int(input())
i,curr=0,0
while True:
curr = 2*curr+1
val = (curr*(curr+1))//2
if x>=val:
x-=val
i+=1
else:break
print(i)
```
| 8,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
ar=[]
r=1
while(1):
x=(r*(r+1))//2
if(x>1e20):
break
ar.append(x)
r=r*2+1
t=int(input())
for a0 in range(t):
x=int(input())
r=0;
while(x>0):
r+=1
x-=ar[r]
print(r)
```
| 8,534 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
def main():
n = int(input())
m = 0
k = 2
while n >= 0:
n -= (k - 1) * ((k - 1)// 2 + 1)
k *= 2
if n < 0:
break
m += 1
print(m)
if __name__ == '__main__':
t = int(input())
for i in range(t):
main()
```
| 8,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
import math
t = int(input())
while (t):
x = int(input())
cnt = 0
i = 1
while (True):
if (i*(i+1)//2 <= x):
cnt += 1
else:
break
x -= i*(i+1)//2
i = i*2 + 1
print(cnt)
t -= 1
```
| 8,536 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Tags: brute force, constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c for j in range(b)] for i in range(a)]
def list3d(a, b, c, d): return [[[d for k in range(c)] for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e for l in range(d)] for k in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10
def rec(sm, cnt, h, x):
if sm > x:
return 0
return rec(sm+cnt*2+(h+1)**2, cnt*2+(h+1)**2, h*2+1, x) + 1
for _ in range(INT()):
x = INT()
ans = rec(1, 1, 1, x)
print(ans)
```
| 8,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
n = int(input())
p = 1
s = 0
ans = 0
while True:
p *= 2
s += (p - 1) * p // 2
if s > n:
break
ans += 1
print(ans)
main()
```
Yes
| 8,538 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
for i in range(int(input())):
x=int(input())
count=0
stair=1
cell=1
while((x-cell)>=0):
x=x-cell
count+=1
stair=(stair*2)+1
cell=(stair*(stair+1))//2
print(count)
```
Yes
| 8,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
import math
t = int(input())
for _ in range(t):
n = int(input())
a = 1
tot = 1
cost = 1
ans = 1
while n>=tot:
a <<= 1
cost = 2*cost + a*a
tot += cost
ans += 1
print(ans-1)
```
Yes
| 8,540 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
wyn = 0
pot = 1
total = 1
while total <= n:
wyn += 1
pot += 1
total += (2**pot-1)*(2**pot)//2
print(wyn)
```
Yes
| 8,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
from math import *
a = int(input())
for x in range(a):
q = int(input())
if q == 1:
print(1)
else:
print(floor(log(q,4)))
```
No
| 8,542 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
arr=[]
for i in range(1,32):
arr.append(2**i-1)
temp=[]
for i in arr:
temp.append(i*(i+1)/2)
res=[temp[0]]
for i in range(1,len(temp)):
res.append(res[-1]+temp[i])
for _ in range(int(input())):
n=int(input())
if n<=6:
print(1)
elif 7<=n<=34:
print(2)
elif 35<=n<=154:
print(3)
else:
for i in range(len(res)):
if n<res[i]:
print(i)
break
```
No
| 8,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
# https://codeforces.com/contest/1419/problem/B
T=int(input())
for t in range(T):
N = int(input())
sum = 0
i=0
cells=0
count=0
while(cells<=N):
sum=sum*2+1
print(sum)
# i+=2
cells+=(sum*(sum+1))/2
count+=1
print(count-1)
```
No
| 8,544 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, β¦, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The description of each test case contains a single integer x (1 β€ x β€ 10^{18}) β the number of cells for building staircases.
Output
For each test case output a single integer β the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
for _ in " "*int(input()):
n,ans=int(input()),0
for j in range(1,40):
d=2**j-1
s=d*(d+1)/2
if n>=s:
n-=s
ans+=1
else:
break
print(ans)
```
No
| 8,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
from math import ceil
for _ in range(int(input())):
n = int(input())
s = input()
ct = 0
for i in range(n-1):
if s[i] == s[i+1]:
ct += 1
print(ceil(ct / 2.0))
```
| 8,546 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for i in range(0, t):
n=int(input())
s=input()
count=0
for j in range(0,n):
if(j!=(n-1)):
if(s[j]==s[j+1]):
count=count+1
else:
if(s[j]==s[0]):
count=count+1
print(count//2)
```
| 8,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
import math
from collections import Counter,defaultdict
LI=lambda:list(map(int,input().split()))
MAP=lambda:map(int,input().split())
IN=lambda:int(input())
S=lambda:input()
def case():
n=IN()
s=S()
a=s.split("0")
b=s.split("1")
x=0
for i in a:
if len(i):
x+=len(i)-1
for i in b:
if len(i):
x+=len(i)-1
print((x+1)//2)
for _ in range(IN()):
case()
```
| 8,548 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
s=list(input())
maxi=0
temp1=0
temp2=0
for i in range(1,n):
if(s[i-1]=='1' and s[i]=='1'):
temp1+=1
maxi=max(maxi,temp1)
if(s[i-1]=='0' and s[i]=='0'):
temp2+=1
maxi=max(maxi,temp2)
print(maxi)
```
| 8,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
t = int(input())
while t > 0:
t -= 1
n = int(input())
s = input()
ans = 0
i = 0
while i+1 < n:
if s[i] == s[i+1]:
ans += 1;
i += 1
print((ans+1)//2)
```
| 8,550 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
s=input()
c=0
for i in range(0,n-1):
if s[i]==s[i+1]:
c+=1
print((c+1)//2)
```
| 8,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
"""
pppppppppppppppppppp
ppppp ppppppppppppppppppp
ppppppp ppppppppppppppppppppp
pppppppp pppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppp pppppppp
ppppppppppppppppppppp ppppppp
ppppppppppppppppppp ppppp
pppppppppppppppppppp
"""
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(2 * (10 ** 5))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
for _ in range(int(data())):
n = int(data())
s = list(data())
answer = 0
# start with 1.
pos = [0] * n
for i in range(n):
if i % 2 != int(s[i]) % 2:
pos[i] = 1
answer = int(pos[0] == 1)
for i in range(1, n):
if pos[i] == pos[i-1] == 1:
continue
if pos[i] == 1:
answer += 1
# start with 0.
pos = [0] * n
temp = 0
for i in range(n):
if i % 2 == int(s[i]) % 2:
pos[i] = 1
temp = int(pos[0] == 1)
for i in range(1, n):
if pos[i] == pos[i-1] == 1:
continue
if pos[i] == 1:
temp += 1
out(min(temp, answer))
```
| 8,552 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Tags: constructive algorithms, greedy
Correct Solution:
```
import math
t=int(input())
for _ in range(t):
n=int(input())
s=str(input())
v1=0;v2=0
p=0
for i in range(len(s)):
if int(s[i])!=i%2:
p+=1
else:
v1+=int(p>0)
p=0
v1+=int(p>0)
p=0
for i in range(len(s)):
if int(s[i])==i%2:
p+=1
else:
v2+=int(p>0)
p=0
v2+=int(p>0)
print(min(v1,v2))
```
| 8,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
for s in[*open(0)][2::2]:print(len(s)//2-sum(map(bool,s.split(s[-2])))+1)
```
Yes
| 8,554 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
t=int(input())
while(t):
zero=0
one=0
size=int(input())
array=list(map(int,input()))
for i in range(len(array)-1):
if(array[i]==1 and array[i+1]==1):
one+=1
if(array[i]==0 and array[i+1]==0):
zero+=1
print(max(one,zero))
t-=1
```
Yes
| 8,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
import sys,math
from heapq import *
from collections import defaultdict as dd
mod = 10**9+7; modd = 998244353
input = lambda: sys.stdin.readline().strip()
inp = lambda: list(map(int,input().split()))
for _ in range(int(input())):
n, = inp()
a = str(input())
ans = 0
for i in range(n):
ans+= 1*(a[i]==a[(i+1)%n])
print(ans//2)
```
Yes
| 8,556 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
if __name__ == '__main__':
for _ in range(int(input())):
n = int(input())
s = input()
c=0
for i in range(len(s)-1):
if s[i] == '0' and s[i+1]=='0':
i +=1
c+=1
k = 0
for i in range(len(s)-1):
if s[i]=='1' and s[i+1] == '1':
i+=1
k+=1
print(max(c,k))
```
Yes
| 8,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
t=int(input())
while(t):
res=0
checker=-1
size=int(input())
array=list(map(int,input()))
for i in range(len(array)-1):
if(array[i]==array[i+1] and checker==-1):
checker=array[i]
if(array[i]==checker and array[i+1]==checker):
res+=1
print(res)
t-=1
```
No
| 8,558 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
for testCase in range(int(input())):
n = int(input())
s = input()
if n == 2:
print(0)
continue
print((s.count("00")+s.count("11"))//2)
```
No
| 8,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*(2*(10**5)+5)
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range((2*(10**5)+5)):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
spf[1]=1
for i in range(2,mx+1):
if spf[i]==mx:
spf[i]=i
for j in range(i*i,mx+1,i):
if i<spf[j]:
spf[j]=i
return
def readTree(n,e): # to read tree
adj=[set() for i in range(n+1)]
for i in range(e):
u1,u2=IP()
adj[u1].add(u2)
return adj
#####################################################################################
mod=10**9+7
def solve():
n=II()
s=input()
cnt=0
ans=0
for i in range(n):
if s[i]=='1':
cnt+=1
else:
if cnt>0:
ans+=(cnt-1)
cnt=0
print(ans)
return
t=II()
for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
# ``````ΒΆ0````1ΒΆ1_```````````````````````````````````````
# ```````ΒΆΒΆΒΆ0_`_ΒΆΒΆΒΆ0011100ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ001_````````````````````
# ````````ΒΆΒΆΒΆΒΆΒΆ00ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0_````````````````
# `````1_``ΒΆΒΆ00ΒΆ0000000000000000000000ΒΆΒΆΒΆΒΆ0_`````````````
# `````_ΒΆΒΆ_`0ΒΆ000000000000000000000000000ΒΆΒΆΒΆΒΆΒΆ1``````````
# ```````ΒΆΒΆΒΆ00ΒΆ00000000000000000000000000000ΒΆΒΆΒΆ_`````````
# ````````_ΒΆΒΆ00000000000000000000ΒΆΒΆ00000000000ΒΆΒΆ`````````
# `````_0011ΒΆΒΆΒΆΒΆΒΆ000000000000ΒΆΒΆ00ΒΆΒΆ0ΒΆΒΆ00000000ΒΆΒΆ_````````
# ```````_ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00000000000ΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆ00000000ΒΆΒΆ1````````
# ``````````1ΒΆΒΆΒΆΒΆΒΆ000000ΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0000000ΒΆΒΆΒΆ````````
# ```````````ΒΆΒΆΒΆ0ΒΆ000ΒΆ00ΒΆ0ΒΆΒΆ`_____`__1ΒΆ0ΒΆΒΆ00ΒΆ00ΒΆΒΆ````````
# ```````````ΒΆΒΆΒΆΒΆΒΆ00ΒΆ00ΒΆ10ΒΆ0``_1111_`_ΒΆΒΆ0000ΒΆ0ΒΆΒΆΒΆ````````
# ``````````1ΒΆΒΆΒΆΒΆΒΆ00ΒΆ0ΒΆΒΆ_ΒΆΒΆ1`_ΒΆ_1_0_`1ΒΆΒΆ_0ΒΆ0ΒΆΒΆ0ΒΆΒΆ````````
# ````````1ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆ0ΒΆ0_0ΒΆ``100111``_ΒΆ1_0ΒΆ0ΒΆΒΆ_1ΒΆ````````
# ```````1ΒΆΒΆΒΆΒΆ00ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ010ΒΆ``1111111_0ΒΆ11ΒΆΒΆΒΆΒΆΒΆ_10````````
# ```````0ΒΆΒΆΒΆΒΆ__10ΒΆΒΆΒΆΒΆΒΆ100ΒΆΒΆΒΆ0111110ΒΆΒΆΒΆ1__ΒΆΒΆΒΆΒΆ`__````````
# ```````ΒΆΒΆΒΆΒΆ0`__0ΒΆΒΆ0ΒΆΒΆ_ΒΆΒΆΒΆ_11````_0ΒΆΒΆ0`_1ΒΆΒΆΒΆΒΆ```````````
# ```````ΒΆΒΆΒΆ00`__0ΒΆΒΆ_00`_0_``````````1_``ΒΆ0ΒΆΒΆ_```````````
# ``````1ΒΆ1``ΒΆΒΆ``1ΒΆΒΆ_11``````````````````ΒΆ`ΒΆΒΆ````````````
# ``````1_``ΒΆ0_ΒΆ1`0ΒΆ_`_``````````_``````1_`ΒΆ1````````````
# ``````````_`1ΒΆ00ΒΆΒΆ_````_````__`1`````__`_ΒΆ`````````````
# ````````````ΒΆ1`0ΒΆΒΆ_`````````_11_`````_``_``````````````
# `````````ΒΆΒΆΒΆΒΆ000ΒΆΒΆ_1```````_____```_1``````````````````
# `````````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0_``````_````_1111__``````````````
# `````````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ01_`````_11____1111_```````````
# `````````ΒΆΒΆ0ΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ1101_______11ΒΆ_```````````
# ``````_ΒΆΒΆΒΆ0000000ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆ0ΒΆΒΆΒΆ1````````````
# `````0ΒΆΒΆ0000000ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ1`````````````
# ````0ΒΆ0000000ΒΆΒΆ0_````_011_10ΒΆ110ΒΆ01_1ΒΆΒΆΒΆ0````_100ΒΆ001_`
# ```1ΒΆ0000000ΒΆ0_``__`````````_`````````0ΒΆ_``_00ΒΆΒΆ010ΒΆ001
# ```ΒΆΒΆ00000ΒΆΒΆ1``_01``_11____``1_``_`````ΒΆΒΆ0100ΒΆ1```_00ΒΆ1
# ``1ΒΆΒΆ00000ΒΆ_``_ΒΆ_`_101_``_`__````__````_0000001100ΒΆΒΆΒΆ0`
# ``ΒΆΒΆΒΆ0000ΒΆ1_`_ΒΆ``__0_``````_1````_1_````1ΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆ0```
# `_ΒΆΒΆΒΆΒΆ00ΒΆ0___01_10ΒΆ_``__````1`````11___`1ΒΆΒΆΒΆ01_````````
# `1ΒΆΒΆΒΆΒΆΒΆ0ΒΆ0`__01ΒΆΒΆΒΆ0````1_```11``___1_1__11ΒΆ000`````````
# `1ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ1_1_01__`01```_1```_1__1_11___1_``00ΒΆ1````````
# ``ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0`__10__000````1____1____1___1_```10ΒΆ0_```````
# ``0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ1___0000000```11___1__`_0111_```000ΒΆ01```````
# ```ΒΆΒΆΒΆ00000ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ01___1___00_1ΒΆΒΆΒΆ`_``1ΒΆΒΆ10ΒΆΒΆ0```````
# ```1010000ΒΆ000ΒΆΒΆ0100_11__1011000ΒΆΒΆ0ΒΆ1_10ΒΆΒΆΒΆ_0ΒΆΒΆ00``````
# 10ΒΆ000000000ΒΆ0________0ΒΆ000000ΒΆΒΆ0000ΒΆΒΆΒΆΒΆ000_0ΒΆ0ΒΆ00`````
# ΒΆΒΆΒΆΒΆΒΆΒΆ0000ΒΆΒΆΒΆΒΆ_`___`_0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00000000000000_0ΒΆ00ΒΆ01````
# ΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ_``_1ΒΆΒΆΒΆ00000000000000000000_0ΒΆ000ΒΆ01```
# 1__```1ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00ΒΆΒΆΒΆΒΆ00000000000000000000ΒΆ_0ΒΆ0000ΒΆ0_``
# ```````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00000000000000000000010ΒΆ00000ΒΆΒΆ_`
# ```````0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00000000000000000000ΒΆ10ΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆ0`
# ````````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ00000000000000000000010ΒΆΒΆΒΆ0011```
# ````````1ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ0000000000000000000ΒΆ100__1_`````
# `````````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ000000000000000000ΒΆ11``_1``````
# `````````1ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ00000000000000000ΒΆ11___1_`````
# ``````````ΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0000000000000000ΒΆ11__``1_````
# ``````````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆ000000000000000ΒΆ1__````__```
# ``````````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0000000000000000__`````11``
# `````````_ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ000ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ000000000000011_``_1ΒΆΒΆΒΆ0`
# `````````_ΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆ000000ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ000000000000100ΒΆΒΆΒΆΒΆ0_`_
# `````````1ΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ000000000ΒΆΒΆΒΆΒΆΒΆΒΆ000000000ΒΆ00ΒΆΒΆ01`````
# `````````ΒΆΒΆΒΆΒΆΒΆ0ΒΆ0ΒΆΒΆΒΆ0000000000000ΒΆ0ΒΆ00000000011_``````_
# ````````1ΒΆΒΆ0ΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆ000000000000000000000ΒΆ11___11111
# ````````ΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆΒΆ00ΒΆΒΆΒΆΒΆΒΆΒΆ000000000000000000ΒΆ011111111_
# ```````_ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ0000000ΒΆ0ΒΆ00000000000000000ΒΆ01_1111111
# ```````0ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ000000000000000000000000000ΒΆ01___`````
# ```````ΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ000000000000000000000000000ΒΆ01___1````
# ``````_ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00000000000000000000000000000011_111```
# ``````0ΒΆΒΆ0ΒΆΒΆΒΆ0ΒΆΒΆ0000000000000000000000000000ΒΆ01`1_11_``
# ``````ΒΆΒΆΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆ0000000000000000000000000000001`_0_11_`
# ``````ΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆΒΆ00000000000000000000000000000ΒΆ01``_0_11`
# ``````ΒΆΒΆΒΆΒΆ0ΒΆΒΆΒΆΒΆ00000000000000000000000000000001```_1_11
```
No
| 8,560 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i β s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 β€ t β€ 1000) β the number of test cases.
The first line of each test case contains a single integer n (2 β€ n β€ 10^5; n is even) β the length of string s.
The second line of each test case contains a binary string s of length n (s_i β {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 β 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 β 10101100;
2. 10101100 β 10101010.
Submitted Solution:
```
#! /usr/bin/python3
import os
import sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n = int(input())
s = list(int(x) for x in input())
zero = 0
one = 0
temp0 = 1
temp1 = 1
for i in range(n-1):
if s[i] == s[i+1]:
if s[i] == 0:
temp0 += 1
else:
temp1 += 1
else:
one = max(one, temp1)
zero = max(zero, temp0)
temp1 = 1
temp0 = 1
print(max(one, zero) - 1)
# 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()
```
No
| 8,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
# Author: yumtam
# Created at: 2020-12-28 23:43
from itertools import groupby, product
def main():
n = int(input())
ar = [int(t) for t in input().split()]
ops = set(input())
if len(ops) == 1:
ans = [ops.pop()] * (n-1)
elif ops == {'+', '-'}:
ans = ['+'] * (n-1)
elif ops == {'*', '-'}:
ans = ['*'] * (n-1)
if 0 in ar:
idx = ar.index(0)
if idx > 0:
ans[idx-1] = '-'
else:
ans = ['?'] * (n-1)
def solve(l, r):
while l < r:
if ar[l] == 1:
ans[l] = '+'
else:
break
l += 1
while l < r:
if ar[r] == 1:
ans[r-1] = '+'
else:
break
r -= 1
if l == r:
return
A = ar[l:r+1]
S = max(sum(A), 2*(r+1-l))
P = 1
for x in A:
P *= x
if P >= S:
for j in range(l, r):
ans[j] = '*'
return
nums = []
conns = []
cl = []
i = l
for ones, it in groupby(A, key=lambda x: x==1):
if ones:
L = len(list(it))
conns.append(L)
cl.append(i)
i += L
else:
p = 1
for x in it:
p *= x
if i < r: ans[i] = '*'
i += 1
nums.append(p)
# print(nums)
# print(conns)
# print(cl)
best_seq = 0
best_val = sum(A)
for seq in range(2**len(conns)):
i = 0
cur = 0
prod = nums[i]
for h in range(len(conns)):
op = seq & (1 << h)
if op:
prod *= nums[i+1]
else:
cur += prod + conns[i]
prod = nums[i+1]
i += 1
cur += prod
if cur > best_val:
best_val = cur
best_seq = seq
for h in range(len(conns)):
op = best_seq & (1 << h)
ch = '*' if op else '+'
for i in range(cl[h]-1, cl[h]+conns[h]):
ans[i] = ch
l = 0
for i in range(n):
if ar[i] == 0:
if i > 0:
ans[i-1] = '+'
if i < n-1:
ans[i] = '+'
if l < i-1:
solve(l, i-1)
l = i+1
if l < n-1:
solve(l, n-1)
res = [None] * (2*n-1)
res[::2] = ar
res[1::2] = ans
print(*res, sep='')
main()
```
| 8,562 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
s = input()
usable = []
if "+" in s:
usable.append(0)
if "*" in s:
usable.append(1)
if "-" in s:
usable.append(2)
usable = tuple(usable)
if len(usable) == 1:
ans = [s] * (n-1)
ansPr = []
for i in range(n - 1):
ansPr.append(str(a[i]))
ansPr.append(ans[i])
ansPr.append(str(a[n - 1]))
print("".join(ansPr))
elif usable == (0,1) or usable == (0,1,2): # +, *
ans = ["+"] * (n - 1)
curIndex = 0
while curIndex < n:
curCpy = curIndex
while curIndex < n and a[curIndex] != 0:
curIndex += 1
left = curCpy
right = curIndex
while left < right and a[left] == 1:
left += 1
while right > left and a[right - 1] == 1:
right -= 1
curCur = left
nonOneProd = []
oneLength = []
while curCur < right:
curProd = 1
while curCur < right and a[curCur] != 1:
if curProd < 2 * (right - left):
curProd *= a[curCur]
curCur += 1
nonOneProd.append(curProd)
if curCur == right:
break
curOneLength = 0
while curCur < right and a[curCur] == 1:
curOneLength += a[curCur]
curCur += 1
oneLength.append(curOneLength)
curAllProd = 1
index = 0
while curAllProd < 2 * (right - left) and index < len(nonOneProd):
curAllProd *= nonOneProd[index]
index += 1
if curAllProd >= 2 * (right - left) or len(nonOneProd) == 1:
for i in range(left, right - 1):
ans[i] = "*"
else:
maskLen = len(oneLength)
bestMask = 0
bestAns = 0
for i in range(1 << maskLen):
curAns = 0
curProd = nonOneProd[0]
for j in range(maskLen):
if i & (1 << j):
curProd *= nonOneProd[j + 1]
else:
curAns += curProd
curAns += oneLength[j]
curProd = nonOneProd[j + 1]
curAns += curProd
if curAns > bestAns:
bestAns = curAns
bestMask = i
curOneCount = 0
for i in range(left, right - 1):
if a[i] != 1 and a[i+1] == 1:
curOneCount += 1
if a[i] != 1 and a[i + 1] != 1:
ans[i] = "*"
elif bestMask & (1 << (curOneCount - 1)):
ans[i] = "*"
while curIndex < n and a[curIndex] == 0:
curIndex += 1
ansPr = []
for i in range(n - 1):
ansPr.append(str(a[i]))
ansPr.append(ans[i])
ansPr.append(str(a[n - 1]))
print("".join(ansPr))
elif usable == (1,2): # -, *
ans = ["-"] * (n - 1)
firstZero = True
for i in range(n - 1):
ans[i] = "*"
if a[i + 1] == 0 and firstZero:
ans[i] = "-"
firstZero = False
ansPr = []
for i in range(n - 1):
ansPr.append(str(a[i]))
ansPr.append(ans[i])
ansPr.append(str(a[n - 1]))
print("".join(ansPr))
elif usable == (0,2): # +, -
ans = ["+"] * (n-1)
ansPr = []
for i in range(n - 1):
ansPr.append(str(a[i]))
ansPr.append(ans[i])
ansPr.append(str(a[n - 1]))
print("".join(ansPr))
```
| 8,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
# n = 10
# A = [random.randint(0, 9) for i in range(10)]
# symbols = ["+", "*"]
n = int(input())
A = list(map(int, input().split()))
symbols = list(input())
def solve1(array):
# "+" or "*"
# 2+1+1+1+1+1+2
# 2*1*1*1*2
j = 0
# we need to break everything into components deprived of 0s
solution = []
first = True
while (j < len(array)):
i = j
if not first:
solution.append("+")
first = False
if array[j] == 0:
j += 1
while (j < len(array) and array[j] == 0):
j += 1
j -= 1
solution.extend(list("+".join(["0"] * (j - i + 1))))
else:
j += 1
while (j < len(array) and array[j] != 0):
j += 1
j -= 1
sol = solve11(array[i:j + 1])
solution.extend(sol)
j += 1
return "".join(list(map(str, solution)))
def solve11(array):
count = 0
first_different_from_one_prefix = 0
for i in range(len(array)):
if array[i] != 1:
break
else:
first_different_from_one_prefix = i + 1
array = array[first_different_from_one_prefix:]
first_part = "+".join(["1"] * (first_different_from_one_prefix))
first_different_from_one_suffix = len(array) - 1
for i in range(len(array) - 1, -1, -1):
if array[i] != 1:
break
else:
first_different_from_one_suffix = i - 1
second_part = "+".join(["1"] * (len(array) - 1 - first_different_from_one_suffix))
array = array[:first_different_from_one_suffix + 1]
SOLUTION = []
if len(array) > 0:
for i in range(0, len(array)):
if array[i] > 2:
count += 1
# Question is : is it better to use a * or a + at given empty space ?
# if subproblem contains more than 20 numbers it's always better to use *
if count >= 20:
SOLUTION.extend(list("*".join(list(map(str, array)))))
SOLUTION = SOLUTION[::-1]
else:
# we need dp here
DP = [0 for i in range(len(array))]
DP_PARENTS = [-1 for i in range(len(array))]
DP[0] = array[0]
for i in range(0, len(array)):
if array[i] == 1:
DP[i] = DP[i - 1] + 1
DP_PARENTS[i] = i - 1
continue
else:
PROD = 1
for k in range(i - 1, -2, -1):
PROD *= array[k + 1]
if k == -1:
if DP[i] < PROD:
DP[i] = PROD
DP_PARENTS[i] = k
elif DP[i] < DP[k] + PROD:
DP[i] = DP[k] + PROD
DP_PARENTS[i] = k
# we need to construct the solution
current = len(array) - 1
while (current != -1):
k = DP_PARENTS[current]
for i in range(current, k, -1):
SOLUTION.append(array[i])
SOLUTION.append("*")
SOLUTION.pop()
SOLUTION.append("+")
current = k
SOLUTION.pop()
RESULT = ""
SOLUTION = list(map(str, SOLUTION[::-1]))
parts = [first_part, "".join(SOLUTION), second_part]
parts = [part for part in parts if part != ""]
return "+".join(parts)
# print(solve11([7, 8, 3, 8, 6, 3, 3, 6, 5, 5, 7, 3, 8, 8, 7, 7, 4, 3, 3, 3]))
# print(solve11([6, 4, 8, 7, 9, 9, 8, 6, 8, 1, 6, 6, 8, 3, 5, 3, 4, 4, 1, 3]))
# print("hum")
def solve2(array):
# "-" or "*"
# 2+1*5-0*2*6*1*1
# when meet zero take -
# after 0 take *
current = 0
solution = []
first_zero = None
array = list(map(str, array))
while (current < len(array) and array[current] != '0'):
current += 1
if current == len(array):
if array[-1] == '0':
return "*".join(array[:-1]) + "-0"
else:
return "*".join(array)
else:
if current != 0:
return "*".join(array[0:current]) + "-" + "*".join(array[current:])
else:
return "*".join(array[current:])
MY_SOLUTION = None
if len(symbols) == 1:
MY_SOLUTION = symbols[0].join(list(map(str, A)))
if len(symbols) == 2:
if "+" in symbols and "-" in symbols:
MY_SOLUTION = "+".join(list(map(str, A)))
if "+" in symbols and "*" in symbols:
MY_SOLUTION = solve1(A)
if "*" in symbols and "-" in symbols:
MY_SOLUTION = solve2(A)
if len(symbols) == 3:
MY_SOLUTION = solve1(A)
print(MY_SOLUTION)
```
| 8,564 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
#!/usr/bin/env python3
from itertools import accumulate, groupby
from functools import reduce
def prod(a):
return reduce(lambda x,y: min(x*y,10**6),a,1)
def solve_positive(a):
if a == '': return '+'
b = [''.join(v) for _,v in groupby(a, key=lambda x: x == '1')]
if b[0][0] == '1':
return '+' * len(b[0]) + solve_positive(a[len(b[0]):])
if b[-1][0] == '1':
return solve_positive(a[:-len(b[-1])]) + '+' * len(b[-1])
p = [prod(map(int,x)) for x in b[::2]]
q = [len(x) for x in b[1::2]]
k = len(p)
if prod(p) >= 10**6:
return '+' + '*' * (len(a)-1) + '+'
dp = [0] * k
go = [k] * k
for i in range(k)[::-1]:
dp[i] = prod(p[i:])
for j in range(i+1,k):
ndp = prod(p[i:j]) + q[j-1] + dp[j]
if ndp > dp[i]:
dp[i], go[i] = ndp, j
offset = [0] + list(accumulate(map(len,b)))
res = ['*'] * (len(a)-1)
i = go[0]
while i < k:
a = offset[2*i-1]-1
b = offset[2*i]
res[a:b] = '+' * (b-a)
i = go[i]
return '+' + ''.join(res) + '+'
def solve(a,ops):
n = len(a)
if len(ops) == 1: return ops * (n-1)
if sorted(ops) == list('+-'): return '+' * (n-1)
if sorted(ops) == list('*-'):
k = a.index('0') if '0' in a else n
if k == 0 or k == n: return '*' * (n-1)
return '*' * (k-1) + '-' + '*' * (n-k-1)
return ''.join(map(solve_positive,a.split('0')))[1:-1]
n = int(input())
a = ''.join(input().split())
ops = input()
b = solve(a,ops) + '\n'
for i in range(n):
print(a[i], end='')
print(b[i], end='')
```
| 8,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int,input().split()))
ops = input().strip()
if len(ops) == 1:
print(ops.join(map(str, a)))
elif '+' in ops and '*' in ops:
seqs = [[a[0]]]
output = []
# split into seqs of all 0's or non 0's
# every seq in seqs is a list of all 0's or non 0's
for i in range(1,n):
if a[i] == 0:
if seqs[-1][-1] == 0:
seqs[-1].append(a[i])
else:
seqs.append([a[i]])
else:
if seqs[-1][-1] == 0:
seqs.append([a[i]])
else:
seqs[-1].append(a[i])
for seq in seqs:
if seq[0] == 0:
# all 0
output.append('+'.join(map(str,seq)))
else:
# if prod >= 2*n, using addition is never optimal
prod = 1
for i in seq:
prod *= i
if prod >= 2 * n:
break
if prod >= 2 * n:
new_seq = ''
l = 0
r = len(seq) - 1
for i in range(len(seq)):
if seq[i] != 1:
l = i
break
for i in range(len(seq)-1,-1,-1):
if seq[i] != 1:
r = i
break
if l != 0:
new_seq += '+'.join('1'*l) + '+'
new_seq += '*'.join(map(str,seq[l:r+1]))
if r != len(seq)-1:
new_seq += '+' + '+'.join('1' * (len(seq) - 1 - r))
output.append(new_seq)
continue
# prod < 2*n so max length of seq after combining 1's is 2*log(2*n)
# use dp to find optimal operations
b = []
lst = -1
for i in seq:
if i == 1:
if lst != 1:
b.append([-1,1])
else:
b[-1][-1] += 1
else:
b.append([0,i])
lst = i
# + -> 0 | * -> 1
last_state = [[None]*2 for i in range(len(b)+1)]
dp = [[-10**9]*2 for i in range(len(b)+1)]
dp[0][0] = 0
dp[0][1] = 0
for i in range(len(b)):
# find state with mx val with i-1 elements used
mx = None
state = None
if dp[i][0] > dp[i][1]:
mx = dp[i][0]
state = [i,0]
else:
mx = dp[i][1]
state = [i,1]
# add
if mx + b[i][1] > dp[i+1][0]:
dp[i+1][0] = mx + b[i][1]
last_state[i+1][0] = ['+', state]
# multiply
prod = 1
for j in range(i,len(b)):
if b[j][0] == 0:
prod *= b[j][1]
if mx + prod > dp[j+1][1]:
dp[j+1][1] = mx + prod
last_state[j+1][1] = ['*', state]
# go in reverse to reconstruct sequence
solved_seq = []
state = None
if dp[len(b)][1] > dp[len(b)][0]:
state = [len(b),1]
else:
state = [len(b),0]
while state[0] != 0:
next_state = last_state[state[0]][state[1]][1]
operation = last_state[state[0]][state[1]][0]
for i in range(state[0] - 1, next_state[0]-1,-1):
# will add extra operation at end of output, but we can remove it later
if b[i][0] == -1:
solved_seq.append(operation.join('1' * b[i][1]) + operation)
else:
solved_seq.append(str(b[i][1]) + operation)
if operation == '*':
solved_seq[-1] = solved_seq[-1][:-1] + '+'
state = next_state
# remove extra operation at beg(was at end but we reversed)
output.append(''.join(solved_seq)[-2::-1])
print('+'.join(output))
elif '+' in ops:
print('+'.join(map(str,a)))
elif '*' in ops:
if 0 in a:
output = []
all_mult = 0
for i in range(n-1):
if a[i+1] == 0 and not all_mult:
output.extend([a[i],'-'])
all_mult = 0
else:
output.extend([a[i],'*'])
output.append(a[-1])
print(*output,sep='')
else:
print('*'.join(map(str, a)))
```
| 8,566 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
s=input()
add_in='+' in s
multi_in='*' in s
sub_in='-' in s
s1=''
s2=''
if multi_in:
if add_in:
for x in range(n):
if l[x]!=0 and l[x]!=1:s1+=str(l[x])+'*'
else:s2+=str(l[x])+'+'
print(s2[:-1]+'+'+s1[:-1])
else:
if sub_in:
for x in range(n):
if l[x]!=0:s1+=str(l[x])+'*'
else:s2+=str(l[x])+'-'
print(s1[:-1]+'-'+s2[:-1])
else:
for x in range(n):s1+=str(l[x])+'*'
print(s1[:-1])
else:
if add_in:
for x in range(n):
s1+=str(l[x])+'+'
print(s1[:-1])
else:
l.sort(reverse=True)
for x in range(n):
s1+=str(l[x])+'-'
print(s1[:-1])
```
No
| 8,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Submitted Solution:
```
n = int(input())
ls = list(map(int, input().split(" ")))
string = input()
ans = ""
mul = False
add = False
sub = False
if "*" in string:
mul = True
if "+" in string:
add = True
if "-" in string:
sub = True
length = len(ls)
for i in range(length-1):
if(ls[i] != 0 and ls[i] != 1 and ls[i+1] != 0 and ls[i+1] != 1 and mul):
ans += str(ls[i])
ans += "*"
elif add:
ans += str(ls[i])
ans += "+"
else:
ans = str(ls[i])
ans += "-"
ans += str(ls[-1])
print(ans)
```
No
| 8,568 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Submitted Solution:
```
import sys,io,os,math
def printlist(n):
sys.stdout.write(" ".join(map(str,n)) + "\n")
def printf(n):
sys.stdout.write(str(n)+"\n")
def printns(n):
sys.stdout.write(str(n))
def intinp():
return int(sys.stdin.readline())
def strinp():
return sys.stdin.readline()
def arrinp():
return list(map(int,sys.stdin.readline().strip().split()))
def mulinp():
return map(int,sys.stdin.readline().strip().split())
def flush():
return stdout.flush()
def solve():
n=intinp()
arr=arrinp()
arr.sort(reverse=True)
s=strinp()
count=[s.count('*'),s.count('+'),s.count('-')]
ans=''+str(arr[0])
for i in range(1,n):
if(count[0]>0 and arr[i]!=1 and arr[i]!=0):
ans=ans+str('*')+str(arr[i])
continue
if(count[1]>0):
ans=ans+str('+')+str(arr[i])
continue
if(count[2]>0):
ans=ans+str('-')+str(arr[i])
print(ans)
def main():
solve()
main()
```
No
| 8,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, β¦, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 β€ n β€ 10^5) β the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 β€ |s| β€ 3) β symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols β a mathematical expression with the greatest result. If there are multiple equally valid results β output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0".
Submitted Solution:
```
def plus(L):
res = ''
for X in L:
if X!=0:
res += str(X) + '+'
return res[:-1]
def mult(L):
res = ''
for X in L:
if X != 0 and X != 1:
res += str(X) + '*'
return res[:-1]
def sub(index, irt):
res = 0
if irt == 0:
return 0
if index[irt] == 0:
return sub(index,irt-1)
print(irt)
res = (str(irt)+'-') *(index[irt])
for i in range(irt-1,0,-1):
res += (str(i)+'-')*index[i]
return res[:-1]
index = {i:0 for i in range(10)}
num = input()
L = list(map(int, input().split()))
exp = input()
for X in L:
index[X] += 1
if exp == '-':
print(sub(index,9))
elif exp == '+':
print(plus(L))
elif exp == '*':
if index[0] == 0:
print("YESS")
print(mult(L))
else:
print(0)
elif exp == '+-' or exp == '-+':
x = '-0' * index[0]
print(plus(L)+x)
elif exp == '-*' or exp == '*-':
x = '*1'*index[1]+'-0'*index[0]
print(mult(L)+x)
else:
x = mult(L)+ '+1'*index[1]+ '+0'*index[0]
print(x)
```
No
| 8,570 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
from pprint import pprint
import sys
input = sys.stdin.readline
def do():
n = int(input())
dat = list(map(int, input().split()))
mix = min(dat)
res = 0
for x in dat:
if x > mix:
res += 1
print(res)
q = int(input())
for _ in range(q):
do()
```
| 8,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
try:
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
a=sorted(a)
pos=-1
for i in range(1,len(a)):
if a[i]!=a[i-1]:
pos=i
break
if pos!=-1:
ans=len(a)-pos
else:
ans=0
print(ans)
except:
pass
```
| 8,572 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
# Problem: A. Arena
# Contest: Codeforces - Educational Codeforces Round 104 (Rated for Div. 2)
# URL: https://codeforces.com/contest/1487/problem/A
# Memory Limit: 256 MB
# Time Limit: 1000 ms
#
# KAPOOR'S
from sys import stdin, stdout
def INI():
return int(stdin.readline())
def INL():
return [int(_) for _ in stdin.readline().split()]
def INS():
return stdin.readline()
def MOD():
return pow(10,9)+7
def OPS(ans):
stdout.write(str(ans)+"\n")
def OPL(ans):
[stdout.write(str(_)+" ") for _ in ans]
stdout.write("\n")
if __name__=="__main__":
for _ in range(INI()):
n=INI()
X=INL()
OPS(n-X.count(min(X)))
```
| 8,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def func(heros):
return len(heros) - sum([1 for hero in heros if hero == min(heros)])
def main():
num_test = int(parse_input())
for _ in range(num_test):
num_hero = int(parse_input())
heros = [int(i) for i in parse_input().split()]
print(func(heros))
# 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)
parse_input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
| 8,574 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
"""
Author - Satwik Tiwari .
7th Feb , 2021 - Sunday
"""
#===============================================================================================
#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,log2
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 testcase(t):
for pp in range(t):
solve(pp)
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()
mn = min(a)
ans = 0
for i in range(n):
if(a[i] != mn): ans+=1
print(ans)
# testcase(1)
testcase(int(inp()))
```
| 8,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
tests = int(input())
for t in range(tests):
n = int(input())
heroes = list(map(int, input().split(' ')))
m = min(heroes)
print(sum([1 for x in heroes if x > m]))
```
| 8,576 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import Counter as cc
from queue import Queue
import math
import itertools
try:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
input = lambda: sys.stdin.buffer.readline().rstrip()
for _ in range(int(input())):
n=int(input())
q=sorted(list(map(int,input().split())))
t=-1
for i in range(n):
if q[i]!=q[0]:
t=i
break
if t==-1:
print(0)
else:
print(n-t)
```
| 8,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Tags: implementation, sortings
Correct Solution:
```
t=int(input())
while t:
n=int(input())
a=list(map(int,input().split()))
b=min(a)
out=0
for i in a:
if i>b:
out+=1
print(out)
t-=1
```
| 8,578 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Mon Feb 15 20:40:17 2021
@author: HP
"""
import sys
def get_arr():
return list (map (int, sys.stdin.readline ().strip ().split ()))
t = int (input ())
for i in range (t):
n = int (input ())
arr = get_arr()
if max (arr) ==min (arr):
print (0)
else:
count = 0
for j in arr:
if j == min (arr):
count += 1
print (len (arr) - count)
```
Yes
| 8,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
t=int(input())
for i in range(0,t):
n=int(input())
a=[int(x) for x in input().split()]
a.sort()
print(n-a.count(a[0]))
```
Yes
| 8,580 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
m = min(a)
print(len([i for i in a if i != m]))
```
Yes
| 8,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def seive(n):
primes = [True]*(n+1)
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
return primes
def factors(n):
factors = []
x = 2
while x*x <= n:
while n % x == 0:
factors.append(x)
n //= x
if n > 1:
factors.append(x)
return factors
# Functions: list of factors, seive of primes, gcd of two numbers
def main():
try:
for _ in range(inp()):
n = inp()
a = inlt()
mn = min(a)
ans = 0
for i in range(n):
if a[i] != mn:
ans += 1
print(ans)
except Exception as e:
print(e)
# 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()
```
Yes
| 8,582 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
n = int(input().strip())
print(n)
```
No
| 8,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
num = int(input())
list = []
for a in range(num):
hero = int(input('ΠΠΎΠ»-Π²ΠΎ Π³Π΅ΡΠΎΠ΅Π²: '))
level = input('Π²Π²Π΅Π΄ΠΈΡΠ΅ ΡΡΠΎΠ²Π½ΠΈ Π³Π΅ΡΠΎΠ΅Π²: ')
for i in level:
if i == ' ':
pass
else:
list.append(int(i))
for run in range(hero-1):
for j in range(hero-1):
if list[j] > list[j+1]:
list[j], list[j+1] = list[j+1], list[j]
else:
pass
w = list.count(list[0])
if w == hero:
print(0)
else:
print(hero - w)
```
No
| 8,584 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
winners = 0
champs = sorted(input().split(" "))
for lvl in champs:
if int(lvl) > int(champs[0]):
winners += 1
print(winners)
```
No
| 8,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 β€ t β€ 500) β the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 β€ n β€ 100) β the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer β the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner.
Submitted Solution:
```
# import math
t = int(input())
for T in range(t):
n = int(input())
l = list(map(int , input().split()))
k = 0
ln = set(l)
# for i in range(1, n):
# if l[i-1]>l[i]:
# k+=1
# l[i-1]+=1
# if l[i]>l[i-1]:
# k+=1
# l[i]+=1
if len(ln)==1:
print(0)
if len(ln)%2==0:
print(len(l) - 1)
else:
print(len(ln))
```
No
| 8,586 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The popular improv website Interpretation Impetus hosts regular improv contests and maintains a rating of the best performers. However, since improv can often go horribly wrong, the website is notorious for declaring improv contests unrated. It now holds a wager before each improv contest where the participants try to predict whether it will be rated or unrated, and they are now more popular than the improv itself.
Izzy and n other participants take part in each wager. First, they each make their prediction, expressed as 1 ("rated") or 0 ("unrated"). Izzy always goes last, so she knows the predictions of the other participants when making her own. Then, the actual competition takes place and it is declared either rated or unrated.
You need to write a program that will interactively play as Izzy. There will be m wagers held in 2021, and Izzy's goal is to have at most 1.3β
b + 100 wrong predictions after all those wagers, where b is the smallest number of wrong predictions that any other wager participant will have after all those wagers.
The number b is not known in advance. Izzy also knows nothing about the other participants β they might somehow always guess correctly, or their predictions might be correlated. Izzy's predictions, though, do not affect the predictions of the other participants and the decision on the contest being rated or not β in other words, in each test case, your program always receives the same inputs, no matter what it outputs.
Interaction
First, a solution must read two integers n (1 β€ n β€ 1000) and m (1 β€ m β€ 10 000). Then, the solution must process m wagers. For each of them, the solution must first read a string consisting of n 0s and 1s, in which the i-th character denotes the guess of the i-th participant. Then, the solution must print Izzy's guess as 0 or 1. Don't forget to flush the output after printing it! Then, the solution must read the actual outcome, also as 0 or 1, and then proceed to the next wager, if this wasn't the last one.
Your solution will be considered correct if it makes at most 1.3β
b + 100 mistakes, where b is the smallest number of mistakes made by any other participant. Note that if a solution outputs anything except 0 or 1 for a wager, it will be considered incorrect even if it made no other mistakes.
There are 200 test cases in this problem.
Example
Input
3 4
000
1
100
1
001
0
111
1
Output
0
0
1
1
Note
In the example, the participants made 1, 2, and 3 mistakes respectively, therefore b=1 (the smallest of these numbers). Izzy made 3 mistakes, which were not more than 1.3β
b + 100=101.3, so these outputs are good enough to pass this test case (as are any other valid outputs).
Submitted Solution:
```
from sys import stdout
from typing import List
def get_guesses() -> List[int]:
inp = input()
return [-1 if x == '0' else 1 for x in inp]
def guess(answer: int) -> None:
print(answer)
stdout.flush()
def get_ans() -> int:
true_ans = int(input())
return true_ans
def solve():
n, m = list(map(int, (input().split(" "))))
lr = .05
weights = [1.0] * n
weights.append(0) # bias
for i in range(m):
g = get_guesses()
g.append(1)
weighted_sum = sum([weights[j] * g[j] for j in range(len(g))])
if weighted_sum > 0:
guess(1)
else:
guess(0)
ans = get_ans()
# Square error
gradient = [g[i] * (2 * weighted_sum - weights[i] * ans) for i in range(len(g))]
weights = [weights[i] - gradient[i] * lr for i in range(len(weights))]
solve()
```
No
| 8,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The popular improv website Interpretation Impetus hosts regular improv contests and maintains a rating of the best performers. However, since improv can often go horribly wrong, the website is notorious for declaring improv contests unrated. It now holds a wager before each improv contest where the participants try to predict whether it will be rated or unrated, and they are now more popular than the improv itself.
Izzy and n other participants take part in each wager. First, they each make their prediction, expressed as 1 ("rated") or 0 ("unrated"). Izzy always goes last, so she knows the predictions of the other participants when making her own. Then, the actual competition takes place and it is declared either rated or unrated.
You need to write a program that will interactively play as Izzy. There will be m wagers held in 2021, and Izzy's goal is to have at most 1.3β
b + 100 wrong predictions after all those wagers, where b is the smallest number of wrong predictions that any other wager participant will have after all those wagers.
The number b is not known in advance. Izzy also knows nothing about the other participants β they might somehow always guess correctly, or their predictions might be correlated. Izzy's predictions, though, do not affect the predictions of the other participants and the decision on the contest being rated or not β in other words, in each test case, your program always receives the same inputs, no matter what it outputs.
Interaction
First, a solution must read two integers n (1 β€ n β€ 1000) and m (1 β€ m β€ 10 000). Then, the solution must process m wagers. For each of them, the solution must first read a string consisting of n 0s and 1s, in which the i-th character denotes the guess of the i-th participant. Then, the solution must print Izzy's guess as 0 or 1. Don't forget to flush the output after printing it! Then, the solution must read the actual outcome, also as 0 or 1, and then proceed to the next wager, if this wasn't the last one.
Your solution will be considered correct if it makes at most 1.3β
b + 100 mistakes, where b is the smallest number of mistakes made by any other participant. Note that if a solution outputs anything except 0 or 1 for a wager, it will be considered incorrect even if it made no other mistakes.
There are 200 test cases in this problem.
Example
Input
3 4
000
1
100
1
001
0
111
1
Output
0
0
1
1
Note
In the example, the participants made 1, 2, and 3 mistakes respectively, therefore b=1 (the smallest of these numbers). Izzy made 3 mistakes, which were not more than 1.3β
b + 100=101.3, so these outputs are good enough to pass this test case (as are any other valid outputs).
Submitted Solution:
```
print("I love 2020-2021 ICPC, NERC, Northern Eurasia so much")
```
No
| 8,588 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The popular improv website Interpretation Impetus hosts regular improv contests and maintains a rating of the best performers. However, since improv can often go horribly wrong, the website is notorious for declaring improv contests unrated. It now holds a wager before each improv contest where the participants try to predict whether it will be rated or unrated, and they are now more popular than the improv itself.
Izzy and n other participants take part in each wager. First, they each make their prediction, expressed as 1 ("rated") or 0 ("unrated"). Izzy always goes last, so she knows the predictions of the other participants when making her own. Then, the actual competition takes place and it is declared either rated or unrated.
You need to write a program that will interactively play as Izzy. There will be m wagers held in 2021, and Izzy's goal is to have at most 1.3β
b + 100 wrong predictions after all those wagers, where b is the smallest number of wrong predictions that any other wager participant will have after all those wagers.
The number b is not known in advance. Izzy also knows nothing about the other participants β they might somehow always guess correctly, or their predictions might be correlated. Izzy's predictions, though, do not affect the predictions of the other participants and the decision on the contest being rated or not β in other words, in each test case, your program always receives the same inputs, no matter what it outputs.
Interaction
First, a solution must read two integers n (1 β€ n β€ 1000) and m (1 β€ m β€ 10 000). Then, the solution must process m wagers. For each of them, the solution must first read a string consisting of n 0s and 1s, in which the i-th character denotes the guess of the i-th participant. Then, the solution must print Izzy's guess as 0 or 1. Don't forget to flush the output after printing it! Then, the solution must read the actual outcome, also as 0 or 1, and then proceed to the next wager, if this wasn't the last one.
Your solution will be considered correct if it makes at most 1.3β
b + 100 mistakes, where b is the smallest number of mistakes made by any other participant. Note that if a solution outputs anything except 0 or 1 for a wager, it will be considered incorrect even if it made no other mistakes.
There are 200 test cases in this problem.
Example
Input
3 4
000
1
100
1
001
0
111
1
Output
0
0
1
1
Note
In the example, the participants made 1, 2, and 3 mistakes respectively, therefore b=1 (the smallest of these numbers). Izzy made 3 mistakes, which were not more than 1.3β
b + 100=101.3, so these outputs are good enough to pass this test case (as are any other valid outputs).
Submitted Solution:
```
from sys import stdout
from typing import List
def get_guesses() -> List[int]:
inp = input()
return [-1 if x == '0' else 1 for x in inp]
def guess(answer: int) -> None:
print(answer)
stdout.flush()
def get_ans() -> int:
true_ans = int(input())
return true_ans
def solve():
n, m = list(map(int, (input().split(" "))))
lr = .005
weights = [1.0] * n
weights.append(0) # bias
for i in range(m):
g = get_guesses()
g.append(1)
weighted_sum = sum([weights[j] * g[j] for j in range(len(g))])
if weighted_sum > 0:
guess(1)
else:
guess(0)
ans = get_ans()
# Square error
gradient = [g[i] * (2 * weighted_sum - weights[i] * ans) for i in range(len(g))]
weights = [weights[i] - gradient[i] * lr for i in range(len(weights))]
solve()
```
No
| 8,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The popular improv website Interpretation Impetus hosts regular improv contests and maintains a rating of the best performers. However, since improv can often go horribly wrong, the website is notorious for declaring improv contests unrated. It now holds a wager before each improv contest where the participants try to predict whether it will be rated or unrated, and they are now more popular than the improv itself.
Izzy and n other participants take part in each wager. First, they each make their prediction, expressed as 1 ("rated") or 0 ("unrated"). Izzy always goes last, so she knows the predictions of the other participants when making her own. Then, the actual competition takes place and it is declared either rated or unrated.
You need to write a program that will interactively play as Izzy. There will be m wagers held in 2021, and Izzy's goal is to have at most 1.3β
b + 100 wrong predictions after all those wagers, where b is the smallest number of wrong predictions that any other wager participant will have after all those wagers.
The number b is not known in advance. Izzy also knows nothing about the other participants β they might somehow always guess correctly, or their predictions might be correlated. Izzy's predictions, though, do not affect the predictions of the other participants and the decision on the contest being rated or not β in other words, in each test case, your program always receives the same inputs, no matter what it outputs.
Interaction
First, a solution must read two integers n (1 β€ n β€ 1000) and m (1 β€ m β€ 10 000). Then, the solution must process m wagers. For each of them, the solution must first read a string consisting of n 0s and 1s, in which the i-th character denotes the guess of the i-th participant. Then, the solution must print Izzy's guess as 0 or 1. Don't forget to flush the output after printing it! Then, the solution must read the actual outcome, also as 0 or 1, and then proceed to the next wager, if this wasn't the last one.
Your solution will be considered correct if it makes at most 1.3β
b + 100 mistakes, where b is the smallest number of mistakes made by any other participant. Note that if a solution outputs anything except 0 or 1 for a wager, it will be considered incorrect even if it made no other mistakes.
There are 200 test cases in this problem.
Example
Input
3 4
000
1
100
1
001
0
111
1
Output
0
0
1
1
Note
In the example, the participants made 1, 2, and 3 mistakes respectively, therefore b=1 (the smallest of these numbers). Izzy made 3 mistakes, which were not more than 1.3β
b + 100=101.3, so these outputs are good enough to pass this test case (as are any other valid outputs).
Submitted Solution:
```
n,m = map(int,input().split())
l=[0 for i in range(n)]
s=input()
print(1)
for f in range(m-1):
x=input()
for i in range(n):
if(s[i]!=x):
l[i]+=1
m=0
for i in range(1,n-1):
if(l[i]<l[m]):
m=i
s=input()
print(s[m])
x=input()
```
No
| 8,590 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
from collections import deque, defaultdict
from math import sqrt, ceil, factorial, floor, inf, log2, sqrt, gcd
import bisect
import copy
from itertools import combinations
import sys
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
from collections import defaultdict
for _ in range(int(input())):
n,l,r=get_ints()
a=get_array()
a.sort()
ans=0
##print(a)
for i in range(len(a)):
ll=l-a[i]
ul=r-a[i]
ind1=bisect.bisect_right(a,ul)
ind2 = bisect.bisect_left(a, ll)
ind1-=1
if ind1<=i:
continue
ind2=bisect.bisect_left(a,ll)
if ind2<=i:
ind2=min(i+1,ind1)
##print(ind1,ind2,ll,ul)
if ind1-ind2+1>=0:
ans+=ind1-ind2+1
##print(ans)
print(ans)
```
| 8,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
import bisect
def solve(arr, n, l, r):
totalCount = 0
for i in range(0, n-1):
start = l - arr[i]
end = r - arr[i]
c1 = bisect.bisect(arr, start-1, i+1, n)
c2 = bisect.bisect(arr, end, i+1, n)
totalCount += (c2 - c1)
return totalCount
testcases = int(input())
while testcases:
n, l, r = map(int, input().split())
arr = list(map(int, input().split()))
arr.sort()
tc = solve(arr, n, l, r)
print(tc)
testcases -= 1
```
| 8,592 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
#ANKIT BEHIND THE KEYBOARD
from collections import defaultdict
t=int(input())
for i in range(0,t):
n,left,right=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
ans = n * (n - 1) // 2
i = 0
j = n - 1
c = 0
while i < j:
if l[j] + l[i] < left:
c += j - i
i += 1
else:
j -= 1
i = 0
j = n - 1
x = 0
while i < j:
if l[j] + l[i] <= right:
x += j - i
i += 1
else:
j -= 1
d = n * (n - 1) // 2 - x
ans -= c
ans -= d
print(ans)
```
| 8,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
from collections import deque, defaultdict, Counter
from itertools import product, groupby, permutations, combinations
from math import gcd, floor
cases = int(input())
for _ in range(cases):
n, l, r = map(int, input().split())
arr = sorted(map(int, input().split()))
a1 =[0] * n
a2 = [0] * n
i, j = 0, n - 1
while i - j <= 0:
if arr[i] + arr[j] >= l:
a1[j] = i
j -= 1
else:
a1[i] = j + 1
i += 1
i, j = 0, n - 1
while i - j <= 0:
if arr[i] + arr[j] <= r:
a2[i] = j
i += 1
else:
a2[j] = i - 1
j -= 1
ans = [ans2 - ans1 + 1 - (ans1 <= i <= ans2) for i, (ans1, ans2) in enumerate(zip(a1, a2))]
print(sum(ans)//2)
```
| 8,594 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
import sys;input=sys.stdin.readline
from bisect import bisect
# from collections import defaultdict
# from itertools import accumulate
# from decimal import *
# import math
# getcontext().prec = 50
# s = input().strip()
# n = int(input())
# lis = list(map(int,input().split()))
# x,y = map(int,input().split())
# chars = 'abcdefghijklmnopqrstuvwxyz'
# import heapq
# def gcd(a,b):
# return gcd (b, a % b) if b else a
def solve():
n,l,r = map(int,input().split())
lis = list(map(int,input().split()))
lis = sorted(lis)
count = 0
for i in range(len(lis)):
start = bisect(lis,l-lis[i]-1)
end = bisect(lis,r-lis[i])
add = 0
if i>=start and i<end:
add = -1
add+=end-start
count+=add
print(count//2)
# solve()
for _ in range(int(input())):
solve()
```
| 8,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
from bisect import *
a=int(input())
total=0
from collections import *
for i in range(a):
n,l,r=map(int,input().split())
total=0
z=list(map(int,input().split()))
al=defaultdict(int)
for i in range(len(z)):
al[z[i]]+=1
z.sort()
for i in range(len(z)):
t=bisect_left(z,l-z[i])
q=bisect_right(z,r-z[i])
if(t==len(z)):
continue;
if(q==len(z)):
q-=1
if(z[q]+z[i]>r):
q-=1
x=0
total+=max(q-t+1,0)
if(l<=z[i]*2<=r):
x-=1
total+=x
print(total//2)
```
| 8,596 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
for _ in range(int(input())):
n,l,r=map(int,input().split())
a=sorted(list(map(int,input().split())))
i,j,q=0,0,0
while i<n:
n-=1
while i<n and a[i]+a[n]<l:i+=1
while j<n and a[j]+a[n]<=r:j+=1
q+=min(j,n)-i
print(q)
```
| 8,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Tags: binary search, data structures, math, two pointers
Correct Solution:
```
def solve(arr,l,r,k,x):
ans=-1
while(l<=r):
m=l+(r-l)//2
if(arr[m]<=k):
ans=m
l=m+1
else:
r=m-1
return ans
def sol(arr,l,r,k,x):
ans=-1
while(l<=r):
m=(r+l)//2
if(arr[m]>=k):
ans=m
r=m-1
else:
l=m+1
return ans
for u in range(int(input())):
n,l,r=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
c=0
for i in range(n):
x,y=l-a[i],r-a[i]
p=sol(a,i+1,n-1,x,y)
q=solve(a,i+1,n-1,y,x)
if(p!=-1 and q!=-1):
c+=(q-p+1)
print(c)
```
| 8,598 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Find the number of pairs (i, j) (1 β€ i < j β€ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l β€ a_i + a_j β€ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 β€ 5 + 1 β€ 7);
* i=1 and j=3 (4 β€ 5 + 2 β€ 7).
Input
The first line contains an integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 β€ n β€ 2 β
10^5, 1 β€ l β€ r β€ 10^9) β the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the number of index pairs (i, j) (i < j), such that l β€ a_i + a_j β€ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1
Submitted Solution:
```
from bisect import bisect_left, bisect_right
for _ in range(int(input())):
n, l, r = map(int, input().split())
a = list(map(int, input().split()))
a.sort(); ans = 0
for i in range(n):
if a[i]>=r: break
lowlim = bisect_left(a, l-a[i], i+1)
highlim = bisect_right(a, r-a[i], i+1)
ans+= highlim - lowlim
print(ans)
```
Yes
| 8,599 |
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