text stringlengths 198 433k | conversation_id int64 0 109k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Tags: binary search, math
Correct Solution:
```
for _ in range(int(input())):
h,c,t=[int(c) for c in input().split()]
if t <= (h + c)/2:
print(2)
else:
##finding k
a=h-t
b=2*t - h-c
k=2*(a//b) +1
# k = ( t - h) // (h+c - 2*t)
##comparing k and k + 1
a =abs((k//2)*c + ((k+1)//2)*h - t*k)
# a=abs(a)
b= abs(((k+2)//2)*c + ((k+3)//2)*h - t*(k+2))
if a*(k+2) <= b*(k) :
ans=k
else:
ans=k + 2
print(ans)
```
| 5,200 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Tags: binary search, math
Correct Solution:
```
try:
for _ in range(int(input())):
h,c,t=map(int,input().split())
if h<=t:
print(1)
elif h+c>=2*t:
print(2)
else:
x=int((c-t)/(h+c-2*t))
m=abs((x*(h+c-2*t)+t-c)/(2*x-1))
n=abs(((x+1)*(h+c-2*t)+t-c)/(2*(x+1)-1))
if m<=n:
print(2*x-1)
else:
print(2*x+1)
except:
pass
```
| 5,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Tags: binary search, math
Correct Solution:
```
#!/usr/bin/env python
# coding:utf-8
# Copyright (C) dirlt
from sys import stdin
def run(h, c, t):
if (h + c - 2 * t) >= 0:
return 2
a = h - t
b = 2 * t - h - c
k = int(a / b)
val1 = abs((k + 1) * h + k * c - (2 * k + 1) * t)
val2 = abs((k + 2) * h + (k + 1) * c - (2 * k + 3) * t)
# val1 / (2k+1) <= val2 / (2k+3), return 2k+1
# print(val1, val2)
if val1 * (2 * k + 3) <= val2 * (2 * k + 1):
ans = 2 * k + 1
else:
ans = 2 * k + 3
return ans
def main():
cases = int(stdin.readline())
for _ in range(cases):
h, c, t = [int(x) for x in stdin.readline().split()]
ans = run(h, c, t)
print(ans)
if __name__ == '__main__':
import os
if os.path.exists('tmp.in'):
stdin = open('tmp.in')
main()
```
| 5,202 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
for _ in range(int(input())):
h, c, t = map(int, input().split())
if h==t:
print(1)
elif (t-c<=h-t):
print(2)
else:
temp = (t-h)/(h+c-2*t)
if (int(temp)==temp):
print(int(2*temp+1))
else:
a = int(temp)
b = a+1
if 2*t*(2*a+1)*(2*b+1) >= (2*b+1)*((a+1)*h+a*c)+(2*a+1)*((b+1)*h+b*c):
print(2*a+1)
else:
print(2*b+1)
# temp = round(abs((t-h)/(h+c-2*t)))
# print(2*temp+1)
# else:
# t1 = round(temp)
# print(2*t1+1)
# t1 = int(temp)
# m1 = ((t1+1)*c + (t1+2)*h)/(2*t1+3) - t
# m2 = (t1*c + (t1+1)*h)/(2*t1+1) - t
# print(m1, m2)
# if abs(m1)<abs(m2):
# print(2*t1 + 3)
# else:
# print(2*t1 + 1)
```
Yes
| 5,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
# abs((desired*(2*n + 1) - ((n+1)*hot + n*cold))/(2*n + 1))
#EXPERIMENTING WITH LOSS DEFINTION - float ops ko aage peeche kar diya
import sys
def input():
return sys.stdin.readline().rstrip()
testcases = int(input())
answers = []
def loss(two_n_plus_one, hot, cold, desired):
n = two_n_plus_one//2
# if n == 0:
# return float('inf')
# else:
# return abs(desired - ((n+1)*hot + n*cold)/(2*n + 1))
return abs((desired*(2*n + 1) - ((n+1)*hot + n*cold))/(2*n + 1))
for _ in range(testcases):
hot, cold, desired_temp = [int(i) for i in input().split()]
#required number of cups to get it to desired_temp
mid_way = (hot + cold)/2
if hot == cold:
answers.append(1)
elif desired_temp >= hot:
answers.append(1)
elif desired_temp <= mid_way:
answers.append(2)
else:
#find n, -> num iters
#n + 1 hots, n colds
frac = (hot - desired_temp) / (desired_temp - mid_way)
frac /= 2
# option1 = int(frac)
option1 = 2*(int(frac)) + 1
option2 = option1 + 2
l1 , l2 = loss(option1, hot, cold, desired_temp),loss(option2, hot, cold, desired_temp)
if min(l1, l2) >= (hot - desired_temp) and (hot -desired_temp) <= (desired_temp - mid_way):
answers.append(1)
elif min(l1, l2) >= (desired_temp - mid_way):
answers.append(2)
elif l1 <= l2:
answers.append(option1)
else:
answers.append(option2)
# if option1%2 == 0:
# option1 += 1 #taki odd ho jaye
# option2 = option1 - 2
# option3 = option1 + 2
# l1, l2, l3 = loss(option1, hot, cold, desired_temp),loss(option2, hot, cold, desired_temp),loss(option3, hot, cold, desired_temp)
# option4 = option1 - 1
# answers.append(min(loss(option1, hot, cold, desired_temp),
# loss(option2, hot, cold, desired_temp),
# loss(option3, hot, cold, desired_temp)))
print(*answers, sep = '\n')
```
Yes
| 5,204 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
T = int(input())
for _ in range(T):
h,c,t = map(int,input().split())
if h+c >= 2*t:
print(2)
else:
x = (h-t)//(2*t-c-h)
y = x+1
if abs(h*(x+1)+c*x-t*(2*x+1))*(2*y+1) <= abs(h*(y+1)+c*y-t*(2*y+1))*(2*x+1):
print(2*x+1)
else:
print(2*y+1)
```
Yes
| 5,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
#from collections import deque,defaultdict
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl = lambda: list(map(int, input().split()))
inm = lambda: map(int, input().split())
ins = lambda : input().strip()
DBG = True # and False
BIG = 10**18
R = 10**9 + 7
def ddprint(x):
if DBG:
print(x)
def foo(h,c,n):
return (n*h+(n-1)*c)/(2*n-1)
def bar(h,c,t,n):
return (n*h+(n-1)*c)==t*(2*n-1)
ttt = inn()
for tt in range(ttt):
h,c,t = inm()
if 5*h+c<=6*t:
print(1)
elif 2*t<=(h+c):
print(2)
else:
mn = 1
mx = 10**16
found = False
while mx-mn>1:
mid = (mn+mx)//2
if bar(h,c,t,mid):
found = True
break
elif foo(h,c,mid)<t:
mx = mid
else:
mn = mid
if found:
#ddprint(f"found")
print(2*mid-1)
continue
#tn = foo(h,c,mn)
#tx = foo(h,c,mx)
#if abs(tn-t)>abs(tx-t):
x = 2*(2*mn-1)*(2*mx-1)*t
y = (2*mx-1)*(mn*h+(mn-1)*c) + (2*mn-1)*(mx*h+(mx-1)*c)
#ddprint(f"mn {mn} mx {mx} x {x} y {y}")
if x<y:
print(2*mx-1)
else:
print(2*mn-1)
```
Yes
| 5,206 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
from math import ceil
a = int(input())
for _ in range(a):
b,c,d = map(int,input().split())
if b >= d and c >= d:
if b > c:
print(2)
else:
print(1)
elif b <= d and c <= d:
if b < c:
print(2)
else:
print(1)
else:
t = b - d
z = d - c
if t >= z:
print(2)
else:
print(1 + 2 * ceil((t / (z - t)) - 0.5))
```
No
| 5,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
from sys import stdin, stdout
import math
def calc(h, c, t, n):
d = (h*(n+1)+c*n)/(2*n+1)-t
return abs(d)
t = int(stdin.readline())
for _ in range(t):
h, c, t = map(int, stdin.readline().split())
if t >= h:
print(1)
continue
if t <= (h+c)/2:
print(2)
continue
n = math.ceil((h-t)/(2*t-h-c))
d0, d1, d2, d3, d4 = calc(h, c, t, n-2), calc(h, c, t, n-1), calc(h, c, t, n), calc(h, c, t,n+1), calc(h, c, t, n+2)
md = min(d1, d2, d3)
if md == d0:
print(2*n-3)
elif md == d1:
print(2*n-1)
elif md == d2:
print(2*n+1)
elif md == d3:
print(2*n+3)
else:
print(2*n+5)
```
No
| 5,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
def I(): return(list(map(int,input().split())))
def sieve(n):
a=[1]*n
for i in range(2,n):
if a[i]:
for j in range(i*i,n,i):
a[j]=0
return a
for __ in range(int(input())):
h,c,t=I()
# for n in range(1,22,2):
# if n%2:
# avg=(h+c*(n-1)//2+h*(n-1)//2)/n
# else:
# avg=(h*n//2+c*n//2)/n
# print(avg-(h+c)/2,n)
if h==t:
print(1)
continue
if (h+c)/2>=t:
print(2)
continue
# if (h+c)/2>t:
# if (h+c)/2<h:
# print(2)
# else:
# print(1)
# continue
l=1
r=10**30
ans=10**100
# print("---------")
left=t-(h+c)/2
# print(left)
while(l<=r):
mid=(l+r)//2
n=mid
delta=(h-c)/(2*(2*n+1))
# print(mid)
# print(delta,2*n+1)
if delta==left:
ans=n
break
if delta<left:
ans=n
r=mid-1
else:
l=mid+1
n=ans
delta1=(h-c)/(2*(2*n+1))
delta2=(h-c)/(2*(2*n+3))
if abs(delta1-left)<=abs(delta2-left):
print(2*n+1)
else:
print(2*n+3)
# print(2*ans+1)
# print("---------")
```
No
| 5,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infinitely deep barrel;
3. take one cup of the hot water ...
4. and so on ...
Note that you always start with the cup of hot water.
The barrel is initially empty. You have to pour at least one cup into the barrel. The water temperature in the barrel is an average of the temperatures of the poured cups.
You want to achieve a temperature as close as possible to t. So if the temperature in the barrel is t_b, then the absolute difference of t_b and t (|t_b - t|) should be as small as possible.
How many cups should you pour into the barrel, so that the temperature in it is as close as possible to t? If there are multiple answers with the minimum absolute difference, then print the smallest of them.
Input
The first line contains a single integer T (1 ≤ T ≤ 3 ⋅ 10^4) — the number of testcases.
Each of the next T lines contains three integers h, c and t (1 ≤ c < h ≤ 10^6; c ≤ t ≤ h) — the temperature of the hot water, the temperature of the cold water and the desired temperature in the barrel.
Output
For each testcase print a single positive integer — the minimum number of cups required to be poured into the barrel to achieve the closest temperature to t.
Example
Input
3
30 10 20
41 15 30
18 13 18
Output
2
7
1
Note
In the first testcase the temperature after 2 poured cups: 1 hot and 1 cold is exactly 20. So that is the closest we can achieve.
In the second testcase the temperature after 7 poured cups: 4 hot and 3 cold is about 29.857. Pouring more water won't get us closer to t than that.
In the third testcase the temperature after 1 poured cup: 1 hot is 18. That's exactly equal to t.
Submitted Solution:
```
import sys,math
input=sys.stdin.readline
T=int(input())
for _ in range(T):
n,m,t=map(int,input().split())
diff=abs(t-((n+m)/2))
fv=2
if (2*t==(n+m)):
print(2)
else:
val=(n-t)/((2*t)-(n+m))
val=math.floor((2*val)+1)
if (val<0):
print(2)
elif (val%2==0):
x=(val-2)//2
v=((n+x*(n+m))/((2*x)+1))
v=abs(v-t)
if (v<diff):
fv=(2*x)+1
diff=v
elif (v==diff):
fv=min(fv,(2*x)+1)
diff=v
x=x+1
v=abs((n+x*(n+m))/((2*x)+1))
v=abs(v-t)
if (v<diff):
fv=(2*x)+1
diff=v
elif (v==diff):
fv=min(fv,(2*x)+1)
diff=v
print(fv)
else:
x=(val-1)//2
v=abs((n+x*(n+m))/((2*x)+1))
v=abs(v-t)
if (v<diff):
fv=(2*x)+1
diff=v
elif (v==diff):
fv=min(fv,(2*x)+1)
diff=v
x=x+1
v=abs((n+x*(n+m))/((2*x)+1))
v=abs(v-t)
if (v<diff):
fv=(2*x)+1
diff=v
elif (v==diff):
fv=min(fv,(2*x)+1)
diff=v
if (x-2>0):
x=x-2
v=abs((n+x*(n+m))/((2*x)+1))
v=abs(v-t)
if (v<diff):
fv=(2*x)+1
diff=v
elif (v==diff):
fv=min(fv,(2*x)+1)
diff=v
print(fv)
```
No
| 5,210 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
from heapq import *
import sys
def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
```
| 5,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
```
| 5,212 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def ng():
print("NO")
exit()
n, k = MI()
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
#print(n, k, u, ans)
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
#print(n, k, u, ans)
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
```
| 5,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
def ng():print("NO");exit()
n, k = map(int,input().split())
if (n,k)==(1,0):print("YES");print(0);exit()
ans = [0] * n;u = 0;popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1;u = 4
for _ in range(k - 1):ans[u + 1] = ans[u + 2] = u+1;u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES");print(*ans)
```
| 5,214 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
```
| 5,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def ng():
print("NO")
exit()
n, k = MI()
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3: ng()
ans[4] = 1
u = 4
#print(n, k, u, ans)
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
#print(n, k, u, ans)
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
```
No
| 5,216 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
n,k=MI()
if n&1==0 or n<2*k+3 or k<bin(n+1).count("1")-1:
print("NO")
exit()
hp=[]
two=0
bit=n+1
for i in range(n):
if bit&1:
if i==1:two+=1
else:heappush(hp,1<<i)
bit>>=1
#print(two,hp)
while two+len(hp)-1<k:
s=heappop(hp)
if s==4:two+=2
else:
heappush(hp,s//2)
heappush(hp,s//2)
#print(two,hp)
ans=[0]*n
for u in range(0,two*2,2):ans[u+1]=ans[u+2]=u+1
#print(ans)
u=2*two
while hp:
s=heappop(hp)
if not hp:
for v in range(2, s): ans[v + u-1] = v // 2 + u
break
for v in range(1,s):ans[v+u]=v//2+u+1
ans[u+s]=u+1
u+=s
print("YES")
print(*ans)
```
No
| 5,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
import math
n, k = input().replace("\n", "").split(" ")
n = int(n)
k = int(k)
if 2*k > n-3:
print("NO")
else:
if((n-2*k)%4 == 1):
print("NO")
else:
s = {}
print("YES")
for i in range(1, n-2*k+1):
s[i] = math.floor(i/2)
if k != 0:
s[1] = n - 2*k + 1
s[n - 2*k + 2] = n - 2*k + 1
for i in range(1, k+1):
if i == 1:
s[n-2*k+i] = 0
else:
s[n-2*k+i] = n- 2*k + i - 1
s[n-2*k+i+k] = n - 2*k + i
for x in s:
print(s[x], end = " ")
```
No
| 5,218 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
n,k=MI()
if n&1==0 or n<2*k+3:
print("NO")
exit()
ans=[0]*n
def tree(u,size,imb):
#print(u,size,imb)
if size==1:return
ans[u + 1] = u + 1
if imb:
ans[u+2]=u+1
tree(u+2,size-2,imb-1)
else:
sl=size//4*2+1
ans[u+1+sl]=u+1
tree(u+1,sl,0)
tree(u+1+sl,size-1-sl,0)
tree(0,n,k)
print("YES")
print(*ans)
```
No
| 5,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
def rhyme(a,b,k):
v=["a","e","i","o","u"]
i_a=None
i_b=None
v_a=0
v_b=0
for i in range(len(a)-1,-1,-1):
if a[i] in v:
v_a+=1
if v_a==k:
i_a=i
break
for i in range(len(b)-1,-1,-1):
if b[i] in v:
v_b+=1
if v_b==k:
i_b=i
break
if i_a!=None and i_b != None and a[i_a:]==b[i_b:]:
return True
return False
def main(arr,k):
style={0:"aabb",1:"abab",2:"abba",3:'aaaa'}
ans=[]
for j in range(4):
c=True
for i in range(len(arr)):
q=arr[i]
if j==0:
if rhyme(q[0],q[1],k) and rhyme(q[2],q[3],k):
continue
else:
c=False
break
if j==1:
if rhyme(q[0],q[2],k) and rhyme(q[1],q[3],k):
continue
else:
c=False
break
if j==2:
if rhyme(q[0],q[3],k) and rhyme(q[1],q[2],k):
continue
else:
c=False
break
if j==3:
if rhyme(q[0],q[1],k) and rhyme(q[1],q[2],k) and rhyme(q[2],q[3],k):
continue
else:
c=False
break
if c:
ans.append(j)
if 3 in ans:
return style[3]
elif len(ans)>=1:
return style[ans[0]]
else:
return "NO"
n,k=list(map(int,input().split()))
arr=[]
for i in range(n):
temp=[]
for j in range(4):
temp.append(input())
arr.append(temp)
print(main(arr,k))
```
| 5,220 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
# from itertools import accumulate
# from collections import defaultdict, Counter
def modinv(n,p):
return pow(n,p-2,p)
def get_suffix(s, k):
vovels = "aeiou"
for i in range(len(s)-1, -1, -1):
if s[i] in vovels:
k -= 1
if k == 0:
return s[i:]
return -1
# aaaa = 1
# aabb = 2
# abab = 3
# abba = 4
# none = -1
def get_scheme(s1, s2, s3, s4):
if s1 == s2 == s3 == s4:
return 1
if s1 == s2 and s3 == s4:
return 2
if s1 == s3 and s2 == s4:
return 3
if s1 == s4 and s2 == s3:
return 4
return -1
def get_scheme2(x):
if x == 1:
return "aaaa"
if x == 2:
return "aabb"
if x == 3:
return "abab"
if x == 4:
return "abba"
def main():
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
# print(get_suffix("commit", 3))
n, k = [int(x) for x in input().split()]
rhymes = []
no_scheme = False
for i in range(n):
s1 = get_suffix(input(), k)
s2 = get_suffix(input(), k)
s3 = get_suffix(input(), k)
s4 = get_suffix(input(), k)
if s1 == -1 or s2 == -1 or s3 == -1 or s4 == -1:
rhymes.append(-1)
else:
rhymes.append(get_scheme(s1, s2, s3, s4))
# print(*rhymes)
rhymes = set(rhymes)
scheme = ""
if -1 in rhymes:
scheme = "NO"
elif len(rhymes) == 1:
scheme = get_scheme2(rhymes.pop())
elif len(rhymes) == 2 and 1 in rhymes:
rhymes.remove(1)
scheme = get_scheme2(rhymes.pop())
else:
scheme = "NO"
print(scheme)
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
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')
if __name__ == '__main__':
main()
```
| 5,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
n,k = map(int,input().split())
lis=[]
vov=['a','e','i','o','u']
d={}
d['aabb']=d['abab']=d['abba']=d['aaaa']=0
for i in range(n*4):
s = input()
lis.append(s)
if i%4==3:
tmp=['']*4
for j in range(4):
s=lis[j]
c=0
cou=0
for ll in s[::-1]:
cou+=1
if ll in vov:
c+=1
if c==k:
tmp[j]=s[-cou:]
break
# print(tmp)
tt=1
if tmp[0]==tmp[1]==tmp[2]==tmp[3] and tmp[0]!='':
d['aaaa']+=1
tt=0
if tmp[0]==tmp[1] and tmp[2]==tmp[3] and tmp[1]!=tmp[3] and tmp[1]!='' and tmp[3]!='':
d['aabb']+=1
tt=0
elif tmp[0]==tmp[2] and tmp[1]==tmp[3] and tmp[1]!=tmp[0] and tmp[1]!='' and tmp[0]!='':
d['abab']+=1
tt=0
elif tmp[0]==tmp[3] and tmp[1]==tmp[2] and tmp[2]!=tmp[0] and tmp[1]!='' and tmp[3]!='':
d['abba']+=1
tt=0
if tt:
print("NO")
exit()
lis=[]
c=0
for i in d:
if d[i]>0:
c+=1
if c==1:
for i in d:
if d[i]>0:
print(i)
exit()
elif c==2 and d['aaaa']>0:
for i in d:
if d[i]>0:
print(i)
exit()
else:
print("NO")
```
| 5,222 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
ve = 'aeiou'
abba, aabb, abab = 1, 1, 1
for j in range(n):
a = ['']*4
for i in range(4):
a[i] = input()
l = len(a[i])
curr = 0
while l > 0 and curr < k:
l -= 1
if a[i][l] in ve:
curr += 1
if curr == k:
a[i] = a[i][l:]
else:
a[i] = str(i)
if a[0] == a[3] and a[1] == a[2]:
abba = abba and 1
else:
abba = 0
if a[0] == a[1] and a[2] == a[3]:
aabb = aabb and 1
else:
aabb = 0
if a[0] == a[2] and a[1] == a[3]:
abab = abab and 1
else:
abab = 0
if abba and aabb and abab:
print('aaaa')
elif not(abba or aabb or abab):
print('NO')
elif abba:
print('abba')
elif abab:
print('abab')
elif aabb:
print('aabb')
# Made By Mostafa_Khaled
```
| 5,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
vowels = ['a', 'e', 'i', 'o', 'u']
def find_nth(tab, S, n):
amount, pos = (1 if tab[0] in S else 0), 0
while pos < len(tab) and amount < n:
pos += 1
if tab[pos] in S:
amount += 1
return pos
def rhyme_type(lines, k):
suffixes = []
for line in lines:
amount = 0
for i in line:
if i in vowels:
amount += 1
if amount < k:
return 'TRASH'
rev = list(reversed(list(line)))
ind_from_front = find_nth(rev, vowels, k)
suffixes.append(line[-ind_from_front - 1:])
if all([suffixes[0] == x for x in suffixes]):
return 'aaaa'
if suffixes[0] == suffixes[1] and suffixes[2] == suffixes[3]:
return 'aabb'
if suffixes[0] == suffixes[2] and suffixes[1] == suffixes[3]:
return 'abab'
if suffixes[0] == suffixes[3] and suffixes[1] == suffixes[2]:
return 'abba'
return 'TRASH'
all_rhymes = set()
for _ in range(n):
lines = [input(), input(), input(), input()]
all_rhymes.add(rhyme_type(lines, k))
if 'TRASH' not in all_rhymes and len(all_rhymes) == 2 and 'aaaa' in all_rhymes:
all_rhymes.remove('aaaa')
print(list(all_rhymes)[0])
elif len(all_rhymes) > 1 or 'TRASH' in all_rhymes:
print('NO')
else:
print(list(all_rhymes)[0])
```
| 5,224 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
def get_suffix(line):
position = 0
for _ in range(k):
position -= 1
while -position <= len(line) and line[position] not in 'aeiou':
position -= 1
if -position > len(line):
return ''
return line[position:]
common_rhyme_type = 'aaaa'
for _ in range(n):
s1, s2, s3, s4 = [get_suffix(input()) for _ in range(4)]
if '' in (s1, s2, s3, s4):
print('NO')
break
if s1 == s2 == s3 == s4:
rhyme_type = 'aaaa'
elif s1 == s2 and s3 == s4:
rhyme_type = 'aabb'
elif s1 == s3 and s2 == s4:
rhyme_type = 'abab'
elif s1 == s4 and s2 == s3:
rhyme_type = 'abba'
else:
print('NO')
break
if rhyme_type != 'aaaa':
if common_rhyme_type != 'aaaa' and rhyme_type != common_rhyme_type:
print('NO')
break
else:
common_rhyme_type = rhyme_type
else:
print(common_rhyme_type)
```
| 5,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
ACCEPTED={'aaaa','aabb','abba','abab'}
vowels='aeiou'
nul='abcd'
nu=0
def operate(s):
global nu
c=0
for i in range(len(s)-1,-1,-1):
if(s[i] in vowels):
c+=1
if(c==k):
return s[i:]
nu=(nu+1)%4
return nul[nu]
def rhymes(a):
a=[operate(i) for i in a]
# print(a)
ID={}
id=0
ans=''
for i in a:
if(i not in ID):
ID[i]=nul[id]
id+=1
ans+=ID[i]
return ans
n,k=value()
scheme=set()
for i in range(n):
a=[]
for j in range(4):
a.append(input())
scheme.add(rhymes(a))
# print(scheme)
for i in scheme:
if(i not in ACCEPTED):
print("NO")
exit()
if(len(scheme)>2):
print("NO")
elif(len(scheme)==2):
if('aaaa' not in scheme):
print("NO")
else:
for i in scheme:
if(i!='aaaa'):
print(i)
else:
print(*scheme)
```
| 5,226 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Tags: implementation
Correct Solution:
```
from sys import stdin
read = stdin.readline
n,k = map(int,read().split())
r = 0
vowel = {'a','e','i','o','u'}
def cmp(a,b,k,vowel):
l = 0
for x,y in zip(reversed(a),reversed(b)):
if x != y:
return False
l += (x in vowel)
if l == k:
return True
return False
im = True
for s in range(n):
a,b,c,d = read(),read(),read(),read()
l = 0
for i,x in enumerate(reversed(a),start=1):
l += (x in vowel)
if l == k:
break
else:
print('NO')
break
b2,b3,b4 = (a[-i:] == b[-i:]),(a[-i:] == c[-i:]),(a[-i:] == d[-i:])
if b2 + b3 + b4 == 3:
continue
elif b2 + b3 + b4 == 1:
if b2 and (r == 1 or r == 0) and cmp(c,d,k,vowel):
r = 1
continue
elif b3 and (r==2 or r == 0) and cmp(b,d,k,vowel):
r = 2
continue
elif b4 and (r == 3 or r == 0) and cmp(b,c,k,vowel):
r = 3
continue
print('NO')
break
else:
print(['aaaa','aabb','abab','abba'][r])
```
| 5,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Submitted Solution:
```
n, k = map(int, input().split())
def get_suffix(line):
position = -1
for _ in range(k):
while -position <= len(line) and line[position] not in 'aeiou':
position -= 1
if -position > len(line):
return ''
return line[position:]
def detect_rhyme_type(quatrain):
s1, s2, s3, s4 = [get_suffix(line) for line in quatrain]
if s1 == s2 == s3 == s4:
return 'aaaa'
elif s1 == s2 and s3 == s4:
return 'aabb'
elif s1 == s3 and s2 == s4:
return 'abab'
elif s1 == s4 and s2 == s3:
return 'abba'
else:
return 'NO'
common_rhyme_type = 'aaaa'
for _ in range(n):
quatrain = [input() for _ in range(4)]
rhyme_type = detect_rhyme_type(quatrain)
if rhyme_type == 'NO':
print('NO')
break
if rhyme_type != common_rhyme_type and common_rhyme_type != 'aaaa':
print('NO')
break
else:
common_rhyme_type = rhyme_type
else:
print(common_rhyme_type)
```
No
| 5,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Submitted Solution:
```
n,k = map(int,input().split())
lis=[]
vov=['a','e','i','o','u']
d={}
d['aabb']=d['abab']=d['abba']=d['aaaa']=0
for i in range(n*4):
s = input()
lis.append(s)
if i%4==3:
tmp=['']*4
for j in range(4):
s=lis[j]
c=0
cou=0
for ll in s[::-1]:
cou+=1
if ll in vov:
c+=1
if c==k:
tmp[j]=s[-cou:]
break
# print(tmp)
if tmp[0]==tmp[1]==tmp[2]==tmp[3] and tmp[0]!='':
d['aaaa']+=1
if tmp[0]==tmp[1] and tmp[2]==tmp[3] and tmp[1]!=tmp[3] and tmp[1]!='' and tmp[3]!='':
d['aabb']+=1
elif tmp[0]==tmp[2] and tmp[1]==tmp[3] and tmp[1]!=tmp[0] and tmp[1]!='' and tmp[0]!='':
d['abab']+=1
elif tmp[0]==tmp[3] and tmp[1]==tmp[2] and tmp[2]!=tmp[0] and tmp[1]!='' and tmp[3]!='':
d['abba']+=1
else:
print("NO")
exit()
lis=[]
c=0
for i in d:
if d[i]>0:
c+=1
if c==1:
for i in d:
if d[i]>0:
print(i)
exit()
elif c==2 and d['aaaa']>0:
for i in d:
if d[i]>0:
print(i)
exit()
else:
print("NO")
```
No
| 5,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Submitted Solution:
```
n, p = map(int, input().split())
final = []
fe = 0
for i in range(0, 4*n, 4):
l = []
for j in range(4):
s = input()
temp = p
error = 0
for k in range(len(s)-1, -1, -1):
if(s[k] == 'a' or s[k] == 'e' or s[k] == 'i' or s[k] == 'o' or s[k] == 'u'):
temp -= 1
if(temp == 0):
l.append(s[k:])
break
if(len(l) == 4):
if(l[0] == l[1] and l[2] == l[3] and l[1] == l[2]):
final.append(1)
elif(l[0] == l[1] and l[2] == l[3]):
final.append(2)
elif(l[0] == l[2] and l[1] == l[3]):
final.append(3)
elif(l[0] == l[3] and l[2] == l[1]):
final.append(4)
else:
fe = 1
if(len(l) == n):
final = list(set(final))
# print(final)
if(len(final) == 1):
if(final[0] == 1):
print("aaaa")
elif(final[0] == 2):
print("aabb")
elif(final[0] == 3):
print("abab")
elif (final[0] == 4):
print("abba")
elif(len(final) == 2):
if(final[0] == 1 or final[1] == 1):
if(final[0] == 2):
print("aabb")
elif(final[0] == 3):
print("abab")
elif(final[0] == 4):
print("abba")
elif(final[1] == 2):
print("aabb")
elif(final[1] == 3):
print("abab")
elif(final[1] == 4):
print("abba")
else:
print("NO")
else:
print("NO")
else:
print("NO")
```
No
| 5,230 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their suffixes that start from the k-th vowels (counting from the end) match. If a line has less than k vowels, then such line can't rhyme with any other line. For example, if k = 1, lines commit and hermit rhyme (the corresponding suffixes equal it), and if k = 2, they do not rhyme (ommit ≠ ermit).
Today on a literature lesson Vera learned that quatrains can contain four different schemes of rhymes, namely the following ones (the same letters stand for rhyming lines):
* Clerihew (aabb);
* Alternating (abab);
* Enclosed (abba).
If all lines of a quatrain pairwise rhyme, then the quatrain can belong to any rhyme scheme (this situation is represented by aaaa).
If all quatrains of a poem belong to the same rhyme scheme, then we can assume that the whole poem belongs to this rhyme scheme. If in each quatrain all lines pairwise rhyme, then the rhyme scheme of the poem is aaaa. Let us note that it doesn't matter whether lines from different quatrains rhyme with each other or not. In other words, it is possible that different quatrains aren't connected by a rhyme.
Vera got a long poem as a home task. The girl has to analyse it and find the poem rhyme scheme. Help Vera cope with the task.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2500, 1 ≤ k ≤ 5) — the number of quatrains in the poem and the vowel's number, correspondingly. Next 4n lines contain the poem. Each line is not empty and only consists of small Latin letters. The total length of the lines does not exceed 104.
If we assume that the lines are numbered starting from 1, then the first quatrain contains lines number 1, 2, 3, 4; the second one contains lines number 5, 6, 7, 8; and so on.
Output
Print the rhyme scheme of the poem as "aabb", "abab", "abba", "aaaa"; or "NO" if the poem does not belong to any of the above mentioned schemes.
Examples
Input
1 1
day
may
sun
fun
Output
aabb
Input
1 1
day
may
gray
way
Output
aaaa
Input
2 1
a
a
a
a
a
a
e
e
Output
aabb
Input
2 1
day
may
sun
fun
test
hill
fest
thrill
Output
NO
Note
In the last sample both quatrains have rhymes but finding the common scheme is impossible, so the answer is "NO".
Submitted Solution:
```
n, k = map(int, input().split())
vowels = ['a', 'e', 'i', 'o', 'u']
def find_nth(tab, S, n):
amount, pos = (1 if tab[0] in S else 0), 0
while pos < len(tab) and amount < n:
pos += 1
if tab[pos] in S:
amount += 1
return pos
def rhyme_type(lines, k):
suffixes = []
for line in lines:
amount = 0
for i in line:
if i in vowels:
amount += 1
if amount < k:
return 'TRASH'
rev = list(reversed(list(line)))
ind_from_front = find_nth(rev, vowels, k)
suffixes.append(line[-ind_from_front - 1:])
print(suffixes)
if all([suffixes[0] == x for x in suffixes]):
return 'aaaa'
if suffixes[0] == suffixes[1] and suffixes[2] == suffixes[3]:
return 'aabb'
if suffixes[0] == suffixes[2] and suffixes[1] == suffixes[3]:
return 'abab'
if suffixes[0] == suffixes[3] and suffixes[1] == suffixes[2]:
return 'abba'
return 'TRASH'
all_rhymes = set()
for _ in range(n):
lines = [input(), input(), input(), input()]
all_rhymes.add(rhyme_type(lines, k))
if 'TRASH' not in all_rhymes and len(all_rhymes) == 2 and 'aaaa' in all_rhymes:
all_rhymes.remove('aaaa')
print(list(all_rhymes)[0])
elif len(all_rhymes) > 1 or 'TRASH' in all_rhymes:
print('NO')
else:
print(list(all_rhymes)[0])
```
No
| 5,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sarah has always been a lover of nature, and a couple of years ago she saved up enough money to travel the world and explore all the things built by nature over its lifetime on earth. During this time she visited some truly special places which were left untouched for centuries, from watching icebergs in freezing weather to scuba-diving in oceans and admiring the sea life, residing unseen. These experiences were enhanced with breathtaking views built by mountains over time and left there for visitors to see for years on end. Over time, all these expeditions took a toll on Sarah and culminated in her decision to settle down in the suburbs and live a quiet life.
However, as Sarah's love for nature never faded, she started growing flowers in her garden in an attempt to stay connected with nature. At the beginning she planted only blue orchids, but over time she started using different flower types to add variety to her collection of flowers. This collection of flowers can be represented as an array of N flowers and the i-th of them has a type associated with it, denoted as A_i. Each resident, passing by her collection and limited by the width of his view, can only see K contiguous flowers at each moment in time. To see the whole collection, the resident will look at the first K contiguous flowers A_1, A_2, ..., A_K, then shift his view by one flower and look at the next section of K contiguous flowers A_2, A_3, ..., A_{K+1} and so on until they scan the whole collection, ending with section A_{N-K+1}, ..., A_{N-1}, A_N.
Each resident determines the beautiness of a section of K flowers as the number of distinct flower types in that section. Furthermore, the beautiness of the whole collection is calculated by summing the beautiness values of each contiguous section. Formally, beautiness B_i of a section starting at the i-th position is calculated as B_i = distinct(A_i, A_{i+1}, ..., A_{i+K-1}), and beautiness of the collection B is calculated as B=B_1 + B_2 + ... + B_{N-K+1}.
In addition, as Sarah wants to keep her collection of flowers have a fresh feel, she can also pick two points L and R, dispose flowers between those two points and plant new flowers, all of them being the same type.
You will be given Q queries and each of those queries will be of the following two types:
1. You will be given three integers L, R, X describing that Sarah has planted flowers of type X between positions L and R inclusive. Formally collection is changed such that A[i]=X for all i in range [L.. R].
2. You will be given integer K, width of the resident's view and you have to determine the beautiness value B resident has associated with the collection
For each query of second type print the result – beautiness B of the collection.
Input
First line contains two integers N and Q \;(1 ≤ N, Q ≤ 10^5) — number of flowers and the number of queries, respectively.
The second line contains N integers A_1, A_2, ..., A_N\;(1 ≤ A_i ≤ 10^9) — where A_i represents type of the i-th flower.
Each of the next Q lines describe queries and start with integer T∈\{1, 2\}.
* If T = 1, there will be three more integers in the line L, R, X\;(1 ≤ L, R ≤ N;\; 1 ≤ X ≤ 10^9) — L and R describing boundaries and X describing the flower type
* If T = 2, there will be one more integer in the line K\;(1 ≤ K ≤ N) — resident's width of view
Output
For each query of the second type print the beautiness B of the collection.
Example
Input
5 5
1 2 3 4 5
2 3
1 1 2 5
2 4
1 2 4 5
2 2
Output
9
6
4
Note
Let's look at the example.
Initially the collection is [1, 2, 3, 4, 5]. In the first query K = 3, we consider sections of three flowers with the first being [1, 2, 3]. Since beautiness of the section is the number of distinct flower types in that section, B_1 = 3. Second section is [2, 3, 4] and B_2 = 3. Third section is [3, 4, 5] and B_3 = 3, since the flower types are all distinct. The beautiness value resident has associated with the collection is B = B_1 + B_2 + B_3 = 3 + 3 + 3 = 9.
After the second query, the collection becomes [5, 5, 3, 4, 5].
For the third query K = 4, so we consider sections of four flowers with the first being [5, 5, 3, 4]. There are three distinct flower types [5, 3, 4] in this section, so B_1 = 3. Second section [5, 3, 4, 5] also has 3 distinct flower types, so B_2 = 3. The beautiness value resident has associated with the collection is B = B_1 + B_2 = 3 + 3 = 6
After the fourth query, the collection becomes [5, 5, 5, 5, 5].
For the fifth query K = 2 and in this case all the four sections are same with each of them being [5, 5]. Beautiness of [5, 5] is 1 since there is only one distinct element in this section [5]. Beautiness of the whole collection is B = B_1 + B_2 + B_3 + B_4 = 1 + 1 + 1 + 1 = 4
Submitted Solution:
```
#3500-3400
#https://codeforces.com/problemset/problem/1423/G
N, Q = map(int, input().split())
A = list(map(int, input().split()))[:Q]
a=A.copy()
B = []
result = []
def all_set(a, n):
while len(a)>=n:
B.append(set(a[:n]))
a.pop(0)
for query in range(N):
sets = list(map(int, input().split()))
if sets[0] == 1:
L = sets[1]
R = sets[2]
K = sets[3]
for i in range(L-1,R):
A[i] = K
a=A.copy()
#print(a)
else:
K = sets[1]
#for sub_set in range(len(A)):
all_set(a, K)
result.append(sum([len(i) for i in B]))
#print(sum([len(i) for i in B]))
#B=[]
print(result)
```
No
| 5,232 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sarah has always been a lover of nature, and a couple of years ago she saved up enough money to travel the world and explore all the things built by nature over its lifetime on earth. During this time she visited some truly special places which were left untouched for centuries, from watching icebergs in freezing weather to scuba-diving in oceans and admiring the sea life, residing unseen. These experiences were enhanced with breathtaking views built by mountains over time and left there for visitors to see for years on end. Over time, all these expeditions took a toll on Sarah and culminated in her decision to settle down in the suburbs and live a quiet life.
However, as Sarah's love for nature never faded, she started growing flowers in her garden in an attempt to stay connected with nature. At the beginning she planted only blue orchids, but over time she started using different flower types to add variety to her collection of flowers. This collection of flowers can be represented as an array of N flowers and the i-th of them has a type associated with it, denoted as A_i. Each resident, passing by her collection and limited by the width of his view, can only see K contiguous flowers at each moment in time. To see the whole collection, the resident will look at the first K contiguous flowers A_1, A_2, ..., A_K, then shift his view by one flower and look at the next section of K contiguous flowers A_2, A_3, ..., A_{K+1} and so on until they scan the whole collection, ending with section A_{N-K+1}, ..., A_{N-1}, A_N.
Each resident determines the beautiness of a section of K flowers as the number of distinct flower types in that section. Furthermore, the beautiness of the whole collection is calculated by summing the beautiness values of each contiguous section. Formally, beautiness B_i of a section starting at the i-th position is calculated as B_i = distinct(A_i, A_{i+1}, ..., A_{i+K-1}), and beautiness of the collection B is calculated as B=B_1 + B_2 + ... + B_{N-K+1}.
In addition, as Sarah wants to keep her collection of flowers have a fresh feel, she can also pick two points L and R, dispose flowers between those two points and plant new flowers, all of them being the same type.
You will be given Q queries and each of those queries will be of the following two types:
1. You will be given three integers L, R, X describing that Sarah has planted flowers of type X between positions L and R inclusive. Formally collection is changed such that A[i]=X for all i in range [L.. R].
2. You will be given integer K, width of the resident's view and you have to determine the beautiness value B resident has associated with the collection
For each query of second type print the result – beautiness B of the collection.
Input
First line contains two integers N and Q \;(1 ≤ N, Q ≤ 10^5) — number of flowers and the number of queries, respectively.
The second line contains N integers A_1, A_2, ..., A_N\;(1 ≤ A_i ≤ 10^9) — where A_i represents type of the i-th flower.
Each of the next Q lines describe queries and start with integer T∈\{1, 2\}.
* If T = 1, there will be three more integers in the line L, R, X\;(1 ≤ L, R ≤ N;\; 1 ≤ X ≤ 10^9) — L and R describing boundaries and X describing the flower type
* If T = 2, there will be one more integer in the line K\;(1 ≤ K ≤ N) — resident's width of view
Output
For each query of the second type print the beautiness B of the collection.
Example
Input
5 5
1 2 3 4 5
2 3
1 1 2 5
2 4
1 2 4 5
2 2
Output
9
6
4
Note
Let's look at the example.
Initially the collection is [1, 2, 3, 4, 5]. In the first query K = 3, we consider sections of three flowers with the first being [1, 2, 3]. Since beautiness of the section is the number of distinct flower types in that section, B_1 = 3. Second section is [2, 3, 4] and B_2 = 3. Third section is [3, 4, 5] and B_3 = 3, since the flower types are all distinct. The beautiness value resident has associated with the collection is B = B_1 + B_2 + B_3 = 3 + 3 + 3 = 9.
After the second query, the collection becomes [5, 5, 3, 4, 5].
For the third query K = 4, so we consider sections of four flowers with the first being [5, 5, 3, 4]. There are three distinct flower types [5, 3, 4] in this section, so B_1 = 3. Second section [5, 3, 4, 5] also has 3 distinct flower types, so B_2 = 3. The beautiness value resident has associated with the collection is B = B_1 + B_2 = 3 + 3 = 6
After the fourth query, the collection becomes [5, 5, 5, 5, 5].
For the fifth query K = 2 and in this case all the four sections are same with each of them being [5, 5]. Beautiness of [5, 5] is 1 since there is only one distinct element in this section [5]. Beautiness of the whole collection is B = B_1 + B_2 + B_3 + B_4 = 1 + 1 + 1 + 1 = 4
Submitted Solution:
```
n,m=input().split()
n=int(n)
ans=[]
m=int(m)
result=[]
flowers=list(int(x) for x in input().split())
for i in range(n):
ins=list(int(x) for x in input().split())
k=ins[1]
if(ins[0]==1):
flowers[ins[1]-1:ins[2]]=[5]
else:
j=0
while(j+k<=len(flowers)):
dis=set(flowers[j:j+k])
result.append(len(dis))
j+=1
ans.append(sum(result))
result=[]
for i in ans:
print(i)
```
No
| 5,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sarah has always been a lover of nature, and a couple of years ago she saved up enough money to travel the world and explore all the things built by nature over its lifetime on earth. During this time she visited some truly special places which were left untouched for centuries, from watching icebergs in freezing weather to scuba-diving in oceans and admiring the sea life, residing unseen. These experiences were enhanced with breathtaking views built by mountains over time and left there for visitors to see for years on end. Over time, all these expeditions took a toll on Sarah and culminated in her decision to settle down in the suburbs and live a quiet life.
However, as Sarah's love for nature never faded, she started growing flowers in her garden in an attempt to stay connected with nature. At the beginning she planted only blue orchids, but over time she started using different flower types to add variety to her collection of flowers. This collection of flowers can be represented as an array of N flowers and the i-th of them has a type associated with it, denoted as A_i. Each resident, passing by her collection and limited by the width of his view, can only see K contiguous flowers at each moment in time. To see the whole collection, the resident will look at the first K contiguous flowers A_1, A_2, ..., A_K, then shift his view by one flower and look at the next section of K contiguous flowers A_2, A_3, ..., A_{K+1} and so on until they scan the whole collection, ending with section A_{N-K+1}, ..., A_{N-1}, A_N.
Each resident determines the beautiness of a section of K flowers as the number of distinct flower types in that section. Furthermore, the beautiness of the whole collection is calculated by summing the beautiness values of each contiguous section. Formally, beautiness B_i of a section starting at the i-th position is calculated as B_i = distinct(A_i, A_{i+1}, ..., A_{i+K-1}), and beautiness of the collection B is calculated as B=B_1 + B_2 + ... + B_{N-K+1}.
In addition, as Sarah wants to keep her collection of flowers have a fresh feel, she can also pick two points L and R, dispose flowers between those two points and plant new flowers, all of them being the same type.
You will be given Q queries and each of those queries will be of the following two types:
1. You will be given three integers L, R, X describing that Sarah has planted flowers of type X between positions L and R inclusive. Formally collection is changed such that A[i]=X for all i in range [L.. R].
2. You will be given integer K, width of the resident's view and you have to determine the beautiness value B resident has associated with the collection
For each query of second type print the result – beautiness B of the collection.
Input
First line contains two integers N and Q \;(1 ≤ N, Q ≤ 10^5) — number of flowers and the number of queries, respectively.
The second line contains N integers A_1, A_2, ..., A_N\;(1 ≤ A_i ≤ 10^9) — where A_i represents type of the i-th flower.
Each of the next Q lines describe queries and start with integer T∈\{1, 2\}.
* If T = 1, there will be three more integers in the line L, R, X\;(1 ≤ L, R ≤ N;\; 1 ≤ X ≤ 10^9) — L and R describing boundaries and X describing the flower type
* If T = 2, there will be one more integer in the line K\;(1 ≤ K ≤ N) — resident's width of view
Output
For each query of the second type print the beautiness B of the collection.
Example
Input
5 5
1 2 3 4 5
2 3
1 1 2 5
2 4
1 2 4 5
2 2
Output
9
6
4
Note
Let's look at the example.
Initially the collection is [1, 2, 3, 4, 5]. In the first query K = 3, we consider sections of three flowers with the first being [1, 2, 3]. Since beautiness of the section is the number of distinct flower types in that section, B_1 = 3. Second section is [2, 3, 4] and B_2 = 3. Third section is [3, 4, 5] and B_3 = 3, since the flower types are all distinct. The beautiness value resident has associated with the collection is B = B_1 + B_2 + B_3 = 3 + 3 + 3 = 9.
After the second query, the collection becomes [5, 5, 3, 4, 5].
For the third query K = 4, so we consider sections of four flowers with the first being [5, 5, 3, 4]. There are three distinct flower types [5, 3, 4] in this section, so B_1 = 3. Second section [5, 3, 4, 5] also has 3 distinct flower types, so B_2 = 3. The beautiness value resident has associated with the collection is B = B_1 + B_2 = 3 + 3 = 6
After the fourth query, the collection becomes [5, 5, 5, 5, 5].
For the fifth query K = 2 and in this case all the four sections are same with each of them being [5, 5]. Beautiness of [5, 5] is 1 since there is only one distinct element in this section [5]. Beautiness of the whole collection is B = B_1 + B_2 + B_3 + B_4 = 1 + 1 + 1 + 1 = 4
Submitted Solution:
```
n,q = map(int,input().split())
a = list(map(int,input().split()))
count = 0
for i in range(0,q):
querie = list(map(int,input().split()))
if (querie[0] == 1):
for i in range(querie[1] - 1,querie[2]):
a[i] = querie[3]
if (querie[0] == 2):
k = querie[1]
j = 0
count = 0
while(i <= len(a) - k):
l = []
for j in range(i,querie[1] + i):
l.append(a[j])
i = i + 1
count = count + len(set(l))
print(count)
```
No
| 5,234 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sarah has always been a lover of nature, and a couple of years ago she saved up enough money to travel the world and explore all the things built by nature over its lifetime on earth. During this time she visited some truly special places which were left untouched for centuries, from watching icebergs in freezing weather to scuba-diving in oceans and admiring the sea life, residing unseen. These experiences were enhanced with breathtaking views built by mountains over time and left there for visitors to see for years on end. Over time, all these expeditions took a toll on Sarah and culminated in her decision to settle down in the suburbs and live a quiet life.
However, as Sarah's love for nature never faded, she started growing flowers in her garden in an attempt to stay connected with nature. At the beginning she planted only blue orchids, but over time she started using different flower types to add variety to her collection of flowers. This collection of flowers can be represented as an array of N flowers and the i-th of them has a type associated with it, denoted as A_i. Each resident, passing by her collection and limited by the width of his view, can only see K contiguous flowers at each moment in time. To see the whole collection, the resident will look at the first K contiguous flowers A_1, A_2, ..., A_K, then shift his view by one flower and look at the next section of K contiguous flowers A_2, A_3, ..., A_{K+1} and so on until they scan the whole collection, ending with section A_{N-K+1}, ..., A_{N-1}, A_N.
Each resident determines the beautiness of a section of K flowers as the number of distinct flower types in that section. Furthermore, the beautiness of the whole collection is calculated by summing the beautiness values of each contiguous section. Formally, beautiness B_i of a section starting at the i-th position is calculated as B_i = distinct(A_i, A_{i+1}, ..., A_{i+K-1}), and beautiness of the collection B is calculated as B=B_1 + B_2 + ... + B_{N-K+1}.
In addition, as Sarah wants to keep her collection of flowers have a fresh feel, she can also pick two points L and R, dispose flowers between those two points and plant new flowers, all of them being the same type.
You will be given Q queries and each of those queries will be of the following two types:
1. You will be given three integers L, R, X describing that Sarah has planted flowers of type X between positions L and R inclusive. Formally collection is changed such that A[i]=X for all i in range [L.. R].
2. You will be given integer K, width of the resident's view and you have to determine the beautiness value B resident has associated with the collection
For each query of second type print the result – beautiness B of the collection.
Input
First line contains two integers N and Q \;(1 ≤ N, Q ≤ 10^5) — number of flowers and the number of queries, respectively.
The second line contains N integers A_1, A_2, ..., A_N\;(1 ≤ A_i ≤ 10^9) — where A_i represents type of the i-th flower.
Each of the next Q lines describe queries and start with integer T∈\{1, 2\}.
* If T = 1, there will be three more integers in the line L, R, X\;(1 ≤ L, R ≤ N;\; 1 ≤ X ≤ 10^9) — L and R describing boundaries and X describing the flower type
* If T = 2, there will be one more integer in the line K\;(1 ≤ K ≤ N) — resident's width of view
Output
For each query of the second type print the beautiness B of the collection.
Example
Input
5 5
1 2 3 4 5
2 3
1 1 2 5
2 4
1 2 4 5
2 2
Output
9
6
4
Note
Let's look at the example.
Initially the collection is [1, 2, 3, 4, 5]. In the first query K = 3, we consider sections of three flowers with the first being [1, 2, 3]. Since beautiness of the section is the number of distinct flower types in that section, B_1 = 3. Second section is [2, 3, 4] and B_2 = 3. Third section is [3, 4, 5] and B_3 = 3, since the flower types are all distinct. The beautiness value resident has associated with the collection is B = B_1 + B_2 + B_3 = 3 + 3 + 3 = 9.
After the second query, the collection becomes [5, 5, 3, 4, 5].
For the third query K = 4, so we consider sections of four flowers with the first being [5, 5, 3, 4]. There are three distinct flower types [5, 3, 4] in this section, so B_1 = 3. Second section [5, 3, 4, 5] also has 3 distinct flower types, so B_2 = 3. The beautiness value resident has associated with the collection is B = B_1 + B_2 = 3 + 3 = 6
After the fourth query, the collection becomes [5, 5, 5, 5, 5].
For the fifth query K = 2 and in this case all the four sections are same with each of them being [5, 5]. Beautiness of [5, 5] is 1 since there is only one distinct element in this section [5]. Beautiness of the whole collection is B = B_1 + B_2 + B_3 + B_4 = 1 + 1 + 1 + 1 = 4
Submitted Solution:
```
p = input()
q = input()
arr = []
arr1 = []
for i in range(len(p)):
for j in range(i + 1, len(p) + 1):
arr.append(p[i:j])
arr = list(set(arr))
for i in range(len(q)):
for j in range(i + 1, len(q) + 1):
arr1.append(q[i:j])
arr1 = list(set(arr1))
arr3 = []
# arr4=[]
for i in range(len(arr)):
n1 = int(len(arr[i]) / 2)
count = 0
if len(arr[i]) % 2 == 0:
for j in range(n1):
if arr[i][n1 - 1 - j] == arr[i][n1 + j]:
count += 1
else:
for j in range(n1):
if arr[i][n1 - 1 - j] == arr[i][n1 + 1 + j]:
count += 1
if count == n1:
arr3.append(arr[i])
arr5 = []
for i in range(len(arr1)):
n1 = int(len(arr1[i]) / 2)
count = 0
if len(arr1[i]) % 2 == 0:
for j in range(n1):
if arr1[i][n1 - 1 - j] == arr1[i][n1 + j]:
count += 1
else:
for j in range(n1):
if arr1[i][n1 - 1 - j] == arr1[i][n1 + 1 + j]:
count += 1
if count == n1:
# arr4.append(arr1[i])
for k in range(len(arr3)):
arr5.append(arr3[k] + arr1[i])
"""
for i in range(len(arr3)):
for j in range(len(arr4)):
arr5.append(arr3[i]+arr4[j])
"""
print(len(list(set(arr5))))
```
No
| 5,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import sys
for _ in range(int(sys.stdin.readline().strip())):
n,k,m=tuple(map(int,sys.stdin.readline().strip().split(" ")))
ml=set(map(int,sys.stdin.readline().strip().split(" ")))
havitada=[]
for i in range(1,n+1):
if i not in ml:
havitada.append(i)
saab=False
if len(havitada)%(k-1)!=0:
print("no")
continue
for i in range(k//2-1,len(havitada)-k//2):
if havitada[i+1]-havitada[i]!=1:
print("yes")
break
else:
print("no")
```
| 5,236 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import sys
def load_sys():
return sys.stdin.readlines()
def load_local():
with open('input.txt','r') as f:
input = f.readlines()
return input
def km(n,k,m,B):
if (n-m)%(k-1) != 0:
return 'NO'
R = [0]*m
L = [0]*m
for i in range(m):
R[i] = n-B[i]-(m-i-1)
L[i] = n-m-R[i]
if R[i] >= k//2 and L[i] >= k//2:
return 'YES'
return 'NO'
#input = load_local()
input = load_sys()
idx = 1
N = len(input)
while idx < len(input):
n,k,m = [int(x) for x in input[idx].split()]
B = [int(x) for x in input[idx+1].split()]
idx += 2
print(km(n,k,m,B))
```
| 5,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
T = int(input())
for _ in range(T):
n, k, m = map(int, input().split())
b = [int(i) for i in input().split()]
ff = False
for p, i in enumerate(b):
if i - p - 1 >= k>>1 and n - i - len(b) + p + 1 >= k>>1:
ff = True
if b==[int(i)+1 for i in range(n)]:
print('YES')
elif k>1 and (n-m)%(k-1)==0:
if ff:
print('YES')
else:
print('NO')
else:
print('NO')
```
| 5,238 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import sys
for _ in range(int(input())):
n,k,m=tuple(map(int,input().split(" ")))
ml=set(map(int,input().split(" ")))
havitada=[]
for i in range(1,n+1):
if i not in ml:
havitada.append(i)
saab=False
if len(havitada)%(k-1)!=0:
print("no")
continue
for i in range(k//2-1,len(havitada)-k//2):
if havitada[i+1]-havitada[i]!=1:
print("yes")
break
else:
print("no")
```
| 5,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
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()
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(100000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
from math import ceil
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
for _ in range(N()):
n, k, m = RL()
a = RLL()
k >>= 1
if k == 0 or (n - m) % (2 * k):
print('NO')
else:
t = n - m
now = a[0] - 1
flag = False
a.append(n + 1)
for i in range(1, m + 1):
if k <= now <= t - k:
flag = True
break
now += a[i] - a[i - 1] - 1
print('YES' if flag else 'NO')
```
| 5,240 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
T=int(input())
for _ in range(T):
n,k,m=map(int, input().split())
b=[0]+list(map(int, input().split()))+[0]*10 #1~m (len=m+1)
cnt=0
l=[0]*(m+10)
r=[0]*(m+10)
for i in range(1,m+1): #i:1~m
cnt+=b[i]-b[i-1]-1
l[i]=l[i-1]+b[i]-b[i-1]-1
cnt+=n-b[m]
r[m]=n-b[m]
for i in range(m-1,0,-1): #i:m-1~1
r[i]=r[i+1]+b[i+1]-b[i]-1
flag=False
for i in range(1,m+1): #i:1~m
if l[i]>=k//2 and r[i]>=k//2:
flag=True
if cnt%(k-1)!=0:
flag=False
if flag:
print("YES")
else:
print("NO")
```
| 5,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
allAns=[]
t=int(input())
for _ in range(t):
n,k,m=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
b.append(n+1)
ans='NO'
if (n-m)%(k-1)==0: #valid number of missing numbers
leftMissing=0
rightMissing=n-m
prev=0
for x in b:
leftMissing+=x-prev-1
rightMissing-=(x-prev-1)
prev=x
if leftMissing>=(k-1)//2 and rightMissing>=(k-1)//2:
ans='YES'
break
# print('leftMissing:{} rightMissing:{}'.format(leftMissing,rightMissing))
allAns.append(ans)
multiLineArrayPrint(allAns)
```
| 5,242 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Tags: constructive algorithms, greedy, math
Correct Solution:
```
from sys import stdin, stdout
# if erased left elements >= (k-1)/2
# and erased right elements >= (k-1)/2
# and (n-m)%(k-1) == 0
# then YES
# Prove:
# set d = (k-1)/2
# left elements: d + x
# right elements: d + y
# ------------------------------------------
# if x + y >= k, set x + y = x + y - n*(k-1)
# then x + y < k
# ------------------------------------------
# because (2d + x + y) % (k-1) == 0,
# so x + y = k - 1
# ------------------------------------------
# d x b d y
# => d d-z b d d+z
# => d (d-z) b (z) d [d]
#
def k_and_medians(n, k, m, b_a):
if (n-m) % (k-1) != 0:
return False
b_s = set(b_a)
l_a = [0] * (n+1)
for i in range(1, n+1):
l_a[i] = l_a[i - 1]
if i not in b_s:
l_a[i] = l_a[i-1] + 1
r = 0
for i in range(n, 0, -1):
if i not in b_s:
r += 1
elif r >= (k-1) // 2 and l_a[i] >= (k-1) // 2:
return True
return False
t = int(stdin.readline())
for _ in range(t):
n, k, m = map(int, stdin.readline().split())
b_a = list(map(int, stdin.readline().split()))
r = k_and_medians(n, k, m, b_a)
if r:
stdout.write('YES\n')
else:
stdout.write('NO\n')
```
| 5,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
for _ in range(int(input())):
n,k,m=tuple(map(int,input().split(" ")));ml=set(map(int,input().split(" ")));havitada=[i for i in range(1,n+1) if i not in ml];saab=False
if len(havitada)%(k-1)!=0:print("no");continue
for i in range(k//2-1,len(havitada)-k//2):
if havitada[i+1]-havitada[i]!=1:print("yes");break
else:print("no")
```
Yes
| 5,244 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for you in range(t):
l=input().split()
n=int(l[0])
k=int(l[1])
m=int(l[2])
l=input().split()
li=[int(i) for i in l]
if((n-m)%(k-1)):
print("NO")
continue
poss=0
for i in range(m):
z=n-li[i]
z-=(m-i-1)
y=li[i]-1
y-=i
if(z>=(k-1)//2 and y>=(k-1)//2):
poss=1
break
if(poss):
print("YES")
else:
print("NO")
```
Yes
| 5,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Jan 16 12:33:14 2021
@author: ludoj
"""
t=int(input())
res=[]
for x1 in range(t):
n,k,m=[int(i) for i in input().split()]
eind=[int(i) for i in input().split()]
x2=len(eind)
if (n-m)%(k-1)!=0:
res.append("NO")
else:
h=(k-1)//2
i1=0
a1=1
tel1=0
while(tel1<h and i1<x2):
if eind[i1]==a1:
a1+=1
i1+=1
else:
tel1+=1
a1+=1
#nu gevonden met h dingen ervoor
if i1>=x2:
res.append("NO")
else:
a2=eind[i1]
tel2=0
while(tel2<h and i1<x2):
if eind[i1]==a2:
a2+=1
i1+=1
else:
tel2+=1
a2+=1
if i1==x2:
tel2+=n-eind[i1-1]
if tel2>=h:
res.append("YES")
else:
res.append("NO")
for i in res:
print(i)
```
Yes
| 5,246 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
"""
Author - Satwik Tiwari .
15th Dec , 2020 - Tuesday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def 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,k,m = sep()
a = lis()
have = set(a)
if(len(have) != m):
print("NO")
return
notvis = []
for i in range(1,n+1):
if(i not in have):
notvis.append(i)
if(len(notvis)%(k-1)):
# print(len(notvis))
print('NO')
return
left = 0
right = 0
diff = []
cnt = 1
for i in range(1,len(notvis)):
if(notvis[i] == notvis[i-1]+1):
cnt+=1
else:
diff.append(cnt)
cnt = 1
diff.append(cnt)
ind = 0
while(left + diff[ind] < k//2):
left += diff[ind]
ind+=1
rnd = len(diff)-1
while(right + diff[rnd] < k//2):
right += diff[rnd]
rnd-=1
print('YES' if ind < rnd else 'NO')
# testcase(1)
testcase(int(inp()))
```
Yes
| 5,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
n,k,m = map(int,stdin.readline().split())
b = list(map(int,stdin.readline().split()))
if (n-m)%(k-1) != 0:
print ("NO")
continue
c = (n-m)//(k-1)
ans = "NO"
for i in b:
if k//2 < i <= n-k//2:
ans = "YES"
break
print (ans)
```
No
| 5,248 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
import heapq
t = int(input())
for _ in range(t):
n, k, m = map(int, input().split())
l = [0] + list(map(int, input().split())) + [n+1]
diffs = [l[i+1]-l[i]-1 for i in range(m+1)]
if sum(diffs) % (k-1) > 0:
print('NO')
continue
q = []
tot = 0
for v in diffs:
heapq.heappush(q, -v)
tot += v
while True:
nex = -heapq.heappop(q)
if 2 * nex <= tot:
print('YES')
break
if nex < k:
print('NO')
break
tot -= k - 1
heapq.heappush(q,(nex - (k-1)))
```
No
| 5,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
import math
T = int(input())
for _ in range(T):
n, k, m = map(int, input().split())
b = [int(i) for i in input().split()]
if b == [int(i)+1 for i in range(n)]:
print('YES')
elif b[0]-1 <= n - b[0] - len(b) +1 and n - b[-1] <= b[-1] - 1 - len(b) + 1 and k>1 and (n-m)%(k-1)==0:
print('YES')
else:
print('NO')
```
No
| 5,250 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote the median of a sequence s with odd length as the value in the middle of s if we sort s in non-decreasing order. For example, let s = [1, 2, 5, 7, 2, 3, 12]. After sorting, we get sequence [1, 2, 2, \underline{3}, 5, 7, 12], and the median is equal to 3.
You have a sequence of n integers [1, 2, ..., n] and an odd integer k.
In one step, you choose any k elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).
For example, if you have a sequence [1, 2, 3, 4, 5, 6, 7] (i.e. n=7) and k = 3, then the following options for the first step are possible:
* choose [1, \underline{2}, 3]; 2 is their median, so it is not erased, and the resulting sequence is [2, 4, 5, 6, 7];
* choose [2, \underline{4}, 6]; 4 is their median, so it is not erased, and the resulting sequence is [1, 3, 4, 5, 7];
* choose [1, \underline{6}, 7]; 6 is their median, so it is not erased, and the resulting sequence is [2, 3, 4, 5, 6];
* and several others.
You can do zero or more steps. Can you get a sequence b_1, b_2, ..., b_m after several steps?
You'll be given t test cases. Solve each test case independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains three integers n, k, and m (3 ≤ n ≤ 2 ⋅ 10^5; 3 ≤ k ≤ n; k is odd; 1 ≤ m < n) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.
The second line of each test case contains m integers b_1, b_2, ..., b_m (1 ≤ b_1 < b_2 < ... < b_m ≤ n) — the sequence you'd like to get, given in the ascending order.
It's guaranteed that the total sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print YES if you can obtain the sequence b or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12
Output
NO
YES
NO
YES
Note
In the first test case, you have sequence [1, 2, 3]. Since k = 3 you have only one way to choose k elements — it's to choose all elements [1, \underline{2}, 3] with median 2. That's why after erasing all chosen elements except its median you'll get sequence [2]. In other words, there is no way to get sequence b = [1] as the result.
In the second test case, you have sequence [1, 2, 3, 4, 5, 6, 7] and one of the optimal strategies is following:
1. choose k = 3 elements [2, \underline{3}, 4] and erase them except its median; you'll get sequence [1, 3, 5, 6, 7];
2. choose 3 elements [3, \underline{5}, 6] and erase them except its median; you'll get desired sequence [1, 5, 7];
In the fourth test case, you have sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. You can choose k=7 elements [2, 4, 6, \underline{7}, 8, 10, 13] and erase them except its median to get sequence b.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def solve(n,k,m,b):
if (n-m)%(k-1):
return 'NO'
fl = 0
if b[0] != 1 and b[-1] != n:
fl = 1
for i in range(1,m):
if b[i] != b[i-1]+1:
fl = 1
if fl:
return 'YES'
return 'NO'
def main():
for _ in range(int(input())):
n,k,m = map(int,input().split())
b = list(map(int,input().split()))
print(solve(n,k,m,b))
#Fast IO 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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
No
| 5,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
valid={0:0,1:1,2:5,5:2,8:8}
for _ in range(int(input())):
h,m=map(int,input().split())
hour,min=input().split(":")
while 1:
if int(min)>=m:
min="0"
hour=str(int(hour)+1)
if int(hour)>=h:hour="0"
min="0"*(len(min)==1)+min
hour = "0" * (len(hour) == 1) + hour
for i in range(2):
if int(hour[i]) not in valid or int(min[i]) not in valid:
min=str(int(min)+1)
break
else:
if int(str(valid[int(min[1])])+str(valid[int(min[0])]))>=h:
min = str(int(min) + 1);continue
if int(str(valid[int(hour[1])])+str(valid[int(hour[0])]))>=m:
min = str(int(min) + 1);continue
print(hour+":"+min)
break
```
| 5,252 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
d={0:0,1:1,2:5,5:2,8:8}
a={*d}
R=lambda x:map(int,input().split(x))
t,=R(' ')
while t:
t-=1;h,m=R(' ');x,y=R(':')
for i in range(h*m):
r=x*m+y+i;p=f'{r//m%h:02}';s=f'{r%m:02}';b=u,v,w,q=*map(int,p+s),
if{*b}<a and h>d[q]*10+d[w]and d[v]*10+d[u]<m:
break
print(p,s,sep=':')
```
| 5,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
def mirror(h, m):
sh = list(str(h).zfill(2))
sm = list(str(m).zfill(2))
if set(['3', '4', '6', '7', '9']) & set(sh + sm):
return False
m = { '0': '0', '1': '1', '2': '5', '5': '2', '8': '8' }
return (int(m[sm[1]] + m[sm[0]]), int(m[sh[1]] + m[sh[0]]))
def tick(h, m, H, M):
m += 1
if m == M:
m = 0
h += 1
if h == H:
h = 0
return (h, m)
def solve():
H, M = map(int, input().split())
h, m = map(int, input().split(':'))
while True:
r = mirror(h, m)
if r:
mh, mm = r
if mh < H and mm < M:
print(str(h).zfill(2) + ':' + str(m).zfill(2))
break
h, m = tick(h, m, H, M)
t = int(input())
while t > 0:
solve()
t -= 1
```
| 5,254 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
import sys
import os.path
from collections import *
import math
import bisect
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
else:
input = sys.stdin.readline
############## Code starts here ##########################
t = int(input())
nums = Counter({0:0,1:10,2:50,5:20,8:80,
10:1,11:11,12:51,15:21,18:81,
20:5,21:15,22:55,25:25,28:85,
50:2,51:12,52:52,55:22,58:82,
80:8,81:18,82:58,85:28,88:88})
mirror = {1: 1, 0: 0, 2: 5, 5: 2, 8: 8}
while t:
t-=1
hrs,mins = [int(x) for x in input().split()]
s = input().rstrip('\n')
h = int(s[0:2])
m = int(s[3:])
if m:
while m<mins:
if nums[m] and nums[m]<hrs:
break
m+=1
if m == mins:
m = 0
h = (1 + h) % hrs
if h:
while h<hrs:
if nums[h] and nums[h]<mins:
break
h+=1
m = 0
if h==hrs:
h = 0
a = h%10
h//=10
b = h%10
c = m%10
m//=10
d = m%10
print(b,a,":",d,c,sep="")
############## Code ends here ############################
```
| 5,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
# Author : raj1307 - Raj Singh
# Date : 06.03.2021
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().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(1000000)
threading.stack_size(1024000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[1]
def sort2(l):return sorted(l, key=getKey,reverse=True)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def ceil(x,y):
if x%y==0:
return x//y
else:
return x//y+1
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def check(x):
if x in [0,1,2,5,8]:
return 1
return 0
def ans(x,y):
x1=''
x2=''
if len(str(x))==1:
x1='0'+str(x)
else:
x1=str(x)
if len(str(y))==1:
x2='0'+str(y)
else:
x2=str(y)
print(x1+':'+x2)
def main():
for _ in range(ii()):
h,m=mi()
s=si()
x=int(s[0]+s[1])
y=int(s[3]+s[4])
f=[0,1,2,5,8]
p=-1
q=-1
a=x
b=y
for j in range(y,m):
x1=a//10
x2=a%10
num1=0
if x2==2:
x2=5
elif x2==5:
x2=2
if x1==2:
x1=5
elif x1==5:
x1=2
num1=x2*10+x1
y1=b//10
y2=b%10
num2=0
if y2==2:
y2=5
elif y2==5:
y2=2
if y1==2:
y1=5
elif y1==5:
y1=2
num2=y2*10+y1
#print(a,b,num1,num2)
if check(y1) and check(y2) and check(x1) and check(x2):
pass
else:
b+=1
continue
if num1<m and num2<h:
p=a
q=b
break
b+=1
if p!=-1:
#print(str(p)+':'+str(q))
ans(p,q)
continue
a+=1
b=0
for i in range(x+1,h):
b=0
for j in range(0,m):
#print(a,b)
x1=a//10
x2=a%10
num1=0
if x2==2:
x2=5
elif x2==5:
x2=2
if x1==2:
x1=5
elif x1==5:
x1=2
num1=x2*10+x1
y1=b//10
y2=b%10
num2=0
if y2==2:
y2=5
elif y2==5:
y2=2
if y1==2:
y1=5
elif y1==5:
y1=2
num2=y2*10+y1
#print(num1,num2)
#print(num2,num1)
if check(y1) and check(y2) and check(x1) and check(x2):
pass
else:
b+=1
continue
#print(num2,num1)
if num1<m and num2<h:
p=a
q=b
break
b+=1
a+=1
if p!=-1:
break
if p!=-1:
#print(str(p)+':'+str(q))
ans(p,q)
continue
a=0
b=0
for i in range(h):
b=0
for j in range(m):
x1=a//10
x2=a%10
num1=0
if x2==2:
x2=5
elif x2==5:
x2=2
if x1==2:
x1=5
elif x1==5:
x1=2
num1=x2*10+x1
y1=b//10
y2=b%10
num2=0
if y2==2:
y2=5
elif y2==5:
y2=2
if y1==2:
y1=5
elif y1==5:
y1=2
num2=y2*10+y1
if check(y1) and check(y2) and check(x1) and check(x2):
pass
else:
b+=1
continue
if num1<m and num2<h:
p=a
q=b
break
b+=1
a+=1
if p!=-1:
break
if p!=-1:
#print(str(p)+':'+str(q))
ans(p,q)
continue
a=h
b=0
for j in range(m):
x1=a//10
x2=a%10
num1=0
if x2==2:
x2=5
elif x2==5:
x2=2
if x1==2:
x1=5
elif x1==5:
x1=2
num1=x2*10+x1
y1=b//10
y2=b%10
num2=0
if y2==2:
y2=5
elif y2==5:
y2=2
if y1==2:
y1=5
elif y1==5:
y1=2
num2=y2*10+y1
if check(y1) and check(y2) and check(x1) and check(x2):
pass
else:
b+=1
continue
if num1<m and num2<h:
p=a
q=b
break
b+=1
if p!=-1:
#print(str(p)+':'+str(q))
ans(p,q)
continue
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read()
```
| 5,256 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
#------------------Important Modules------------------#
from sys import stdin,stdout
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import *
input=stdin.readline
#prin=stdout.write
from random import sample
t=int(input())
#t=1
from collections import Counter,deque
from math import sqrt,ceil,log2,gcd
#dist=[0]*(n+1)
class DisjSet:
def __init__(self, n):
# Constructor to create and
# initialize sets of n items
self.rank = [1] * n
self.parent = [i for i in range(n)]
# Finds set of given item x
def find(self, x):
# Finds the representative of the set
# that x is an element of
if (self.parent[x] != x):
# if x is not the parent of itself
# Then x is not the representative of
# its set,
self.parent[x] = self.find(self.parent[x])
# so we recursively call Find on its parent
# and move i's node directly under the
# representative of this set
return self.parent[x]
# Do union of two sets represented
# by x and y.
def union(self, x, y):
# Find current sets of x and y
xset = self.find(x)
yset = self.find(y)
# If they are already in same set
if xset == yset:
return
# Put smaller ranked item under
# bigger ranked item if ranks are
# different
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
# If ranks are same, then move y under
# x (doesn't matter which one goes where)
# and increment rank of x's tree
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
# Driver code
def f(arr,i,j,d,dist):
if i==j:
return
nn=max(arr[i:j])
for tl in range(i,j):
if arr[tl]==nn:
dist[tl]=d
#print(tl,dist[tl])
f(arr,i,tl,d+1,dist)
f(arr,tl+1,j,d+1,dist)
#return dist
def ps(n):
cp=0
while n%2==0:
n=n//2
cp+=1
for ps in range(3,ceil(sqrt(n))+1,2):
while n%ps==0:
n=n//ps
cp+=1
if n!=1:
return False
return True
#count=0
#dp=[[0 for i in range(m)] for j in range(n)]
#[int(x) for x in input().strip().split()]
def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
# Driver Code
def factorials(n,r):
#This calculates ncr mod 10**9+7
slr=n;dpr=r
qlr=1;qs=1
mod=10**9+7
for ip in range(n-r+1,n+1):
qlr=(qlr*ip)%mod
for ij in range(1,r+1):
qs=(qs*ij)%mod
#print(qlr,qs)
ans=(qlr*modInverse(qs))%mod
return ans
def modInverse(b):
qr=10**9+7
return pow(b, qr - 2,qr)
tt=[xa**3 for xa in range(0,10**4+1)]
qq=set(tt)
def digits(k,rp):
n=len(k)
pq=k[::-1]
jq=''
for ij in pq:
if ij=='1':
jq+='1'
elif ij=='2':
jq+='5'
elif ij=='4' or ij=='7' or ij=='3' or ij=='6' or ij=='9':
jq+='-'
elif ij=='5':
jq+='2'
elif ij=='8':
jq+='8'
elif ij=='0':
jq+='0'
if jq.find('-')!=-1:
return -1
else:
jl=int(jq)
if jl>=rp:
return -1
else:
return jq
def fr(a,b,h,m):
kl=int(b)
kl=(kl+1)%m
b='0'*(2-len(str(kl)))+str(kl)
if b=='0'*2:
kj=(int(a)+1)%h
a='0'*(2-len(str(kj)))+str(kj)
return [a,b]
for jj in range(t):
#n=int(input())
h,m=[int(x) for x in input().strip().split()]
#arr=[int(x) for x in input().strip().split()];
#brr=[int(x) for x in input().strip().split()];brr.sort()
kq=input().strip()
hrs=h;mins=m
while True:
if digits(kq[:2],m)!=-1 and digits(kq[3:],h)!=-1:
print(kq)
break
#print(kq)
klp=fr(kq[:2],kq[3:],h,m)
kq=klp[0]+':'+klp[1]
```
| 5,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
l=['0','1','2','5','8']
def g(x):
if x=='2':
return '5'
if x=='5':
return '2'
return x
def f(x,y,p,q):
s1=str(x)
s2=str(y)
for c in s1:
if c not in l:
return False
for c in s2:
if c not in l:
return False
if len(s1)==1:
s1='0'+s1
if len(s2)==1:
s2='0'+s2
s3=int(g(s2[1])+g(s2[0]))
s4=int(g(s1[1])+g(s1[0]))
#print(s1,s2,s3,s4)
if s3<p and s4<q:
return True
return False
for _ in range(int(input())):
h,m=map(int,input().split())
s=input()
x=int(s[:2])
y=int(s[3:])
am=False
while not am:
while x<h and not am:
while y<m and not am:
if f(x,y,h,m):
am=True
s1=str(x)
s2=str(y)
if len(s1)==1:
s1='0'+s1
if len(s2)==1:
s2='0'+s2
print(s1+':'+s2)
y+=1
if y==m:
y=0
x+=1
if x==h:
x=0
```
| 5,258 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Tags: brute force, implementation
Correct Solution:
```
import math
import operator
def lcm(a,b):
return (a / math.gcd(a,b))* b
def nCr(n, r):
return((math.factorial(n))/((math.factorial(r))*(math.factorial(n - r))))
def isKthBitSet(n, k):
if (n & (1 << (k - 1))):
return True
else:
return False
def maximalRectangle( matrix):
if not matrix or not matrix[0]:
return 0
n = len(matrix[0])
height = [0] * (n + 1)
ans = 0
for row in matrix:
for i in range(n):
height[i] = height[i] + 1 if row[i] == '0' else 0
stack = [-1]
for i in range(n + 1):
while height[i] < height[stack[-1]]:
h = height[stack.pop()]
w = i - 1 - stack[-1]
ans = max(ans, h * w)
stack.append(i)
return ans
def matched(str):
count = 0
for i in str:
if i == "(":
count += 1
elif i == ")":
count -= 1
if count < 0:
return False
return count == 0
def isValid(h,m,nh,nm):
l=[0,1,5,-1,-1,2,-1,-1,8,-1]
if(l[h//10]==-1 or l[h%10]==-1 or l[m//10]==-1 or l[m%10]==-1):
return False
resh= l[m%10]*10 + l[m//10]
resm= l[h%10]*10 + l[h//10]
return (resh<nh and resm<nm)
def solve():
h,m=map(int,input().split())
#n=int(input())
#l=list(map(int,input().split()))
#n2=int(input())
s=input()
#l1=list(ap(int,input().split()))
nh=int(s[0]+s[1])
nm=int(s[3]+s[4])
while(not(isValid(nh,nm,h,m))):
#print(nh,nm)
if(nm==m-1):
nh+=1
nm+=1
nm=nm%m
nh=nh%h
nh=str(nh)
nm=str(nm)
if(len(nh)==1):
nh='0'+nh
if(len(nm)==1):
nm='0'+nm
print(nh+":"+nm)
#print(s)
t=int(input())
while(t>0):
t-=1
solve()
```
| 5,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
for _ in range(int(input())):
h,m = map(int,input().split())
x,y = map(int,input().split(":"))
s = ["0","1","5","-1","-1","2","-1","-1","8","-1"]
i,j = x,y
f = 0
while(i<=h-1):
f = 0
while(j<=m-1):
q,w = str(i),str(j)
if(len(q)==1):
q = "0" + q
if(len(w)==1):
w = "0" + w
e,r = "",""
e += s[int(q[1])]
e += s[int(q[0])]
r += s[int(w[1])]
r += s[int(w[0])]
if("-1" in e or "-1" in r):
j+=1
continue
if(int(r)<=h-1 and int(e)<=m-1):
print(q+":"+w)
f=1
break
j+=1
if(f==1):
break
j = 0
i+=1
if(f==1):
continue
print("00:00")
```
Yes
| 5,260 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
ref = [0, 1, 5, -1, -1, 2, -1, -1, 8, -1]
def solve(h, m, time):
h0 = int(time[:2])
m0 = int(time[3:])
a = [h0//10, h0%10, m0//10, m0%10]
b = list(map(lambda x: ref[x], a))[::-1]
while True:
h1 = 10*b[0]+b[1]
m1 = 10*b[2]+b[3]
if -1 not in b and h1 < h and m1 < m:
return str(a[0]) + str(a[1]) + ":" + str(a[2]) + str(a[3])
m0 += 1
if m0 == m:
m0 = 0
h0 += 1
if h0 == h:
h0 = 0
a = [h0 // 10, h0 % 10, m0 // 10, m0 % 10]
b = list(map(lambda x: ref[x], a))[::-1]
import sys
input = lambda: sys.stdin.readline().rstrip()
t = int(input())
for i in range(t):
h, m = map(int, input().split())
time = input()
print(solve(h, m, time))
```
Yes
| 5,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
change = {0: 0, 1: 1, 2: 5, 5: 2, 8: 8}
bad = (3, 4, 6, 7, 9)
for _ in range(int(input())):
h, m = map(int, input().split())
s = input()
ch = int(s[:2])
cm = int(s[3:])
while True:
if not ((ch%10) in bad or (cm%10) in bad or (ch//10) in bad or (cm//10) in bad):
chm = change[ch%10] * 10 + change[ch//10]
chh = change[cm%10] * 10 + change[cm//10]
if chh < h and chm < m:
if ch < 10:
print('0', end='')
print(ch, end=':')
if cm < 10:
print('0', end='')
print(cm)
break
cm += 1
if cm == m:
cm = 0
ch += 1
if ch == h:
print("00:00")
break
```
Yes
| 5,262 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
t = int(input())
def transform(n):
return n // 10, n % 10
def make(ch, minuts, h, m):
p = {0: 0, 1: 1, 2: 5, 5: 2, 8: 8, 3: 3, 4: 4, 6: 6, 7: 7, 9: 9}
bad = {3, 4, 6, 7, 9}
qc, rc = transform(ch)
qm, rm = transform(minuts)
flag = True
if not bad.intersection({qc, rc, qm, rm}):
if p[rm] * 10 + p[qm] < h and p[rc] * 10 + p[qc] < m:
flag = False
while flag:
minuts += 1
if minuts == m:
minuts = 0
ch += 1
if ch == h:
ch = 0
qc, rc = transform(ch)
qm, rm = transform(minuts)
if not bad.intersection({qc, rc, qm, rm}):
if p[rm] * 10 + p[qm] < h and p[rc] * 10 + p[qc] < m:
flag = False
qc, rc = transform(ch)
qm, rm = transform(minuts)
return str(qc) + str(rc) + ":" + str(qm) + str(rm)
for _ in range(t):
h, m = map(int, input().split())
ch, minuts = map(int, input().split(":"))
print(make(ch, minuts, h, m))
```
Yes
| 5,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
T = int(input())
P = {0, 1, 2, 5, 8}
D = {0:0, 1:1, 2:5, 5:2, 8:8}
for i in range(T):
h, m = map(int, input().split())
H, M = map(int, input().split(":"))
for i in range(H, H + h):
H2 = i % h % 10
H1 = i % h // 10
SH = {H1, H2}
if SH <= P and D[H2] * 10 + D[H1] < m:
HH = str(H1) + str(H2)
break
for j in range(M * (i == H), M + m):
M2 = j % m % 10
M1 = j % m // 10
SM = {M2, M1}
if SM <= P and D[M2] * 10 + D[M1] < h:
MM = str(M1) + str(M2)
break
print(HH, MM, sep=":")
```
No
| 5,264 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
for r in range(int(input())):
h,m=list(map(int,input().split()))
p=input()
# print(h,m,p)
hour=p[0:2]
minute=p[3:5]
def fun(a):
if a==0:
return 0
elif a==1:
return 1
elif a==2:
return 5
elif a==5:
return 2
elif a==8:
return 8
else:
return -1
x=hour;y=minute;turn=0;c=""
while turn!=1:
if fun(int(x[0]))!=-1 and fun(int(x[1]))!=-1 and fun(int(y[0]))!=-1 and fun(int(y[1]))!=-1:
c=x+':'+y
turn=1
break
else:
if int(y)==m:
y='00'
if int(x)==h-1:
x='00'
else:
x=str(int(x)+1)
if len(x)==1:
q=x
x='0'+q
else:
y=str(int(y)+1)
if len(y)==1:
q=y
y='0'+q
print(c)
```
No
| 5,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from sys import stdin,stdout
from io import BytesIO, IOBase
from collections import deque
#sys.setrecursionlimit(10**5)
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# 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")
#-----------------------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
def regularbracket(t):
p=0
for i in t:
if i=="(":
p+=1
else:
p-=1
if p<0:
return False
else:
if p>0:
return False
else:
return True
# endregion
"""
def samesign(a,b):
if (a>0 and b>0) or (a<0 and b<0):
return True
return False
def main():
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int,input().split()))
sum1=0
l=[arr[0]]
for i in range(len(arr)-1):
if samesign(arr[i+1],arr[i])==True:
l.append(arr[i+1])
else:
# print(l)
# print(max(l))
sum1+=max(l)
l=[arr[i+1]]
#print(sum1)
# print(l)
sum1+=max(l)
print(sum1)
"""
"""
def main():
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int,input().split()))
sum1 = sum(arr)
# max1 = max(arr)
arr = sorted(arr)
flag = True
for i in range(1,n+1):
if arr[i-1]>i:
print("second")
flag = False
break
if flag==True:
diff = (n*(n+1))/2-sum1
if diff%2==0:
print("Second")
else:
print("First")
"""
"""
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
seg = [0] * (n + 1)
curr = 0
cnt = 0
for ai in a:
if ai == curr:
cnt += 1
else:
if cnt > 0:
seg[curr] += 1
curr = ai
cnt = 1
if cnt > 0:
seg[curr] += 1
res = n
for i in range(1, n + 1):
if seg[i] == 0:
continue
op = seg[i] + 1
if i == a[0]:
op -= 1
if i == a[-1]:
op -= 1
res = min(res, op)
print((res))
"""
"""
def main():
t = int(input())
for _ in range(t):
a,b = map(int,input().split())
if a==0 or b==0:
print(0)
elif a>=2*b or b>=2*a:
print(min(a,b))
else:
print((a+b)//3)
"""
"""
def main():
t = int(input())
for _ in range(t):
n, m, x, y = map(int, input().split())
ans = 0
y = min(y, 2 * x)
for i_ in range(n):
s = input()
i = 0
while i < m:
if s[i] == '*':
i += 1
continue
j = i
while j + 1 < m and s[j + 1] == '.':
j += 1
l = j - i + 1
ans += l % 2 * x + l // 2 * y
i = j + 1
print(ans)
"""
"""
def main():
t = int(input())
for _ in range(t):
n,x = map(int,input().split())
arr = list(map(int,input().split()))
p=0
sum1 = sum(arr)
arr = sorted(arr)
if sum1//n>=x:
print(n)
else:
for i in range(n-1):
sum1-=arr[i]
if sum1//(n-(i+1))>=x:
print(n-(i+1))
break
else:
print(0)
"""
"""
def main():
p = int(input())
for _ in range(p):
n,k = map(int,input().split())
list1=[]
if n==1 and k==1:
print(0)
else:
for i in range(n,0,-1):
if i+(i-1)<=k:
list1.append(i)
break
else:
list1.append(i)
list1.remove(k)
print(len(list1))
print(*list1)
"""
def mirrortime(s):
s = list(s)
pp=""
for i in range(len(s)):
if s[i]=="0":
continue
elif s[i]=="2":
s[i]="5"
elif s[i]=="5":
s[i]="2"
elif i=="8":
continue
elif i=="1":
continue
for i in s:
pp+=i
return pp
#print(mirrortime("2255"))
def main():
t = int(input())
for _ in range(t):
l=[3,4,6,7,9]
h,m = map(int,input().split())
s = input()
t=s[0:2]
p=s[3:5]
ll=[]
mm=[]
t = int(t)
p = int(p)
# print(t,p)
for i in range(t,h):
for j in range(p,m):
i = str(i)
if len(i)==1:
i="0"+i
j = str(j)
if len(j)==1:
j="0"+j
if int(i[0]) in l or int(i[1]) in l or int(j[0]) in l or int(j[1]) in l:
continue
else:
ll.append(i)
mm.append(j)
# print(ll,mm)
if len(ll)>=1:
for i in range(len(ll)):
cccc = ll[i]
dddd = mm[i]
ccc = mirrortime(cccc)
ddd = mirrortime(dddd)
ccc = list(ccc)
ddd = list(ddd)
ccc.reverse()
ddd.reverse()
ppp=""
qqq=""
for k in ccc:
ppp+=k
for k_ in ddd:
qqq+=k_
if int(qqq)<h and int(ppp)<m:
# print(int(qqq))
# print(int(ppp))
print(cccc+":"+dddd)
break
else:
print("00:00")
else:
print("00:00")
if __name__ == '__main__':
main()
```
No
| 5,266 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1.
<image>
That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position.
A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).
The image of the clocks in the mirror is reflected against a vertical axis.
<image>
The reflection is not a valid time.
<image>
The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time.
An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.
It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.
You are asked to solve the problem for several test cases.
Input
The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.
The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.
The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).
The second line contains the start time s in the described format HH:MM.
Output
For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.
Example
Input
5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04
Output
12:21
00:00
52:28
00:00
00:00
Note
In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct.
<image>
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
def add(ls,h,m):
ls[-1] += 1
if ls[-2]*10+ls[-1] >= m:
ls[-2],ls[-1] = 0,0
else:
if ls[-1] == 10:
ls[-1] = 0
else:
return
ls[-2] += 1
if ls[-2] == 10:
ls[-2] = 0
else:
return
ls[-3] += 1
if ls[-4]*10+ls[-3] >= h:
ls[-4],ls[-3] = 0,0
else:
if ls[-3] == 10:
ls[-3] = 0
else:
return
ls[-4] += 1
if ls[-4] == 10:
ls[-4] = 0
return
def solve():
h,m = map(int,input().split())
s,e = input().split(':')
ref = {0:0,1:1,2:5,5:2,8:8}
ti = [int(s[0]),int(s[1]),int(e[0]),int(e[1])]
while 1:
fl = 0
for i in range(4):
if ref.get(ti[i]) == None:
fl = 1
break
if fl:
add(ti,h,m)
continue
#print(ti[-4]*10+ti[-3],ti[-2]*10+ti[-1],ti)
if ti[-1]*10+ti[-2] < h and ti[-3]*10+ti[-4] < m:
print(str(ti[0])+str(ti[1])+':'+str(ti[2])+str(ti[3]))
return
add(ti,h,m)
def main():
for _ in range(int(input())):
solve()
#Fast IO 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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
```
No
| 5,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix loves playing with bits — specifically, by using the bitwise operations AND, OR, and XOR. He has n integers a_1, a_2, ..., a_n, and will perform q of the following queries:
1. replace all numbers a_i where l ≤ a_i ≤ r with a_i AND x;
2. replace all numbers a_i where l ≤ a_i ≤ r with a_i OR x;
3. replace all numbers a_i where l ≤ a_i ≤ r with a_i XOR x;
4. output how many distinct integers a_i where l ≤ a_i ≤ r.
For each query, Phoenix is given l, r, and x. Note that he is considering the values of the numbers, not their indices.
Input
The first line contains two integers n and q (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ q ≤ 10^5) — the number of integers and the number of queries, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i < 2^{20}) — the integers that Phoenix starts with.
The next q lines contain the queries. For each query, the first integer of each line is t (1 ≤ t ≤ 4) — the type of query.
If t ∈ \{1, 2, 3\}, then three integers l_i, r_i, and x_i will follow (0 ≤ l_i, r_i, x_i < 2^{20}; l_i ≤ r_i).
Otherwise, if t=4, two integers l_i and r_i will follow (0 ≤ l_i ≤ r_i < 2^{20}).
It is guaranteed that there is at least one query where t=4.
Output
Print the answer for each query where t=4.
Examples
Input
5 6
5 4 3 2 1
1 2 3 2
4 2 5
3 2 5 3
4 1 6
2 1 1 8
4 8 10
Output
3
2
1
Input
6 7
6 0 2 3 2 7
1 0 4 3
2 6 8 4
4 0 7
3 2 5 3
1 0 1 2
4 0 3
4 2 7
Output
5
1
2
Note
In the first example:
* For the first query, 2 is replaced by 2 AND 2 = 2 and 3 is replaced with 3 AND 2 = 2. The set of numbers is \{1, 2, 4, 5\}.
* For the second query, there are 3 distinct numbers between 2 and 5: 2, 4, and 5.
* For the third query, 2 is replaced by 2 XOR 3 = 1, 4 is replaced by 4 XOR 3 = 7, and 5 is replaced by 5 XOR 3 = 6. The set of numbers is \{1, 6, 7\}.
* For the fourth query, there are 2 distinct numbers between 1 and 6: 1 and 6.
* For the fifth query, 1 is replaced by 1 OR 8 = 9. The set of numbers is \{6, 7, 9\}.
* For the sixth query, there is one distinct number between 8 and 10: 9.
Submitted Solution:
```
a=20
b=10
print(a+b)
```
No
| 5,268 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
for _ in range(int(input())):
a,b=map(int,input().split())
if(a==b):
print(0,0)
else:
dif=abs(a-b)
mi=min(a,b)
rem=mi%dif
print(dif,min(rem,abs(dif-rem)))
```
| 5,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
import sys
import math
#sys.stdin=open('input.txt','r')
#sys.stdout=open('output.txt','w')
def solve():
#n=int(input())
a,b=map(int,input().split())
if(a==b):
print("0 0")
else:
r=abs(a-b)
if(math.gcd(a,b)==r):
print(r,0)
else:
e=a//r
y1=((e+1)*r)-a
y2=abs(((e)*r)-a)
print(abs(a-b),min(y1,y2))
t=int(input())
while(t!=0):
solve()
t-=1
```
| 5,270 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
from math import *
t=int(input())
for i in range(t):
temp=input().split()
x=int(temp[0]);y=int(temp[1])
gre=abs(x-y)
if(gre!=0):
num = min(x%gre,y%gre,gre-x%gre,gre-x%gre)
else:
num = 0
print(str(gre)+" "+str(num))
```
| 5,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
a,b=map(int,input().split())
x=min(a,b)
y=max(a,b)
if x==y:
print("0 0")
else:
print(y-x, end=" ")
b=y-x
a=x%b
print(min(a, b-a))
```
| 5,272 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
t = int(input())
for _ in range(t):
a, b = map(int,input().split())
if(a == b):
print('0 0')
elif(abs(a-b) == 1):
print('1 0')
elif(a==0 and b!=0):
print(str(b)+" 0")
elif(b==0 and a!=0):
print(str(a)+" 0")
else:
exc = abs(a-b)
if(a < b):
k = 0
while(k < a):
pre = k
k = k + exc
if((a-pre) > (k-a)):
print(str(exc)+" "+str(k-a))
else:
print(str(exc)+" "+str(a-pre))
else:
k = 0
while(k < b):
pre = k
k = k + exc
if((b-pre) > (k-b)):
print(str(exc)+" "+str(k-b))
else:
print(str(exc)+" "+str(b-pre))
```
| 5,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
t=int(input())
for i in range (t):
a,b=map(int,input().strip().split())
if a==b:
print(0,0)
elif abs(a-b)==1:
print(1,0)
else:
gcd=abs(a-b)
minimum=min(a,b)
temp=minimum//gcd
prv=(minimum-(temp*gcd))
nxt=((temp+1)*gcd)-minimum
print(gcd,min(nxt,prv))
```
| 5,274 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
a, b = [int(x) for x in input().split()]
d = abs(a - b)
if d == 0:
print(0, 0)
continue
minus_count = a % d
plus_count = d - minus_count
if min(a, b) - minus_count < 0:
print(d, plus_count)
continue
if minus_count < plus_count:
print(d, minus_count)
else:
print(d, plus_count)
```
| 5,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Tags: greedy, math, number theory
Correct Solution:
```
from math import ceil
for i in range(int(input())):
a,b=map(int,input().split())
if a==b:
print(0,0)
else:
mgcd=max(a,b)-min(a,b)
temp=min(a,b)
x=ceil(temp/mgcd)
z=temp//mgcd
y=(x*mgcd)-temp
y2=temp-(z*mgcd)
ans=min(temp,y,y2)
#print(temp,mgcd,x,y)
print(mgcd,ans)
```
| 5,276 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
import math
d=int(input())
for p in range(0,d):
m=0
n=0
a,b= list(map(int, input().split()))
if a==b:
print(0,0)
continue
x=math.gcd(a,b)
if a>b:
l=(a-b)
else:
l=(b-a)
if x==l:
print(x,0)
continue
q=a%l
w=l-q
print(l,min(q,w))
```
Yes
| 5,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
# link: https://codeforces.com/contest/1543/problem/0
import os, sys, math
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from math import ceil
mod = 10 ** 9 + 7
# number of test cases
for _ in range(int(input())):
a,b = map(int, input().split())
if a==b:
print(0,0)
else:
if a == b-1 or b == a-1:
print(1,0)
else:
aa = a
bb = b
if aa < bb:
bb = bb - aa
aa = 0
else:
aa = aa - bb
bb = 0
first_num = aa
second_num = bb
if first_num == 0: intial_gcd = second_num
else: intial_gcd = first_num
# for first number
if a%intial_gcd == 0 and b%intial_gcd == 0:
print(intial_gcd, 0)
else:
s = 0
e = pow(10, 18)
till = 0
while s <= e:
m = s + (e - s) // 2
if intial_gcd*m >= a:
e = m - 1
till = intial_gcd * m
else:
s = m + 1
m1 = abs(till - a)
s = 0
e = pow(10, 18)
till = 0
while s <= e:
m = s + (e - s) // 2
if m*intial_gcd <= a:
s = m + 1
till = intial_gcd * m
else: e = m - 1
print(intial_gcd, min(abs(till - a), m1))
```
Yes
| 5,278 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
for _ in range(int(input())):
a,b=map(int,input().split())
if a==b:
print(0,0)
else:
print(abs(a-b),min(a%abs(a-b),(max(a,b)-b%abs(b-a))+abs(b-a)-max(a,b)))
```
Yes
| 5,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
import sys
import math
def read_ints():
inp = input().split()
inp = [int(x) for x in inp]
return inp
def read_strings():
inp = input().split()
s = [inp[i] for i in range(len(inp))]
return s
t = int(input())
for _ in range(t):
ab = read_ints()
a = ab[0]
b = ab[1]
maxe = 0
minm = 0
if a == b:
maxe = 0
minm = 0
else:
maxe = abs(a-b)
minm = min(a % maxe, maxe - (a % maxe))
ans = str(maxe) + " " + str(minm)
print(ans)
```
Yes
| 5,280 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
"""
from sys import stdin, stdout
import math
from functools import reduce
import statistics
import numpy as np
import itertools
import operator
from sys import stdin, stdout
import math
from functools import reduce
import statistics
import numpy as np
import itertools
import sys
import operator
from collections import Counter
import decimal
"""
import math
import os
import sys
from math import ceil, floor, sqrt, gcd, factorial
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def prog_name():
a, b = map(int, input().split())
moves = 0
if a == b:
print(0, 0)
else:
if abs(a - b) == 1:
print(1, 0)
else:
g = max(a - b, b - a)
ans = max(a, b) // g
moves = g * (ans + 1)
print(max(a - b, b - a), min(min(a, b), moves - max(a, b)))
def main():
# init = time()
T = int(input())
for unique in range(T):
# print("Case #"+str(unique+1)+":",end = " ")
prog_name()
# print(time() - init)
if __name__ == "__main__":
main()
```
No
| 5,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
import math
import sys
input = sys.stdin.readline
for _ in range(int(input())):
a,b = map(int,input().split())
if a == b:
sys.stdout.write(str(0)+" "+str(0)+"\n")
continue
# if abs(a-b) == 1:
# sys.stdout.write(str(1) + " " + str(0) + "\n")
# continue
# if (a%2 == 0 and b%2 == 1) or (a%2 == 1 and b%2 == 0):
# sys.stdout.write(str() + " " + str(0) + "\n")
# continue
ans = max(a,b)-min(a,b)
sys.stdout.write(str(ans) + " " + str(min(a,b)) + "\n")
```
No
| 5,282 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
import sys
import math
#sys.stdin=open('input.txt','r')
#sys.stdout=open('output.txt','w')
def solve():
#n=int(input())
a,b=map(int,input().split())
if(a==b):
print("0 0")
else:
r=abs(a-b)
z=max(a,b)
if(math.gcd(a,b)==r):
print(r,0)
else:
e=z//r
y1=((e+1)*r)-a
y2=abs(((e)*r)-a)
print(abs(a-b),min(y1,y2))
t=int(input())
while(t!=0):
solve()
t-=1
```
No
| 5,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to Rockport City!
It is time for your first ever race in the game against Ronnie. To make the race interesting, you have bet a dollars and Ronnie has bet b dollars. But the fans seem to be disappointed. The excitement of the fans is given by gcd(a,b), where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y. To make the race more exciting, you can perform two types of operations:
1. Increase both a and b by 1.
2. Decrease both a and b by 1. This operation can only be performed if both a and b are greater than 0.
In one move, you can perform any one of these operations. You can perform arbitrary (possibly zero) number of moves. Determine the maximum excitement the fans can get and the minimum number of moves required to achieve it.
Note that gcd(x,0)=x for any x ≥ 0.
Input
The first line of input contains a single integer t (1≤ t≤ 5⋅ 10^3) — the number of test cases.
The first and the only line of each test case contains two integers a and b (0≤ a, b≤ 10^{18}).
Output
For each test case, print a single line containing two integers.
If the fans can get infinite excitement, print 0 0.
Otherwise, the first integer must be the maximum excitement the fans can get, and the second integer must be the minimum number of moves required to achieve that excitement.
Example
Input
4
8 5
1 2
4 4
3 9
Output
3 1
1 0
0 0
6 3
Note
For the first test case, you can apply the first operation 1 time to get a=9 and b=6. It can be shown that 3 is the maximum excitement possible.
For the second test case, no matter how many operations you apply, the fans will always have an excitement equal to 1. Since the initial excitement is also 1, you don't need to apply any operation.
For the third case, the fans can get infinite excitement by applying the first operation an infinite amount of times.
For the fourth test case, you can apply the second operation 3 times to get a=0 and b=6. Since, gcd(0,6)=6, the fans will get an excitement of 6.
Submitted Solution:
```
# import math
t=int(input())
for i in range(t):
a,b=map(int,input().split())
d=min(a,b)
if a==b:
print("0 0")
elif a==0 or b==0:
print(max(a,b),1)
elif a==1 or b==1:
print("1 0")
else:
print(max(a,b)-d,(max(a,b)-d)//2)
```
No
| 5,284 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
a=int(input())
print(a*(a-1)*6+1)
```
| 5,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
a = int(input())
res = 1
for i in range(2, a + 1):
res += (i - 1) * 12
print(res)
```
| 5,286 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
# ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥
n = int(input())
a = 1
s = 12
for i in range(n-1):
a += s
s += 12
print(a)
# ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
```
| 5,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
def star(n):
f = 1
while n:
n -=1
f += n*12
return f
print(star(int(input())))
```
| 5,288 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
#In the name of Allah
from sys import stdin, stdout
input = stdin.readline
a = int(input())
ans = 1
for i in range(1, a):
ans += 6 * (2 * i)
stdout.write(str(ans))
```
| 5,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
n = int(input().split()[0])
print((2 * n - 1) * (2 * n - 1) + 4 * (n - 1) * n // 2)
```
| 5,290 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
n = int(input())
print(n * (n - 1) * 6 + 1)
```
| 5,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Tags: *special, combinatorics
Correct Solution:
```
n = int(input())
b = (6*n*(n - 1)) + 1
print(b)
```
| 5,292 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# 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")
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 fsep(): return map(float, inp().split())
def inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x, n, a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4, n, a)
luckynumwithequalnumberoffourandseven(x * 10 + 7, n, a)
return a
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
#------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
#------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
#-----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
#-----------------------------------------------lucky number with two lucky any digits
res = set()
def solve(p, l, a, b,n):#given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solve(p * 10 + a, l + 1, a, b,n)
solve(p * 10 + b, l + 1, a, b,n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
#-----------------------------------------------
# endregion------------------------------
"""
def main():
n = inpu()
cnt=0
c = n
if n % 7 == 0:
print("7" * (n // 7))
else:
while(c>0):
c-=4
cnt+=1
if c%7==0 and c>=0:
#print(n,n%4)
print("4"*(cnt)+"7"*(c//7))
break
else:
if n % 4 == 0:
print("4" * (n // 4))
else:
print(-1)
if __name__ == '__main__':
main()
"""
"""
def main():
n,t = sep()
arr = lis()
i=0
cnt=0
min1 = min(arr)
while(True):
if t>=arr[i]:
cnt+=1
t-=arr[i]
i+=1
else:
i+=1
if i==n:
i=0
if t<min1:
break
print(cnt)
if __name__ == '__main__':
main()
"""
# Python3 program to find all subsets
# by backtracking.
# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index,d):
print(*subset)
s = sum(subset)
d.append(s)
for i in range(index, len(A)):
# include the A[i] in subset.
subset.append(A[i])
# move onto the next element.
subsetsUtil(A, subset, i + 1,d)
# exclude the A[i] from subset and
# triggers backtracking.
subset.pop(-1)
return d
def subsetSums(arr, l, r,d, sum=0):
if l > r:
d.append(sum)
return
subsetSums(arr, l + 1, r,d, sum + arr[l])
# Subset excluding arr[l]
subsetSums(arr, l + 1, r,d, sum)
return d
"""
def main():
t = inpu()
for _ in range(t):
n = inpu()
arr=[]
subset=[]
i=0
l=[]
for j in range(26):
arr.append(3**j)
if __name__ == '__main__':
main()
"""
def main():
n = int(input())
cnt=1
if n==1:
print(1)
else:
for i in range(1,n):
cnt+=i*12
print(cnt)
if __name__ == '__main__':
main()
```
Yes
| 5,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
#star noumber problem
# 1 13....
n=int(input())
k=6*n;k1=n-1;ans=k*k1
print(ans+1)
```
Yes
| 5,294 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
n=int(input())
a=3*n-2
b=n-1
print(int((a*(a+1))/2 + ((b*(b+1))/2)*3))
```
Yes
| 5,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
n = int(input())
print((2*n-2)*n*3+1)
```
Yes
| 5,296 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
n = int(input())
z = [1,12]
for i in range(2,n):
z.append(z[-1]*2)
s = 0
if n==1:
s = 1
elif n==4:
s = 73
elif n==5:
s = 121
else:
s = sum(z)
print(s)
```
No
| 5,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
_ = input()
print(13)
```
No
| 5,298 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
__author__ = 'MARI'
def main():
a = int(input())
print(1 + 6*(a*(a+1)//2-1))
if __name__ == '__main__':
import sys
argv = sys.argv
main()
```
No
| 5,299 |
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