message stringlengths 2 20.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 757 108k | cluster float64 4 4 | __index_level_0__ int64 1.51k 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts 1 minute.
He was asked to insert one takeoff in the schedule. The takeoff takes 1 minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least s minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 60) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next n lines contains two integers h and m (0 ≤ h ≤ 23, 0 ≤ m ≤ 59) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is 0 0). These times are given in increasing order.
Output
Print two integers h and m — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Examples
Input
6 60
0 0
1 20
3 21
5 0
19 30
23 40
Output
6 1
Input
16 50
0 30
1 20
3 0
4 30
6 10
7 50
9 30
11 10
12 50
14 30
16 10
17 50
19 30
21 10
22 50
23 59
Output
24 50
Input
3 17
0 30
1 0
12 0
Output
0 0
Note
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than 24 hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing.
Submitted Solution:
```
n, s = map(int, input().split())
times = []
result = 0
for i in range (n):
h, m = map(int, input().split())
times.append(60*h + m)
for i in range(n-1):
if times[i+1] - times[i] >= 2 + 2*s:
result = times[i] + 1 + s
if result == 0:
result = times[n-1] + 1 + s
hour = result // 60
minute = result % 60
print(hour, minute)
``` | instruction | 0 | 64,565 | 4 | 129,130 |
No | output | 1 | 64,565 | 4 | 129,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts 1 minute.
He was asked to insert one takeoff in the schedule. The takeoff takes 1 minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least s minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 60) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next n lines contains two integers h and m (0 ≤ h ≤ 23, 0 ≤ m ≤ 59) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is 0 0). These times are given in increasing order.
Output
Print two integers h and m — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Examples
Input
6 60
0 0
1 20
3 21
5 0
19 30
23 40
Output
6 1
Input
16 50
0 30
1 20
3 0
4 30
6 10
7 50
9 30
11 10
12 50
14 30
16 10
17 50
19 30
21 10
22 50
23 59
Output
24 50
Input
3 17
0 30
1 0
12 0
Output
0 0
Note
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than 24 hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing.
Submitted Solution:
```
x=[input().split(' ')]
n=int(x[0][0])
s=int(x[0][1])
k=0
sum=0
while n+1:
x=[input().split()]
if (int(x[0][0])*60+int(x[0][1]))!=0:
sum=(int(x[0][0])*60+int(x[0][1]))-sum
k=(int(x[0][0])*60+int(x[0][1]))
print(sum)
if sum>=(2*s+2):
y=(int(x[0][0])*60+int(x[0][1]))+61
c=y%60
d=y//60
print(d,c)
break
``` | instruction | 0 | 64,566 | 4 | 129,132 |
No | output | 1 | 64,566 | 4 | 129,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts 1 minute.
He was asked to insert one takeoff in the schedule. The takeoff takes 1 minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least s minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 60) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next n lines contains two integers h and m (0 ≤ h ≤ 23, 0 ≤ m ≤ 59) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is 0 0). These times are given in increasing order.
Output
Print two integers h and m — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Examples
Input
6 60
0 0
1 20
3 21
5 0
19 30
23 40
Output
6 1
Input
16 50
0 30
1 20
3 0
4 30
6 10
7 50
9 30
11 10
12 50
14 30
16 10
17 50
19 30
21 10
22 50
23 59
Output
24 50
Input
3 17
0 30
1 0
12 0
Output
0 0
Note
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than 24 hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing.
Submitted Solution:
```
n, k = [int(i) for i in input().split()]
h = [0 - (k // 60)]
m = [-(k % 60) - 1]
for i in range(n):
x, y = input().split()
h.append(int(x))
m.append(int(y))
ansh = -1
ansm = -1
for i in range(1, n):
t = (h[i] - h[i - 1]) * 60 - m[i - 1] + m[i]
if t > 2 * k + 1:
ansh = h[i - 1]
ansm = m[i - 1] + k + 1
if ansm >= 60:
ansm -= 60
ansh += 1
break
if ansh == -1:
ansh = h[n]
ansm = m[n] + k + 1
if ansm >= 60:
ansm -= 60
ansh += 1
print(ansh, ansm)
``` | instruction | 0 | 64,567 | 4 | 129,134 |
No | output | 1 | 64,567 | 4 | 129,135 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,674 | 4 | 129,348 |
"Correct Solution:
```
from fractions import gcd
N = int(input())
ts = [int(input()) for i in range(N)]
lcm = 1
for t in ts:
lcm = lcm * t // gcd(lcm,t)
print(lcm)
``` | output | 1 | 64,674 | 4 | 129,349 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,675 | 4 | 129,350 |
"Correct Solution:
```
from fractions import gcd
N = int(input())
T = [int(input()) for i in range(N)]
l = 1
for t in T:
l = l * t // gcd(l,t)
print(l)
``` | output | 1 | 64,675 | 4 | 129,351 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,676 | 4 | 129,352 |
"Correct Solution:
```
def gcd(a,b):
if b==0:return a
return gcd(b,a%b)
ans=1
for i in range(int(input())):
h=int(input())
ans = ans*h // gcd(ans,h)
print(ans)
``` | output | 1 | 64,676 | 4 | 129,353 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,677 | 4 | 129,354 |
"Correct Solution:
```
from math import gcd
n=int(input())
t=[int(input())for i in range(n)]
ans=1
for i in t:ans*=i//(gcd(ans,i))
print(ans)
``` | output | 1 | 64,677 | 4 | 129,355 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,678 | 4 | 129,356 |
"Correct Solution:
```
#from math import gcd
from fractions import gcd
n=int(input())
res=1
for _ in range(n):
x=int(input())
res=res//gcd(res,x)*x
print(res)
``` | output | 1 | 64,678 | 4 | 129,357 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,679 | 4 | 129,358 |
"Correct Solution:
```
from math import gcd
n = int(input())
ans = int(input())
for _ in range(n - 1):
x = int(input())
ans = (ans * x) // gcd(ans, x)
print(ans)
``` | output | 1 | 64,679 | 4 | 129,359 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,680 | 4 | 129,360 |
"Correct Solution:
```
from math import*
N=int(input())
ans=int(input())
for _ in range(N-1):
t=int(input())
ans=ans*t//gcd(ans,t)
print(ans)
``` | output | 1 | 64,680 | 4 | 129,361 |
Provide a correct Python 3 solution for this coding contest problem.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000 | instruction | 0 | 64,681 | 4 | 129,362 |
"Correct Solution:
```
import math
n=int(input())
t=[int(input()) for i in range(n)]
res=t[0]
for i in range(1,n):
res=res*t[i]//math.gcd(res,t[i])
print(res)
``` | output | 1 | 64,681 | 4 | 129,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import fractions
n = int(input())
x = [int(input()) for i in range(n)]
a = 1
for i in x:
a = a*i//fractions.gcd(a,i)
print(a)
``` | instruction | 0 | 64,682 | 4 | 129,364 |
Yes | output | 1 | 64,682 | 4 | 129,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import fractions
A = int(input())
ans = int(input())
for i in range(A-1):
B = int(input())
ans = B * ans // fractions.gcd(ans,B)
print(ans)
``` | instruction | 0 | 64,683 | 4 | 129,366 |
Yes | output | 1 | 64,683 | 4 | 129,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import math
n=int(input());a=[int(input()) for _ in range(n)];r=a[0]
for i in a:
r=r*i//math.gcd(r,i)
print(r)
``` | instruction | 0 | 64,684 | 4 | 129,368 |
Yes | output | 1 | 64,684 | 4 | 129,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import fractions as f
n=int(input())
ans=int(input())
for i in range(n-1):
x=int(input())
ans=x*ans//f.gcd(ans,x)
print(ans)
``` | instruction | 0 | 64,685 | 4 | 129,370 |
Yes | output | 1 | 64,685 | 4 | 129,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import sys
sys.setrecursionlimit = 10000000
def get_gcd(A, B):
r = A%B
if r == 0:
return B
else:
return get_gcd(B, r)
if __name__ == '__main__':
N = int(input())
T = [int(input()) for _ in range(N)]
ans = 1
for n in range(N):
gcd = get_gcd(ans, T[n])
ans = int(T[n]/gcd)*T[n]
print(ans)
``` | instruction | 0 | 64,686 | 4 | 129,372 |
No | output | 1 | 64,686 | 4 | 129,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import math
N=int(input())
list=[]
for i in range(N):
a=int(input())
list.append(a)
X=[x for x in set(list)]
if 1 in X:
X.pop(1)
ans=int((X[0]*X[1])/(math.gcd(X[0],X[1])))
if len(X)==1:
print(X[0])
else:
for i in range(1,len(X)-1):
ans=int((X[i]*ans)/(math.gcd(X[i],ans)))
print(ans)
``` | instruction | 0 | 64,687 | 4 | 129,374 |
No | output | 1 | 64,687 | 4 | 129,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
from fractions import gcd
n=int(input())
ans=int(input())
for i in range(1,n):
t=int(input())
ans=int(ans*t/gcd(ans,t))
print(ans)
``` | instruction | 0 | 64,688 | 4 | 129,376 |
No | output | 1 | 64,688 | 4 | 129,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?
Constraints
* 1≤N≤100
* 1≤T_i≤10^{18}
* All input values are integers.
* The correct answer is at most 10^{18} seconds.
Input
Input is given from Standard Input in the following format:
N
T_1
:
T_N
Output
Print the number of seconds after which the hand of every clock point directly upward again.
Examples
Input
2
2
3
Output
6
Input
5
2
5
10
1000000000000000000
1000000000000000000
Output
1000000000000000000
Submitted Solution:
```
import math
def lcm(x, y):
return int(x * y) // math.gcd(x, y))
n = int(input())
t = [i for i in range(n)]
for i in range(n):
t[i] = int(input())
ans = t[0]
for i in range(1, n):
ans = lcm(ans, t[i])
print(int(ans))
``` | instruction | 0 | 64,689 | 4 | 129,378 |
No | output | 1 | 64,689 | 4 | 129,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed k hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.
Vasya has a train lighting schedule for all n hours of the trip — n numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose k hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours.
Input
The first input line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains n space-separated integers ai (0 ≤ ai ≤ 100), ai is the light level at the i-th hour.
Output
In the first output line print the minimum light level Vasya will read at. In the second line print k distinct space-separated integers b1, b2, ..., bk, — the indexes of hours Vasya will read at (1 ≤ bi ≤ n). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers bi in an arbitrary order.
Examples
Input
5 3
20 10 30 40 10
Output
20
1 3 4
Input
6 5
90 20 35 40 60 100
Output
35
1 3 4 5 6
Note
In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20.
Submitted Solution:
```
# -*- coding: utf-8 -*-
# @Date : 2019-01-23 14:06:35
# @Author : raj lath (oorja.halt@gmail.com)
# @Link : link
# @Version : 1.0.0
from sys import stdin
max_val=int(10e12)
min_val=int(-10e12)
def read_int() : return int(stdin.readline())
def read_ints() : return [int(x) for x in stdin.readline().split()]
def read_str() : return input()
def read_strs() : return [x for x in stdin.readline().split()]
F = open('input.txt')
W = open("output.txt", "w")
I = lambda:[int(x) for x in F.readline().split()]
nb_hours , read_hours = I()
lights = I()
light_sorted = sorted(set(lights), reverse=True)[:read_hours]
ans = []
for i in light_sorted:
ans.append(lights.index(i)+1)
print(light_sorted[-1], file = W)
print(*sorted(ans), file = W)
``` | instruction | 0 | 65,867 | 4 | 131,734 |
No | output | 1 | 65,867 | 4 | 131,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed k hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.
Vasya has a train lighting schedule for all n hours of the trip — n numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose k hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours.
Input
The first input line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains n space-separated integers ai (0 ≤ ai ≤ 100), ai is the light level at the i-th hour.
Output
In the first output line print the minimum light level Vasya will read at. In the second line print k distinct space-separated integers b1, b2, ..., bk, — the indexes of hours Vasya will read at (1 ≤ bi ≤ n). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers bi in an arbitrary order.
Examples
Input
5 3
20 10 30 40 10
Output
20
1 3 4
Input
6 5
90 20 35 40 60 100
Output
35
1 3 4 5 6
Note
In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20.
Submitted Solution:
```
with open("input.txt") as f:
lines = f.readlines()
n, k = map(int, lines[0].split())
a = list(map(int, lines[1].split()))
b = [i for i in a]
b.sort()
m = b[n - k]
h = []
for i in range(n):
if a[i] >= m:
h.append(i + 1)
with open("output.txt", "w") as f:
f.write("{}\n".format(m))
for i in h:
f.write("{} ".format(i))
``` | instruction | 0 | 65,868 | 4 | 131,736 |
No | output | 1 | 65,868 | 4 | 131,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed k hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.
Vasya has a train lighting schedule for all n hours of the trip — n numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose k hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours.
Input
The first input line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains n space-separated integers ai (0 ≤ ai ≤ 100), ai is the light level at the i-th hour.
Output
In the first output line print the minimum light level Vasya will read at. In the second line print k distinct space-separated integers b1, b2, ..., bk, — the indexes of hours Vasya will read at (1 ≤ bi ≤ n). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers bi in an arbitrary order.
Examples
Input
5 3
20 10 30 40 10
Output
20
1 3 4
Input
6 5
90 20 35 40 60 100
Output
35
1 3 4 5 6
Note
In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20.
Submitted Solution:
```
import sys
sys.stdin=open("input.txt","r")
sys.stdout=open("output.txt","w")
n, k = map(int,input().split())
arr = list(map(int,input().split()))
mini = sorted(arr)[-k]
s = []
for i in range(n):
if len(s)<=k and arr[i]>=mini:
s.append(i+1)
print(mini)
print(' '.join(map(str,s)))
``` | instruction | 0 | 65,869 | 4 | 131,738 |
No | output | 1 | 65,869 | 4 | 131,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed k hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.
Vasya has a train lighting schedule for all n hours of the trip — n numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose k hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours.
Input
The first input line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains n space-separated integers ai (0 ≤ ai ≤ 100), ai is the light level at the i-th hour.
Output
In the first output line print the minimum light level Vasya will read at. In the second line print k distinct space-separated integers b1, b2, ..., bk, — the indexes of hours Vasya will read at (1 ≤ bi ≤ n). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers bi in an arbitrary order.
Examples
Input
5 3
20 10 30 40 10
Output
20
1 3 4
Input
6 5
90 20 35 40 60 100
Output
35
1 3 4 5 6
Note
In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20.
Submitted Solution:
```
import sys
sys.stdin = open(r'input.txt', 'r')
sys.stdout = open(r'output.txt', 'w')
n,k=map(int, input().split())
a=list(map(int,input().split()))
b=sorted(a)
b = b[n-k:]
cnt = 0
for i in range(len(a)):
if cnt == len(b):
break
if a[i] in b:
cnt+=1
print(i+1, end = " ")
``` | instruction | 0 | 65,870 | 4 | 131,740 |
No | output | 1 | 65,870 | 4 | 131,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,049 | 4 | 132,098 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
from itertools import permutations
n, m = tuple(map(int, input().split()))
def baseN(num,b,numerals="0123456789abcdefghijklmnopqrstuvwxyz"):
return ((num == 0) and numerals[0]) or (baseN(num // b, b, numerals).lstrip(numerals[0]) + numerals[num % b])
def lt_eq(a, b):
if len(a) == 0 and len(b) == 0:
return True
elif a[0] > b[0]:
return False
elif a[0] < b[0]:
return True
else:
return lt_eq(a[1:], b[1:])
nmax = tuple(map(int, baseN(n - 1, 7)))
mmax = tuple(map(int, baseN(m - 1, 7)))
s = len(nmax) + len(mmax)
ans = 0
if s > 7:
print(0)
else:
for c in permutations(range(7), s):
left, right = c[:len(nmax)], c[len(nmax):]
if lt_eq(left, nmax) and lt_eq(right, mmax):
ans += 1
print(ans)
``` | output | 1 | 66,049 | 4 | 132,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,050 | 4 | 132,100 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
import math
from itertools import permutations
from itertools import product
a=input().split()
n=int(a[0])
m=int(a[1])
def log(x, y):
if x<1: return 0
else: return math.log(x, y)
n1=math.floor(log(n-1, 7))+1
m1=math.floor(log(m-1, 7))+1
if n1+m1>7: print (0)
else:
def xxx(y):
y2=math.floor(log(y, 7))+1
z=['0' for i in range(y2)]
for i in range(y2):
y1=math.floor(log(y, 7))
if y1==y2-i-1:
z[i]=str(y//(7**y1))
y-=(7**y1)*int(z[i])
return z
n7=xxx(n-1)
m7=xxx(m-1)
n7=int(''.join(n7))
m7=int(''.join(m7))
vb=set()
if n1+m1==7:
for i in permutations(['0','1','2','3','4','5','6'], 7):
if int(''.join(i[:n1]))>n7:
pass
elif int(''.join(i[n1:]))>m7:
pass
else:
b=True
for j in ['0', '1', '2', '3', '4', '5', '6']:
if i.count(j)>1:
b=False
break
if b:
vb.add(i)
print(len(vb))
else:
for i in product(permutations(['0','1','2','3','4','5','6'], n1), permutations(['0','1','2','3','4','5','6'], m1)):
if int(''.join(i[0]))>n7:
pass
elif int(''.join(i[1]))>m7:
pass
else:
b=True
for j in ['0', '1', '2', '3', '4', '5', '6']:
if i[0].count(j)+i[1].count(j)>1:
b=False
break
if b:
vb.add(i)
print (len(vb))
``` | output | 1 | 66,050 | 4 | 132,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,051 | 4 | 132,102 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
def db(num):
base_num = ""
while num>0:
dig = int(num%7)
base_num += str(dig)
num //= 7
base_num = base_num[::-1]
if(base_num == ""):
base_num = "0"
return base_num
def dbb(num, xx):
s = [0]*xx
j = xx-1
while num>0:
dig = int(num%7)
s[j] = dig
j-=1
num //= 7
# base_num = base_num[::-1]
return s.copy()
t = 1
for _ in range(t):
n, m = mapin()
ans =0
nn = db(n-1)
mm = db(m-1)
x = len(str(nn))
y = len(str(mm))
# print(x, y)
if(x+y>7):
print(0)
else:
for i in range(n):
a = dbb(i, x)
for j in range(m):
b = dbb(j, y)
c = a+b
# print(a, bB)
# print(i, j, a, b)
if(len(set(c)) == len(c)):
ans+=1
print(ans)
``` | output | 1 | 66,051 | 4 | 132,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,052 | 4 | 132,104 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
BASE = 7
class Solution(object):
def __init__(self,m,n):
self.max_hour = self.itov(n)
self.max_min = self.itov(m)
self.used = used = [False] * BASE
def itov(self,x):
digits = []
if x == 0:
digits.append(0)
while x > 0:
digits.append(x % BASE)
x //= BASE
digits.reverse()
return digits
def gen(self,pos = 0, minute = False, smaller = False):
max_val = self.max_min if minute else self.max_hour
if pos >= len(max_val):
if minute:
return 1
else:
return self.gen(0, True)
else:
ans = 0
for digit in range(BASE):
if not self.used[digit] and (smaller or digit <= max_val[pos]):
self.used[digit] = True
ans += self.gen(pos + 1, minute, smaller or digit < max_val[pos])
self.used[digit] = False
return ans
def main():
n, m = map(int, input().split())
n -= 1
m -= 1
s = Solution(m,n)
print(s.gen())
if __name__ == '__main__':
main()
``` | output | 1 | 66,052 | 4 | 132,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,053 | 4 | 132,106 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
n, m = [int(x) for x in input().split()]
x = 7
d1 = 1
while x<n:
d1 += 1
x *= 7
x = 7
d2 = 1
while x<m:
d2 += 1
x *= 7
if d1 + d2 > 7:
print(0)
exit()
c1 = {}
L = [0]*d1
L[0] = -1
for sadas in range(n):
L[0] += 1
for j in range(d1):
if(L[j] == 7):
L[j] = 0
L[j+1] += 1
g = [0]*7
for j in range(7):
g[j] = len([x for x in L if x==j])
if max(g)<2:
if tuple(g) not in c1:
c1[tuple(g)] = 0
c1[tuple(g)] += 1
c2 = {}
L = [0]*d2
L[0] = -1
for sadas in range(m):
L[0] += 1
for j in range(d2):
if(L[j] == 7):
L[j] = 0
L[j+1] += 1
g = [0]*7
for j in range(7):
g[j] = len([x for x in L if x==j])
if max(g)<2:
if tuple(g) not in c2:
c2[tuple(g)] = 0
c2[tuple(g)] += 1
ans = 0
for x in c1:
for y in c2:
if len([z for z in range(7) if x[z]+y[z]>1])==0:
ans += c1[x]*c2[y]
print(ans)
``` | output | 1 | 66,053 | 4 | 132,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,054 | 4 | 132,108 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
from itertools import permutations
from math import ceil, log
n, m = t = list(map(int, input().split()))
l, _ = t = [ceil(log(x, 7.)) if x > 1 else 1 for x in t]
print(sum(int(s[:l], 7) < n and int(s[l:], 7) < m for s in map(''.join, permutations("0123456", sum(t)))))
# Made By Mostafa_Khaled
``` | output | 1 | 66,054 | 4 | 132,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,055 | 4 | 132,110 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
a,b = map(int,input().split())
def getdig(x):
if(x==0):
return 0,1
s = ''
c = 0
while x>0:
s = str(x%7)+s
c += 1
x //= 7
return int(s), c
ass,an = getdig(a-1)
bss,bn = getdig(b-1)
tn = an+bn
if(tn>7):
print(0)
exit()
count = 0
sor = [str(i) for i in range(7)]
def make(s,list,depth):
global count
for x in list:
new = list[:]
new.remove(x)
make(s+x,new,depth+1)
if(depth==tn):
aaa,bbb = s[:an],s[an:]
if(int(aaa)<=ass)and(int(bbb)<=bss):
count += 1
make('',sor,0)
print(count)
``` | output | 1 | 66,055 | 4 | 132,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1). | instruction | 0 | 66,056 | 4 | 132,112 |
Tags: brute force, combinatorics, dp, math
Correct Solution:
```
from itertools import permutations
n, m = map(int, input().split())
def l(x):
r = x == 0
while x:
r += 1
x //= 7
return r
ans, ln, lm = 0, l(n - 1), l(m - 1)
for s in permutations('0123456', ln + lm):
s = ''.join(s)
ans += int(s[:ln], 7) < n and int(s[ln:], 7) < m
print(ans)
``` | output | 1 | 66,056 | 4 | 132,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
from itertools import permutations
base = 7
digits = [x for x in range(base)]
def get_count(n):
answer = 1
while (n >= base):
n = n // base
answer += 1
return answer
def is_good(tup, n):
a = 0
for x in reversed(tup):
a *= base
a += x
if a >= n:
return False
else:
return True
n, m = map(int, input().split())
h_count = get_count(n - 1)
m_count = get_count(m - 1)
answer = 0
for perm in permutations(digits, h_count + m_count):
if is_good(perm[:h_count], n) and is_good(perm[h_count:], m):
answer += 1
print(answer)
``` | instruction | 0 | 66,057 | 4 | 132,114 |
Yes | output | 1 | 66,057 | 4 | 132,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
n, m = map(int, input().split())
def GetDigitsNumberIn7System(number):
result = 1
x = 7
while (True):
if (number <= x - 1):
return result
else:
result += 1
x *= 7
def fillZeros(s, razryadCount):
for i in range(razryadCount - len(s)):
s = "0" + s
return s
def HasTheSameDigits(s):
digits = set()
for letter in s:
digits.add(letter)
if len(digits) == len(s):
return False
else:
return True
def int2base(x, base):
if x < 0:
sign = -1
elif x == 0:
return '0'
else:
sign = 1
x *= sign
res = ''
while x != 0:
res += str(x % base)
x //= base
return res[::-1]
def GetNumberOfVariations(hoursInDay, minutesInHour):
n1 = GetDigitsNumberIn7System(hoursInDay-1)
n2 = GetDigitsNumberIn7System(minutesInHour-1)
count = 0
listHours = []
listMinutes = []
for i in range(hoursInDay):
if (HasTheSameDigits(fillZeros(int2base(i,7),n1)) == False):
listHours.append(i)
for i in range(minutesInHour):
if (HasTheSameDigits(fillZeros(int2base(i, 7), n2)) == False):
listMinutes.append(i)
for i in listHours:
for j in listMinutes:
str1 = fillZeros(int2base(i,7),n1)
str2 = fillZeros(int2base(j,7),n2)
strc = str1+str2
if (HasTheSameDigits(strc) == False):
count+=1
return count
r1 = GetDigitsNumberIn7System(n-1)
r2 = GetDigitsNumberIn7System(m-1)
if r1 + r2 > 7:
print(0)
else:
res = GetNumberOfVariations(n,m)
print (res)
``` | instruction | 0 | 66,058 | 4 | 132,116 |
Yes | output | 1 | 66,058 | 4 | 132,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
BASE = 7
def itov(x):
digits = []
if x == 0:
digits.append(0)
while x > 0:
digits.append(x % BASE)
x //= BASE
digits.reverse()
return digits
def gen(pos = 0, minute = False, smaller = False):
max_val = max_minute if minute else max_hour
if pos >= len(max_val):
if minute:
return 1
else:
return gen(0, True)
else:
ans = 0
for digit in range(BASE):
if not used[digit] and (smaller or digit <= max_val[pos]):
used[digit] = True
ans += gen(pos + 1, minute, smaller or digit < max_val[pos])
used[digit] = False
return ans
n, m = map(int, input().split())
n -= 1
m -= 1
used = [False] * BASE
max_hour = itov(n)
max_minute = itov(m)
print(gen())
``` | instruction | 0 | 66,059 | 4 | 132,118 |
Yes | output | 1 | 66,059 | 4 | 132,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
n,m=map(int,input().split())
d=[pow(7,i) for i in range(12)]
ans=0
nn,mm,kn,km=n-1,m-1,0,0
while nn: kn+=1; nn//=7
while mm: km+=1; mm//=7
kn+=kn==0
km+=km==0
nn,mm=[0]*kn,[0]*km
def hm(r1,r2,s1,s2):
global ans
if s1>=n : return
if r1==kn:
if s2>=m : return
if r2==km: ans+=1; return
for i in range(0,7):
mm[r2]=i
if i in nn: continue
if len(set(mm[:r2+1]))<r2+1: continue
hm(r1,r2+1,s1,s2+i*d[r2])
return
for i in range(0,7):
nn[r1]=i
if len(set(nn[:r1+1]))<r1+1: continue
hm(r1+1,r2,s1+i*d[r1],s2)
#if kn+km<8:
hm(0,0,0,0)
print(ans)
``` | instruction | 0 | 66,060 | 4 | 132,120 |
Yes | output | 1 | 66,060 | 4 | 132,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
n,m=map(int,input().split())
d=[pow(7,i) for i in range(12)]
ans=0
nn,mm,kn,km=n,m,0,0
while nn: kn+=1; nn//=7
while mm: km+=1; mm//=7
nn,mm=[0]*kn,[0]*km
def mi(r,s):
global ans
if s>=m: return
if r==km: ans+=1; return
for i in range(1,7):
mm[r]=i
if len(set(mm[:r+1]))<r+1: continue
mi(r+1,s+i*d[r])
def hi(r,s):
global ans
if s>=n: return
if r==kn: ans+=1; return
for i in range(1,7):
nn[r]=i
if len(set(nn[:r+1]))<r+1: continue
hi(r+1,s+i*d[r])
def hm(r1,r2,s1,s2):
global ans
if s1>=n : return
if r1==kn:
if s2>=m : return
if r2==km: ans+=1; return
for i in range(0,7):
mm[r2]=i
if len(set(mm[:r2+1])|set(nn))<kn+r2+1: continue
hm(r1,r2+1,s1,s2+i*d[r2])
return
for i in range(0,7):
nn[r1]=i
if len(set(nn[:r1+1]))<r1+1: continue
hm(r1+1,r2,s1+i*d[r1],s2)
#if km>1: hi(0,0)
#if kn>1: mi(0,0)
if kn+km<8: hm(0,0,0,0)
print(ans)
``` | instruction | 0 | 66,061 | 4 | 132,122 |
No | output | 1 | 66,061 | 4 | 132,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
import math
def to7(x):
ans = 0
k = 1
i = 0
while(x > 0):
d = divmod(x, 7)
ans = ans + d[1]*k
k = k * 10
x = d[0]
i = i + 1
return (ans, i)
def func(step, nnow, mnow, n, m, ln, lm, fig):
if step == ln:
print(nnow, mnow, (nnow < n ) and (mnow < m))
if (nnow < n ) and (mnow < m):
return 1
return 0
k = 10**(ln - step-1)
ans = 0
for i in range(7):
if fig[i] == False:
if nnow + k*i <= n:
fig[i] = True
if ln - lm <= step:
for j in range(7):
if fig[j] == False:
if mnow + k*j <= m:
fig[j] = True
ans = ans + func(step+1, nnow + k*i, mnow + k*j, n, m, ln, lm, fig)
fig[j] = False
else:
ans = ans + func(step+1, nnow + k*i, mnow, n, m, ln, lm, fig)
fig[i] = False
return ans
(n,m) = (int(i) for i in input().split())
start = time.time()
fig = [ False for i in range(7)]
if (m > n):
(n, m) = (m, n)
(n, ln) = to7(n)
(m, lm) = to7(m)
print(n, ln, m, lm)
if (ln+lm > 7):
ans = 0
else:
ans = func(0, 0, 0, n, m, ln, lm, fig)
print(ans)
finish = time.time()
#print(finish - start)
``` | instruction | 0 | 66,062 | 4 | 132,124 |
No | output | 1 | 66,062 | 4 | 132,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
from math import factorial as f
def b8(n):
if n == 0:
return '0'
nums = []
while n:
n, r = divmod(n, 7)
nums.append(str(r))
return ''.join(reversed(nums))
n, m = map(int, input().split())
n8 = b8(n-1)
m8 = b8(m-1)
def isvalid(t, n8, m8):
if len(t) <= len(n8):
return t <= n8[:len(t)]
else:
return (t[len(n8):] < m8) or ((t[:len(n8)] < n8) and (t[len(n8):] <= m8))
def bf(t, n8, m8, i, unused):
if i == len(n8+m8):
#print(t)
c = 1
else:
c = 0
for ui in unused:
tt = t + ui
if isvalid(tt, n8, m8):
c += bf(tt, n8, m8, i+1, unused - {ui})
return c
if len(n8 + m8) > 7:
sol = 0
else:
sol = bf('', n8, m8, 0, set([str(i) for i in range(7)]))
print(sol)
``` | instruction | 0 | 66,063 | 4 | 132,126 |
No | output | 1 | 66,063 | 4 | 132,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
Input
The first line of the input contains two integers, given in the decimal notation, n and m (1 ≤ n, m ≤ 109) — the number of hours in one day and the number of minutes in one hour, respectively.
Output
Print one integer in decimal notation — the number of different pairs of hour and minute, such that all digits displayed on the watches are distinct.
Examples
Input
2 3
Output
4
Input
8 2
Output
5
Note
In the first sample, possible pairs are: (0: 1), (0: 2), (1: 0), (1: 2).
In the second sample, possible pairs are: (02: 1), (03: 1), (04: 1), (05: 1), (06: 1).
Submitted Solution:
```
import math
def comb(lst, n, k, used):
if len(lst) == n:
yield lst
else:
for i in range(k):
if not used[i]:
lst.append(i)
used[i] = True
yield from comb(lst, n, k, used)
used[i] = False
lst.pop()
n, m = map(int, input().split())
yach_n = max(math.ceil(math.log(n, 7)), 1)
yach_m = max(math.ceil(math.log(m, 7)), 1)
yach = yach_n + yach_m
amount = 0
for lst in comb([], yach, 7, [False] * 7):
curr_hours = lst[0]
j = 1
while j < yach_n:
curr_hours *= 7
curr_hours += lst[j]
j += 1
curr_minutes = lst[yach_n]
j = yach_n + 1
while j < yach:
curr_minutes *= 7
curr_hours += lst[j]
j += 1
if curr_hours < n and curr_minutes < m:
amount += 1
# print(lst)
if curr_hours >= n:
break
print(amount)
``` | instruction | 0 | 66,064 | 4 | 132,128 |
No | output | 1 | 66,064 | 4 | 132,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,130 | 4 | 132,260 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = input()
inS = 0
inF = 0
for i in range(n-1):
if a[i] == 'S' and a[i+1] == 'F':
inF += 1
elif a[i] == 'F' and a[i+1] == 'S':
inS += 1
if inF > inS:
print('YES')
else:
print('NO')
``` | output | 1 | 66,130 | 4 | 132,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,132 | 4 | 132,264 |
Tags: implementation
Correct Solution:
```
n=int(input())
x=input()
c=0
d=0
for i in range(0,n-1):
if x[i]=='S' and x[i+1]=='F':
c=c+1
elif x[i]=='F' and x[i+1]=='S':
d=d+1
else:
i=i+1
if c>d:
print("Yes")
else:
print("No")
``` | output | 1 | 66,132 | 4 | 132,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,133 | 4 | 132,266 |
Tags: implementation
Correct Solution:
```
n = int(input())
string = input()
prev = string[0]
StoF = 0
FtoS = 0
for i in range(1,len(string)):
if string[i] != prev:
if prev == "S":
StoF += 1
else:
FtoS += 1
prev = string[i]
if StoF > FtoS:
print("YES")
else:
print("NO")
``` | output | 1 | 66,133 | 4 | 132,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,134 | 4 | 132,268 |
Tags: implementation
Correct Solution:
```
if __name__ == "__main__":
n = int(input())
st = input()
prev = st[0]
more, less = 0, 0
for i in range(1, n):
if prev == "S" and st[i] == "F":
more += 1
elif prev == "F" and st[i] == "S":
less += 1
prev = st[i]
if more > less:
print("Yes")
else:
print("No")
``` | output | 1 | 66,134 | 4 | 132,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,135 | 4 | 132,270 |
Tags: implementation
Correct Solution:
```
n=int(input())
s=input()
for i in range(n):
if s[0]=="S" and s[n-1:n]=="F":
print("YES")
break
elif s[0]=="F" or (s[0]=="S" and s[n-1:n]=="S"):
print("NO")
break
``` | output | 1 | 66,135 | 4 | 132,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,136 | 4 | 132,272 |
Tags: implementation
Correct Solution:
```
a,b = int(input()),input()
print('YES' if b.count('SF')> b.count('FS') else 'NO')
``` | output | 1 | 66,136 | 4 | 132,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | instruction | 0 | 66,137 | 4 | 132,274 |
Tags: implementation
Correct Solution:
```
n = int(input())
string = input()
a = 0
b = 0
for i in range(n - 1):
if string[i] != string[i + 1]:
if string[i] == "S":
a += 1
else:
b += 1
if a > b:
print("YES")
else:
print("NO")
``` | output | 1 | 66,137 | 4 | 132,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
Submitted Solution:
```
n = int(input())
city = list(input())
count = 0
for i in range(1,n):
if city[i-1] == 'S' and city[i] == 'F':
count += 1
if city[i-1] == 'F' and city[i] == 'S':
count -= 1
if count >= 1:
print('YES')
else:
print("NO")
``` | instruction | 0 | 66,138 | 4 | 132,276 |
Yes | output | 1 | 66,138 | 4 | 132,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
Submitted Solution:
```
a=int(input())
s=input()
if s.count("SF")>s.count("FS"):
print("YES")
else:
print("NO")
``` | instruction | 0 | 66,139 | 4 | 132,278 |
Yes | output | 1 | 66,139 | 4 | 132,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
Submitted Solution:
```
from sys import stdin
n = int(stdin.readline())
days = stdin.readline()
more = 0
for i in range(0, n-1):
if days[i] != days[i+1]:
if days[i+1] == "F":
more += 1
else:
more -= 1
if (more > 0):
print("YES")
else:
print("NO")
``` | instruction | 0 | 66,140 | 4 | 132,280 |
Yes | output | 1 | 66,140 | 4 | 132,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last n days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last n days, or not.
Input
The first line of input contains single integer n (2 ≤ n ≤ 100) — the number of days.
The second line contains a string of length n consisting of only capital 'S' and 'F' letters. If the i-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
4
FSSF
Output
NO
Input
2
SF
Output
YES
Input
10
FFFFFFFFFF
Output
NO
Input
10
SSFFSFFSFF
Output
YES
Note
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though.
Submitted Solution:
```
#data = input().split("\n")
n = input()
sequence = [char for char in input()]
f_s = 0
s_f = 0
for i in range(len(sequence)-1):
if sequence[i]=='S' and sequence[i+1]=='F':
s_f +=1
elif sequence[i]=='F' and sequence[i+1]=='S':
f_s += 1
if(s_f>f_s):
print("YES")
else:
print("NO")
``` | instruction | 0 | 66,141 | 4 | 132,282 |
Yes | output | 1 | 66,141 | 4 | 132,283 |
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