message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Doodle Jump the aim is to guide a four-legged creature called "The Doodler" up a never-ending series of platforms without falling. — Wikipedia.
It is a very popular game and xiaodao likes it very much. One day when playing the game she wondered whether there exists a platform that the doodler couldn't reach due to the limits of its jumping ability. Consider the following problem.
There are n platforms. The height of the x-th (1 ≤ x ≤ n) platform is a·x mod p, where a and p are positive co-prime integers. The maximum possible height of a Doodler's jump is h. That is, it can jump from height h1 to height h2 (h1 < h2) if h2 - h1 ≤ h. Initially, the Doodler is on the ground, the height of which is 0. The question is whether it can reach the highest platform or not.
For example, when a = 7, n = 4, p = 12, h = 2, the heights of the platforms are 7, 2, 9, 4 as in the picture below. With the first jump the Doodler can jump to the platform at height 2, with the second one the Doodler can jump to the platform at height 4, but then it can't jump to any of the higher platforms. So, it can't reach the highest platform.
<image>
User xiaodao thought about the problem for a long time but didn't solve it, so she asks you for help. Also, she has a lot of instances of the problem. Your task is solve all of these instances.
Input
The first line contains an integer t (1 ≤ t ≤ 104) — the number of problem instances. Each of the next t lines contains four integers a, n, p and h (1 ≤ a ≤ 109, 1 ≤ n < p ≤ 109, 0 ≤ h ≤ 109). It's guaranteed that a and p are co-prime.
Output
For each problem instance, if the Doodler can reach the highest platform, output "YES", otherwise output "NO".
Examples
Input
3
7 4 12 2
7 1 9 4
7 4 12 3
Output
NO
NO
YES
Submitted Solution:
```
def nerf():
for i in range(1,n+1):
if h<(a*i)%p:
return "NO"
return "YES"
for j in range(int(input())):
a,n,p,h=[int(i) for i in input().split()]
print(nerf())
``` | instruction | 0 | 318 | 8 | 636 |
No | output | 1 | 318 | 8 | 637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Doodle Jump the aim is to guide a four-legged creature called "The Doodler" up a never-ending series of platforms without falling. — Wikipedia.
It is a very popular game and xiaodao likes it very much. One day when playing the game she wondered whether there exists a platform that the doodler couldn't reach due to the limits of its jumping ability. Consider the following problem.
There are n platforms. The height of the x-th (1 ≤ x ≤ n) platform is a·x mod p, where a and p are positive co-prime integers. The maximum possible height of a Doodler's jump is h. That is, it can jump from height h1 to height h2 (h1 < h2) if h2 - h1 ≤ h. Initially, the Doodler is on the ground, the height of which is 0. The question is whether it can reach the highest platform or not.
For example, when a = 7, n = 4, p = 12, h = 2, the heights of the platforms are 7, 2, 9, 4 as in the picture below. With the first jump the Doodler can jump to the platform at height 2, with the second one the Doodler can jump to the platform at height 4, but then it can't jump to any of the higher platforms. So, it can't reach the highest platform.
<image>
User xiaodao thought about the problem for a long time but didn't solve it, so she asks you for help. Also, she has a lot of instances of the problem. Your task is solve all of these instances.
Input
The first line contains an integer t (1 ≤ t ≤ 104) — the number of problem instances. Each of the next t lines contains four integers a, n, p and h (1 ≤ a ≤ 109, 1 ≤ n < p ≤ 109, 0 ≤ h ≤ 109). It's guaranteed that a and p are co-prime.
Output
For each problem instance, if the Doodler can reach the highest platform, output "YES", otherwise output "NO".
Examples
Input
3
7 4 12 2
7 1 9 4
7 4 12 3
Output
NO
NO
YES
Submitted Solution:
```
n = int(input())
for j in range(n):
WR = []
f = True
a = list(map(int, input().split(' ')))
for i in range(1,a[1]+1):
WR.append(a[0]*i%a[2])
WR = sorted(WR)
#print(WR)
#======================================#
if len(WR)>1:
for i in range(len(WR)-1):
if WR[i+1]-WR[i]>a[3]:
f = False
break
elif len(WR) == 1:
if WR[0]>a[3]:
f = False
if f == False:
print('NO')
else:
print('YES')
``` | instruction | 0 | 319 | 8 | 638 |
No | output | 1 | 319 | 8 | 639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Doodle Jump the aim is to guide a four-legged creature called "The Doodler" up a never-ending series of platforms without falling. — Wikipedia.
It is a very popular game and xiaodao likes it very much. One day when playing the game she wondered whether there exists a platform that the doodler couldn't reach due to the limits of its jumping ability. Consider the following problem.
There are n platforms. The height of the x-th (1 ≤ x ≤ n) platform is a·x mod p, where a and p are positive co-prime integers. The maximum possible height of a Doodler's jump is h. That is, it can jump from height h1 to height h2 (h1 < h2) if h2 - h1 ≤ h. Initially, the Doodler is on the ground, the height of which is 0. The question is whether it can reach the highest platform or not.
For example, when a = 7, n = 4, p = 12, h = 2, the heights of the platforms are 7, 2, 9, 4 as in the picture below. With the first jump the Doodler can jump to the platform at height 2, with the second one the Doodler can jump to the platform at height 4, but then it can't jump to any of the higher platforms. So, it can't reach the highest platform.
<image>
User xiaodao thought about the problem for a long time but didn't solve it, so she asks you for help. Also, she has a lot of instances of the problem. Your task is solve all of these instances.
Input
The first line contains an integer t (1 ≤ t ≤ 104) — the number of problem instances. Each of the next t lines contains four integers a, n, p and h (1 ≤ a ≤ 109, 1 ≤ n < p ≤ 109, 0 ≤ h ≤ 109). It's guaranteed that a and p are co-prime.
Output
For each problem instance, if the Doodler can reach the highest platform, output "YES", otherwise output "NO".
Examples
Input
3
7 4 12 2
7 1 9 4
7 4 12 3
Output
NO
NO
YES
Submitted Solution:
```
t = int(input())
j = 0
l = 0
flag = True
while(j<t):
a,n,p,h = input().split()
a=int(a)
n=int(n)
p=int(p)
h=int(h)
if(h==0):
flag=False
else:
if(n==1):
x = a*1%p
if(x>h):
flag=False
else:
flag=True
else:
mas = [int((a*(i+1))%p) for i in range(n)]
mas = sorted(mas)
while (l<n-1):
if(mas[l+1]-mas[l]>h):
flag = False
break
l+=1
l=0
if(flag == False):
print('NO')
else:
print('YES')
flag=True
j+=1
``` | instruction | 0 | 320 | 8 | 640 |
No | output | 1 | 320 | 8 | 641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Doodle Jump the aim is to guide a four-legged creature called "The Doodler" up a never-ending series of platforms without falling. — Wikipedia.
It is a very popular game and xiaodao likes it very much. One day when playing the game she wondered whether there exists a platform that the doodler couldn't reach due to the limits of its jumping ability. Consider the following problem.
There are n platforms. The height of the x-th (1 ≤ x ≤ n) platform is a·x mod p, where a and p are positive co-prime integers. The maximum possible height of a Doodler's jump is h. That is, it can jump from height h1 to height h2 (h1 < h2) if h2 - h1 ≤ h. Initially, the Doodler is on the ground, the height of which is 0. The question is whether it can reach the highest platform or not.
For example, when a = 7, n = 4, p = 12, h = 2, the heights of the platforms are 7, 2, 9, 4 as in the picture below. With the first jump the Doodler can jump to the platform at height 2, with the second one the Doodler can jump to the platform at height 4, but then it can't jump to any of the higher platforms. So, it can't reach the highest platform.
<image>
User xiaodao thought about the problem for a long time but didn't solve it, so she asks you for help. Also, she has a lot of instances of the problem. Your task is solve all of these instances.
Input
The first line contains an integer t (1 ≤ t ≤ 104) — the number of problem instances. Each of the next t lines contains four integers a, n, p and h (1 ≤ a ≤ 109, 1 ≤ n < p ≤ 109, 0 ≤ h ≤ 109). It's guaranteed that a and p are co-prime.
Output
For each problem instance, if the Doodler can reach the highest platform, output "YES", otherwise output "NO".
Examples
Input
3
7 4 12 2
7 1 9 4
7 4 12 3
Output
NO
NO
YES
Submitted Solution:
```
def main(a,n,p,h):
heights = []
for x in range(1, n % p + 1):
heights.append(a*x % p)
heights.sort()
if 0 not in heights:
heights = [0]+heights
for i in range(1, n+1):
d = heights[i]-heights[i-1]
if d>h:
return 'NO'
return 'YES'
def init():
t = int(input())
s=''
for i in range(t):
a,n,p,h = list(map(int, input().split()))
if h==0 or (h==1 and n%p!=0):
s+='NO\n'
else:
s+=main(a,n,p,h)+'\n'
print(s[:-1])
init()
``` | instruction | 0 | 321 | 8 | 642 |
No | output | 1 | 321 | 8 | 643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 426 | 8 | 852 |
Tags: implementation, math
Correct Solution:
```
#!/usr/bin/env python3
N, S = map(int, input().split())
stuff = [map(int, input().split()) for i in range(N)]
answer = S
for a, k in stuff:
answer = max(answer, a + k)
print(answer)
``` | output | 1 | 426 | 8 | 853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 427 | 8 | 854 |
Tags: implementation, math
Correct Solution:
```
n, m = map(int, input().split())
fl = []
ps = []
fl.append(0)
ps.append(0)
for i in range(0, n):
a, b = map(int, input().split())
fl.append(a)
ps.append(b)
lv = m
time = 0
for i in range(n, -1, -1):
time+=lv-fl[i]
if ps[i]>time:
time+=ps[i]-time
lv = fl[i]
print(time)
``` | output | 1 | 427 | 8 | 855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 428 | 8 | 856 |
Tags: implementation, math
Correct Solution:
```
import sys
import math
import collections
from pprint import pprint
mod = 1000000007
def vector(size, val=0):
vec = [val for i in range(size)]
return vec
def matrix(rowNum, colNum, val=0):
mat = []
for i in range(rowNum):
collumn = [val for j in range(colNum)]
mat.append(collumn)
return mat
n, s = map(int, input().split())
a = []
for i in range(n):
temp = list(map(int, input().split()))
a.append(temp)
a.sort(reverse=True)
t = 0
for x in a:
t += s - x[0]
s = x[0]
t += max(0, x[1] - t)
if s != 0:
t += s
print(t)
``` | output | 1 | 428 | 8 | 857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 429 | 8 | 858 |
Tags: implementation, math
Correct Solution:
```
n, s = map(int, input().split())
l = []
for i in range(n):
l.append(tuple(map(int, input().split())))
res = 0
for p, t in l[::-1]:
res += s - p
s = p
if t - res > 0:
res += t - res
res += l[0][0]
print(res)
``` | output | 1 | 429 | 8 | 859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 430 | 8 | 860 |
Tags: implementation, math
Correct Solution:
```
import sys
import math
from functools import reduce
import bisect
def getN():
return int(input())
def getNM():
return map(int, input().split())
def getList():
return list(map(int, input().split()))
def input():
return sys.stdin.readline().rstrip()
# input = sys.stdin.buffer.readline
def index(a, x):
i = bisect.bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
return -1
#############
# MAIN CODE #
#############
n, s = getNM()
pasengers = []
for i in range(n):
pasengers.append(list(getNM()))
pasengers.sort(reverse=True)
ans = s
t = 0
for a, b in pasengers:
t += s - a
ans += b - t if b - t > 0 else 0
t += b - t if b - t > 0 else 0
s = a
print(ans)
``` | output | 1 | 430 | 8 | 861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 431 | 8 | 862 |
Tags: implementation, math
Correct Solution:
```
n, s = list(map(int, input().split()))
lp = list()
for i in range(n):
lp.append(tuple(map(int, input().split())))
lp.sort(reverse=True)
time = 0
current_floor = s
for i in range(n):
predicted_time = time + current_floor-lp[i][0]
passenger_atime = lp[i][1]
if passenger_atime > predicted_time:
time = passenger_atime
else:
time = predicted_time
current_floor = lp[i][0]
time += current_floor
print(time)
``` | output | 1 | 431 | 8 | 863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 432 | 8 | 864 |
Tags: implementation, math
Correct Solution:
```
if __name__ == '__main__':
n, s = map(int, input().split())
arr = [0 for _ in range(s + 1)]
for _ in range(n):
f, t = map(int, input().split())
arr[f] = max(arr[f], t)
arr.reverse()
cnt = arr[0]
for x in arr[1:]:
cnt = max(x, cnt + 1)
print(cnt)
``` | output | 1 | 432 | 8 | 865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | instruction | 0 | 433 | 8 | 866 |
Tags: implementation, math
Correct Solution:
```
n, m = map(int, input().split())
ans = m
for x in range(n):
a, b = map(int, input().split())
ans = max(ans, a+b)
print(ans)
``` | output | 1 | 433 | 8 | 867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n,s=map(int,input().split())
ip=[]
count=0
for i in range(n):
a,b=map(int,input().split())
ip.append((a,b))
ip=sorted(ip,key=lambda l:l[0], reverse=True)
for i in range(n):
a,b=ip[i]
count+=s-a
if count<b:
count+=(b-count)
s=a
count+=a
print(count)
``` | instruction | 0 | 434 | 8 | 868 |
Yes | output | 1 | 434 | 8 | 869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n, s = map(int, input().split())
ans = s
for _ in range(n):
f, t = map(int, input().split())
ans = max(ans, f + t)
print(ans)
``` | instruction | 0 | 435 | 8 | 870 |
Yes | output | 1 | 435 | 8 | 871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n=list(map(int,input().split()))
lst=[]
for i in range(n[0]):
n1=list(map(int,input().split()))
lst.append(n1)
lst.sort()
lst.reverse()
w=n[1]-lst[0][0]
for j in range(len(lst)):
if(j==0):
if(w<lst[j][1]):
w=w+(lst[j][1]-w)
else:
w=w+(lst[j-1][0]-lst[j][0])
if(w<lst[j][1]):
w=w+(lst[j][1]-w)
m=lst[-1][0]-0
print(w+m)
``` | instruction | 0 | 436 | 8 | 872 |
Yes | output | 1 | 436 | 8 | 873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n, up = map(int, input().split())
res = 0
for i in range(n):
fl, t = map(int, input().split())
res = max(res, max(t, up - fl) + fl)
print(res)
``` | instruction | 0 | 437 | 8 | 874 |
Yes | output | 1 | 437 | 8 | 875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n,s = map(int,input().split())
a = []
z = []
d = {}
for i in range(n):
f,t = map(int,input().split())
d[f] = t
z.append(f)
s1 = 0
z.sort(reverse=True)
k = 0
for i in range(s,-1,-1):
if i in z:
if d[z[k]] > s1:
s1 = d[z[k]]
else:
s1 = s1 + 1
k = k + 1
else:
s1 = s1 + 1
print(s1)
``` | instruction | 0 | 438 | 8 | 876 |
No | output | 1 | 438 | 8 | 877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n,m=map(int,input().split())
ls=[]
for i in range(n) :
f,t=map(int,input().split())
tp=f,t
ls.append(tp)
time=0
ls.sort()
ls=ls[::-1]
time=m-ls[0][0]
m=ls[0][1]
for i in ls:
time=time+m-i[0]
if time>=i[1]:
pass
else:
time=i[1]
m=i[0]
time=time+ls[n-1][0]
print(time)
``` | instruction | 0 | 439 | 8 | 878 |
No | output | 1 | 439 | 8 | 879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n, s = map(int, input().split())
data = []
elapsed = 0
latest_floor = s
for i in range(n):
f, t = map(int, input().split())
data.append([f,t])
data.sort(reverse=True)
#print(data)
for pair in data:
elapsed += (latest_floor - pair[0]) # 2
if pair[1] <= elapsed: # y
latest_floor = pair[0] # 5
continue
else:
elapsed += (pair[1]-elapsed)
print(elapsed)
``` | instruction | 0 | 440 | 8 | 880 |
No | output | 1 | 440 | 8 | 881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input
The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.
The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.
Output
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.
Submitted Solution:
```
n, s = [int(x) for x in input().split()]
lst = []
maxi =0
for i in range(n):
f, t = [int(x) for x in input().split()]
if t>maxi:
maxi = t
m1 = f
lst.append([f,t])
print(m1+maxi)
``` | instruction | 0 | 441 | 8 | 882 |
No | output | 1 | 441 | 8 | 883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,212 | 8 | 2,424 |
Tags: greedy, implementation
Correct Solution:
```
l1 = [int(x) for x in input().split()]
n,k = l1[0],l1[1]
# 1 2 3 4 5 6 7 8
#
l2 = list(input())
if k > (n+1)/2:
i=k-1
while i<n-1:
print("RIGHT")
i+=1
l2.reverse()
i=0
while i<n:
print("PRINT",l2[i])
if i!=n-1:print("LEFT")
i+=1
else:
i = k-1
while i>0:
print("LEFT")
i-=1
i=0
while i<n:
print("PRINT",l2[i])
if i!=n-1:print("RIGHT")
i+=1
``` | output | 1 | 1,212 | 8 | 2,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,213 | 8 | 2,426 |
Tags: greedy, implementation
Correct Solution:
```
n,k = map(int,input().split())
s = input()
if n//2 >= k:
for i in range(k-1):
print("LEFT")
for i in range(n-1):
print("PRINT",s[i])
print("RIGHT")
print("PRINT",s[n-1])
else:
for i in range(n-k):
print("RIGHT")
for i in range(1,n):
print("PRINT",s[-i])
print("LEFT")
print("PRINT",s[0])
``` | output | 1 | 1,213 | 8 | 2,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,214 | 8 | 2,428 |
Tags: greedy, implementation
Correct Solution:
```
def dir_print(word, dir):
for i in range(len(word)):
print("PRINT", word[i])
if i < len(word)-1 and dir != "":
print(dir)
n, k = map(int, input().split())
word = input().strip()
if n == 1:
dir_print(word, "")
elif k == n:
dir_print(word[::-1], "LEFT")
elif k == 1:
dir_print(word, "RIGHT")
else:
r = n-k
l = k - 1
if l < r:
for i in range(l):
print("LEFT")
dir_print(word, "RIGHT")
else:
for i in range(r):
print("RIGHT")
dir_print(word[::-1], "LEFT")
``` | output | 1 | 1,214 | 8 | 2,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,215 | 8 | 2,430 |
Tags: greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
line = input()
if k <= n // 2:
for i in range(k - 1):
print('LEFT')
print('PRINT', line[0])
for i in range(n - 1):
print('RIGHT')
print('PRINT', line[i + 1])
else:
for i in range(n - k):
print('RIGHT')
print('PRINT', line[-1])
for i in range(n - 1):
print('LEFT')
print('PRINT', line[n - i - 2])
``` | output | 1 | 1,215 | 8 | 2,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,216 | 8 | 2,432 |
Tags: greedy, implementation
Correct Solution:
```
n,k=map(int,input().split())
s=input()
if(n<2*k):
print("RIGHT\n"*(n-k))
l=len(s)
for i in range(l-1,0,-1):
print("PRINT", s[i])
print("LEFT")
print("PRINT", s[0])
else:
print("LEFT\n"*(k-1))
l=len(s)
for i in range(0,l-1):
print("PRINT",s[i])
print("RIGHT")
print("PRINT", s[l-1])
``` | output | 1 | 1,216 | 8 | 2,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,217 | 8 | 2,434 |
Tags: greedy, implementation
Correct Solution:
```
n,k = map(int,input().split())
s = list(input())
if n-k < k-1:
for i in range(k-1,n-1):
print("RIGHT")
i = n-1
while i > 0:
print("{} {}".format("PRINT",s[i]))
print("{}".format("LEFT"))
i-=1
print("{} {}".format("PRINT",s[i]))
else:
i = k-1
while i > 0 :
print("LEFT")
i-=1
for i in range(0,n-1):
print("{} {}".format("PRINT", s[i]))
print("{}".format("RIGHT"))
print("{} {}".format("PRINT", s[n-1]))
``` | output | 1 | 1,217 | 8 | 2,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,218 | 8 | 2,436 |
Tags: greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
s = input()
if n//2 < k:
for _ in range(n-k):
print('RIGHT')
for i in range(1, n):
print('PRINT', s[-i])
print('LEFT')
print('PRINT', s[0])
else:
for _ in range(k-1):
print('LEFT')
for i in range(n-1):
print('PRINT', s[i])
print('RIGHT')
print('PRINT', s[-1])
``` | output | 1 | 1,218 | 8 | 2,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | instruction | 0 | 1,219 | 8 | 2,438 |
Tags: greedy, implementation
Correct Solution:
```
n, k=input().split(" ")
n=int(n)
k=int(k)
s=input()
if k>n//2:
i=k-1
while i<n-1:
print("RIGHT")
i+=1
for i in reversed(range(1, n)):
print("PRINT {}".format(s[i]))
print("LEFT")
print("PRINT {}".format(s[0]))
else:
i=k-1
while i>0:
print("LEFT")
i-=1
for i in range(n-1):
print("PRINT {}".format(s[i]))
print("RIGHT")
print("PRINT {}".format(s[n-1]))
``` | output | 1 | 1,219 | 8 | 2,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
l, k = list(map(int, input().split()))
phrase = input()
def print_sw(word: str) -> str:
arr = ["PRINT " + i for i in word]
return "\nRIGHT\n".join(arr)
def print_bw(word: str) -> str:
arr = ["PRINT " + i for i in word[::-1]]
return "\nLEFT\n".join(arr)
if k > l / 2:
while k != l:
print("RIGHT")
k += 1
print(print_bw(phrase))
else:
while k != 1:
print("LEFT")
k -= 1
print(print_sw(phrase))
``` | instruction | 0 | 1,220 | 8 | 2,440 |
Yes | output | 1 | 1,220 | 8 | 2,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
[n, k] = [int(i) for i in input().split()]
s = input()
if (n-k > k-1):
i = 0
d = 1
for j in range(k-1):
print('LEFT')
else:
i = n-1
d = -1
for j in range(n-k):
print('RIGHT')
for j in range(n-1):
print('PRINT', s[i])
print('LEFT' if d<0 else 'RIGHT')
i += d
print('PRINT', s[i])
``` | instruction | 0 | 1,221 | 8 | 2,442 |
Yes | output | 1 | 1,221 | 8 | 2,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
a = list(map(int,input().split()))
b = input()
if a[1] -1 < a[0] - a[1]:
for x in range(0,a[1]-1):
print("LEFT")
for x in range(0,a[0]):
print("PRINT " + b[x])
if x != a[0] - 1:
print("RIGHT")
else:
for x in range(0,a[0] - a[1]):
print("RIGHT")
for x in range(1,a[0] + 1):
print("PRINT " + b[-x])
if x != a[0]:
print("LEFT")
``` | instruction | 0 | 1,222 | 8 | 2,444 |
Yes | output | 1 | 1,222 | 8 | 2,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
no_sq , pos = map(int,input().split())
slogan = input()
left = pos - 1
right = no_sq - pos
if left > right:
print('RIGHT\n'*(right))
flag = 0
for i in slogan[::-1]:
print('PRINT',i)
if flag == no_sq-1:
break
print('LEFT')
flag += 1
else:
print('LEFT\n'*(left))
flag = 0
for i in slogan:
print('PRINT', i)
if flag == no_sq - 1:
break
print('RIGHT')
flag += 1
``` | instruction | 0 | 1,223 | 8 | 2,446 |
Yes | output | 1 | 1,223 | 8 | 2,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
n, k = map(int, input().split())
S = input()
k -= 1
if k >= (n - 1) / 2:
print('PRINT', S[k])
for i in range(k+1, n):
print('RIGHT')
print('PRINT', S[i])
for i in range(k+1, n):
print('LEFT')
for i in range(k-1, -1, -1):
print('LEFT')
print('PRINT', S[i])
else:
print('RRINT', S[k])
for i in range(k-1, -1, -1):
print('LEFT')
print('PRINT', S[i])
for i in range(k-1, -1, -1):
print('LEFT')
for i in range(k+1, n):
print('RIGHT')
print('PRINT', S[i])
``` | instruction | 0 | 1,224 | 8 | 2,448 |
No | output | 1 | 1,224 | 8 | 2,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(input())
n, k = n-1, k-1
if k == n:
a = a[::-1]
for i in range(len(a) - 1):
print('PRINT', a[i])
print('LEFT')
print('PRINT', a[-1])
elif k == 0:
for i in range(len(a) - 1):
print('PRINT', a[i])
print('RIGHT')
print('PRINT', a[-1])
elif k < n/2:
for i in range(k):
print('LEFT')
for i in range(len(a) - 1):
print('PRINT', a[i])
print('RIGHT')
print('PRINT', a[-1])
else:
for i in range(n - k):
print('RIGHT')
a = a[::-1]
for i in range(len(a)):
print('PRINT', a[i])
print('LEFT')
print('PRINT', a[-1])
``` | instruction | 0 | 1,225 | 8 | 2,450 |
No | output | 1 | 1,225 | 8 | 2,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
n,k = list(map(int, input().split(" ")))
s=input()
print("PRINT "+s[k-1])
if k>n//2:
for i in range(k,n):
print("RIGHT")
print("PRINT "+s[i])
for i in range(k-2,-1,-1):
print("LEFT")
print("PRINT "+s[i])
else:
for i in range(0,k-1):
print("LEFT")
print("PRINT " + s[i])
for i in range(k,n):
print("RIGHT")
print("PRINT " + s[i])
``` | instruction | 0 | 1,226 | 8 | 2,452 |
No | output | 1 | 1,226 | 8 | 2,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of n characters, so the decorators hung a large banner, n meters wide and 1 meter high, divided into n equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the k-th square of the poster. To draw the i-th character of the slogan on the poster, you need to climb the ladder, standing in front of the i-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the i-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input
The first line contains two integers, n and k (1 ≤ k ≤ n ≤ 100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as n characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output
In t lines, print the actions the programmers need to make. In the i-th line print:
* "LEFT" (without the quotes), if the i-th action was "move the ladder to the left";
* "RIGHT" (without the quotes), if the i-th action was "move the ladder to the right";
* "PRINT x" (without the quotes), if the i-th action was to "go up the ladder, paint character x, go down the ladder".
The painting time (variable t) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Examples
Input
2 2
R1
Output
PRINT 1
LEFT
PRINT R
Input
2 1
R1
Output
PRINT R
RIGHT
PRINT 1
Input
6 4
GO?GO!
Output
RIGHT
RIGHT
PRINT !
LEFT
PRINT O
LEFT
PRINT G
LEFT
PRINT ?
LEFT
PRINT O
LEFT
PRINT G
Note
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
Submitted Solution:
```
n , k = map(int, input().split(' '))
slogan = input()
def begin():
print('PRINT ' + slogan[0])
for i in range(1, len(slogan)):
print('RIGHT')
print('PRINT ' + slogan[i])
def end():
print('PRINT ' + slogan[-1])
for i in range(2, len(slogan)+1):
print('LEFT')
print('PRINT ' + slogan[-i])
mid = n //2
if k == 1:
begin()
elif k == n:
end()
elif k > mid:
diff = n-k
for i in range(diff):
print('RIGHT')
end()
elif k < mid:
diff = mid - k
for i in range(diff):
print('LEFT')
begin()
elif mid == k:
for i in range(mid -1):
print('LEFT')
begin()
``` | instruction | 0 | 1,227 | 8 | 2,454 |
No | output | 1 | 1,227 | 8 | 2,455 |
Provide a correct Python 3 solution for this coding contest problem.
The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.
In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.
Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.
Input
The input is a sequence of datasets. Each dataset is given in the following format:
n
x1 y1 dir1
...
xn yn dirn
n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .
It is guaranteed that no two futons in the input overlap each other.
The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.
Output
For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.
Example
Input
4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0
Output
Yes
No
Yes | instruction | 0 | 1,628 | 8 | 3,256 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
move = [(0,1),(1,0)]
def solve(n):
f = defaultdict(lambda : 0)
v = defaultdict(list)
l = []
for i in range(n):
a,b,dir = input().split()
a = int(a)
b = int(b)
f[(a,b)] = 1
dir = 0 if dir == "x" else 1
na,nb = a+1-dir,b+dir
f[(na,nb)] = 1
l.append((a,b,na,nb))
l.append((na,nb,a,b))
v[(a,b)].append(((na,nb),1))
v[(na,nb)].append(((a,b),1))
for a,b,c,d in l:
for dx,dy in move:
na,nb = a+dx,b+dy
if f[(na,nb)] and (c,d) != (na,nb):
v[(a,b)].append(((na,nb),0))
v[(na,nb)].append(((a,b),0))
bfs = defaultdict(lambda : -1)
q = deque()
for a,b,c,d in l:
if bfs[(a,b)] < 0:
q.append((a,b))
bfs[(a,b)] = 0
while q:
x,y = q.popleft()
for node,k in v[(x,y)]:
nx,ny = node
if k:
nb = 1-bfs[(x,y)]
else:
nb = bfs[(x,y)]
if bfs[(nx,ny)] >= 0:
if bfs[(nx,ny)] != nb:
print("No")
return
else:
bfs[(nx,ny)] = nb
q.append((nx,ny))
print("Yes")
return
#Solve
if __name__ == "__main__":
while 1:
n = I()
if n == 0:
break
solve(n)
``` | output | 1 | 1,628 | 8 | 3,257 |
Provide a correct Python 3 solution for this coding contest problem.
The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.
In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.
Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.
Input
The input is a sequence of datasets. Each dataset is given in the following format:
n
x1 y1 dir1
...
xn yn dirn
n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .
It is guaranteed that no two futons in the input overlap each other.
The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.
Output
For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.
Example
Input
4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0
Output
Yes
No
Yes | instruction | 0 | 1,629 | 8 | 3,258 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [LS() for _ in range(n)]
d = collections.defaultdict(int)
ed = collections.defaultdict(lambda: None)
for i in range(n):
x, y, di = a[i]
x = int(x)
y = int(y)
d[(x,y)] = i + 1
if di == 'x':
d[(x+1,y)] = i + 1
ed[(x,y)] = (x+1,y)
ed[(x+1,y)] = (x,y)
else:
d[(x,y+1)] = i + 1
ed[(x,y)] = (x,y+1)
ed[(x,y+1)] = (x,y)
ee = collections.defaultdict(set)
dka = list(d.keys())
for x,y in list(d.keys()):
dt = d[(x,y)]
for di,dj in dd:
ni = x + di
nj = y + dj
if d[(ni,nj)] > 0 and d[(ni,nj)] != dt:
ee[(x,y)].add((ni,nj))
v = collections.defaultdict(bool)
f = True
for k in dka:
if v[k]:
continue
s1 = set()
s2 = set()
ns1 = set([k])
ns2 = set()
while ns1:
na = list(ns1)
s1 |= ns1
ns1 = set()
for k in na:
ns1 |= ee[k]
ns2.add(ed[k])
ns2 -= s2
while ns2:
na = list(ns2)
s2 |= ns2
ns2 = set()
for k in na:
ns2 |= ee[k]
ns1.add(ed[k])
ns2 -= s2
ns1 -= s1
if s1 & s2:
f = False
# print('k', k)
# print('s1', s1)
# print('s2', s2)
if f:
for k in s1:
v[k] = 1
for k in s2:
v[k] = 1
else:
break
if f:
rr.append('Yes')
else:
rr.append('No')
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 1,629 | 8 | 3,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.
In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.
Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.
Input
The input is a sequence of datasets. Each dataset is given in the following format:
n
x1 y1 dir1
...
xn yn dirn
n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .
It is guaranteed that no two futons in the input overlap each other.
The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.
Output
For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.
Example
Input
4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0
Output
Yes
No
Yes
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [LS() for _ in range(n)]
d = collections.defaultdict(int)
for i in range(n):
x, y, di = a[i]
x = int(x)
y = int(y)
d[(x,y)] = i + 1
if di == 'x':
d[(x+1,y)] = i + 1
else:
d[(x,y+1)] = i + 1
v = collections.defaultdict(bool)
f = True
for k in list(d.keys()):
if v[k]:
continue
s1 = set()
s2 = set()
ns1 = set([k])
ns2 = set()
while ns1 or ns2:
for ak in list(ns1):
for di,dj in dd:
nk = (ak[0]+di, ak[1]+dj)
if d[nk] == 0:
continue
if d[nk] == d[ak]:
ns2.add(nk)
else:
ns1.add(nk)
ns2 -= s2
for ak in list(ns2):
for di,dj in dd:
nk = (ak[0]+di, ak[1]+dj)
if d[nk] == 0:
continue
if d[nk] == d[ak]:
ns1.add(nk)
else:
ns2.add(nk)
ns2 -= s2
ns1 -= s1
s2 |= ns2
s1 |= ns1
if s1 & s2:
f = False
break
if not f:
break
for c in s1:
v[c] = 1
for c in s2:
v[c] = 1
if f:
rr.append('Yes')
else:
rr.append('No')
return '\n'.join(map(str,rr))
print(main())
``` | instruction | 0 | 1,630 | 8 | 3,260 |
No | output | 1 | 1,630 | 8 | 3,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.
In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.
Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.
Input
The input is a sequence of datasets. Each dataset is given in the following format:
n
x1 y1 dir1
...
xn yn dirn
n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .
It is guaranteed that no two futons in the input overlap each other.
The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.
Output
For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.
Example
Input
4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0
Output
Yes
No
Yes
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [LS() for _ in range(n)]
d = collections.defaultdict(int)
for i in range(n):
x, y, di = a[i]
x = int(x)
y = int(y)
d[(x,y)] = i + 1
if di == 'x':
d[(x+1,y)] = i + 1
else:
d[(x,y+1)] = i + 1
v = collections.defaultdict(bool)
f = True
for k in list(d.keys()):
if v[k]:
continue
s1 = set()
s2 = set()
ns1 = set([k])
ns2 = set()
while ns1 or ns2:
for ak in list(ns1):
for di,dj in dd:
nk = (ak[0]+di, ak[1]+dj)
if d[nk] == 0:
continue
if d[nk] == d[ak]:
ns1.add(nk)
else:
ns2.add(nk)
ns2 -= s2
for ak in list(ns2):
for di,dj in dd:
nk = (ak[0]+di, ak[1]+dj)
if d[nk] == 0:
continue
if d[nk] == d[ak]:
ns2.add(nk)
else:
ns1.add(nk)
ns2 -= s2
ns1 -= s1
s2 |= ns2
s1 |= ns1
if s1 & s2:
f = False
break
if not f:
break
for c in s1:
v[c] = 1
for c in s2:
v[c] = 1
if f:
rr.append('Yes')
else:
rr.append('No')
return '\n'.join(map(str,rr))
print(main())
``` | instruction | 0 | 1,631 | 8 | 3,262 |
No | output | 1 | 1,631 | 8 | 3,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.
In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.
Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.
Input
The input is a sequence of datasets. Each dataset is given in the following format:
n
x1 y1 dir1
...
xn yn dirn
n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .
It is guaranteed that no two futons in the input overlap each other.
The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.
Output
For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.
Example
Input
4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0
Output
Yes
No
Yes
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [LS() for _ in range(n)]
d = collections.defaultdict(int)
ed = collections.defaultdict(lambda: None)
for i in range(n):
x, y, di = a[i]
x = int(x)
y = int(y)
d[(x,y)] = i + 1
if di == 'x':
d[(x+1,y)] = i + 1
ed[(x,y)] = (x+1,y)
ed[(x+1,y)] = (x,y)
else:
d[(x,y+1)] = i + 1
ed[(x,y)] = (x,y+1)
ed[(x,y+1)] = (x,y)
ee = collections.defaultdict(set)
dka = list(d.keys())
for x,y in list(d.keys()):
dt = d[(x,y)]
for di,dj in dd:
ni = x + di
nj = y + dj
if d[(ni,nj)] > 0 and d[(ni,nj)] != dt:
ee[(x,y)].add((ni,nj))
v = collections.defaultdict(bool)
f = True
for k in dka:
if v[k]:
continue
s1 = set()
s2 = set()
ns1 = set([k])
ns2 = set()
while ns1 or ns2:
s1 |= ns1
na = list(ns1)
ns1 = set()
for k in na:
ns1 |= ee[k]
ns2.add(ed[k])
ns2 -= s2
na = list(ns2)
s2 |= ns2
ns2 = set()
for k in na:
ns2 |= ee[k]
ns1.add(ed[k])
ns1 -= s1
ns2 -= s2
s1 |= ns1
s2 |= ns2
if s1 & s2:
f = False
break
# print('k', k)
# print('s1', s1)
# print('s2', s2)
if f:
for k in s1:
v[k] = 1
for k in s2:
v[k] = 1
else:
break
if f:
rr.append('Yes')
else:
rr.append('No')
return '\n'.join(map(str,rr))
print(main())
``` | instruction | 0 | 1,632 | 8 | 3,264 |
No | output | 1 | 1,632 | 8 | 3,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.
In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.
Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.
Input
The input is a sequence of datasets. Each dataset is given in the following format:
n
x1 y1 dir1
...
xn yn dirn
n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .
It is guaranteed that no two futons in the input overlap each other.
The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.
Output
For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.
Example
Input
4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0
Output
Yes
No
Yes
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [LS() for _ in range(n)]
d = collections.defaultdict(int)
for i in range(n):
x, y, di = a[i]
x = int(x)
y = int(y)
d[(x,y)] = i + 1
if di == 'x':
d[(x+1,y)] = i + 1
else:
d[(x,y+1)] = i + 1
v = collections.defaultdict(bool)
f = True
for k in list(d.keys()):
if v[k]:
continue
s1 = set()
s2 = set()
ns1 = set([k])
ns2 = set()
while ns1:
while ns1:
ns1 -= s1
s1 |= ns1
for ak in list(ns1):
for di,dj in dd:
nk = (ak[0]+di, ak[1]+dj)
if d[nk] == 0:
continue
if d[nk] == d[ak]:
ns2.add(nk)
else:
ns1.add(nk)
if s1 & s2:
f = False
break
while ns2:
ns2 -= s2
s2 |= ns2
for ak in list(ns2):
for di,dj in dd:
nk = (ak[0]+di, ak[1]+dj)
if d[nk] == 0:
continue
if d[nk] == d[ak]:
ns1.add(nk)
else:
ns2.add(nk)
if s1 & s2:
f = False
break
if s1 & s2:
f = False
break
if not f:
break
for c in s1:
v[c] = 1
for c in s2:
v[c] = 2
if f:
rr.append('Yes')
else:
rr.append('No')
return '\n'.join(map(str,rr))
print(main())
``` | instruction | 0 | 1,633 | 8 | 3,266 |
No | output | 1 | 1,633 | 8 | 3,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,833 | 8 | 3,666 |
Tags: math, sortings
Correct Solution:
```
t=int(input())
while t>0:
t-=1
n=int(input())
A=list(map(int,input().split(' ')))
B=[i for i in A]
B.sort(reverse=True)
if B==A and len(list(set(B)))==len(B):
print("NO")
else:
print("YES")
``` | output | 1 | 1,833 | 8 | 3,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,834 | 8 | 3,668 |
Tags: math, sortings
Correct Solution:
```
# Author : -pratyay-
import sys
inp=sys.stdin.buffer.readline
wrl=print
inar=lambda: list(map(int,inp().split()))
inin=lambda: int(inp())
inst=lambda: inp().decode().strip()
_T_=inin()
for _t_ in range(_T_):
n=inin()
a=inar()
ans=True
for i in range(n-1):
if a[i]<=a[i+1]:
ans=False
if not ans:
wrl('YES')
else:
wrl('NO')
``` | output | 1 | 1,834 | 8 | 3,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,835 | 8 | 3,670 |
Tags: math, sortings
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
lst = [int(x) for x in input().split()]
f=False
for i in range(1,n):
if lst[i]>=lst[i-1]:
f=True
print('YES' if f == True else 'NO')
``` | output | 1 | 1,835 | 8 | 3,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,836 | 8 | 3,672 |
Tags: math, sortings
Correct Solution:
```
from collections import defaultdict
from queue import deque
def arrinp():
return [*map(int, input().split(' '))]
def mulinp():
return map(int, input().split(' '))
def intinp():
return int(input())
def solution():
n = intinp()
a = arrinp()
count = 0
for i in range(n-1):
if a[i] > a[i+1]:
count += 1
if count == n-1:
print('NO')
else:
print('YES')
testcases = 1
testcases = int(input())
for _ in range(testcases):
solution()
``` | output | 1 | 1,836 | 8 | 3,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,837 | 8 | 3,674 |
Tags: math, sortings
Correct Solution:
```
def sol():
n = int(input())
arr = [int(x) for x in input().split()]
for i, val in enumerate(arr):
if i == n - 1:
print("NO")
elif val <= arr[i+1]:
print("YES")
break
def main():
t = int(input())
for _ in range(t):
sol()
if __name__ == "__main__":
main()
``` | output | 1 | 1,837 | 8 | 3,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,838 | 8 | 3,676 |
Tags: math, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
def test():
n = int(input())
c = list(map(int, input().split()))
s = 0
p = sorted(c)
m = sorted(c, reverse=True)
o = set(c)
if c == p:
return 'YES'
elif m == c and len(o) == n:
return 'NO'
else:
return 'YES'
t = int(input())
for _ in range(t):
print(test())
``` | output | 1 | 1,838 | 8 | 3,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,839 | 8 | 3,678 |
Tags: math, sortings
Correct Solution:
```
t = int(input())
while t:
t -= 1
n = int(input())
arr =[int(x) for x in input().split()]
flag = True
for i in range(1,n):
if arr[i] < arr[i-1]:
continue
else:
flag = False
break
if flag:
print("NO")
else:
print("YES")
``` | output | 1 | 1,839 | 8 | 3,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For god's sake, you're boxes with legs! It is literally your only purpose! Walking onto buttons! How can you not do the one thing you were designed for?
Oh, that's funny, is it? Oh it's funny? Because we've been at this for twelve hours and you haven't solved it either, so I don't know why you're laughing. You've got one hour! Solve it!
Wheatley decided to try to make a test chamber. He made a nice test chamber, but there was only one detail absent — cubes.
For completing the chamber Wheatley needs n cubes. i-th cube has a volume a_i.
Wheatley has to place cubes in such a way that they would be sorted in a non-decreasing order by their volume. Formally, for each i>1, a_{i-1} ≤ a_i must hold.
To achieve his goal, Wheatley can exchange two neighbouring cubes. It means that for any i>1 you can exchange cubes on positions i-1 and i.
But there is a problem: Wheatley is very impatient. If Wheatley needs more than (n ⋅ (n-1))/(2)-1 exchange operations, he won't do this boring work.
Wheatly wants to know: can cubes be sorted under this conditions?
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 1000), denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains one positive integer n (2 ≤ n ≤ 5 ⋅ 10^4) — number of cubes.
The second line contains n positive integers a_i (1 ≤ a_i ≤ 10^9) — volumes of cubes.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a word in a single line: "YES" (without quotation marks) if the cubes can be sorted and "NO" (without quotation marks) otherwise.
Example
Input
3
5
5 3 2 1 4
6
2 2 2 2 2 2
2
2 1
Output
YES
YES
NO
Note
In the first test case it is possible to sort all the cubes in 7 exchanges.
In the second test case the cubes are already sorted.
In the third test case we can make 0 exchanges, but the cubes are not sorted yet, so the answer is "NO". | instruction | 0 | 1,840 | 8 | 3,680 |
Tags: math, sortings
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
b = list(a)
b.sort(reverse = True)
sa = set(a)
if(len(sa) == len(a) and a == b):
print("NO")
else:
print("YES")
``` | output | 1 | 1,840 | 8 | 3,681 |
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