message stringlengths 2 20.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 757 108k | cluster float64 4 4 | __index_level_0__ int64 1.51k 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
while True:
N = int(input())
if not N:
break
D = [[] for _ in range(30)]
for i in range(N):
for x in map(int, input().split()[1:]):
D[x - 1].append(i)
C = [1 << i for i in range(N)]
for d in range(30):
for i in D[d]:
for j in D[d]:
C[i] = C[i] | C[j]
if any(x == (1 << N) - 1 for x in C):
print(d + 1)
break
else:
print(-1)
``` | instruction | 0 | 96,026 | 4 | 192,052 |
Yes | output | 1 | 96,026 | 4 | 192,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
import itertools
while True:
N = int(input())
if not N:
break
D = [[] for _ in range(30)]
for i in range(N):
for x in map(int, input().split()[1:]):
D[x - 1].append(i)
C = [1 << i for i in range(N)]
for d in range(30):
for i, j in itertools.permutations(D[d], 2):
C[i] = C[i] | C[j]
if any(x == (1 << N) - 1 for x in C):
print(d + 1)
break
else:
print(-1)
``` | instruction | 0 | 96,027 | 4 | 192,054 |
Yes | output | 1 | 96,027 | 4 | 192,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
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 = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-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 = [LI()[1:] for _ in range(n)]
d = collections.defaultdict(list)
for i in range(n):
for c in a[i]:
d[c].append(i)
r = -1
s = [set([i]) for i in range(n)]
for k in sorted(list(d.keys())):
ts = set()
for c in d[k]:
ts |= s[c]
if len(ts) == n:
r = k
break
for c in d[k]:
s[c] |= ts
rr.append(r)
return '\n'.join(map(str,rr))
print(main())
``` | instruction | 0 | 96,028 | 4 | 192,056 |
Yes | output | 1 | 96,028 | 4 | 192,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
def solve():
from sys import stdin
f_i = stdin
ans = []
while True:
n = int(f_i.readline())
if n == 0:
break
last_day = 1
schedule = [set() for i in range(31)]
for i in range(n):
line = map(int, f_i.readline().split())
f = next(line)
for day in line:
schedule[day].add(i)
if day > last_day:
last_day = day
groups = [set() for i in range(n)]
for day in range(1, last_day + 1):
member1 = schedule[day]
member2 = member1.copy()
for person in member1:
member2 |= groups[person]
if len(member2) == n:
ans.append(day)
break
for person in member1:
groups[person] |= member2
else:
ans.append(-1)
print('\n'.join(map(str, ans)))
solve()
``` | instruction | 0 | 96,029 | 4 | 192,058 |
Yes | output | 1 | 96,029 | 4 | 192,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
while(True):
n = int(input())
if n == 0:
break
d = []
for i in range(n):
ll = input().split()
l = [int(_) for _ in ll]
f = l[0]
tmp = []
for j in range(1, f+1):
tmp.append(l[j])
d += [tmp]
mp = [[0 for i in range(31)] for j in range(n)]
for member in range(n):
for day in d[member]:
mp[member][day] = 1
clr = [0 for i in range(n)]
ans = [1 for i in range(n)]
flg = 0
for day in range(31):
tmp = []
s = 0
for member in range(n):
s += mp[member][day]
if mp[member][day] == 1:
tmp.append(member)
if s > 1:
for member in tmp:
clr[member] = 1
if clr == ans and flg == 0:
print(day)
flg = 1
if flg == 0:
print(-1)
``` | instruction | 0 | 96,030 | 4 | 192,060 |
No | output | 1 | 96,030 | 4 | 192,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
from itertools import combinations
from decimal import Decimal
def get_point(line1, line2):
m1, k1 = line1
m2, k2 = line2
if m1 == m2:
return None
elif m1 is None and m2 is None:
return None
elif m1 is None:
x = k1
y = m2 * x + k2
elif m2 is None:
x = k2
y = m1 * x + k1
else:
x = (k2-k1)/(m2-m1)
y = m1*x+k
if -100 < x < 100 and -100 < y < 100:
return (x, y)
else:
return None
while True:
N = int(input())
if not N:
break
lines = []
for i in range(N):
x1, y1, x2, y2 = [Decimal(int(i)) for i in input().split()]
if x1 == x2:
lines.append(None, x1)
else:
m = (y2-y1) / (x2-x1)
k = y1 - m * x1
lines.append((m, k))
iterated_lines = []
ans = 1
for line1 in lines:
pt_set = set()
for line2 in iterated_lines:
pt = get_point(line1, line2)
# print(pt)
if pt:
pt_set.add(pt)
ans += len(pt_set) + 1
iterated_lines.append(line1)
print(ans)
``` | instruction | 0 | 96,031 | 4 | 192,062 |
No | output | 1 | 96,031 | 4 | 192,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
while(True):
n = int(input())
if n == 0:
break
d = []
for i in range(n):
ll = input().split()
l = [int(_) for _ in ll]
f = l[0]
tmp = []
for j in range(1, f+1):
tmp.append(l[j])
d += [tmp]
mp = [[0 for i in range(30)] for j in range(n)]
for member in range(n):
for day in d[member]:
mp[member][day] = 1
clr = [0 for i in range(n)]
ans = [1 for i in range(n)]
flg = 0
for day in range(30):
tmp = []
s = 0
for member in range(n):
s += mp[member][day]
if mp[member][day] == 1:
tmp.append(member)
if s > 1:
for member in tmp:
clr[member] = 1
if clr == ans and flg == 0:
print(day)
flg = 1
if flg == 0:
print(-1)
``` | instruction | 0 | 96,032 | 4 | 192,064 |
No | output | 1 | 96,032 | 4 | 192,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is said that a legendary treasure left by Mr. Yao is sleeping somewhere in Hachioji long ago. The treasure map, which is said to show its whereabouts, has been handed down by Yao's n descendants, divided into several pieces.
Now, the descendants of Mr. Yao were trying to cooperate to obtain the treasure. However, the treasure cannot be found only by a part of the treasure map that points to the location of the treasure. Therefore, all the descendants of Mr. Yao gathered and tried to collect the map in one place. However, even if I tried to put it into practice, I couldn't get together because I couldn't meet the schedule. However, this information about the treasure is valuable information that has been secretly passed down in the clan. Considering the risk of leakage, exchanging maps using public communication means is out of the question.
Therefore, I decided to collect the map for one descendant by repeating the process of meeting the descendants in person and handing over the map. There is no limit to the number of people that one person can meet in a day, but it is necessary that there is a schedule for each other.
Your job is to write a program that asks for at least how many days it will take to collect a map from a list of open days on schedule for each offspring.
By the way, the unity of the Yao clan is very tight. If the descendants who finally get the entire map betray the other descendants and take away the treasure, they will be sanctioned by the clan. The sanctions are so horrifying that it is virtually impossible for their descendants to actually carry away the treasure.
Input
The input consists of multiple datasets.
Each dataset consists of multiple rows. The first line contains the integer n (1 <n <= 50), which represents the number of people with a piece of the map. The next n lines contain the schedule for each descendant. Line i represents the schedule of the i-th descendant, with some integers separated by a single character space. The first integer fi (0 <= fi <= 30) is an integer that represents the number of days that the descendant's schedule is free. The following fi integers represent dates when the schedule is free. These dates differ from each other and are all greater than or equal to 1 and less than or equal to 30.
There is one line containing only 0 at the end of the input.
Output
Print one integer on one line for each dataset. If you can collect the map within 30 days, output the minimum number of days required to collect the map, otherwise output -1.
Addendum: The above "minimum number of days required to collect maps" means the date when all maps are collected earliest starting from one day.
Example
Input
4
1 1
2 2 3
2 1 2
3 3 4 5
0
Output
3
Submitted Solution:
```
import heapq
while True:
N = int(input())
if not N:
break
f = [0] * N
for i in range(N):
f[i] = list(map(int,input().split()[1:]))
dp = [ [0] * 51 for i in range(51) ]
for i in range(N):
for j in range(N):
dp[j][i] = 1<<i
for i in range(1,31):
ds = [ j for j in range(N) if i in f[j]]
for d1 in ds:
for d2 in ds:
dp[i][d1] |= dp[i-1][d2]
ans = 40
for i in range(31):
for j in range(N):
if dp[i][j] == (1<<N)-1:
ans = min( (ans,i) )
if ans > 30:
print(-1)
else:
print(ans)
``` | instruction | 0 | 96,033 | 4 | 192,066 |
No | output | 1 | 96,033 | 4 | 192,067 |
Provide a correct Python 3 solution for this coding contest problem.
JOI has a stove in your room. JOI himself is resistant to the cold, so he doesn't need to put on the stove when he is alone in the room, but he needs to put on the stove when there are visitors.
On this day, there are N guests under JOI. The ith (1 \ leq i \ leq N) visitor arrives at time T_i and leaves at time T_i + 1. There can never be more than one visitor at the same time.
JOI can turn the stove on and off at any time. However, each time you turn on the stove, you will consume one match. JOI has only K matches, so he can only stove up to K times. The stove is gone at the beginning of the day.
When the stove is on, fuel is consumed by that amount, so I want to set the time when the stove is on and off and minimize the total time when the stove is on.
Example
Input
3 2
1
3
6
Output
4 | instruction | 0 | 96,896 | 4 | 193,792 |
"Correct Solution:
```
n,k=map(int,input().split())
guest=[]
interval=[]
flag=0
for i in range(n):
guest.append(int(input()))
if flag==1:
interval.append(guest[-1]-guest[-2]-1)
flag=1
interval.sort()
ans=n
if n>k:
for i in range(n-k):
ans+=interval[i]
print(ans)
``` | output | 1 | 96,896 | 4 | 193,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156 | instruction | 0 | 98,212 | 4 | 196,424 |
Tags: math
Correct Solution:
```
from fractions import Fraction
import sys
sys.setrecursionlimit(1000*100)
A=int(input())
p=[]
c=1
for _ in range(600):
p.append(c)
c*=12
r=[]
for i in range(600):
for j in range(i+1):
if p[j]+p[i-j]==A:
r.append(i+1)
break
s=set()
for i in r:
for j in range(i):
v=p[j]+p[i-1-j]
if v!=A:
s.add(v)
if len(r)==0:
print('NO')
else:
print('YES')
print(min(len(r),1000))
print(' '.join(map(str,r[:1000])))
print(min(len(s),1000))
if len(s)>0:
print(' '.join(map(str,sorted(s)[:1000])))
#c=int(input())
#a,b=tuple(map(int,input().split()))
#edges=dict((i,[]) for i in range(1,c+1))
#children=filter(lambda x: x != p, edges[r])
#cs.sort(key=lambda x:Fraction(x[0],x[1]),reverse=True)
#if dp[r] is not None:
#chr(ord('a')+1)
#''.join(['a','b','c'])
#sys.exit()
# Made By Mostafa_Khaled
``` | output | 1 | 98,212 | 4 | 196,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156 | instruction | 0 | 98,213 | 4 | 196,426 |
Tags: math
Correct Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 2000):
d[j] = set()
big = 10**302
for i in range(1, 2000):
ok = 0
for a, b in d[i]:
c = 12*a
if a <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if a <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
break
n = int(input())
g = set()
for i in range(1, 2000):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
``` | output | 1 | 98,213 | 4 | 196,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156 | instruction | 0 | 98,214 | 4 | 196,428 |
Tags: math
Correct Solution:
```
from fractions import Fraction
import sys
sys.setrecursionlimit(1000*100)
A=int(input())
p=[]
c=1
for _ in range(600):
p.append(c)
c*=12
r=[]
for i in range(600):
for j in range(i+1):
if p[j]+p[i-j]==A:
r.append(i+1)
break
s=set()
for i in r:
for j in range(i):
v=p[j]+p[i-1-j]
if v!=A:
s.add(v)
if len(r)==0:
print('NO')
else:
print('YES')
print(min(len(r),1000))
print(' '.join(map(str,r[:1000])))
print(min(len(s),1000))
if len(s)>0:
print(' '.join(map(str,sorted(s)[:1000])))
#c=int(input())
#a,b=tuple(map(int,input().split()))
#edges=dict((i,[]) for i in range(1,c+1))
#children=filter(lambda x: x != p, edges[r])
#cs.sort(key=lambda x:Fraction(x[0],x[1]),reverse=True)
#if dp[r] is not None:
#chr(ord('a')+1)
#''.join(['a','b','c'])
#sys.exit()
``` | output | 1 | 98,214 | 4 | 196,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156 | instruction | 0 | 98,215 | 4 | 196,430 |
Tags: math
Correct Solution:
```
A = int(input())
T = A
x = 0
while (A % 12 == 0):
A = A // 12
x = x + 1
b = [0] * 1000
b[1] = 2
b[2] = 13
for i in range(3, 606):
b[i] = 13 * b[i - 1] - 12 * b[i - 2]
y = 1
while (b[y] < A):
y = y + 1
if (b[y] != A):
print("NO")
else:
print("YES\n1")
t = 2 * x + y
print(t)
c = [0] * 1000
d = 0
for i in range((t - 1) // 2 + 1):
if (i != x and t - 2 * i != 0):
c[d] = b[t - 2 * i] * 12 ** i
d = d + 1
print(d)
for i in range(d):
print(c[d - i - 1])
``` | output | 1 | 98,215 | 4 | 196,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156 | instruction | 0 | 98,216 | 4 | 196,432 |
Tags: math
Correct Solution:
```
A=int(input())
res=[]
for i in range(300):
for j in range(300):
if(pow(12,i)+pow(12,j)==A):
res.append([i,j])
if(len(res)==0):
print("NO")
quit()
print("YES")
print(1)
sm=sum(res[0])
print(sm+1)
print(sm//2)
for i in range(sm//2,-1,-1):
V=pow(12,i)+pow(12,sm-i)
if(V!=A):
print(V)
``` | output | 1 | 98,216 | 4 | 196,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156 | instruction | 0 | 98,217 | 4 | 196,434 |
Tags: math
Correct Solution:
```
a = int(input())
ans = set()
def work(i, z, y):
z.add(y)
# print(">>", i, y)
if y == a:
ans.add(i)
if len(z) > 1000:
z.remove(max(z))
pos = [set(), set([2])]
for i in range(2):
for x in pos[i]:
if x == a:
ans.add(i)
def dfs(i, last, cur):
if i > 988:
return
while len(pos) - 1 < i:
pos.append(set())
if len(pos[i]) == 0 and cur > a:
return
if cur in pos[i]:
return
work(i, pos[i], cur)
dfs(i + 1, cur, last * 12)
dfs(i + 1, cur, cur * 13 - last * 12)
dfs(2, 2, 13)
if len(ans) == 0:
print("NO")
else:
print("YES")
count = 0
print(min(1000, len(ans)))
z = set()
for i in sorted(ans):
if count < 1000:
print(i)
count += 1
for y in pos[i]:
if y != a:
z.add(y)
if len(z) > 1000:
z.remove(max(z))
count = 0
print(min(1000, len(z)))
for i in sorted(z):
print(i)
count += 1
if count >= 1000:
break
``` | output | 1 | 98,217 | 4 | 196,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
A = int(input())
T = A
x = 0
while (A % 12 == 0):
A = A / 12
x = x + 1
b = [0] * 1000
b[1] = 2
b[2] = 13
for i in range(3, 603):
b[i] = 13 * b[i - 1] - 12 * b[i - 2]
y = 1
while (b[y] < A):
y = y + 1
if (b[y] != A):
print("NO")
else:
print("YES\n1")
t = 2 * x + y
print(t)
c = [0] * 1000
d = 0
for i in range((t - 1) // 2 + 1):
if (i != x and t - 2 * i != 0):
c[d] = b[t - 2 * i] * 12 ** i
d = d + 1
print(d)
for i in range(d):
print(c[d - i - 1])
``` | instruction | 0 | 98,218 | 4 | 196,436 |
No | output | 1 | 98,218 | 4 | 196,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 2000):
d[j] = set()
big = 10**302
for i in range(1, 2000):
ok = 0
for a, b in d[i]:
c = 12*a
if b <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if b <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
break
n = int(input())
g = set()
for i in range(1, 2000):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
``` | instruction | 0 | 98,219 | 4 | 196,438 |
No | output | 1 | 98,219 | 4 | 196,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 2000):
d[j] = set()
big = 10**333
for i in range(1, 2000):
ok = 0
for a, b in d[i]:
c = 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
break
n = int(input())
g = set()
for i in range(1, 2000):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
``` | instruction | 0 | 98,220 | 4 | 196,440 |
No | output | 1 | 98,220 | 4 | 196,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 600):
d[j] = set()
big = 10**302
h = -1
for i in range(1, 600):
ok = 0
for a, b in d[i]:
c = 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
h = i
break
n = int(input())
g = set()
for i in range(1, h):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
``` | instruction | 0 | 98,221 | 4 | 196,442 |
No | output | 1 | 98,221 | 4 | 196,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
from math import sqrt
def criba(n):
primos = [True] * n #Inicialmente todos los valores hasta n son marcados como primos
primos[0] = primos[1] = False #Desmarco el 0 y el 1
for i in range(2, int(sqrt(n))+1): #Recorrer solo hasta la raiz de n
if primos[i]:
for j in range(i * i, n, i): #Desmarcar todos los multiplos del primo i
primos[j] = False
return [index for index, item in enumerate(primos) if item] #retornar una lista solo con los primos
a,b = map(int,input().split())
resp = 0
primos = criba(b)
for i in primos:
if (i==2 or i%4==1)and i>=a: #Solo incremento el contador cuando cumplen la condicion de exprsarse como suma de dos cuadrados y son >= que el limite inferior
resp+=1
print(resp)
``` | instruction | 0 | 98,657 | 4 | 197,314 |
No | output | 1 | 98,657 | 4 | 197,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def isprime(n):
if n <= 100:
return n in primes_under_100
if n % 2 == 0 or n % 3 == 0:
return False
for f in range(5, int(n ** .5), 6):
if n % f == 0 or n % (f + 2) == 0:
return False
return True
s=input()
a=[int(i) for i in s.split()]
b=a[1]
a=a[0]
t=0
for i in range(a,b+1):
if (i-1)%4==0:
if isprime(i):
t=t+1
if a<=2<=b:
t=t+1
print(t)
``` | instruction | 0 | 98,658 | 4 | 197,316 |
No | output | 1 | 98,658 | 4 | 197,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def isprime(n):
if n <= 100:
return n in primes_under_100
if n % 2 == 0 or n % 3 == 0:
return False
for f in range(5, int(n ** .5), 6):
if n % f == 0 or n % (f + 2) == 0:
return False
return True
'''
t=0
for i in range(n):
if n%(i+1)==0:
t=t+1
if t==2:
return True
else:
return False
for i in range(5000):
if isprime(i+1):
print(i+1)'''
s=input()
a=[int(i) for i in s.split()]
b=a[1]
a=a[0]
t=0
if a<=2<=b:
t.append(2)
for i in range(a,b+1):
if (i-1)%4==0:
if isprime(i):
t=t+1
print(t)
``` | instruction | 0 | 98,659 | 4 | 197,318 |
No | output | 1 | 98,659 | 4 | 197,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def isprime(n):
if n <= 100:
return n in primes_under_100
if n % 2 == 0 or n % 3 == 0:
return False
for f in range(5, int(n ** .5), 6):
if n % f == 0 or n % (f + 2) == 0:
return False
return True
s=input()
a=[int(i) for i in s.split()]
b=a[1]
a=a[0]
t=0
for i in range(a,b+1):
if (i-1)%4==0:
if isprime(i):
t=t+1
print(t)
``` | instruction | 0 | 98,660 | 4 | 197,320 |
No | output | 1 | 98,660 | 4 | 197,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,693 | 4 | 197,386 |
Tags: implementation
Correct Solution:
```
n,x,y=map(int,input().split())
a=*map(int,input().split()),
for k in range(n):
v=1
for l in range(max(0,k-x),k):v&=a[k]<a[l]
for l in range(k+1,min(k+y+1,n)):v&=a[k]<a[l]
if v:print(k+1);break
``` | output | 1 | 98,693 | 4 | 197,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,694 | 4 | 197,388 |
Tags: implementation
Correct Solution:
```
n,x,y = map(int,input().split())
l = list(map(int,input().split()))
for i in range(n):
if l[i] <= min(l[max(0,i-x):i+y+1]):
break
print(i+1)
``` | output | 1 | 98,694 | 4 | 197,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,695 | 4 | 197,390 |
Tags: implementation
Correct Solution:
```
def solve(days, x, y, n):
window = dict()
win_beg = 0
win_end = 0
res = 0
pre = 0
while pre < n:
while win_beg < pre - x:
del window[days[win_beg]]
win_beg += 1
while win_end < min(pre + y + 1, n):
window[days[win_end]] = True
if days[win_end] < days[res]:
res = win_end
win_end += 1
if pre == res:
return pre + 1
pre = res
return -1
if __name__ == "__main__":
n, x, y = [int(ele) for ele in input().split()]
days = [int(ele) for ele in input().split()]
print(solve(days, x, y, n))
``` | output | 1 | 98,695 | 4 | 197,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,696 | 4 | 197,392 |
Tags: implementation
Correct Solution:
```
n,x,y=map(int,input().split())
l=list(map(int,input().split()))
for i in range(n):
j=i-x
k=i+y
if j<0:
j=0
if k>=n:
k=n-1
if all(l[i]<arr for arr in l[j:i]) and all(l[i]<brr for brr in l[i+1:k+1]):
print(i+1)
break
``` | output | 1 | 98,696 | 4 | 197,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,697 | 4 | 197,394 |
Tags: implementation
Correct Solution:
```
N, BEFORE, AFTER = map(int, input().split())
days = list(map(int, input().split()))
def goodbefore(j):
for diff in range(1, BEFORE+1):
k = j - diff
if k < 0:
continue
if days[k] <= days[j]:
return False
return True
def goodafter(j):
for diff in range(1, AFTER+1):
k = j + diff
if k >= N:
continue
if days[k] <= days[j]:
return False
return True
for i in range(N):
if goodbefore(i) and goodafter(i):
print(i+1)
break
``` | output | 1 | 98,697 | 4 | 197,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,698 | 4 | 197,396 |
Tags: implementation
Correct Solution:
```
n, x, y = map(int, input().split())
inf = 10**10
a = [inf]*7 + list(map(int, input().split())) + [inf]*7
for i in range(7, n+7):
if a[i] < (min(a[i-x:i]) if x else inf) and a[i] < (min(a[i+1:i+y+1]) if y else inf):
print(i-6)
exit()
``` | output | 1 | 98,698 | 4 | 197,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,699 | 4 | 197,398 |
Tags: implementation
Correct Solution:
```
n, x, y = map(int, input().split());
arr = list(map(int, input().split()));
for i in range(n):
if ((not arr[max(i-1, 0):i] or min(arr[i-x:i] + [2e9]) > arr[i]) and (not arr[i+1: min(n, i+y+1)] or min(arr[i+1:i+y+1]+[2e9]) > arr[i])):
print(i+1)
exit()
``` | output | 1 | 98,699 | 4 | 197,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer. | instruction | 0 | 98,700 | 4 | 197,400 |
Tags: implementation
Correct Solution:
```
n,x,y=map(int,input().split(" "))
a=list(map(int,input().split()))
i=f1=0
while i<n:
if i-x<0:
start=0
else:
start=i-x
if i+y>n-1:
end=n-1
else:
end=i+y
f=0
j=start
while j<=end:
if i!=j:
if a[i]<a[j]:
j+=1
else:
f=1
break
else:
j+=1
if f==0:
f1=1
break
i+=1
if f1==1:
print(i+1)
``` | output | 1 | 98,700 | 4 | 197,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
n, x, y = mapin()
fl = 0
a = [int(x) for x in input().split()]
for i in range(n):
s = min(a[max(0,i-x):min(n,i+y+1)])
if(a[i] == s):
fl = i
break
print(fl+1)
``` | instruction | 0 | 98,701 | 4 | 197,402 |
Yes | output | 1 | 98,701 | 4 | 197,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
n,x,y=map(int,input().split())
arr=list(map(int,input().split()))
for i in range(n):
ol=all(t>arr[i] for t in arr[i+1:min(i+y+1,n)])
uf=all(t>arr[i] for t in arr[max(i-x,0):i])
if ol==True and uf==True:
print(i+1)
break
``` | instruction | 0 | 98,702 | 4 | 197,404 |
Yes | output | 1 | 98,702 | 4 | 197,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
from collections import deque
dq = deque()
n, x, y = map(int, input().split())
arr = list(map(int, input().split()))
for i in range(y):
while dq and arr[i] < arr[dq[-1]]:
dq.pop()
dq.append(i)
ans = -1
for i in range(y, n + y):
if i < n:
while dq and dq[0] < i - y - x:
dq.popleft()
while dq and arr[i] < arr[dq[-1]]:
dq.pop()
dq.append(i)
if dq[0] == i - y:
ans = dq[0] + 1
break
else:
while dq and dq[0] < i - y - x:
dq.popleft()
if dq[0] == i - y:
ans = dq[0] + 1
break
print(ans)
# h, l = 3, 5
# print((l ** 2 - h ** 2) / (2 * h))
``` | instruction | 0 | 98,703 | 4 | 197,406 |
Yes | output | 1 | 98,703 | 4 | 197,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
a,b,c=map(int,input().split())
arr=list(map(int,input().split()))
for i in range(0,a):
flag=0
j=i+1
while(j<i+c+1 and j<a):
if(arr[i]>arr[j]):
flag=1
break
j+=1
if(flag==1):
continue
k=i-1
while(k>i-b-1 and k>=0):
if(arr[i]>arr[k]):
flag=1
break
k-=1
if(flag==0):
print(i+1)
break
``` | instruction | 0 | 98,704 | 4 | 197,408 |
Yes | output | 1 | 98,704 | 4 | 197,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
n, x, y = map(int, input().split())
array = list(map(int, input().split()))
inf = float('inf')
ar_x = [float('inf')]*x
ar_y = []
for i in range(1, min(y+1, n)):
ar_y.append(array[i])
if not ar_y or array[0] == min(ar_y):
print(1)
exit(0)
for i in range(1, n):
ar_x.append(array[i-1])
if i + y < n:
ar_y.append(array[i+y])
ar_x.pop(0)
ar_y.pop(0)
if not ar_y and not ar_x:
print(i+1)
exit(0)
elif not ar_y and array[i] < min(ar_x):
print(i+1)
exit(0)
elif not ar_x and array[i] < min(ar_y):
print(i+1)
exit(0)
elif array[i] < min(ar_y) and array[i] < min(ar_x):
print(i+1)
exit(0)
``` | instruction | 0 | 98,705 | 4 | 197,410 |
No | output | 1 | 98,705 | 4 | 197,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
n,x,y=list(map(int,input().split()))
a=list(map(int,input().split()))
d=[]
for i in range(n-1):
d.append(a[i]-a[i+1])
#print(*d)
t=n-1
for i in range(n-1):
if d[i]<0:
#print('i=',i)
f=0;g=0
if x>i:
r=i
else:
r=x
if y>n-1-i:
s=n-1-i
else:
s=y
for j in range(r):
#print('i-j-1=',i-j-1)
if d[i-j-1]<0:
f=1
break
#print(d[i-j-1])
for j in range(s-1):
#print('i+j+1=',i+j+1)
if d[i+j+1]>0:
g=1
break
#print(d[i+j+1])
if f==0 and g==0:
#print(a[i],d[i])
t=i
break
print(t+1)
``` | instruction | 0 | 98,706 | 4 | 197,412 |
No | output | 1 | 98,706 | 4 | 197,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
from collections import deque
n, x, y = map(int, input().split())
arr = list(map(int, input().split()))
left_dq = deque()
right_dq = deque()
for i in range(1, y):
while right_dq and arr[right_dq[-1]] > arr[i]:
right_dq.pop()
right_dq.append(i)
ans = -1
for i in range(n):
while left_dq and left_dq[0] < i-x:
left_dq.popleft()
while right_dq and right_dq[0] <= i:
right_dq.popleft()
if i > 0:
while left_dq and arr[left_dq[-1]] > arr[i-1]:
left_dq.pop()
left_dq.append(i-1)
min_left = arr[left_dq[0]]
else:
min_left = float('inf')
if i+y < n:
while right_dq and arr[right_dq[-1]] > arr[i+y]:
right_dq.pop()
right_dq.append(i+y)
min_right = arr[right_dq[0]]
else:
min_right = arr[right_dq[0]] if right_dq else float('inf')
if arr[i] < min_left and arr[i] < min_right:
ans = i+1
break
print(ans)
``` | instruction | 0 | 98,707 | 4 | 197,414 |
No | output | 1 | 98,707 | 4 | 197,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Submitted Solution:
```
'''
tìm vị trí sớm nhất mà lượng mưa nhỏ hơn x ngày trước và y ngày sau
'''
n,x,y=map(int,input().split(' '))
rainingday=list(map(int,input().split(' ')))
stack=[0]
left=[0 for i in range(n)]
right=[n-1 for i in range(n)]
for i in range(1,n):
while(len(stack)!=0 and rainingday[stack[-1]]>=rainingday[i]):
right[stack[-1]]=i-1
stack.pop()
if(len(stack)!=0):
left[i]=stack[-1]+1
else:
left[i]=0
stack.append(i)
for i in range(n):
if(i-left[i]>=x and right[i]-i>=y):
print(i+1)
break
elif((left[i]==0 or i-left[i]>x) and (right[i]==n-1 or right[i]-i>y)):
print(i+1)
break
``` | instruction | 0 | 98,708 | 4 | 197,416 |
No | output | 1 | 98,708 | 4 | 197,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,022 | 4 | 198,044 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
n=int(input())
a,ans,b=[],[0]*n,[0]*10000001
for i in range(n):
L,R=map(int,input().split())
a.append([R,L,i])
a.sort()
for z in a:
R,L,i=z
for j in range(L,R+1):
if b[j]==0:
b[j]=1
ans[i]=j
break
print(*ans)
``` | output | 1 | 99,022 | 4 | 198,045 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,023 | 4 | 198,046 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
n = int(input())
intervals = [list(map(int, input().split())) for i in range(n)]
intervals = list(enumerate(intervals))
intervals.sort(key=lambda x: (x[1][1], x[1][0]))
ans = [0] * n
vis = set()
for idx, (l, _) in intervals:
while l in vis:
l += 1
vis.add(l)
ans[idx] = l
print(' '.join(list(map(str, ans))))
``` | output | 1 | 99,023 | 4 | 198,047 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,024 | 4 | 198,048 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
n = int(input())
arr = [0]*10000001
ans = [0]*100
a2=[]
for i in range(n):
l,r = map(int,input().split())
tmp = (r,l,i)
a2.append(tmp)
a2.sort()
for i in range(n):
m=a2[i]
r=m[0]
l=m[1]
v=m[2]
dif=r-l+1
for j in range(dif):
if(arr[l+j]==0):
arr[l+j]=1
ans[v]=l+j
break
ans=ans[:n]
print(*ans)
``` | output | 1 | 99,024 | 4 | 198,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,025 | 4 | 198,050 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
n=int(input())
a=[[list(map(int, input().split())),i] for i in range(n)]
a.sort()
now=1
avail=[1 for i in range(int(1e7+1)+1)]
ans=[-1 for i in range(n)]
#print(a)
for k in range(n):
i=a[k]
for j in range(i[0][0], i[0][1]+1):
if avail[j]:
avail[j]=0
ans[i[1]]=j
break
for t in range(k+1, n):
if a[t][0][0]==a[k][0][0]:
a[t][0][0]+=1
else:
break
a.sort()
#now+=1
print(*ans)
return
if __name__ == "__main__":
main()
``` | output | 1 | 99,025 | 4 | 198,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,026 | 4 | 198,052 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
# sort by the end
n = int(input())
intervals = [list(map(int, input().split())) for i in range(n)]
intervals = list(enumerate(intervals))
intervals.sort(key=lambda x: (x[1][1], x[1][0]))
ans = [0] * n
vis = set()
for idx, (l, _) in intervals:
while l in vis:
l += 1
vis.add(l)
ans[idx] = l
print(' '.join(list(map(str, ans))))
``` | output | 1 | 99,026 | 4 | 198,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,027 | 4 | 198,054 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
import random
class Event:
def __init__(self, id, start, end):
self.start = start
self.end = end
self.id = id
self.pick = None
def key(self):
return (self.start, self.end)
n = int(input())
events = [ Event(i, *map(int, input().split())) for i in range(n) ]
limit = events[0].end
for event in events:
limit = max(limit, event.end)
current_events = set()
for i in range(n):
if len(current_events) == 0:
pos = limit + 1
for event in events:
if event.pick == None and event.start < pos:
pos = event.start
for event in events:
if event.pick == None and event.start == pos:
current_events.add(event)
best_event = random.sample(current_events, 1)[0]
for event in current_events:
if event.end < best_event.end:
best_event = event
best_event.pick = pos
current_events.remove(best_event)
pos += 1
result = [ events[i].pick for i in range(n) ]
print(' '.join(map(str, result)))
``` | output | 1 | 99,027 | 4 | 198,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,028 | 4 | 198,056 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
n = int(input())
freq = [0] * 10000001
ans = [0] * 100
a = []
for i in range(n):
l , r = map(int, input().split())
temp = (r, l, i)
a.append(temp)
a.sort()
for i in range(n):
r, l, v = a[i][0], a[i][1], a[i][2]
for j in range(l, r+1):
if freq[j] == 0:
freq[j] = 1
ans[v] = j
break
for i in range(n):
print(ans[i], " ", end = "")
``` | output | 1 | 99,028 | 4 | 198,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2 | instruction | 0 | 99,029 | 4 | 198,058 |
Tags: greedy, meet-in-the-middle, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
n = int(input())
freq = [0] * 10000001
ans = [0] * 100
a = []
for i in range(n):
l , r = map(int, input().split())
temp = (r, l, i)
a.append(temp)
a.sort()
for i in range(n):
r, l, v = a[i][0], a[i][1], a[i][2]
for j in range(l, r+1):
if freq[j] == 0:
freq[j] = 1
ans[v] = j
break
for i in range(n):
print(ans[i], " ", end = "")
# Made By Mostafa_Khaled
``` | output | 1 | 99,029 | 4 | 198,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2
Submitted Solution:
```
# from dust i have come, dust i will be
n = int(input())
seg = []
for i in range(n):
l, r = map(int, input().split())
seg.append((l, r, i))
seg.sort(key=lambda tup: (tup[1], tup[0], tup[2]))
assigned = set()
ans = [0] * n
for i in range(n):
# this loop won't run more than 100 times as n <= 100
for j in range(seg[i][0], seg[i][1] + 1, 1):
if j not in assigned:
ans[seg[i][2]] = j
assigned.add(j)
break
for i in range(n):
print(ans[i], end=' ')
'''
10
1 10
3 4
8 10
3 4
5 9
1 4
7 10
1 9
1 8
4 10
'''
``` | instruction | 0 | 99,030 | 4 | 198,060 |
Yes | output | 1 | 99,030 | 4 | 198,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2
Submitted Solution:
```
# http://codeforces.com/contest/45/problem/D
from sys import stdin
inFile = stdin
tokens = []
tokens_next = 0
def next_str():
global tokens, tokens_next
while tokens_next >= len(tokens):
tokens = inFile.readline().split()
tokens_next = 0
tokens_next += 1
return tokens[tokens_next - 1]
def nextInt():
return int(next_str())
n = nextInt()
intervals = [(nextInt(), nextInt(), i) for i in range(n)]
intervals.sort(key=lambda x: (x[1], x[0]))
# for i in intervals:
# print(i[0], i[1])
seen = set()
for i in range(n):
cur = intervals[i][0]
while cur in seen:
cur += 1
seen.add(cur)
intervals[i] = (intervals[i][0], intervals[i][1], intervals[i][2], cur)
intervals.sort(key=lambda x: x[2])
for i in range(n):
print(intervals[i][3], end=' ')
print()
``` | instruction | 0 | 99,031 | 4 | 198,062 |
Yes | output | 1 | 99,031 | 4 | 198,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Output
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
Examples
Input
3
1 2
2 3
3 4
Output
1 2 3
Input
2
1 3
1 3
Output
1 2
Submitted Solution:
```
n = int(input())
arr = [0]*10000001
ans = [0]*100
print()
a2=[]
for i in range(n):
l,r = map(int,input().split())
tmp = (r,l,i)
a2.append(tmp)
a2.sort()
for i in range(n):
m=a2[i]
r=m[0]
l=m[1]
v=m[2]
dif=r-l+1
for j in range(dif):
if(arr[l+j]==0):
arr[l+j]=1
ans[v]=l+j
break
for x in range(n):
print(ans[x], " ",end="")
``` | instruction | 0 | 99,032 | 4 | 198,064 |
Yes | output | 1 | 99,032 | 4 | 198,065 |
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