message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
from collections import defaultdict
n, q = map(int, input().split())
S = [0] * (n + 1)
S[0] = 1
s = n
L = defaultdict(list)
pre = -1
for i in range(q):
t, k, d = map(int, input().split())
for j in range(pre, t):
for l1 in L[j]:
for l2 in l1:
S[l2] = 0
s += 1
if s >= k:
tmp = []
for j in range(1, n + 1):
if S[j] == 0:
S[j] = 1
tmp.append(j)
if len(tmp) == k:
break
s -= k
L[t + d - 1].append(tmp)
print(sum(tmp))
pre = t
else:
print(-1)
pre = t
``` | instruction | 0 | 58,541 | 14 | 117,082 |
Yes | output | 1 | 58,541 | 14 | 117,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
n, q = [int(i) for i in input().split()]
available = [1 for i in range(n)]
for task in range(q):
t, k , d = [int(i) for i in input().split()]
assign = []
len_assign = 0
for i in range(n):
if available[i] <= t:
assign.append(i)
len_assign += 1
if len_assign >= k:
break
if len_assign >= k:
for j in range(k):
available[assign[j]] = t+d
print(sum(assign)+k)
else:
print(-1)
``` | instruction | 0 | 58,542 | 14 | 117,084 |
Yes | output | 1 | 58,542 | 14 | 117,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
n, q = map(int, input().split())
a = [0]*n
for _ in range(q):
t, k, d = map(int, input().split())
c = 0
test = 0
val = []
for i in range(n):
if a[i]<=t:
val.append(i)
c+=(i+1)
k-=1
if k==0:
test = 1
break
if test:
for i in val:
a[i]=t+d
print(c)
else:
print(-1)
``` | instruction | 0 | 58,543 | 14 | 117,086 |
Yes | output | 1 | 58,543 | 14 | 117,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
for t in range(1):
n,q=map(int,input().split())
time=[0]*n
for i in range(q):
t,k,d=map(int,input().split())
count=0
pos=0
for j in range(n):
if time[j]<=t:
pos+=j+1
count+=1
if count==k:
break
if count<k:
print(-1)
continue
changed=0
for j in range(n):
if time[j]<=t:
time[j]=t+d
changed+=1
if changed==count:
break
print(pos)
``` | instruction | 0 | 58,544 | 14 | 117,088 |
Yes | output | 1 | 58,544 | 14 | 117,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
import heapq
n, q = [int(i) for i in input().split(' ')]
server = [i for i in range(1, n+1)]
running = []
for _ in range(q):
t, k, d = [int(i) for i in input().split(' ')]
if running:
not_yet = []
while running:
r = running.pop()
if t - r['t'] >= r['d']:
server += r['occupied']
else:
not_yet.append(r)
running = not_yet
heapq.heapify(server)
if len(server) < k:
print(-1)
continue
else:
# run_server = [heapq.heappop(server) for _ in range(k)]
running.append({'t': t, 'd': d, 'occupied': server[0:k].copy()})
print(sum(server[0:k]))
server = server[k:]
``` | instruction | 0 | 58,545 | 14 | 117,090 |
No | output | 1 | 58,545 | 14 | 117,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
f = lambda: map(int, input().split())
n, q = f()
s = [0] * n
for i in range(q):
t, k, d = f()
p = [j for j in range(n) if s[j] <= t][:k]
p = [] if len(p) < k else p[:k]
for j in p: s[j] = t + d
print(-1 if p else sum(p) + k)
``` | instruction | 0 | 58,546 | 14 | 117,092 |
No | output | 1 | 58,546 | 14 | 117,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
n, q = map(int, input().split())
E = []
for i in range(q):
t, k, d = map(int, input().split())
E.append([t, 1, k, i])
E.append([t + d, -1, k, i])
E.sort()
b = 0
ans = []
get_it = [0] * (q + 1)
for t, typ, k, i in E:
if typ == -1 and get_it[i]:
b -= k
elif typ == 1:
if b + k <= n:
get_it[i] = True
ans.append((b + k) * (b + k + 1) // 2 - b * (b + 1) // 2)
b += k
else:
ans.append(-1)
print('\n'.join(list(map(str, ans))))
``` | instruction | 0 | 58,547 | 14 | 117,094 |
No | output | 1 | 58,547 | 14 | 117,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.
It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.
To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.
Write the program that determines which tasks will be performed and which will be ignored.
Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.
Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.
Output
Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.
Examples
Input
4 3
1 3 2
2 2 1
3 4 3
Output
6
-1
10
Input
3 2
3 2 3
5 1 2
Output
3
3
Input
8 6
1 3 20
4 2 1
6 5 5
10 1 1
15 3 6
21 8 8
Output
6
9
30
-1
15
36
Note
In the first example in the second 1 the first task will come, it will be performed on the servers with ids 1, 2 and 3 (the sum of the ids equals 6) during two seconds. In the second 2 the second task will come, it will be ignored, because only the server 4 will be unoccupied at that second. In the second 3 the third task will come. By this time, servers with the ids 1, 2 and 3 will be unoccupied again, so the third task will be done on all the servers with the ids 1, 2, 3 and 4 (the sum of the ids is 10).
In the second example in the second 3 the first task will come, it will be performed on the servers with ids 1 and 2 (the sum of the ids is 3) during three seconds. In the second 5 the second task will come, it will be performed on the server 3, because the first two servers will be busy performing the first task.
Submitted Solution:
```
a,b=map(int,input().split())
l=[a]*1000001
for i in range(b):
t,k,d=map(int,input().split())
h=a
r=0
g=True
for i in range(t,t+d):
l[i]=l[i]-k
if l[i]<0:
print(-1)
g=False
break
else:
h=min(l[i],h)
if g:
for i in range(k):
r+=a-i-h
print(r)
``` | instruction | 0 | 58,548 | 14 | 117,096 |
No | output | 1 | 58,548 | 14 | 117,097 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,666 | 14 | 117,332 |
"Correct Solution:
```
N=int(input())
*A,=map(int, input().split())
B=[(i+1,a) for i,a in enumerate(A)]
from operator import itemgetter
B.sort(reverse=True, key=itemgetter(1))
dp = [[-1] * (N+1) for _ in range(N+1)]
dp[0][0] = 0
for i in range(1,N+1):
idx, act = B[i-1]
for j in range(i+1):
k = i-j
if 0<j: dp[j][k] = max(dp[j][k], dp[j-1][k] + act * abs(idx-j))
if 0<k: dp[j][k] = max(dp[j][k], dp[j][k-1] + act * abs(idx-(N-k+1)))
ans=0
for j in range(N+1):
ans = max(ans, dp[j][N-j])
print(ans)
``` | output | 1 | 58,666 | 14 | 117,333 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,667 | 14 | 117,334 |
"Correct Solution:
```
def solve(n, aaa):
dp = [0]
aaa_with_idx = list(zip(aaa, range(n)))
aaa_with_idx.sort(reverse=True)
for k in range(1, n + 1):
ndp = [0] * (k + 1)
a, i = aaa_with_idx[k - 1]
for l in range(k):
# 左に配置
ndp[l + 1] = max(ndp[l + 1], dp[l] + a * abs(i - l))
# 右に配置
r = n - (k - l)
ndp[l] = max(ndp[l], dp[l] + a * abs(i - r))
dp = ndp
return max(dp)
n = int(input())
aaa = list(map(int, input().split()))
print(solve(n, aaa))
``` | output | 1 | 58,667 | 14 | 117,335 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,668 | 14 | 117,336 |
"Correct Solution:
```
e=enumerate
n,a=open(0)
d=[0]+[-1e18]*2001
for j,(a,i)in e(sorted((-int(a),i)for i,a in e(a.split()))):d=[max(t-a*abs(~i-j+k+int(n)),d[k-1]-a*abs(~i+k))for k,t in e(d)]
print(max(d))
``` | output | 1 | 58,668 | 14 | 117,337 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,669 | 14 | 117,338 |
"Correct Solution:
```
N = int(input())
def solve(a, i, prev):
r = N - len(prev) - i + 1
p = -i*a
for j, s in enumerate(prev):
yield p+abs(j-i)*a, s+abs(j+r)*a
p = s
yield s+abs(len(prev)-i)*a,
pd = [0]
A = map(int, input().split())
for a,i in sorted(((a, i) for i, a in enumerate(A, 1)), reverse=True):
pd = [*map(max, solve(a,i, pd))]
print(max(pd))
``` | output | 1 | 58,669 | 14 | 117,339 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,670 | 14 | 117,340 |
"Correct Solution:
```
e=enumerate
n,*a=map(int,open(0).read().split())
dp=[0]
for j,(a,i)in e(sorted((a,i)for i,a in e(a))[::-1]):dp=[dp[0]+a*abs(n-j-i-1)]+[max(dp[k]+a*abs(n-j+k-i-1),dp[k-1]+a*abs(i-k+1))for k in range(1,j+1)]+[dp[j]+a*abs(i-j)]
print(max(dp))
``` | output | 1 | 58,670 | 14 | 117,341 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,671 | 14 | 117,342 |
"Correct Solution:
```
n = int(input())
A = [(a, i) for i, a in enumerate(map(int, input().split()))]
A.sort(reverse=True)
dp = [[0]*(n+1) for _ in range(n+1)]
ans = 0
for L in range(n+1):
for R in range(n+1):
if n <= L+R-1:
continue
a, i = A[L+R-1]
if 0 <= R-1:
dp[L][R] = max(dp[L][R], dp[L][R-1]+abs((n-R)-i)*a)
if 0 <= L-1:
dp[L][R] = max(dp[L][R], dp[L-1][R]+abs(i-(L-1))*a)
ans = max(ans, dp[L][R])
print(ans)
``` | output | 1 | 58,671 | 14 | 117,343 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,672 | 14 | 117,344 |
"Correct Solution:
```
N = int(input())
A = map(int, input().split())
AI = sorted(((a, i) for i, a in enumerate(A, 1)), reverse=True)
def solve(a, i, prev):
pl, pr, ps = i, 0, 0
for l, r, s in prev:
yield l, r-1, max(s+abs(r-i)*a, ps+abs(i-pl)*a)
pl, pr, ps = l, r, s
yield pl+1, pr, ps+abs(i-pl)*a
prev = [(1, N, 0)]
for a,i in AI:
prev = [*solve(a,i, prev)]
print(max(s for l, r, s in prev))
``` | output | 1 | 58,672 | 14 | 117,345 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85 | instruction | 0 | 58,673 | 14 | 117,346 |
"Correct Solution:
```
from heapq import heappush, heappop
n = int(input())
A = list(map(int, input().split()))
f_inf = float('inf')
dp = [[0 for _ in range(n+1)] for _ in range(n+1)]
dp[0][0] = 0
hq = []
for i in range(n):
heappush(hq, (-A[i], i))
ans = 0
for i in range(n):
x, pi = heappop(hq)
for l in range(i+1):
r = i-l
dp[i+1][l+1] = max(dp[i+1][l+1], dp[i][l] + (pi-l)*A[pi])
dp[i+1][l] = max(dp[i+1][l], dp[i][l] + ((n-r-1)-pi)*A[pi])
ans = 0
for i in range(n+1):
ans = max(ans, dp[n][i])
print(ans)
``` | output | 1 | 58,673 | 14 | 117,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
# かつっぱさん見て
from collections import defaultdict
n=int(input())
a=list(map(int,input().split()))
babies=[(q,i) for i,q in enumerate(a)]
babies.sort(reverse=True)
dp=defaultdict(int)
ans=0
for l in range(n):
for r in range(n):
if l+r==n:
ans=max(ans,dp[(l,r)])
break
act,ind=babies[l+r]
dp[(l+1,r)]=max(dp[(l+1,r)],dp[(l,r)]+act*abs(ind-l))
dp[(l,r+1)]=max(dp[(l,r+1)],dp[(l,r)]+act*abs(ind-(n-1-r)))
print(ans)
``` | instruction | 0 | 58,674 | 14 | 117,348 |
Yes | output | 1 | 58,674 | 14 | 117,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
def solve():
N = int(input())
A = [(a, i+1) for i, a in enumerate(map(int, input().split()))]
A.sort(reverse=True)
INF = float('inf')
dp = [[-INF] * (N+1) for _ in range(N+1)]
dp[0][0] = 0
for s in range(1, N+1):
for l in range(s+1):
r = s - l
dp[l][r] = max(dp[l-1][r] + A[s-1][0] * abs(A[s-1][1]-l), dp[l][r-1] + A[s-1][0] * abs(N-r+1-A[s-1][1]))
ans = 0
for m in range(N):
if dp[m][N-m] > ans:
ans = dp[m][N-m]
print(ans)
solve()
``` | instruction | 0 | 58,675 | 14 | 117,350 |
Yes | output | 1 | 58,675 | 14 | 117,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
from heapq import heappush, heappop
n = int(input())
A = list(map(int, input().split()))
MX = 2005
f_inf = float('inf')
dp = [[0 for _ in range(MX)] for _ in range(MX)]
dp[0][0] = 0
hq = []
for i in range(n):
heappush(hq, (-A[i], i))
ans = 0
for i in range(n):
x, pi = heappop(hq)
for l in range(i+1):
r = i-l
dp[i+1][l+1] = max(dp[i+1][l+1], dp[i][l] + (pi-l)*A[pi])
dp[i+1][l] = max(dp[i+1][l], dp[i][l] + ((n-r-1)-pi)*A[pi])
ans = 0
for i in range(n+1):
ans = max(ans, dp[n][i])
print(ans)
``` | instruction | 0 | 58,676 | 14 | 117,352 |
Yes | output | 1 | 58,676 | 14 | 117,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
n=int(input())
d=[0]+[-10**18]*n
for j,(a,i)in enumerate(sorted((-a,i)for i,a in enumerate(map(int,input().split())))):
d=[max(t-a*abs(~i-j+k+n),d[k-1]-a*abs(~i+k))for k,t in enumerate(d)]
print(max(d))
``` | instruction | 0 | 58,677 | 14 | 117,354 |
Yes | output | 1 | 58,677 | 14 | 117,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
n = int(input())
a = [(A, i) for i, A in enumerate(map(int, input().split()))]
a.sort(reverse=True)
dp = [[-1]*(n+1) for _ in range(n+1)]
dp[0][0] = 0
for num, (A, pos) in enumerate(a):
for i in range(num+1):
j = num - i
dp[i+1][j] = max(dp[i+1][j], dp[i][j] + A*abs(pos-i)) # 左からi番目の時
dp[i][j+1] = max(dp[i][j+1], dp[i][j] + A*abs((n-1-j)-pos)) # 右からj番目の時
ans = 0
for i in range(n+1):
j = n-i
ans = max(ans, dp[i][j])
print(ans)
``` | instruction | 0 | 58,678 | 14 | 117,356 |
No | output | 1 | 58,678 | 14 | 117,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
N=int(input())
A=list(map(int,input().split()))
B=[(j,i) for i,j in enumerate(A)]
B.sort(reverse=True)
dp=[[0]*(N+1) for _ in range(N+1)]
c=0
for v,k in B:
for i in range(c+1):
dp[i+1][c-i]=max(dp[i+1][c-i],dp[i][c-i]+abs(v*(k-i)))
dp[i][c-i+1]=max(dp[i][c-i+1],dp[i][c-i]+abs(v*(N-(c-i)-k-1)))
c+=1
ans=0
for i in range(N):
ans=max(ans,dp[i][N-i])
print(ans)
``` | instruction | 0 | 58,679 | 14 | 117,358 |
No | output | 1 | 58,679 | 14 | 117,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
e=enumerate
n,a=open(0)
d=[0]+[-1e18]*2000
for j,(a,i)in e(sorted((-int(a),i)for i,a in e(a.split()))):d=[max(t-a*(~i-j+k+int(n)),d[k-1]-a*abs(~i+k))for k,t in e(d)]
print(max(d))
``` | instruction | 0 | 58,680 | 14 | 117,360 |
No | output | 1 | 58,680 | 14 | 117,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.
You can rearrange these children just one time in any order you like.
When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.
Find the maximum total happiness points the children can earn.
Constraints
* 2 \leq N \leq 2000
* 1 \leq A_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the maximum total happiness points the children can earn.
Examples
Input
4
1 3 4 2
Output
20
Input
6
5 5 6 1 1 1
Output
58
Input
6
8 6 9 1 2 1
Output
85
Submitted Solution:
```
def main():
n = int(input())
active = [int(i) for i in input().split()]
moved = [1 for _ in range(n)]
visited = [1 for _ in range(n)]
score = []
for i, a in enumerate(active):
for j in range(n):
score.append((a * abs(i-j), a , i, j))
score = sorted(score, key=lambda x: x[0] + 1/x[1], reverse=True)
res = 0
for sc in score:
infant = sc[2]
pos = sc[3]
if visited[pos] and moved[infant]:
res += sc[0]
visited[pos] = 0
moved[infant] = 0
print(res)
if __name__ == '__main__':
main()
``` | instruction | 0 | 58,681 | 14 | 117,362 |
No | output | 1 | 58,681 | 14 | 117,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,164 | 14 | 118,328 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
weights = map(int, input())
weights = [cnt for cnt, x in enumerate(weights, 1) if x]
m = int(input())
state = [(0, 0, 0, [])]
res = "NO"
while state:
w, b, k, l = state.pop()
if k == m:
res = 'YES\n' + ' '.join(map(str, l))
break
for wt in reversed(weights):
if wt != w and wt > b:
state.append((wt, wt-b, k+1, l+[wt]))
print(res)
``` | output | 1 | 59,164 | 14 | 118,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,165 | 14 | 118,330 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit(2000)
weights = input()
m=int(input())
l = [x+1 for x in range(len(weights)) if weights[x]=='1']
sol = []
def solve(x, m, diff):
if m==0:
return True
for p in l:
if p>diff and p!=x:
sol.append(p)
if solve(p, m-1, p-diff):
return True
sol.pop()
return False
if solve(0,m,0):
print("YES")
for x in sol:
print(x, end=' ')
else:
print("NO")
``` | output | 1 | 59,165 | 14 | 118,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,166 | 14 | 118,332 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
ans=[]
def dfs(bal,m,last,flag):
x=bal
#print(bal,m,last)
if(m==te):
return 1
for i in w:
if(i==last):
continue
if(flag): x+=i
else: x-=i
if((x>0 and flag) or (x<0 and not flag)) and dfs(x,m+1,i,not flag):
ans.append(i)
return 1
if(flag): x-=i
else: x+=i
return 0
s=input()
te=int(input())
w=[]
for i in range(len(s)):
if(s[i]=='1'):
w.append(i+1)
dfs(0,0,0,1)
if(ans):
print("YES")
print(*ans[::-1])
else:
print("NO")
``` | output | 1 | 59,166 | 14 | 118,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,167 | 14 | 118,334 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
# Target - Expert on CF
# Be Humblefool
import sys
# inf = float("inf")
# sys.setrecursionlimit(10000000)
# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
# mod, MOD = 1000000007, 998244353
# words = {1:'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine',10:'ten',11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',15:'quarter',16:'sixteen',17:'seventeen',18:'eighteen',19:'nineteen',20:'twenty',21:'twenty one',22:'twenty two',23:'twenty three',24:'twenty four',25:'twenty five',26:'twenty six',27:'twenty seven',28:'twenty eight',29:'twenty nine',30:'half'}
# vow=['a','e','i','o','u']
# dx,dy=[-1,1,0,0],[0,0,1,-1]
# import random
# from collections import deque, Counter, OrderedDict,defaultdict
# from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# from math import ceil,floor,log,sqrt,factorial,pi,gcd
# from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
string = input()
p = [i for i,j in enumerate(string, 1) if j=='1']
m = int(input())
queue = [(-1,0,0,[])] # prev, position, till_sum, list
while queue:
prev,position,till_sum,lst = queue.pop()
if position==m:
print('YES')
print(*lst)
exit()
for q in p:
if q!=prev and q>till_sum:
queue.append((q, position+1, q-till_sum,lst+[q]))
print('NO')
``` | output | 1 | 59,167 | 14 | 118,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,168 | 14 | 118,336 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
p, m = [i for i, x in enumerate(input(), 1) if x == '1'], int(input())
r, q = [(-1, 0, 0, [])], 'NO'
while r:
x, d, s, t = r.pop()
if s == m:
q = 'YES\n' + ' '.join(map(str, t))
break
for y in p:
if y != x and y > d: r.append((y, y - d, s + 1, t + [y]))
print(q)
# Made By Mostafa_Khaled
``` | output | 1 | 59,168 | 14 | 118,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,169 | 14 | 118,338 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
def main():
alpha = 'abcdefghijklmnopqrstuvwxyz'
ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
inf = 1e17
mod = 10 ** 9 + 7
# Max = 10 ** 1
# primes = []
# prime = [True for i in range(Max + 1)]
# p = 2
# while (p * p <= Max + 1):
#
# # If prime[p] is not
# # changed, then it is a prime
# if (prime[p] == True):
#
# # Update all multiples of p
# for i in range(p * p, Max + 1, p):
# prime[i] = False
# p += 1
#
# for p in range(2, Max + 1):
# if prime[p]:
# primes.append(p)
#
# print(primes)
def factorial(n):
f = 1
for i in range(1, n + 1):
f = (f * i) % mod # Now f never can
# exceed 10^9+7
return f
def ncr(n, r):
# initialize numerator
# and denominator
num = den = 1
for i in range(r):
num = (num * (n - i)) % mod
den = (den * (i + 1)) % mod
return (num * pow(den,
mod - 2, mod)) % mod
def solve(s,m):
#print(s)
queue = [[0,0,0,[]]]# (diff,last,taken,answer)
while queue:
diff,last,taken,answer = queue.pop()
if taken == m:
print('YES')
print(" ".join(list(map(str,answer))))
return
for opt in s:
if opt != last and opt - diff > 0:
queue.append([opt-diff,opt,taken+1,answer+[opt]])
#print(queue)
print('NO')
pass
t = 1#int(input())
ans = []
for _ in range(t):
# n = int(input())
#c, m, p, v = map(float, input().split())
s = input()
s_ = []
for i in range(len(s)):
if s[i] == '1':
s_.append(i+1)
m = int(input())
#arr = [int(x) for x in input().split()]
# a2 = [int(x) for x in input().split()]
# grid = []
# for i in range(n):
# grid.append([int(x) for x in input().split()][::-1])
ans.append(solve(s_,m))
# for answer in ans:
# print(answer)
if __name__ == "__main__":
import sys, threading
import bisect
import math
import itertools
from sys import stdout
# Sorted Containers
import heapq
from queue import PriorityQueue
# Tree Problems
# sys.setrecursionlimit(2 ** 32 // 2 - 1)
# threading.stack_size(1 << 27)
# fast io
input = sys.stdin.readline
thread = threading.Thread(target=main)
thread.start()
thread.join()
``` | output | 1 | 59,169 | 14 | 118,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,170 | 14 | 118,340 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
if __name__=='__main__':
inp = input()
S = []
for i in range(10):
if inp[i]=='1':
S.append(i+1)
m = int(input())
L = []
ba = 0
bb = 0
ta = True
for x in range(len(S)):
lt=[]
lt.append(S[x])
lst = S[x]
ta = False
ba = S[x]
bb = 0
for i in range(m-1):
fnd = False
for s in range(len(S)):
if lst!=S[s]:
if ta:
if (ba+S[s])>bb:
ba+=S[s]
lt.append(S[s])
lst = S[s]
fnd = True
break
else:
if (bb+S[s])>ba:
bb+=S[s]
lt.append(S[s])
lst = S[s]
fnd = True
break
if fnd:
ta=not(ta)
else:
break
if len(lt)==m:
L=lt[:]
break
if len(L)<m :
print("NO")
else:
print("YES")
ret = str(L[0])
for i in range(1,m):
ret += ' '+str(L[i])
print(ret)
``` | output | 1 | 59,170 | 14 | 118,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | instruction | 0 | 59,171 | 14 | 118,342 |
Tags: constructive algorithms, dfs and similar, dp, graphs, greedy, shortest paths
Correct Solution:
```
a=[i for i,j in enumerate(input(),1) if j=='1']
m=int(input())
q=[(-1,0,0,[])]
p='NO'
while q:
x,y,z,t=q.pop()
if z==m:
p='YES\n'+' '.join(map(str,t))
break
for i in a:
if i!=x and i>y:
q.append((i,i-y,z+1,t+[i]))
print(p)
``` | output | 1 | 59,171 | 14 | 118,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
import traceback
import math
from collections import defaultdict
from functools import lru_cache
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def geti():
return int(input())
def gets():
return input()
def getil():
return list(map(int, input().split()))
def getsl():
return input().split()
def get2d(nrows, ncols, n=0):
return [[n] * ncols for r in range(nrows)]
inf = float('inf')
mod = 10 ** 9 + 7
def main():
w = gets()
m = geti()
W = [i+1 for i, k in enumerate(w) if k=='1']
ans = []
def do(tot, prv, k):
if k == m:
return True
for w in W:
if w == prv or w <= abs(tot): continue
if do(tot+(w if k%2==0 else -w), w, k+1):
ans.append(w)
return True
return False
return do(0, -1, 0), ans[::-1]
try:
ans, arr = main()
ans = 'YES' if ans else 'NO'
print(ans)
if arr:
print(*arr)
except Exception as e:
traceback.print_exc()
``` | instruction | 0 | 59,172 | 14 | 118,344 |
Yes | output | 1 | 59,172 | 14 | 118,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
__author__ = 'ratnesh.mishra'
weights = map(int, input())
weights = [cnt for cnt, x in enumerate(weights, 1) if x]
m = int(input())
state = [(0, 0, 0, [])]
res = "NO"
while state:
w, b, k, l = state.pop()
if k == m:
res = 'YES\n' + ' '.join(map(str, l))
break
for wt in weights:
if wt != w and wt > b:
state.append((wt, wt-b, k+1, l+[wt]))
print(res)
``` | instruction | 0 | 59,173 | 14 | 118,346 |
Yes | output | 1 | 59,173 | 14 | 118,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
import sys
sys.setrecursionlimit(2000)
n = input()
m = int(input())
arr = [i+1 for i in range(10) if n[i]=='1']
if len(arr)<1:
print('NO')
exit(0)
a = []
def dfs(x, y, k):
if k>m:
return 1
f = 0
for i in arr:
if x>=0 and y!=i and (x-i)<0:
f = dfs(x-i, i, k+1)
elif x<0 and y!=i and (x+i)>0:
f = dfs(x+i, i, k+1)
if f==1:
a.append(i)
return 1
return 0
result = dfs(0, 0, 1)
if result==1:
print("YES")
for i in a[::-1]:
print(i, end=' ')
else:
print("NO")
``` | instruction | 0 | 59,174 | 14 | 118,348 |
Yes | output | 1 | 59,174 | 14 | 118,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
R = lambda: map(int, input().split())
ws = [i + 1 for i, x in enumerate(map(int, input())) if x]
m = int(input())
acc = [0, 0]
seq = []
idx = 0
def dfs(acc, seq, idx):
if len(seq) == m:
print('YES')
print(' '.join(map(str, seq)))
return 1
for w in ws:
if (not seq or w != seq[-1]) and acc[idx] + w > acc[1 - idx]:
acc[idx] += w
seq.append(w)
idx = 1 - idx
if not dfs(acc, seq, idx):
idx = 1 - idx
acc[idx] -= w
seq.pop(-1)
else:
return 1
if not dfs(acc, seq, idx):
print('NO')
``` | instruction | 0 | 59,175 | 14 | 118,350 |
Yes | output | 1 | 59,175 | 14 | 118,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
weight=[x for x,y in enumerate(input(),1) if y=='1']
l,r=0,0
prev=0
c=True
a=[]
for i in range(int(input())):
z=0
for j in weight:
if c and j!=prev and l+j>r:
a.append(j)
l+=j
prev=j
c=not c
z=1
break
elif not c and j!=prev and r+j>l:
a.append(j)
r+=j
prev=j
c=not c
z=1
break
if not z:
print('NO')
exit(0)
print('YES')
print(*a)
``` | instruction | 0 | 59,176 | 14 | 118,352 |
No | output | 1 | 59,176 | 14 | 118,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
__author__ = 'ratnesh.mishra'
weights = map(int, input())
weights = [cnt+1 for cnt, x in enumerate(weights) if x]
m = int(input())
if m == 1 or len(weights) == 1:
print('NO')
else:
ll = [str(weights[0])]
print('YES')
print(' '.join([str(weights[-1]) if ant else str(weights[0]) for ant in range(m) ]))
``` | instruction | 0 | 59,177 | 14 | 118,354 |
No | output | 1 | 59,177 | 14 | 118,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
x = input()
n = int( input() )
L = []
for i in range( len(x) ):
if x[i] == '1':
L.append( i+1 )
if n == 1 and len(L) == 1:
print( "YES" )
print( L[0] )
exit(0)
if len(L) <= 1:
print( "NO" )
exit(0)
#print( L )
fl = True
S = []
s1, s2 = 0, 0
last = -1
if L[0] == 1:
del L[0]
for i in range( n ):
if fl:
ok = False
for i in L:
if i != last and s1+i > s2:
ok = True
S.append(i)
last = i
s1 += i
break
if ok == False:
print( "NO" )
exit(0)
else:
ok = False
for i in L:
if i != last and s2+i > s1:
ok = True
S.append(i)
last = i
s2 += i
break
if ok == False:
print( "NO" )
exit(0)
fl = not fl
print( "YES" )
for i in S:
print( i, end=' ' )
print()
``` | instruction | 0 | 59,178 | 14 | 118,356 |
No | output | 1 | 59,178 | 14 | 118,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
# from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
s=input()
m=L()[0]
A=[]
for i in range(10):
if s[i]=="1":
A.append(i+1)
if len(A)==1:
print("NO")
exit()
a=-1
b=-1
for i in range(len(A)-1):
if A[i]*2>A[i+1]:
a=A[i]
b=A[i+1]
break
if a==-1:
print("NO")
exit()
print("YES")
for i in range(m):
if i%2==0:
print(a,end=" ")
else:
print(b,end=" ")
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | instruction | 0 | 59,179 | 14 | 118,358 |
No | output | 1 | 59,179 | 14 | 118,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4. | instruction | 0 | 59,203 | 14 | 118,406 |
Tags: implementation, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
arr = list(map(int, input().split()))
price = sum(abs(a - b) for a, b in zip(arr, arr[1:]))
from collections import defaultdict
adj = defaultdict(list)
for a,b in zip(arr, arr[1:]):
if a != b:
adj[a].append(b)
adj[b].append(a)
def median(xs):
return sorted(xs)[len(xs)//2]
res = 0
for i, xs in adj.items():
m = median(xs)
a = sum(abs(i - x) for x in xs)
b = sum(abs(m - x) for x in xs)
res = max(res, a - b)
print(price - res)
``` | output | 1 | 59,203 | 14 | 118,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4. | instruction | 0 | 59,204 | 14 | 118,408 |
Tags: implementation, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
s, p = 0, [[] for i in range(n + 1)]
t = list(map(int, input().split()))
for i in range(m - 1):
if t[i] != t[i + 1]:
p[t[i + 1]].append(t[i])
p[t[i]].append(t[i + 1])
for i, q in enumerate(p):
if q:
q.sort()
k = q[len(q) // 2]
d = sum(abs(i - j) - abs(k - j) for j in q)
if d > s: s = d
print(sum(abs(t[i + 1] - t[i]) for i in range(m - 1)) - s)
``` | output | 1 | 59,204 | 14 | 118,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4. | instruction | 0 | 59,205 | 14 | 118,410 |
Tags: implementation, math, sortings
Correct Solution:
```
import base64,zlib
exec(zlib.decompress(base64.b85decode(b'c${@mOKt-p4Bc~z*PT)6B<(U%dJ)8eYDh&g5Poz2noK66M5-4c+wWsLDg+!6Y$6|<f{$r#B!`+N&%YE^Au(XDptZs2THqOkV&ku)h<AKSBN`(nUmN|aBgG3f;Df0L0S<pc)863i*#E9+mPU`H*P0n@N-mDn!SWgK)V!hZjtaS(x;O_{Y;%-0)4aLlp<W9Q`Y2p@rK~3xDNF2j_%e^y$gMgDvOkqPuW^?A^yIQ6-S^y0^Crh`Bfe1&sLUQkDtsndRo0oAf7|W#&dag7O_hH&R&{3')))
``` | output | 1 | 59,205 | 14 | 118,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4. | instruction | 0 | 59,206 | 14 | 118,412 |
Tags: implementation, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = {}
for ai in a:
b[ai] = []
amt = 0
for i in range(1, len(a)):
amt += abs(a[i]-a[i-1])
if a[i]-a[i-1]:
b[a[i]] += [a[i-1]]
b[a[i-1]] += [a[i]]
diff = 0
for bi in b:
if b[bi]:
center = sorted(b[bi])[len(b[bi])//2]
sum = 0
for page in b[bi]:
sum += abs(page-center)-abs(page-bi)
diff = min(diff, sum)
print(amt+diff)
``` | output | 1 | 59,206 | 14 | 118,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4. | instruction | 0 | 59,207 | 14 | 118,414 |
Tags: implementation, math, sortings
Correct Solution:
```
def median(a):
if len(a) == 0:
return 0
if len(a) % 2 == 1:
return a[len(a) // 2]
else:
return (a[len(a) // 2] + a[(len(a) // 2) - 1]) // 2
def profit(a, old_val):
a.sort()
med = median(a)
sum_old = 0
sum_new = 0
for i in a:
sum_old += abs(i - old_val)
sum_new += abs(i - med)
return sum_old - sum_new
n, m = [int(c) for c in input().split()]
pages = [int(c) for c in input().split()]
count = {pages[0]: []}
current_page_switches = 0
for i in range(1, len(pages)):
cur_i = pages[i]
prev_i = pages[i - 1]
if not(cur_i in count):
count[cur_i] = []
if cur_i != prev_i:
count[cur_i].append(prev_i)
count[prev_i].append(cur_i)
current_page_switches += abs(cur_i - prev_i)
max_profit = 0
for i in count:
if len(count[i]) > 0:
tmp = profit(count[i], i)
if tmp > max_profit:
max_profit = tmp
print(current_page_switches - max_profit)
``` | output | 1 | 59,207 | 14 | 118,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4.
Submitted Solution:
```
n, m = map(int, input().split())
s, p = 0, [[] for i in range(n + 1)]
t = list(map(int, input().split()))
for i in range(m - 1):
if t[i] != t[i + 1]:
p[t[i + 1]].append(t[i])
p[t[i]].append(t[i + 1])
for i, q in enumerate(p):
if q:
q.sort()
k = q[len(q) // 2]
d = sum(abs(i - j) - abs(k - j) for j in q)
if d > s: s = d
print(sum(abs(t[i + 1] - t[i]) for i in range(m - 1)) - d)
``` | instruction | 0 | 59,208 | 14 | 118,416 |
No | output | 1 | 59,208 | 14 | 118,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4.
Submitted Solution:
```
n, m = [int(c) for c in input().split()]
pages = [int(c) for c in input().split()]
count = {pages[0]: []}
current_page_switches = 0
def median(a):
if len(a) == 0:
return 0
if len(a) % 2 == 1:
return a[len(a) // 2]
else:
return (a[len(a) // 2] + a[(len(a) // 2) - 1]) // 2
def profit(a, old_val):
med = median(a)
sum_old = 0
sum_new = 0
for i in a:
sum_old += abs(i - old_val)
sum_new += abs(i - med)
return sum_old - sum_new
for i in range(1, len(pages)):
cur_i = pages[i]
prev_i = pages[i - 1]
if not(cur_i in count):
count[cur_i] = []
count[cur_i].append(prev_i)
count[prev_i].append(cur_i)
current_page_switches += abs(cur_i - prev_i)
max_profit = 0
for i in count:
if len(count[i]) > 0:
tmp = profit(count[i], i)
if tmp > max_profit:
max_profit = tmp
print(current_page_switches - max_profit)
``` | instruction | 0 | 59,209 | 14 | 118,418 |
No | output | 1 | 59,209 | 14 | 118,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4.
Submitted Solution:
```
n, m = [int(c) for c in input().split()]
pages = [int(c) for c in input().split()]
def count_switch(a, b):
res = 0
for i in range(1,len(pages)):
f = pages[i-1] if pages[i-1] != a else b
t = pages[i] if pages[i] != a else b
res += abs(f - t)
return res
#
#
#
# res = []
# counted = {}
#
# for i in range(1,len(pages)):
# if not((pages[i-1], pages[i]) in counted):
# f1 = min(count_switch(pages[i-1], pages[i]), count_switch(pages[i], pages[i-1]))
# counted[(pages[i-1], pages[i])] = True
# counted[(pages[i], pages[i-1])] = True
# if len(res) == 0 or f1 < res[0]:
# res.insert(0, f1)
count = {}
res = -1
def add_count (a,b):
f = min(a,b)
s = max(a,b)
cur = 0 if not((f, s) in count) else count[(f, s)]
count[(f,s)] = cur + (s - f)
for i in range(1, len(pages)):
add_count(pages[i - 1],pages[i])
max_c = max(count.values())
for i in count.keys():
if count[i] == max_c:
f1 = min(count_switch(i[0], i[1]), count_switch(i[1], i[0]))
if res == -1 or f1 < res:
res = f1
print(res)
``` | instruction | 0 | 59,210 | 14 | 118,420 |
No | output | 1 | 59,210 | 14 | 118,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4.
Submitted Solution:
```
def median(a):
if len(a) == 0:
return 0
if len(a) % 2 == 1:
return a[len(a) // 2]
else:
return (a[len(a) // 2] + a[(len(a) // 2) - 1]) // 2
def profit(a, old_val):
a.sort()
med = median(a)
sum_old = 0
sum_new = 0
for i in a:
sum_old += abs(i - old_val)
sum_new += abs(i - med)
return sum_old - sum_new
n, m = [int(c) for c in input().split()]
pages = [int(c) for c in input().split()]
count = {pages[0]: []}
current_page_switches = 0
for i in range(1, len(pages)):
cur_i = pages[i]
prev_i = pages[i - 1]
if not(cur_i in count):
count[cur_i] = []
count[cur_i].append(prev_i)
count[prev_i].append(cur_i)
current_page_switches += abs(cur_i - prev_i)
max_profit = 0
for i in count:
if len(count[i]) > 0:
tmp = profit(count[i], i)
if tmp > max_profit:
max_profit = tmp
print(current_page_switches - max_profit)
``` | instruction | 0 | 59,211 | 14 | 118,422 |
No | output | 1 | 59,211 | 14 | 118,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a social website with n fanpages, numbered 1 through n. There are also n companies, and the i-th company owns the i-th fanpage.
Recently, the website created a feature called following. Each fanpage must choose exactly one other fanpage to follow.
The website doesn’t allow a situation where i follows j and at the same time j follows i. Also, a fanpage can't follow itself.
Let’s say that fanpage i follows some other fanpage j0. Also, let’s say that i is followed by k other fanpages j1, j2, ..., jk. Then, when people visit fanpage i they see ads from k + 2 distinct companies: i, j0, j1, ..., jk. Exactly ti people subscribe (like) the i-th fanpage, and each of them will click exactly one add. For each of k + 1 companies j0, j1, ..., jk, exactly <image> people will click their ad. Remaining <image> people will click an ad from company i (the owner of the fanpage).
The total income of the company is equal to the number of people who click ads from this copmany.
Limak and Radewoosh ask you for help. Initially, fanpage i follows fanpage fi. Your task is to handle q queries of three types:
* 1 i j — fanpage i follows fanpage j from now. It's guaranteed that i didn't follow j just before the query. Note an extra constraint for the number of queries of this type (below, in the Input section).
* 2 i — print the total income of the i-th company.
* 3 — print two integers: the smallest income of one company and the biggest income of one company.
Input
The first line of the input contains two integers n and q (3 ≤ n ≤ 100 000, 1 ≤ q ≤ 100 000) — the number of fanpages and the number of queries, respectively.
The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1012) where ti denotes the number of people subscribing the i-th fanpage.
The third line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n). Initially, fanpage i follows fanpage fi.
Then, q lines follow. The i-th of them describes the i-th query. The first number in the line is an integer typei (1 ≤ typei ≤ 3) — the type of the query.
There will be at most 50 000 queries of the first type. There will be at least one query of the second or the third type (so, the output won't be empty).
It's guaranteed that at each moment a fanpage doesn't follow itself, and that no two fanpages follow each other.
Output
For each query of the second type print one integer in a separate line - the total income of the given company. For each query of the third type print two integers in a separate line - the minimum and the maximum total income, respectively.
Example
Input
5 12
10 20 30 40 50
2 3 4 5 2
2 1
2 2
2 3
2 4
2 5
1 4 2
2 1
2 2
2 3
2 4
2 5
3
Output
10
36
28
40
36
9
57
27
28
29
9 57
Note
<image>
In the sample test, there are 5 fanpages. The i-th of them has i·10 subscribers.
On drawings, numbers of subscribers are written in circles. An arrow from A to B means that A follows B.
The left drawing shows the initial situation. The first company gets income <image> from its own fanpage, and gets income <image> from the 2-nd fanpage. So, the total income is 5 + 5 = 10. After the first query ("2 1") you should print 10.
The right drawing shows the situation after a query "1 4 2" (after which fanpage 4 follows fanpage 2). Then, the first company still gets income 5 from its own fanpage, but now it gets only <image> from the 2-nd fanpage. So, the total income is 5 + 4 = 9 now.
Submitted Solution:
```
import fileinput
N = 100000
n, q = [int(x) for x in input().split()]
T = [None] + [int(x) for x in input().split()]
F = [None] + [int(x) for x in input().split()]
E = [None] + [set()] * n
for i in range(1, n+1):
E[F[i]] = E[F[i]].union([i])
def income(i):
k = len(E[i])
value = T[i] - (k+1) * (T[i] // (k+2))
for j in E[i] | set([F[i]]):
k = len(E[j])
value += T[j] // (k+2)
return value
V = [None] + [income(x) for x in range(1, n+1)]
minv, maxv = min(V[1:]), max(V[1:])
for qi in range(q):
Q = [int(x) for x in input().split()]
if Q[0] == 1:
i, j = Q[1], Q[2]
fi = F[i]
E[F[i]] = E[F[i]].difference([i])
E[j] = E[j].union([i])
F[i] = j
for j in E[i] | E[F[i]] | set([fi, F[i], i, j]):
V[j] = income(j)
minv, maxv = min(minv, V[j]), max(maxv, V[j])
elif Q[0] == 2:
i = Q[1]
print(V[i])
elif Q[0] == 3:
print(minv, maxv)
#print(n, q, T, F)
#print(E)
``` | instruction | 0 | 59,298 | 14 | 118,596 |
No | output | 1 | 59,298 | 14 | 118,597 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.