message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000 | instruction | 0 | 25,505 | 10 | 51,010 |
"Correct Solution:
```
import sys
f = sys.stdin
values = [None,6000,4000,3000,2000]
for line in f:
t, n = map(int, line.split())
print(values[t] * n)
``` | output | 1 | 25,505 | 10 | 51,011 |
Provide a correct Python 3 solution for this coding contest problem.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000 | instruction | 0 | 25,506 | 10 | 51,012 |
"Correct Solution:
```
L = [6000,4000,3000,2000]
for i in range(4):
t,n = map(int,input().split())
print(L[t-1]*n)
``` | output | 1 | 25,506 | 10 | 51,013 |
Provide a correct Python 3 solution for this coding contest problem.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000 | instruction | 0 | 25,507 | 10 | 51,014 |
"Correct Solution:
```
c = 0
while c <= 3:
t, n = map(int,input().split())
c = c + 1
if t == 1:
n = n*6000
elif t == 2:
n = n*4000
elif t == 3:
n = n*3000
elif t == 4:
n = n*2000
print(n)
``` | output | 1 | 25,507 | 10 | 51,015 |
Provide a correct Python 3 solution for this coding contest problem.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000 | instruction | 0 | 25,508 | 10 | 51,016 |
"Correct Solution:
```
a=[0,6000,4000,3000,2000]
for x in range(4):
t,n=list(map(int,input().split()))
print(a[t]*n)
``` | output | 1 | 25,508 | 10 | 51,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000
Submitted Solution:
```
p=[0,6000,4000,3000,2000]
for i in range(4):
t,n=map(int,input().split())
print(p[t]*n)
``` | instruction | 0 | 25,509 | 10 | 51,018 |
Yes | output | 1 | 25,509 | 10 | 51,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000
Submitted Solution:
```
a,n1=map(int,input().split())
b,n2=map(int,input().split())
c,n3=map(int,input().split())
d,n4=map(int,input().split())
t1=6000
t2=4000
t3=3000
t4=2000
if a==1:
if b==2:
if c==3:
print(t1*n1)
print(t2*n2)
print(t3*n3)
print(t4*n4)
else:
print(t1*n1)
print(t2*n2)
print(t4*n3)
print(t3*n4)
elif b==3:
if c==2:
print(t1*n1)
print(t3*n2)
print(t2*n3)
print(t4*n4)
else:
print(t1*n1)
print(t3*n2)
print(t4*n3)
print(t2*n4)
else:
if c==2:
print(t1*n1)
print(t4*n2)
print(t2*n3)
print(t3*n4)
else:
print(t1*n1)
print(t4*n2)
print(t3*n3)
print(t2*n4)
elif a==2:
if b==1:
if c==3:
print(t2*n1)
print(t1*n2)
print(t3*n3)
print(t4*n4)
else:
print(t2*n1)
print(t1*n2)
print(t4*n3)
print(t3*n4)
elif b==3:
if c==1:
print(t2*n1)
print(t3*n2)
print(t1*n3)
print(t4*n4)
else:
print(t2*n1)
print(t3*n2)
print(t4*n3)
print(t1*n4)
else:
if c==1:
print(t2*n1)
print(t4*n2)
print(t1*n3)
print(t3*n4)
else:
print(t2*n1)
print(t4*n2)
print(t3*n3)
print(t*n4)
elif a==3:
if b==1:
if c==2:
print(t3*n1)
print(t1*n2)
print(t2*n3)
print(t4*n4)
else:
print(t3*n1)
print(t1*n2)
print(t4*n3)
print(t2*n4)
elif b==2:
if c==1:
print(t3*n1)
print(t2*n2)
print(t1*n3)
print(t4*n4)
else:
print(t3*n1)
print(t2*n2)
print(t4*n3)
print(t1*n4)
else:
if c==1:
print(t3*n1)
print(t4*n2)
print(t1*n3)
print(t2*n4)
else:
print(t3*n1)
print(t4*n2)
print(t2*n3)
print(t1*n4)
else:
if b==1:
if c==2:
print(t4*n1)
print(t1*n2)
print(t2*n3)
print(t3*n4)
else:
print(t4*n1)
print(t1*n2)
print(t3*n3)
print(t2*n4)
elif b==2:
if c==1:
print(t4*n1)
print(t2*n2)
print(t1*n3)
print(t3*n4)
else:
print(t4*n1)
print(t2*n2)
print(t3*n3)
print(t1*n4)
else:
if c==1:
print(t4*n1)
print(t3*n2)
print(t1*n3)
print(t2*n4)
else:
print(t3*n1)
print(t3*n2)
print(t2*n3)
print(t1*n4)
``` | instruction | 0 | 25,510 | 10 | 51,020 |
Yes | output | 1 | 25,510 | 10 | 51,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000
Submitted Solution:
```
tickets= {1: 6000, 2: 4000, 3: 3000, 4: 2000}
for _ in range(4):
t, n= map(int, input().split())
print(tickets[t]*n)
``` | instruction | 0 | 25,511 | 10 | 51,022 |
Yes | output | 1 | 25,511 | 10 | 51,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today is the ticket release date for Aizu Entertainment's recommended idol group "Akabeko & Koboushi". There are four types of tickets:
S seat 6000 yen
A seat 4000 yen
B seat 3000 yen
C seat 2000 yen
You, the sales manager, are excitedly waiting for the launch. Finally on sale. It's selling very well!
Shortly after the launch, I received a table summarizing the orders up to that point. Each row of the table shows the type and number of tickets sold so far. However, the ticket types do not always appear in the order of S, A, B, C. Create a program to find the sales amount for each row in this table.
input
Input data is given in the following format.
t1 n1
t2 n2
t3 n3
t4 n4
The input consists of 4 lines. Line i is given the integer ti (1 ≤ ti ≤ 4) for the ticket type and the integer ni (0 ≤ ni ≤ 10000) for the number of tickets. The integers 1, 2, 3, and 4 representing the ticket types represent S seats, A seats, B seats, and C seats, respectively. Numbers from 1 to 4 always appear once as values for t1, t2, t3, and t4, but they are not always given in the order of 1, 2, 3, 4.
output
Output the sales amount for each line.
Example
Input
3 10
1 4
4 1
2 5
Output
30000
24000
2000
20000
Submitted Solution:
```
i=0
m=0
for i in range(4):
a,b=map(int,input().split())
if a==1:
print(6000*b)
elif a==2:
print(4000*b)
elif a==3:
print(3000*b)
else:
print(2000*b)
i+=1
``` | instruction | 0 | 25,512 | 10 | 51,024 |
Yes | output | 1 | 25,512 | 10 | 51,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n coins labeled from 1 to n. Initially, coin c_i is on position i and is facing upwards ((c_1, c_2, ..., c_n) is a permutation of numbers from 1 to n). You can do some operations on these coins.
In one operation, you can do the following:
* Choose 2 distinct indices i and j.
* Then, swap the coins on positions i and j.
* Then, flip both coins on positions i and j. (If they are initially faced up, they will be faced down after the operation and vice versa)
Construct a sequence of at most n+1 operations such that after performing all these operations the coin i will be on position i at the end, facing up.
Note that you do not need to minimize the number of operations.
Input
The first line contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of coins.
The second line contains n integers c_1,c_2,...,c_n (1 ≤ c_i ≤ n, c_i ≠ c_j for i≠ j).
Output
In the first line, output an integer q (0 ≤ q ≤ n+1) — the number of operations you used.
In the following q lines, output two integers i and j (1 ≤ i, j ≤ n, i ≠ j) — the positions you chose for the current operation.
Examples
Input
3
2 1 3
Output
3
1 3
3 2
3 1
Input
5
1 2 3 4 5
Output
0
Note
Let coin i facing upwards be denoted as i and coin i facing downwards be denoted as -i.
The series of moves performed in the first sample changes the coins as such:
* [~~~2,~~~1,~~~3]
* [-3,~~~1,-2]
* [-3,~~~2,-1]
* [~~~1,~~~2,~~~3]
In the second sample, the coins are already in their correct positions so there is no need to swap. | instruction | 0 | 25,833 | 10 | 51,666 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
Let's find the first number that is in the wrong place. Keep swapping the
I need to partition the cycles, and then make sure I link each cycle
"""
def solve():
N = getInt()
A = [0]+getInts()
used = [False]*(N+1)
ans = []
prev_ = -1
def swap_(i,j):
ans.append((i,j))
A[i], A[j] = -A[j], -A[i]
return
"""
[2,3,1,5,6,4]
swap_(1,4)
[-5,3,1,-2,6,4]
[-6,3,1,-2,5,4]
[-4,3,1,-2,5,6]
[-4,2,1,-3,5,6]
[-4,2,3,-1,5,6]
idx = 1
eg A[-A[idx]] = 6, so swap 1 and 5
once we hit a negative number, swap to that index
"""
def join_cycles(i,j):
swap_(i,j)
idx = i
while A[-A[idx]] > 0: swap_(idx,-A[idx]) #A[idx] is negative because we swapped i with j (this is the first 'part' of the cycle, up to the artificially induced element
idx = -A[idx]
while A[-A[idx]] > 0: swap_(idx,-A[idx]) #this is the second part of the cycle
swap_(idx,-A[idx]) # the final two elements
for i in range(1,N+1):
if A[i] == i: used[i] = True
if used[i]: continue
idx = i
while True:
used[idx] = True
idx = A[idx]
if idx == i: break # cycle has ended
if prev_ == -1:
prev_ = i # we don't have a cycle to join it with yet
else:
join_cycles(prev_,i) # we do have a cycle to join it with
prev_ = -1
if prev_ > -1: # we have one remaining unjoined cycle
j = 1
while j <= N and A[j] != j: j += 1
if j <= N:
# if possible join the cycle with a correct element - this only adds one extra move. This treats the swapping of 2 case.
join_cycles(prev_,j)
else:
# if this isn't possible (eg [2,3,1]) then breaking any pair will suffice, eg [-3,-2,1]. Then we join with any other element in the permutation.
i0, i1 = prev_, A[prev_]
swap_(i0,i1)
j = 1
while j in [i0,i1]: j += 1 # make sure j is different to the elements we just swapped
join_cycles(i1,j)
print(len(ans))
for a, b in ans: print(a,b)
return
#for _ in range(getInt()):
#print(solve())
solve()
#print(time.time()-start_time)
``` | output | 1 | 25,833 | 10 | 51,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n coins labeled from 1 to n. Initially, coin c_i is on position i and is facing upwards ((c_1, c_2, ..., c_n) is a permutation of numbers from 1 to n). You can do some operations on these coins.
In one operation, you can do the following:
* Choose 2 distinct indices i and j.
* Then, swap the coins on positions i and j.
* Then, flip both coins on positions i and j. (If they are initially faced up, they will be faced down after the operation and vice versa)
Construct a sequence of at most n+1 operations such that after performing all these operations the coin i will be on position i at the end, facing up.
Note that you do not need to minimize the number of operations.
Input
The first line contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of coins.
The second line contains n integers c_1,c_2,...,c_n (1 ≤ c_i ≤ n, c_i ≠ c_j for i≠ j).
Output
In the first line, output an integer q (0 ≤ q ≤ n+1) — the number of operations you used.
In the following q lines, output two integers i and j (1 ≤ i, j ≤ n, i ≠ j) — the positions you chose for the current operation.
Examples
Input
3
2 1 3
Output
3
1 3
3 2
3 1
Input
5
1 2 3 4 5
Output
0
Note
Let coin i facing upwards be denoted as i and coin i facing downwards be denoted as -i.
The series of moves performed in the first sample changes the coins as such:
* [~~~2,~~~1,~~~3]
* [-3,~~~1,-2]
* [-3,~~~2,-1]
* [~~~1,~~~2,~~~3]
In the second sample, the coins are already in their correct positions so there is no need to swap. | instruction | 0 | 25,835 | 10 | 51,670 |
Tags: constructive algorithms, graphs, math
Correct Solution:
```
def swap(i, j):
graph[i], graph[j] = -graph[j], -graph[i]
ans.append((i, j))
def solve_pair(x, y):
swap(x, y)
i = x
while (graph[-graph[i]] > 0):
swap(i, -graph[i])
i = -graph[i]
while (graph[-graph[i]] > 0):
swap(i, -graph[i])
swap(i, -graph[i])
n = int(input())
j = 1
graph = [0]
solved = set()
for i in input().split():
graph.append(int(i))
if (int(i) == j):
solved.add(j)
j+=1
ans = []
cycles = set()
for x in range(1, n+1):
if not (x in solved):
i = x
solved.add(i)
i = graph[i]
while (i != x):
solved.add(i)
i = graph[i]
cycles.add(i)
while (len(cycles) > 1):
solve_pair(cycles.pop(), cycles.pop())
if (len(cycles) == 1):
j = 1
while (not (j == n+1 or j == graph[j])):
j+=1
if (j != n+1):
solve_pair(j, cycles.pop())
else:
i, j = graph[1], graph[graph[1]]
swap(1, graph[1])
solve_pair(i, j)
print(len(ans))
for i in ans:
print(i[0], i[1])
``` | output | 1 | 25,835 | 10 | 51,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n coins labeled from 1 to n. Initially, coin c_i is on position i and is facing upwards ((c_1, c_2, ..., c_n) is a permutation of numbers from 1 to n). You can do some operations on these coins.
In one operation, you can do the following:
* Choose 2 distinct indices i and j.
* Then, swap the coins on positions i and j.
* Then, flip both coins on positions i and j. (If they are initially faced up, they will be faced down after the operation and vice versa)
Construct a sequence of at most n+1 operations such that after performing all these operations the coin i will be on position i at the end, facing up.
Note that you do not need to minimize the number of operations.
Input
The first line contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of coins.
The second line contains n integers c_1,c_2,...,c_n (1 ≤ c_i ≤ n, c_i ≠ c_j for i≠ j).
Output
In the first line, output an integer q (0 ≤ q ≤ n+1) — the number of operations you used.
In the following q lines, output two integers i and j (1 ≤ i, j ≤ n, i ≠ j) — the positions you chose for the current operation.
Examples
Input
3
2 1 3
Output
3
1 3
3 2
3 1
Input
5
1 2 3 4 5
Output
0
Note
Let coin i facing upwards be denoted as i and coin i facing downwards be denoted as -i.
The series of moves performed in the first sample changes the coins as such:
* [~~~2,~~~1,~~~3]
* [-3,~~~1,-2]
* [-3,~~~2,-1]
* [~~~1,~~~2,~~~3]
In the second sample, the coins are already in their correct positions so there is no need to swap.
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
Let's find the first number that is in the wrong place. Keep swapping the
I need to partition the cycles, and then make sure I link each cycle
"""
def solve():
N = getInt()
A = [0]+getInts()
used = [False]*(N+1)
ans = []
prev_ = -1
def swap_(i,j):
ans.append((i,j))
A[i], A[j] = -A[j], -A[i]
return
"""
[2,3,1,5,6,4]
swap_(1,4)
[-5,3,1,-2,6,4]
[-6,3,1,-2,5,4]
[-4,3,1,-2,5,6]
[-4,2,1,-3,5,6]
[-4,2,3,-1,5,6]
idx = 1
eg A[-A[idx]] = 6, so swap 1 and 5
once we hit a negative number, swap to that index
"""
def join_cycles(i,j):
swap_(i,j)
idx = i
while A[-A[idx]] > 0: swap_(idx,-A[idx]) #A[idx] is negative because we swapped i with j (this is the first 'part' of the cycle, up to the artificially induced element
idx = -A[idx]
while A[-A[idx]] > 0: swap_(idx,-A[idx]) #this is the second part of the cycle
swap_(idx,-A[idx]) # the final two elements
for i in range(1,N+1):
if A[i] == i: used[i] = True
if used[i]: continue
idx = i
while True:
used[idx] = True
idx = A[idx]
if idx == i: break # cycle has ended
if prev_ == -1:
prev_ = i # we don't have a cycle to join it with yet
else:
join_cycles(prev_,i) # we do have a cycle to join it with
prev_ = -1
if prev_ > -1: # we have one remaining unjoined cycle
j = 1
while j <= N and A[j] != j: j += 1
if j <= N:
# if possible join the cycle with a correct element - this only adds one extra move. This treats the swapping of 2 case.
join_cycles(prev_,j)
else:
# if this isn't possible (eg [2,3,1]) then breaking any pair will suffice, eg [-1,3,-2]. Then we join with any other element in the permutation.
i0, i1 = prev_, A[prev_]
swap_(i0,i1)
j = 1
while j in [i0,i1]: j += 1 # make sure j is different to the elements we just swapped
join_cycles(i0,j)
print(len(ans))
for a, b in ans: print(a,b)
return
#for _ in range(getInt()):
#print(solve())
solve()
#print(time.time()-start_time)
``` | instruction | 0 | 25,838 | 10 | 51,676 |
No | output | 1 | 25,838 | 10 | 51,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n coins labeled from 1 to n. Initially, coin c_i is on position i and is facing upwards ((c_1, c_2, ..., c_n) is a permutation of numbers from 1 to n). You can do some operations on these coins.
In one operation, you can do the following:
* Choose 2 distinct indices i and j.
* Then, swap the coins on positions i and j.
* Then, flip both coins on positions i and j. (If they are initially faced up, they will be faced down after the operation and vice versa)
Construct a sequence of at most n+1 operations such that after performing all these operations the coin i will be on position i at the end, facing up.
Note that you do not need to minimize the number of operations.
Input
The first line contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of coins.
The second line contains n integers c_1,c_2,...,c_n (1 ≤ c_i ≤ n, c_i ≠ c_j for i≠ j).
Output
In the first line, output an integer q (0 ≤ q ≤ n+1) — the number of operations you used.
In the following q lines, output two integers i and j (1 ≤ i, j ≤ n, i ≠ j) — the positions you chose for the current operation.
Examples
Input
3
2 1 3
Output
3
1 3
3 2
3 1
Input
5
1 2 3 4 5
Output
0
Note
Let coin i facing upwards be denoted as i and coin i facing downwards be denoted as -i.
The series of moves performed in the first sample changes the coins as such:
* [~~~2,~~~1,~~~3]
* [-3,~~~1,-2]
* [-3,~~~2,-1]
* [~~~1,~~~2,~~~3]
In the second sample, the coins are already in their correct positions so there is no need to swap.
Submitted Solution:
```
import sys
n = int(input())
c = list(map(int,sys.stdin.readline().split()))
swap = []
for i in range(n):
for j in range(n-i-1):
if c[j]>c[j+1]:
c[j],c[j+1] = c[j+1],c[j]
swap.append((j,j+1))
print(len(swap)*3)
def how_to_swap(n1,n2):
temp = max(n1,n2)+1
if temp>=n:
temp=min(n1,n2)-1
if temp<0:
temp = 1
print(n1+1,temp+1)
print(temp+1,n2+1)
print(temp+1,n1+1)
for u,v in swap:
how_to_swap(u,v)
``` | instruction | 0 | 25,839 | 10 | 51,678 |
No | output | 1 | 25,839 | 10 | 51,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n coins labeled from 1 to n. Initially, coin c_i is on position i and is facing upwards ((c_1, c_2, ..., c_n) is a permutation of numbers from 1 to n). You can do some operations on these coins.
In one operation, you can do the following:
* Choose 2 distinct indices i and j.
* Then, swap the coins on positions i and j.
* Then, flip both coins on positions i and j. (If they are initially faced up, they will be faced down after the operation and vice versa)
Construct a sequence of at most n+1 operations such that after performing all these operations the coin i will be on position i at the end, facing up.
Note that you do not need to minimize the number of operations.
Input
The first line contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of coins.
The second line contains n integers c_1,c_2,...,c_n (1 ≤ c_i ≤ n, c_i ≠ c_j for i≠ j).
Output
In the first line, output an integer q (0 ≤ q ≤ n+1) — the number of operations you used.
In the following q lines, output two integers i and j (1 ≤ i, j ≤ n, i ≠ j) — the positions you chose for the current operation.
Examples
Input
3
2 1 3
Output
3
1 3
3 2
3 1
Input
5
1 2 3 4 5
Output
0
Note
Let coin i facing upwards be denoted as i and coin i facing downwards be denoted as -i.
The series of moves performed in the first sample changes the coins as such:
* [~~~2,~~~1,~~~3]
* [-3,~~~1,-2]
* [-3,~~~2,-1]
* [~~~1,~~~2,~~~3]
In the second sample, the coins are already in their correct positions so there is no need to swap.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Mar 3 03:29:35 2021
@author: NA
"""
def f(a):
x=a[:]
l=len(x)
y=[]
for i in range(l):
for j in range(i+1,l):
x[i],x[j]=-1*x[j],-1*x[i]
y.append([x,(abs(a[i]),abs(a[j]))])
x=a[:]
return y
n=int(input())
x=list(map(int,input().split(" ")))
y=list(range(1,n+1))
if x==y:
print("0")
else:
z=f(x)
d={}
for i in z:
d[tuple(i[0])]=[(i[1])]
for i in z:
temp=f(i[0])
for j in temp:
d[tuple(j[0])]=d[tuple(i[0])]+[tuple(j[1])]
z+=temp
if tuple(y) in d:
break
ans=d[tuple(y)]
print (len(ans))
for i in ans:
print (*i,sep=" ")
``` | instruction | 0 | 25,841 | 10 | 51,682 |
No | output | 1 | 25,841 | 10 | 51,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera runs a 24/7 fast food cafe. He magically learned that next day n people will visit his cafe. For each person we know the arrival time: the i-th person comes exactly at hi hours mi minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all n customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), that is the number of cafe visitors.
Each of the following n lines has two space-separated integers hi and mi (0 ≤ hi ≤ 23; 0 ≤ mi ≤ 59), representing the time when the i-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Examples
Input
4
8 0
8 10
8 10
8 45
Output
2
Input
3
0 12
10 11
22 22
Output
1
Note
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
Submitted Solution:
```
#uff ki prblm.at last parsi. Tbe builtin function chara solve hbe na. TLE KHABA. amre tw dkso koto bar tle kahisi. huhh
import statistics
from statistics import mode
a=int(input())
d=[]
for i in range(0,a):
b=str(input())
d.append(b)
e=mode(d)
print(d.count(e))
``` | instruction | 0 | 25,906 | 10 | 51,812 |
Yes | output | 1 | 25,906 | 10 | 51,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera runs a 24/7 fast food cafe. He magically learned that next day n people will visit his cafe. For each person we know the arrival time: the i-th person comes exactly at hi hours mi minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all n customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), that is the number of cafe visitors.
Each of the following n lines has two space-separated integers hi and mi (0 ≤ hi ≤ 23; 0 ≤ mi ≤ 59), representing the time when the i-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Examples
Input
4
8 0
8 10
8 10
8 45
Output
2
Input
3
0 12
10 11
22 22
Output
1
Note
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
Submitted Solution:
```
n = int(input())
s1 = input()
m = 1
ans = 1
for f in range(1,n):
s2 = input()
if s2 == s1 :
m += 1
if m > ans : ans = m
else : m = 1
s1 = s2
print(ans)
``` | instruction | 0 | 25,909 | 10 | 51,818 |
Yes | output | 1 | 25,909 | 10 | 51,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera runs a 24/7 fast food cafe. He magically learned that next day n people will visit his cafe. For each person we know the arrival time: the i-th person comes exactly at hi hours mi minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all n customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), that is the number of cafe visitors.
Each of the following n lines has two space-separated integers hi and mi (0 ≤ hi ≤ 23; 0 ≤ mi ≤ 59), representing the time when the i-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Examples
Input
4
8 0
8 10
8 10
8 45
Output
2
Input
3
0 12
10 11
22 22
Output
1
Note
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
Submitted Solution:
```
n = int(input())
print(len(set(input() for _ in range(n))))
``` | instruction | 0 | 25,910 | 10 | 51,820 |
No | output | 1 | 25,910 | 10 | 51,821 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,383 | 10 | 52,766 |
"Correct Solution:
```
for i in range(1, 100):
W = int(input())
if not W:
break
N = int(input())
data = [tuple(map(int, input().split(','))) for _ in range(N)]
dp = [[[0, 0] for j in range(W + 1)] for i in range(N + 1)]
print('Case {0}:'.format(i))
for i in range(1, N + 1):
for j in range(W + 1):
if data[i - 1][1] <= j:
pt = dp[i - 1][j - data[i - 1][1]][:]
for k in range(2):
pt[k] += data[i - 1][k]
dpt = dp[i - 1][j][:]
if pt[0] > dpt[0]:
dp[i][j] = pt
elif dpt[0] > pt[0]:
dp[i][j] = dpt
else:
dp[i][j] = pt if pt[1] < dpt[1] else dpt
else:
dp[i][j] = dp[i - 1][j]
for i in range(2):
print(dp[N][W][i])
``` | output | 1 | 26,383 | 10 | 52,767 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,384 | 10 | 52,768 |
"Correct Solution:
```
c=0
for W in iter(input,'0'):
c+=1
W=int(W)
dp=[0]*-~W
for _ in[0]*int(input()):
v,w=map(int,input().split(','))
for i in range(W,w-1,-1):
if dp[i]<dp[i-w]+v:dp[i]=dp[i-w]+v
print(f'Case {c}:\n{dp[W]}\n{dp.index(dp[W])}')
``` | output | 1 | 26,384 | 10 | 52,769 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,385 | 10 | 52,770 |
"Correct Solution:
```
caseN = 0
while True:
caseN += 1
W = int(input())
if W == 0: break
N = int(input())
tr = [list(map(int, input().split(','))) for _ in range(N)]
dp = [0 for x in range(W+1)]
for x in range(N):
for y in range(W,tr[x][1]-1,-1):
dp[y] = max(dp[y], dp[y-tr[x][1]] + tr[x][0])
sum_weight = 0
sum_value = 0
ind = N%2
for x in range(W+1):
if dp[x] > sum_value:
sum_weight = x
sum_value = dp[x]
print("Case ", caseN, ":", sep='')
print(sum_value)
print(sum_weight)
``` | output | 1 | 26,385 | 10 | 52,771 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,386 | 10 | 52,772 |
"Correct Solution:
```
c=0
for W in iter(input,'0'):
c+=1
W=int(W)
dp=[0]*-~W
for _ in[0]*int(input()):
v,w=map(int,input().split(','))
for i in range(W,w-1,-1):
if dp[i]<dp[i-w]+v:dp[i]=dp[i-w]+v
for i in range(W+1):
if dp[W]==dp[i]:print(f'Case {c}:\n{dp[W]}\n{i}');break
``` | output | 1 | 26,386 | 10 | 52,773 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,387 | 10 | 52,774 |
"Correct Solution:
```
ctr = 1
while True:
w = int(input())
if w==0: break
dp = [0]*(w+1)
n = int(input())
for i in range(n):
vi, wi = map(int, input().split(","))
for j in range(w,wi-1,-1):
dp[j] = max(dp[j],dp[j-wi]+vi)
print("Case "+str(ctr)+":")
print(max(dp))
print(dp.index(max(dp)))
ctr += 1
``` | output | 1 | 26,387 | 10 | 52,775 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,388 | 10 | 52,776 |
"Correct Solution:
```
mininf = -10**8
inf = 10**8
count = 0
while(1):
count += 1
W = int(input())
if W == 0:
break
N = int(input())
dp = [0 for j in range(W+1)]
for _ in range(N):
v,w = [int(i) for i in input().split(",")]
for i in range(w,W+1)[::-1]:
dp[i] = max(dp[i], dp[i-w]+v)
value = 0
weight = 0
for i in range(W+1):
if value < dp[i]:
value = dp[i]
weight = i
print("Case {}:".format(count))
print(value)
print(weight)
``` | output | 1 | 26,388 | 10 | 52,777 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,389 | 10 | 52,778 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve(k):
W = int(readline())
if W == 0:
return False
INF = 10**18
dp = [-INF]*(W+1)
dp[0] = 0
N = int(readline())
for i in range(N):
v, w = map(int, readline().split(","))
for j in range(W-w, -1, -1):
dp[j+w] = max(dp[j+w], dp[j] + v)
vm = max(dp)
wm = dp.index(vm)
write("Case %d:\n%d\n%d\n" % (k, vm, wm))
return True
i = 1
while solve(i):
i += 1
``` | output | 1 | 26,389 | 10 | 52,779 |
Provide a correct Python 3 solution for this coding contest problem.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49 | instruction | 0 | 26,390 | 10 | 52,780 |
"Correct Solution:
```
l=1
for W in iter(input,'0'):
W=int(W)
D=[0,0]*W
for p in[0]*int(input()):
v,w=map(int,input().split(','))
for i in range(W,w-1,-1):D[i]=max(D[i],D[i-w]+v)
print(f'Case {l}:\n{D[W]}\n{D.index(D[W])}')
l+=1
``` | output | 1 | 26,390 | 10 | 52,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
if __name__ == '__main__':
W = int(input())
v = [0]
w = [0]
d = []
m = []
counter = 1
result = []
while W != 0:
N = int(input())
v[:] = [0]
w[:] = [0]
d[:] = []
m[:] = []
"""the initialization of array d"""
for x in range(W + 1):
temp = [0] * (N + 1)
d.append(temp)
for x in range(W + 1):
temp = [0] * (N + 1)
m.append(temp)
"""prepare the dataset"""
for x in range(N):
i = input().split(',')
v.append(int(i[0]))
w.append(int(i[1]))
"""processing"""
for weight in range(W + 1):
for kind in range(N + 1):
if weight != 0 and kind != 0:
if w[kind] > weight:
d[weight][kind] = d[weight][kind - 1]
m[weight][kind] = m[weight][kind - 1]
else:
if d[weight - w[kind]][kind - 1] + v[kind] > d[weight][kind - 1]:
d[weight][kind] = d[weight - w[kind]][kind - 1] + v[kind]
m[weight][kind] = m[weight - w[kind]][kind - 1] + w[kind]
elif d[weight - w[kind]][kind - 1] + v[kind] == d[weight][kind - 1]:
d[weight][kind] = d[weight][kind - 1]
if m[weight - w[kind]][kind - 1] + w[kind] < m[weight][kind - 1]:
m[weight][kind] = m[weight - w[kind]][kind- 1] + w[kind]
else:
m[weight][kind] = m[weight][kind - 1]
else:
d[weight][kind] = d[weight][kind - 1]
m[weight][kind] = m[weight][kind - 1]
result.append("Case " + counter.__str__() + ":")
result.append(d[W][N])
result.append(m[W][N])
W = int(input())
counter += 1
for line in result:
print(line)
``` | instruction | 0 | 26,391 | 10 | 52,782 |
Yes | output | 1 | 26,391 | 10 | 52,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
nCase = 0
while True:
W = int(input())
if W == 0: break
nCase += 1
N = int(input())
# [(value, weight), ... ]
items = [tuple(map(int, input().split(','))) for _ in range(N)]
bestValueFor = [0] * (W + 1) # item -> best value, idx -> capacity
for item in items:
for capa in reversed(range(W-item[1]+1)):
# if taken wins not taken
if bestValueFor[capa] + item[0] > bestValueFor[capa + item[1]]:
bestValueFor[capa + item[1]] = bestValueFor[capa] + item[0]
bestValue = max(bestValueFor)
bestWeight = bestValueFor.index(bestValue)
print('Case ',str(nCase),':',sep='')
print(bestValue)
print(bestWeight)
``` | instruction | 0 | 26,392 | 10 | 52,784 |
Yes | output | 1 | 26,392 | 10 | 52,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
case = 0
while True:
W = int(input())
if W == 0:
break
N, case = int(input()), case + 1
items = [[0,0]] + [list(map(int, input().split(','))) for _ in range(N)]
C = [[0 for _ in range(W+1)] for _ in range(N+1)]
for i in range(1, N+1):
for w in range(1, W+1):
if items[i][1] <= w:
C[i][w] = max(items[i][0]+C[i-1][w-items[i][1]], C[i-1][w])
else:
C[i][w] = C[i-1][w]
maxv = max(C[-1])
print("Case {0}:\n{1}\n{2}".format(case, maxv, C[-1].index(maxv)))
``` | instruction | 0 | 26,393 | 10 | 52,786 |
Yes | output | 1 | 26,393 | 10 | 52,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
cnt=0
while True:
cnt+=1
W=int(input())
if W==0:
break
n=int(input())
dp=[0]*(W+1)
for i in range(n):
v,w=map(int,input().split(','))
for j in range(W,w-1,-1):
dp[j]=max(dp[j],dp[j-w]+v)
wgt=0
md=max(dp)
for i in range(W+1):
if md==dp[i]:
wgt=i
break
print("Case {0}:".format(cnt))
print(md)
print(wgt)
``` | instruction | 0 | 26,394 | 10 | 52,788 |
Yes | output | 1 | 26,394 | 10 | 52,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
c=1
while 1:
W=int(input())
if W==0:break
dp=[0]*(W+1)
for i in range(int(input())):
v,w = map(int, input().split(','))
for j in range(W,w-1,-1):
if dp[j]<dp[j-w]:dp[j]=dp[j-w]
for i in range(W+1):
if dp[W]==dp[i]:break
print('Case %d:\n%d\n%d'%(c,dp[W],i))
c+=1
``` | instruction | 0 | 26,395 | 10 | 52,790 |
No | output | 1 | 26,395 | 10 | 52,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import itertools
def solve(W, N, v):
cache = [[None for _ in range(W + 1)] for _ in range(N)]
def loop(i, w):
if i == N:
return (0, 0)
if cache[i][w] is not None:
return cache[i][w]
cache[i][w] = loop(i + 1, w)
if v[i][1] <= w:
t = loop(i + 1, w - v[i][1])
cache[i][w] = max(cache[i][w], (t[0] + v[i][0], t[1] + v[i][1]), key=lambda u: u[0])
return cache[i][w]
return loop(0, W)
for dn in itertools.count(1):
W = int(input())
if W == 0: break
N = int(input())
v = [tuple(map(int, input().split(','))) for _ in range(N)]
optim = solve(W, N, v)
print("Case %d:\n%d\n%d" % (dn, optim[0], optim[1]))
``` | instruction | 0 | 26,396 | 10 | 52,792 |
No | output | 1 | 26,396 | 10 | 52,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
def rec(i, j):
if dp[i][j] != -1:
return dp[i][j]
if i == N:
ans = 0, W - j
elif j < wl[i]:
ans = rec(i + 1, j)
else:
v1, w1 = rec(i + 1, j)
v2, w2 = rec(i + 1, j - wl[i])
v2 += vl[i]
if v1 > v2:
ans = v1, w1
elif v1 < v2:
ans = v2, w2
elif v1 == v2:
if w1 < w2:
ans = v1, w1
else:
ans = v2, w2
dp[i][j] = ans
return ans
case = 1
while True:
W = int(input())
if W == 0: break
N = int(input())
vl = []
wl = []
for _ in range(N):
v, w = map(int, input().split(','))
vl.append(v)
wl.append(w)
dp = [[-1 for j in range(1100)] for i in range(1100)]
v, w = rec(0, W)
print("Case {}:".format(case))
print(v)
print(w)
case += 1
``` | instruction | 0 | 26,397 | 10 | 52,794 |
No | output | 1 | 26,397 | 10 | 52,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A thief sneaked into a museum with a lot of treasures with only one large furoshiki. There are many things I want to steal, but the weight that the furoshiki can withstand is limited, and if it exceeds this, the furoshiki will tear. Therefore, the thief must consider a combination of treasures that will not break the prepared furoshiki and will be the most valuable.
The weight W that the bath room can withstand, and the value and weight of each treasure in the museum are read, and the total value of the treasure is the maximum when the total value does not exceed W. Create a program that outputs the total weight. However, if there are multiple combinations that maximize the sum of values, the one with the smallest sum of weights will be output.
Input
Given multiple datasets. Each dataset is given in the following format:
W
N
v1, w1
v2, w2
::
vN, wN
The first line gives the integer W (W ≤ 1,000), which represents the weight that the furoshiki can bear, and the second line gives the number of treasures N (1 ≤ N ≤ 1,000). The next N lines are given a set of the integer vi (0 ≤ vi ≤ 10,000) representing the value of the i-th treasure and the integer wi (0 ≤ wi ≤ W) representing its weight.
When W is 0, it is the last input. The number of datasets does not exceed 50.
Output
Output as follows for each data set.
Case dataset number:
Sum of the value of the treasure in the furoshiki
Sum of the weight of the treasure at that time
Example
Input
50
5
60,10
100,20
120,30
210,45
10,4
50
5
60,10
100,20
120,30
210,45
10,4
0
Output
Case 1:
220
49
Case 2:
220
49
Submitted Solution:
```
count = 0
while True:
count += 1
W = int(input())
if W == 0:
break
N = int(input())
value = []
weight = []
for i in range(N):
a, b = map(int, input().split())
value.append(a)
weight.append(b)
dp = [[-float("inf") for i in range(W+1)] for j in range(N+1)]
for i in range(W+1):
dp[0][i] = 0
for i in range(N):
for w in range(W+1):
if w >= weight[i]:
dp[i+1][w] = max(dp[i][w-weight[i]]+value[i], dp[i][w])
else:
dp[i+1][w] = dp[i][w]
print("Case {0}:\n{1}\n{2}".format(count, dp[N][W], dp[N].index(dp[N][W])))
``` | instruction | 0 | 26,398 | 10 | 52,796 |
No | output | 1 | 26,398 | 10 | 52,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m = map(int,input().split())
dg = 10**10
s = [0]
t = [0]
tank = []
for _ in [0]*n:
X,d = map(int,input().split())
tank.append((X+d)*dg + X)
tank.sort()
for e in tank:
X_d = e//dg
X = e%dg
d = X_d-X
s.append(max(1,X_d-d*2))
t.append(min(m,X_d))
n += 1
dp = [[0]*(m+1) for i in range(2)]
for j in range(1,t[0]+1):
dp[0][j] = s[0]-1
for j in range(t[0]+1,m+1):
dp[0][j] = max(s[0]-1,j-t[0])
for i in range(1,n):
for j in range(1,m+1):
tmp = dp[(i+1)%2][j]
if j <= t[i]:
tmp = min(tmp,dp[(i+1)%2][s[i]-1])
else:
tmp = min(tmp,dp[(i+1)%2][max(0,s[i]+t[i]-j-1)]+j-t[i])
dp[i%2][j] = tmp
print(dp[(n+1)%2][-1])
``` | instruction | 0 | 26,665 | 10 | 53,330 |
Yes | output | 1 | 26,665 | 10 | 53,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
def dp(ind, max_covered):
max_covered = min(m, max_covered)
if ind not in cache:
cache[ind] = {}
d = cache[ind]
if max_covered in d:
return d[max_covered]
ans = blah(ind, max_covered)
d[max_covered] = ans
return ans
# path = {}
class Node:
def __init__(self, key, val, next=None):
self.key = key
self.val = val
self.next = next
def blah(ind, max_covered):
x, s = antenna[ind]
# key = (ind, max_covered)
if max_covered >= m:
# path[key] = Node(key, 0)
return 0
if ind == len(antenna) - 1:
if max_covered < x - s - 1:
left_needed = x - s - (max_covered + 1)
right_needed = max(m - (x + s), 0)
ans = max(left_needed, right_needed)
# path[key] = Node(key, ans)
return ans
else:
right_boundary = max(max_covered, x + s)
ans = max(0, m - right_boundary)
# path[key] = Node(key, ans)
return ans
if max_covered < x - s - 1:
num_needed = x - s - (max_covered + 1)
new_boundary = min(x + s + num_needed, m)
use_i = num_needed + dp(ind + 1, new_boundary)
dont_use_i = dp(ind + 1, max_covered)
# if use_i < dont_use_i:
# path[key] = Node(key, num_needed, path[(ind + 1, new_boundary)])
# else:
# path[key] = Node(key, 0, path[(ind + 1, max_covered)])
return min(use_i, dont_use_i)
else:
new_boundary = min(max(max_covered, x + s), m)
ans = dp(ind + 1, new_boundary)
# path[key] = Node(key, 0, path[(ind + 1, new_boundary)])
return ans
import sys
cache = {}
n, m = [int(x) for x in sys.stdin.readline().split(" ")]
antenna = []
for i in range(n):
x, s = [int(x) for x in sys.stdin.readline().split(" ")]
antenna.append((x, s))
antenna.sort(key=lambda a: a[0])
print(dp(0, 0))
``` | instruction | 0 | 26,666 | 10 | 53,332 |
Yes | output | 1 | 26,666 | 10 | 53,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
A = []
n, m = map(int, input().split())
for i in range(1, n+1):
x, s = map(int, input().split())
A.append((max(0, x-s), min(m, x+s)))
dp = [9999999] * (m+1)
dp[0] = 0
for p in range(1, m+1):
for i in range(n):
a, b = A[i]
if p >= a and p <= b:
dp[p] = dp[p-1]
elif p < a:
pwr = a - p
dp[p] = min(dp[p], pwr + dp[max(0,a-pwr-1)], p)
elif p > b:
pwr = p - b
dp[p] = min(dp[p], pwr + dp[max(a-pwr-1,0)], p)
print(dp[m])
``` | instruction | 0 | 26,667 | 10 | 53,334 |
Yes | output | 1 | 26,667 | 10 | 53,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
A = [(0, 0)]
inf = 999999999999
n, m = map(int, input().split())
for i in range(1, n+1):
x, s = map(int, input().split())
A.append((max(0, x-s), min(m, x+s)))
dp = [inf] * (m+1)
dp[0] = 0
A = sorted(A)
for p in range(1, m+1):
for i in range(1, n+1):
a, b = A[i]
if p >= a and p <= b:
dp[p] = dp[p-1]
elif p < a:
pwr = a - p
dp[p] = min(dp[p], pwr + dp[max(0,a-pwr-1)], p)
elif p > b:
pwr = p - b
dp[p] = min(dp[p], pwr + dp[max(a-pwr-1,0)], p)
print(dp[m])
``` | instruction | 0 | 26,668 | 10 | 53,336 |
Yes | output | 1 | 26,668 | 10 | 53,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
#n = int(input())
n, m = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
#a = list(map(int, input().split()))
#b = list(map(int, input().split()))
#s = input()
a = []
for i in range(n):
x, s = map(int, input().split())
a += [[x-s, x+s]]
ans = 0
segc = [0]*len(a)
for i in range(1, m+1):
found = False
gap = 6969696969
gap2 = gap
index = -1
index2 = -1
for j in range(len(a)):
alpha, omega = a[j]
if alpha <= i <= omega:
found = True
break
if alpha >= i:
gap = min(gap, alpha - i)
if alpha-i == gap:
index = j
if i >= omega:
gap2 = min(gap2, i - omega)
if i-omega == gap2:
index2 = j
if not found:
if index != -1:
a[index][0] -= gap
a[index][1] += gap
ans += gap
segc[index] += gap
elif index2 != -1:
a[index2][1] += gap2
a[index2][0] -= gap2
ans += gap2
segc[index2] += gap2
while 1:
#print(a)
exists = False
for i in range(len(a)):
for j in range(i+1, len(a)):
if not segc[j]:continue
a1, o1 = a[i]
a2, o2 = a[j]
if a1 <= a2 <= o2 <= o1:
# segment j is inside i
a[j][0] += segc[j]
a[j][1] -= segc[j]
ans -= segc[j]
segc[j] = 0
exists = True
break
if exists:break
if not exists:break
print(ans)
``` | instruction | 0 | 26,669 | 10 | 53,338 |
No | output | 1 | 26,669 | 10 | 53,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
A = [(0, 0)]
inf = 999999999999
n, m = map(int, input().split())
for i in range(1, n+1):
x, s = map(int, input().split())
A.append((max(0, x-s), min(m, x+s)))
dp = [inf] * (m+1)
dp[0] = 0
A = sorted(A)
for p in range(1, m+1):
for i in range(1, n+1):
a, b = A[i]
if p >= a and p <= b:
dp[p] = dp[p-1]
elif p < a:
pwr = a - p
dp[p] = min(dp[p], pwr + dp[max(0,a-pwr-1)], a)
elif p > b:
pwr = p - b
dp[p] = min(dp[p], pwr + dp[max(a-pwr-1,0)])
print(dp[m])
``` | instruction | 0 | 26,670 | 10 | 53,340 |
No | output | 1 | 26,670 | 10 | 53,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
import sys
infile = sys.stdin
# infile = open("ein")
A = [(0, 0)]
inf = 9999
n, m = map(int, infile.readline().split())
for i in range(1, n+1):
x, s = map(int, infile.readline().split())
A.append((max(0, x-s), min(m, x+s)))
dp = [inf] * (m+1)
dp[0] = 0
A = sorted(A)
for p in range(1, m+1):
for i in range(1, n+1):
a, b = A[i]
if p >= a and p <= b:
dp[p] = dp[p-1]
elif p < a:
pwr = a - p
dp[p] = min(dp[p], pwr + dp[max(0,a-pwr-1)], p)
elif p > b:
pwr = p - b
dp[p] = min(dp[p], pwr + dp[max(a-pwr-1,0)], p)
print(dp[m])
``` | instruction | 0 | 26,671 | 10 | 53,342 |
No | output | 1 | 26,671 | 10 | 53,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The mayor of the Central Town wants to modernize Central Street, represented in this problem by the (Ox) axis.
On this street, there are n antennas, numbered from 1 to n. The i-th antenna lies on the position x_i and has an initial scope of s_i: it covers all integer positions inside the interval [x_i - s_i; x_i + s_i].
It is possible to increment the scope of any antenna by 1, this operation costs 1 coin. We can do this operation as much as we want (multiple times on the same antenna if we want).
To modernize the street, we need to make all integer positions from 1 to m inclusive covered by at least one antenna. Note that it is authorized to cover positions outside [1; m], even if it's not required.
What is the minimum amount of coins needed to achieve this modernization?
Input
The first line contains two integers n and m (1 ≤ n ≤ 80 and n ≤ m ≤ 100\ 000).
The i-th of the next n lines contains two integers x_i and s_i (1 ≤ x_i ≤ m and 0 ≤ s_i ≤ m).
On each position, there is at most one antenna (values x_i are pairwise distinct).
Output
You have to output a single integer: the minimum amount of coins required to make all integer positions from 1 to m inclusive covered by at least one antenna.
Examples
Input
3 595
43 2
300 4
554 10
Output
281
Input
1 1
1 1
Output
0
Input
2 50
20 0
3 1
Output
30
Input
5 240
13 0
50 25
60 5
155 70
165 70
Output
26
Note
In the first example, here is a possible strategy:
* Increase the scope of the first antenna by 40, so that it becomes 2 + 40 = 42. This antenna will cover interval [43 - 42; 43 + 42] which is [1; 85]
* Increase the scope of the second antenna by 210, so that it becomes 4 + 210 = 214. This antenna will cover interval [300 - 214; 300 + 214], which is [86; 514]
* Increase the scope of the third antenna by 31, so that it becomes 10 + 31 = 41. This antenna will cover interval [554 - 41; 554 + 41], which is [513; 595]
Total cost is 40 + 210 + 31 = 281. We can prove that it's the minimum cost required to make all positions from 1 to 595 covered by at least one antenna.
Note that positions 513 and 514 are in this solution covered by two different antennas, but it's not important.
—
In the second example, the first antenna already covers an interval [0; 2] so we have nothing to do.
Note that the only position that we needed to cover was position 1; positions 0 and 2 are covered, but it's not important.
Submitted Solution:
```
n, m = [int(x) for x in input().split(" ")]
antenna = []
for i in range(n):
x, s = [int(x) for x in input().split(" ")]
antenna.append((x, s))
antenna.sort(key=lambda a: a[0])
cache = {}
def dp(ind, max_covered):
max_covered = min(m, max_covered)
key = (ind, max_covered)
if key in cache:
return cache[key]
ans = blah(ind, max_covered)
cache[key] = ans
return ans
def blah(ind, max_covered):
x, s = antenna[ind]
if max_covered >= x - 1:
if ind == len(antenna) - 1:
right_boundary = max(max_covered, x + s)
right_boundary = min(m, right_boundary)
return m - right_boundary
else:
return dp(ind + 1, max_covered)
else:
num_needed = max(x - s - max_covered, 0)
use_i = num_needed + dp(ind + 1, x + s + num_needed)
dont_use_i = dp(ind + 1, max_covered)
return min(use_i, dont_use_i)
``` | instruction | 0 | 26,672 | 10 | 53,344 |
No | output | 1 | 26,672 | 10 | 53,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,495 | 10 | 54,990 |
Tags: implementation
Correct Solution:
```
t = int(input())
for q in range(t):
n, k, d = map(int, input().split())
a = []
a = [int(s) for s in input().split()]
#print(*a)
ans = n
for i in range(n-d+1):
p = set(a[i:i+d])
#print(*p)
ans = min(ans, len(p))
print(ans)
``` | output | 1 | 27,495 | 10 | 54,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,496 | 10 | 54,992 |
Tags: implementation
Correct Solution:
```
for x in range(int(input())):
n,k,d = input().split();l1=[];min=int(k)
l = list(map(int,input().split()))
for i in range((int(n)-int(d))+1):
for j in range(i,i+int(d)):
if l[j] not in l1:l1.append(l[j])
if len(l1)<min:min = len(l1)
l1.clear()
print(min)
``` | output | 1 | 27,496 | 10 | 54,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,497 | 10 | 54,994 |
Tags: implementation
Correct Solution:
```
t = int(input())
for i in range(t):
n, k, d = [int(r) for r in input().split()]
s = input()
a = []
for i in s.split():
a.append(int(i))
ans = 0
for j in range(0, d):
if j == a[0 : d].index(a[j]):
ans += 1
mn = ans
for i in range(1, n - d + 1):
ans = 0
for j in range(i, i + d):
if j - i == a[i : i + d].index(a[j]):
ans += 1
if ans < mn:
mn = ans
print(mn)
``` | output | 1 | 27,497 | 10 | 54,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,498 | 10 | 54,996 |
Tags: implementation
Correct Solution:
```
for x in range(int(input())):
n,k,d=map(int,input().split())
a,l=d,list(map(int,input().split()))
for y in range(n-d+1):
a=min(a,len(set(l[y:y+d])))
if a==1:break
print(a)
``` | output | 1 | 27,498 | 10 | 54,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,499 | 10 | 54,998 |
Tags: implementation
Correct Solution:
```
for i in range(int(input())):
n = input().split()
d = int(n[2])
k = int(n[1])
n = int(n[0])
maxim = []
count = 0
c2 = k + 1
days = input().split()
try:
for i in range(n):
row = days[i:i+d]
if len(row) == d:
for j in range(k + 1):
if str(j) in days[i:i+d]:
count += 1
maxim.append(count)
count = 0
except Exception:
d = 0
print(min(maxim))
``` | output | 1 | 27,499 | 10 | 54,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,500 | 10 | 55,000 |
Tags: implementation
Correct Solution:
```
#Winners never quit, quiters never win............................................................................
from collections import deque as de
import math
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
from itertools import groupby as gb
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
class My_stack():
def __init__(self):
self.data = []
def my_push(self, x):
return (self.data.append(x))
def my_pop(self):
return (self.data.pop())
def my_peak(self):
return (self.data[-1])
def my_contains(self, x):
return (self.data.count(x))
def my_show_all(self):
return (self.data)
def isEmpty(self):
return len(self.data)==0
arrStack = My_stack()
def decimalToBinary(n):
return bin(n).replace("0b", "")
def isPrime(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def get_prime_factors(number):
prime_factors = []
while number % 2 == 0:
prime_factors.append(2)
number = number / 2
for i in range(3, int(math.sqrt(number)) + 1, 2):
while number % i == 0:
prime_factors.append(int(i))
number = number / i
if number > 2:
prime_factors.append(int(number))
return prime_factors
def get_frequency(list):
dic={}
for ele in list:
if ele in dic:
dic[ele] += 1
else:
dic[ele] = 1
return dic
def Log2(x):
return (math.log10(x) /
math.log10(2));
def isPowerOfTwo(n):
return (math.ceil(Log2(n)) == math.floor(Log2(n)));
#here we go......................
#Winners never quit, Quitters never win
t=int(input())
while t:
t-=1
n,k,d=map(int,input().split())
a=list(map(int,input().split()))
if d==n:
print(len(set(a)))
else:
mn=k
for i in range(n-d+1):
#print(a[i:i+d])
if mn > len(set(a[i:i+d])):
#print(mn,"mn")
#print(a[i:i+d])
mn=len(set(a[i:i+d]))
if mn==1:
break
print(mn)
``` | output | 1 | 27,500 | 10 | 55,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,501 | 10 | 55,002 |
Tags: implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n, k, d = map(int, input().split())
data = [x for x in input().split()]
data1 = []
if len(data) != d:
for i in range(0, len(data) - d + 1):
data1.append(len(set(data[i:i + d])))
print(min(data1))
else:
print(len(set(data)))
``` | output | 1 | 27,501 | 10 | 55,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days. | instruction | 0 | 27,502 | 10 | 55,004 |
Tags: implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n, k, d = list(map(int, input().split()))
s = input().split()
x = []
for i in range(n-d+1):
x.append(len(set(s[i:i+d])))
print(min(x))
``` | output | 1 | 27,502 | 10 | 55,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
for i in range(int(input())):
n, k, d = map(int, input().split())
a = list(map(int, input().split()))
mn = 1000000000000
for i in range(n - d + 1):
b = set(a[i : i + d])
if len(b) < mn:
mn = len(b)
print(mn)
``` | instruction | 0 | 27,503 | 10 | 55,006 |
Yes | output | 1 | 27,503 | 10 | 55,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show, the episode of which will be shown in i-th day.
The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.
How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows d (1 ≤ d ≤ n) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n, k and d (1 ≤ n ≤ 100, 1 ≤ k ≤ 100, 1 ≤ d ≤ n). The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 100.
Output
Print t integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
Input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
Output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. So the answer is two.
In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.
In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.
In the fourth test case, you can buy subscriptions to shows 3,5,7,8,9, and you will be able to watch shows for the last eight days.
Submitted Solution:
```
t=int(input())
y=list(range(t))
for i in range(t):
n, k, d = map(int, input().split())
a=list(range(n))
l=[[] for q in range(n-d+1)]
a=list(map(int, input().split()))
for p in range (n-d+1):
for j in range(1, d+1):
if l[p].count(a[p-1+j])==0:
l[p].append(a[p-1+j])
for p in range (n-d+1):
l[p]=len(l[p])
y[i]=min(l)
for i in range(t):
print(y[i])
``` | instruction | 0 | 27,504 | 10 | 55,008 |
Yes | output | 1 | 27,504 | 10 | 55,009 |
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