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 tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,630 | 10 | 91,260 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import sys
input = sys.stdin.readline
from random import shuffle
n, m, p = map(int, input().split())
A = [input()[:-1] for _ in range(n)]
order = list(range(n))
shuffle(order)
best = 0
sub = '0' * m
for it in range(min(n, 10)):
x = order[it]
bits = [j for j in range(m) if A[x][j] == '1']
sz = len(bits)
cnt = [0] * (1 << sz)
for i in range(n):
cur = sum(1 << j for j in range(sz) if A[i][bits[j]] == '1')
cnt[cur] += 1
for i in range(sz):
for mask in range(1 << sz):
if mask & (1 << i):
cnt[mask ^ (1 << i)] += cnt[mask]
for mask in range(1 << sz):
if 2 * cnt[mask] >= n and bin(mask).count('1') > best:
best = bin(mask).count('1')
sub = ['0'] * m
for i in range(sz):
if mask >> i & 1: sub[bits[i]] = '1'
sub = "".join(sub)
print(sub)
``` | output | 1 | 45,630 | 10 | 91,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,631 | 10 | 91,262 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import sys
input = sys.stdin.readline
from random import shuffle
n, m, p = map(int, input().split())
A = [input().strip() for _ in range(n)]
order = list(range(n))
shuffle(order)
best = 0
sub = '0' * m
for it in range(min(n, 10)):
x = order[it]
bits = [j for j in range(m) if A[x][j] == '1']
sz = len(bits)
cnt = [0] * (1 << sz)
for i in range(n):
cur = sum(1 << j for j in range(sz) if A[i][bits[j]] == '1')
cnt[cur] += 1
for i in range(sz):
for mask in range(1 << sz):
if mask & (1 << i):
cnt[mask ^ (1 << i)] += cnt[mask]
for mask in range(1 << sz):
if 2 * cnt[mask] >= n and bin(mask).count('1') > best:
best = bin(mask).count('1')
sub = ['0'] * m
for i in range(sz):
if mask >> i & 1: sub[bits[i]] = '1'
sub = "".join(sub)
print(sub)
``` | output | 1 | 45,631 | 10 | 91,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
import random
import sys
from sys import stdin
def popcnt(x):
ret = 0
while x > 0:
ret += x % 2
x //= 2
return ret
n,m,p = map(int,stdin.readline().split())
s = [stdin.readline()[:-1] for i in range(n)]
ans = 0
anslis = ["0"] * m
for loop in range(30):
v = random.randint(0,n-1)
vbits = []
for i in range(m):
if s[v][i] == "1":
vbits.append(i)
dp = [0] * (2**len(vbits))
vbits.reverse()
for i in range(n):
now = 0
for j in vbits:
now *= 2
if s[i][j] == "1":
now += 1
dp[now] += 1
vbits.reverse()
#print (v,dp)
for j in range(len(vbits)):
for i in range(len(dp)):
if i & 2**j == 0:
dp[i] += dp[i | 2**j]
#print (v,dp)
for i in range(len(dp)):
if dp[i] >= (n+1)//2:
pc = popcnt(i)
if ans < pc:
#print ("change!")
ans = pc
anslis = [0] * m
for j in range(len(vbits)):
if 2**j & i > 0:
anslis[vbits[j]] = 1
print ("".join(map(str,anslis)))
``` | instruction | 0 | 45,632 | 10 | 91,264 |
Yes | output | 1 | 45,632 | 10 | 91,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
def popCount(a):
cnt = 0
for i in range(60):
if a & (1 << i):
cnt += 1
return cnt
n,m,p = map(int,input().split())
person = []
for _ in range(n):
person.append(int(input(),base = 2))
base = 0
for i in range(m):
cnt = 0
for j in range(n):
if person[j] & (1 << i):
cnt += 1
if cnt >= (n + 1) // 2:
base |= 1 << i
for i in range(n):
person[i] &= base
person.sort(reverse = True)
keys = []
cnt = {}
for elem in person:
if elem in cnt:
cnt[elem] += 1
else:
cnt[elem] = 1
keys.append(elem)
for b in range(60):
if base & (1 << b) == 0:
continue
toAdd = []
toAddVal = []
for key in cnt:
if key & (1 << b):
if key ^ (1 << b) in cnt:
cnt[key ^ (1 << b)] += cnt[key]
else:
toAdd.append(key ^ (1 << b))
toAddVal.append(cnt[key])
for i in range(len(toAdd)):
cnt[toAdd[i]] = toAddVal[i]
maxAns = 0
for key in cnt:
if cnt[key] < (n + 1) // 2:
continue
if popCount(maxAns) < popCount(key):
maxAns = key
print("0" * (m - len(bin(maxAns)) + 2) + bin(maxAns)[2:])
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 45,633 | 10 | 91,266 |
Yes | output | 1 | 45,633 | 10 | 91,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
import random
import sys
from sys import stdin
def popcnt(x):
ret = 0
while x > 0:
ret += x % 2
x //= 2
return ret
n,m,p = map(int,stdin.readline().split())
s = [stdin.readline()[:-1] for i in range(n)]
ans = 0
anslis = ["0"] * m
for loop in range(10):
v = random.randint(0,n-1)
vbits = []
for i in range(m):
if s[v][i] == "1":
vbits.append(i)
dp = [0] * (2**len(vbits))
vbits.reverse()
for i in range(n):
now = 0
for j in vbits:
now *= 2
if s[i][j] == "1":
now += 1
dp[now] += 1
vbits.reverse()
#print (v,dp)
for j in range(len(vbits)):
for i in range(len(dp)):
if i & 2**j == 0:
dp[i] += dp[i | 2**j]
#print (v,dp)
for i in range(len(dp)):
if dp[i] >= (n+1)//2:
pc = popcnt(i)
if ans < pc:
#print ("change!")
ans = pc
anslis = [0] * m
for j in range(len(vbits)):
if 2**j & i > 0:
anslis[vbits[j]] = 1
print ("".join(map(str,anslis)))
``` | instruction | 0 | 45,634 | 10 | 91,268 |
Yes | output | 1 | 45,634 | 10 | 91,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
from itertools import combinations
n, m, p = map(int, input().split())
kols, valutes = dict(), dict()
for i in range(m):
kols[i] = 0
valutes[i] = set()
for friend in range(n):
s = input()
for i in range(m):
if s[i] == '1':
kols[i] += 1
valutes[i].add(friend)
valkeys = list(valutes.keys())
for i in valkeys:
if len(valutes[i]) < n // 2 + n % 2:
valutes.pop(i)
valkeys = list(valutes.keys())
ans = set()
for i in range(1, p + 1):
for attempt in combinations(valkeys, i):
res = set(range(m))
for valute in attempt:
res = res & valutes[valute]
if len(res) >= n // 2 + n % 2:
ans = attempt
break
else:
break
strans = ['0'] * m
for i in ans:
strans[i] = '1'
print(''.join(strans))
``` | instruction | 0 | 45,635 | 10 | 91,270 |
No | output | 1 | 45,635 | 10 | 91,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
import math
import sys
input = sys.stdin.readline
a,b,c = map(int,input().split())
z = [0 for i in range(b)]
like = [[] for i in range(b)]
for i in range(a):
k = input()
for p in range(len(k)):
if k[p] == '1':
z[p]+=1
like[p].append(i)
ans = []
for i in range(len(z)):
if z[i]>=math.ceil(a/2):
ans.append(i)
mx = [0 for i in range(b)]
mxval = 0
for cur in range(2**(len(ans))):
temp = []
for i in range(len(ans)):
if cur&(2**i):
temp.append(ans[i])
if len(temp) == 0:
continue
vis = set(like[temp[0]])
stat = True
for i in range(1,len(temp)):
vis = vis.intersection(set(like[temp[i]]))
if len(list(vis))<math.ceil(a/2):
stat = False
break
if stat == False:
continue
if stat == True:
print(temp)
if len(temp) > mxval:
for i in range(len(mx)):
if i in temp:
mx[i] = 1
mxval = len(temp)
print(*mx,sep='')
``` | instruction | 0 | 45,636 | 10 | 91,272 |
No | output | 1 | 45,636 | 10 | 91,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
import sys
IS_INTERACTIVE = False
input = input # type: ignore
# input: function
if not IS_INTERACTIVE:
*data, = sys.stdin.read().split("\n")[::-1]
def input(): # type: ignore
return data.pop()
def fprint(*args, **kwargs):
print(*args, **kwargs, flush=True)
def eprint(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
n, m, p = map(int, input().split())
likes = []
for _ in range(n):
likes.append(int(input(), base=2))
ans_bits = 0
ans = 0
def popcnt(i):
assert 0 <= i < 0x100000000
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
dp = [0] * (1 << (p + 1))
dp[0] = 1
for i in range(1, 1 << (p + 1)):
popped = popcnt(i)
if popped <= ans_bits:
dp[i] = True
continue
should = False
for j in range(60):
if (1 << j) & i and dp[(1 << j) ^ i]:
should = True
break
cnt = 0
if should:
for v in likes:
if (v & i) == i:
cnt += 1
if cnt * 2 >= n:
dp[i] = True
ans_bits = popcnt(i)
ans = i
print(bin(ans)[2:])
``` | instruction | 0 | 45,637 | 10 | 91,274 |
No | output | 1 | 45,637 | 10 | 91,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.
Submitted Solution:
```
fr, cn, _ = map(int, input().split())
n = (fr % 2) + (fr // 2)
res, ss = [], []
for _ in range(fr):
s = input()
s = set([str(i) for i in range(cn) if s[i]=='1'])
ss.append(s)
del s
ss = sorted(ss, key=len, reverse=True)
for s1 in ss:
cnt, sa = 0, []
for s2 in ss:
sa.append(s1 & s2)
if s1.issubset(s2):
cnt +=1
if cnt>=n:
res.append(''.join(list(s1)))
for sa1 in sa:
cnt = 0
for s2 in ss:
if sa1.issubset(s2):
cnt +=1
if cnt>=n:
res.append(''.join(list(sa1)))
res = sorted(res, key=len, reverse=True)
result = ['0']*cn
for i in res[0]:
result[int(i)] = '1'
print(''.join(result))
``` | instruction | 0 | 45,638 | 10 | 91,276 |
No | output | 1 | 45,638 | 10 | 91,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,785 | 10 | 91,570 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
ans = 0
if c >= n:
ans = n // a
else:
ans += (n - c) // min(b - c, a) + ((n - c) % min(b - c, a) + c) // a
print(ans)
``` | output | 1 | 45,785 | 10 | 91,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,786 | 10 | 91,572 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
if n >= b:
if b - c >= a:
print(n // a)
else:
answer = (n - b) // (b - c)
n = b + (n - b) % (b - c)
cnt = 1
cnt = max(cnt + (n - b + c) // a, n // a)
print(answer + cnt)
else:
print(n // a)
``` | output | 1 | 45,786 | 10 | 91,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,787 | 10 | 91,574 |
Tags: implementation, math
Correct Solution:
```
import math
n=int(input())
a=int(input())
b=int(input())
c=int(input())
g = max(0, (n-c)//(b-c))
print(max(n//a, g+(n-g*(b-c))//a))
``` | output | 1 | 45,787 | 10 | 91,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,788 | 10 | 91,576 |
Tags: implementation, math
Correct Solution:
```
from math import factorial as fact
def gcd(a, b):
while a != 0:
a, b = b % a, a
return b
def lcm(a, b):
return a * b // gcd(a, b)
def popcount(x):
r = 0
while x > 0:
r += 1
x &= (x - 1)
return r
def bit(m, i):
return (m >> i) & 1
def cnk(n, k):
r = 1
for i in range(n - k + 1, n + 1):
r *= i
return r // fact(k)
n = int(input())
a = int(input())
b = int(input())
c = int(input())
'''
1000000000000000000
2
10
9
'''
assert b > c
ans = 0
ans = max(ans, n // a)
if n >= b:
k = 1 + (n - b) // (b - c)
ans = max(ans, k)
ans = max(ans, 1 + (n - b) // a)
ans = max(ans, k + (n - k * (b - c)) // a)
print(ans)
``` | output | 1 | 45,788 | 10 | 91,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,789 | 10 | 91,578 |
Tags: implementation, math
Correct Solution:
```
n, a, b, c = int(input()), int(input()), int(input()), int(input())
x = (n - c) // (b - c)
if c > n:
x = 0
y = n - x * (b - c)
x += y // a
print(max(x, n // a))
``` | output | 1 | 45,789 | 10 | 91,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,790 | 10 | 91,580 |
Tags: implementation, math
Correct Solution:
```
budget = int(input())
plastic = int(input())
glass = int(input())
refund = int(input())
if glass - refund < plastic:
ans = max((budget - refund) // (glass - refund), 0)
budget -= ans * (glass - refund)
ans += budget // plastic
print(ans)
else:
print(budget // plastic)
``` | output | 1 | 45,790 | 10 | 91,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,791 | 10 | 91,582 |
Tags: implementation, math
Correct Solution:
```
n,a,b,c = [int( input()) for i in range(4)]
g = max(0,(n-c)//(b-c))
print (max(n//a,g+(n-g*(b-c))//a))
``` | output | 1 | 45,791 | 10 | 91,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | instruction | 0 | 45,792 | 10 | 91,584 |
Tags: implementation, math
Correct Solution:
```
I = lambda: int(input())
n, a, b, c = I(), I(), I(), I()
m = max(n // a, ((n - b) // (b - c) + 1) if b <= n else 0)
print(m + max(0, ((n - m * (b - c)) // a) if b <= n else 0))
``` | output | 1 | 45,792 | 10 | 91,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
[n,a,b,c] = list(map(int,[input() for _ in range(4)]))
s = 0
if n > c and a > b - c:
n -= c
s += n//(b-c)
n = n%(b-c) + c
print(s+n//a)
``` | instruction | 0 | 45,793 | 10 | 91,586 |
Yes | output | 1 | 45,793 | 10 | 91,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
n, a, b, c = int(input()), int(input()), int(input()), int(input())
l = 0
if b - c < a and n >= c:
l = (n - c)//(b - c)
n -= l*(b - c)
l += n//a
print(l)
``` | instruction | 0 | 45,794 | 10 | 91,588 |
Yes | output | 1 | 45,794 | 10 | 91,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
n,a,b,c=int(input()),int(input()),int(input()),int(input())
plas=a
glas=b-c
if(plas<=glas):
print(int(n//plas))
else:
if(n>=b):
k=int((n-b)//(b-c))+1
n-=((b-c)*k)
if(n>=0):
print(k+int(n//plas))
else:
print(k)
else:
print(int(n//plas))
``` | instruction | 0 | 45,795 | 10 | 91,590 |
Yes | output | 1 | 45,795 | 10 | 91,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
s = int(input())
a = int(input())
b = int(input())
c = int(input())
d = b-c
l = 0
if d < a:
if s >= b:
l = (s-b) // d + 1
s -= l * d
else:
l = s // a
s -= l * a
m = min(a,b)
if s >= m:
l += s // m
print(l)
``` | instruction | 0 | 45,796 | 10 | 91,592 |
Yes | output | 1 | 45,796 | 10 | 91,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
tot=int(input())
plas=int(input())
glass=int(input())
rem=int(input())
c=0
if(glass-rem>=plas):
c=tot//plas+(tot%plas)//glass
else:
tot-=rem
glass-=rem
c+=tot//glass
c+=(tot%glass)//plas
print(c)
``` | instruction | 0 | 45,797 | 10 | 91,594 |
No | output | 1 | 45,797 | 10 | 91,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
d = b - c
if n < a and n < b:
print(0)
elif a < d:
print(n // a)
else:
ans = 0
x = (n - b) // d + 1
ans += x
n -= d * x
ans += (n // a)
print(ans)
``` | instruction | 0 | 45,798 | 10 | 91,596 |
No | output | 1 | 45,798 | 10 | 91,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
if b - c < a:
if n < b:
print(0)
else:
count = n // (b - c) + 1 - b // (b - c)
if a <= b:
if a < n:
print(count + (n - (b - c) * count) // a)
else:
print(count)
else:
print(count)
else:
if n < a:
print(0)
else:
print(n // a)
``` | instruction | 0 | 45,799 | 10 | 91,598 |
No | output | 1 | 45,799 | 10 | 91,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.
Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input
First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Examples
Input
10
11
9
8
Output
2
Input
10
5
6
1
Output
2
Note
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
Submitted Solution:
```
import math
grana = int(input())
a = int(input())
b = int(input())
c = int(input())
if(a<(b-c)):
total = math.ceil(grana/a)
else:
total = math.ceil((grana-b)/(b-c))
grana = grana-total*(b-c)
if(grana >= a):
total = total + math.floor(grana/a)
print(total)
``` | instruction | 0 | 45,800 | 10 | 91,600 |
No | output | 1 | 45,800 | 10 | 91,601 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,148 | 10 | 92,296 |
"Correct Solution:
```
#!python3
iim = lambda: map(int, input().rstrip().split())
def resolve():
N, W = iim()
S = [list(iim()) for i in range(N)]
NV = N * 100 + 1
Inf = float("inf")
dp = [Inf] * (NV+1)
dp[0] = 0
ans = 0
for v, w in S:
for i in range(ans, -1, -1):
j = i + v
w2 = dp[i] + w
if dp[i] == Inf: continue
if w2 > W: continue
dp[j] = min(dp[j], w2)
ans = max(ans, j)
print(ans)
if __name__ == "__main__":
resolve()
``` | output | 1 | 46,148 | 10 | 92,297 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,149 | 10 | 92,298 |
"Correct Solution:
```
import sys, bisect
input = lambda: sys.stdin.readline().rstrip()
sys.setrecursionlimit(10**7)
INF = 10**10
def I(): return int(input())
def F(): return float(input())
def SS(): return input()
def LI(): return [int(x) for x in input().split()]
def LI_(): return [int(x)-1 for x in input().split()]
def LF(): return [float(x) for x in input().split()]
def LSS(): return input().split()
def resolve():
N, W = LI()
vw = [LI() for _ in range(N)]
v_sum = sum([i[0] for i in vw])
# dp[i][j]: [0, i)を使って価値j以上になる場合の重さの最小値
dp = [[INF] * (v_sum + 1) for _ in range(N + 1)]
for i in range(N + 1):
dp[i][0] = 0
for i in range(N):
for j in range(v_sum):
v_p = max(j + 1 - vw[i][0], 0)
if v_p >= 0:
dp[i+1][j+1] = min(dp[i][j+1], dp[i][v_p] + vw[i][1])
else:
dp[i+1][j+1] = dp[i][j+1]
# for i in dp:
# print(i)
ans = bisect.bisect_right(dp[-1], W) - 1
print(ans)
if __name__ == '__main__':
resolve()
``` | output | 1 | 46,149 | 10 | 92,299 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,150 | 10 | 92,300 |
"Correct Solution:
```
def knapsack(n, w, info):
sum_v = 0
for i in range(n):
value = info[i][0]
sum_v += value
# dp[i][j] := i番目までの品物の中から価値の総和がjとなるように選んだときの、
# 重さの総和の最小値
dp = [[float("inf")]*(sum_v + 1) for i in range(n + 1)]
dp[0][0] = 0
for i in range(n):
for j in range(sum_v + 1):
weight = info[i][1]
value = info[i][0]
if j < value:
dp[i + 1][j] = dp[i][j]
else:
dp[i + 1][j] = min(dp[i][j], dp[i][j - value] + weight)
ans = 0
for j in range(sum_v + 1):
if dp[n][j] <= w:
ans = j
return ans
n, w = map(int, input().split())
info = [list(map(int, input().split())) for i in range(n)]
print(knapsack(n, w, info))
``` | output | 1 | 46,150 | 10 | 92,301 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,151 | 10 | 92,302 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
from bisect import *
from collections import *
from heapq import *
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def SI(): return sys.stdin.readline()[:-1]
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def MF(): return map(float, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LF(): return list(map(float, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
dij = [(0, 1), (1, 0), (0, -1), (-1, 0)]
def main():
inf=10**9+5
n,wn=MI()
dp=[inf]*10005
dp[0]=0
for _ in range(n):
v,w=MI()
for i in range(10004,-1,-1):
if i-v<0:break
dp[i]=min(dp[i],dp[i-v]+w)
for i in range(10004,-1,-1):
if dp[i]<=wn:
print(i)
break
main()
``` | output | 1 | 46,151 | 10 | 92,303 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,152 | 10 | 92,304 |
"Correct Solution:
```
from math import isfinite
from itertools import dropwhile
n, knapsack = map(int, input().split())
dpn = n * 100 # MAX VALUE
dp = [float('inf')] * (dpn + 1)
dp[0] = 0
for i in range(n):
value, weight = map(int, input().split())
for j in range(dpn - value, 0, -1):
if not isfinite(dp[j]):
continue
new_weight = dp[j] + weight
if new_weight <= knapsack and dp[j + value] > new_weight:
dp[j + value] = new_weight
if weight <= knapsack and dp[value] > weight:
dp[value] = weight
print(next(dropwhile(lambda i: not isfinite(dp[i]), range(dpn, -1, -1))))
``` | output | 1 | 46,152 | 10 | 92,305 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,153 | 10 | 92,306 |
"Correct Solution:
```
N, W = map(int, input().split())
I = [list(map(int, input().split())) for _ in range(N)]
MAX_V = sum(v for v, _ in I)
dp = [W+1] * (MAX_V + 1)
dp[0] = 0
for v, w in I:
for j in range(MAX_V, v-1, -1):
dp[j] = min(dp[j-v] + w, dp[j])
for i in range(MAX_V, -1, -1):
if dp[i] <= W:
print(i)
break
``` | output | 1 | 46,153 | 10 | 92,307 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,154 | 10 | 92,308 |
"Correct Solution:
```
INF = 10 ** 20
n, w = map(int, input().split())
value = []
weight = []
for _ in range(n):
vi, wi = map(int, input().split())
value.append(vi)
weight.append(wi)
sum_value = sum(value)
dp = [INF for _ in range(sum_value + 1)]
dp[0] = 0
for i in range(n):
for j in range(sum_value, value[i] - 1, -1):
dp[j] = min(dp[j], dp[j - value[i]] + weight[i])
for j in range(sum_value, -1, -1):
if dp[j] <= w:
print(j)
break
``` | output | 1 | 46,154 | 10 | 92,309 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 46,155 | 10 | 92,310 |
"Correct Solution:
```
import heapq
from collections import deque
from enum import Enum
import sys
import math
from _heapq import heappush, heappop
BIG_NUM = 2000000000
MOD = 1000000007
EPS = 0.000000001
NUM = 10000
N,W = map(int,input().split())
dp = [W+1]*(NUM+1)
dp[0] = 0
for loop in range(N):
value,weight = map(int,input().split())
for v in range(NUM,0,-1):
if v-value < 0:
break
dp[v] = min(dp[v],dp[v-value]+weight) #価値あたりの最小の重さを求める
ans = 0
for i in range(NUM,0,-1):
if dp[i] <= W:
ans = i
break
print("%d"%(ans))
``` | output | 1 | 46,155 | 10 | 92,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
N,W=map(int,input().split())
v=[0]*N
w=[0]*N
for i in range(N):
v[i],w[i]=map(int,input().split())
V=max(v)*N
dp=[-1]*(V+1)
dp[0]=0
for i in range(N):
for j in range(V+1):
if dp[V-j]!=-1 and -j+v[i]<=0:
if dp[V-j+v[i]]==-1:
dp[V-j+v[i]]=dp[V-j]+w[i]
else:
dp[V-j+v[i]]=min(dp[V-j+v[i]],dp[V-j]+w[i])
for i in range(V+1):
if dp[V-i]!=-1 and dp[V-i]<=W:
print(V-i)
break
``` | instruction | 0 | 46,156 | 10 | 92,312 |
Yes | output | 1 | 46,156 | 10 | 92,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
n, m = map(int, input().split())
VW = [tuple(map(int, input().split())) for i in range(n)]
V = [v for v, w in VW]
W = [w for v, w in VW]
# n, m = 4, 5
# V = [4, 5, 2, 8]
# W = [2, 2, 1, 3]
# DP[i][j]=i個の品物で価値j以上で最小の重さ
sv = sum(V)
inf = 10**10
DP = [[inf for j in range(sv+1)] for i in range(n+1)]
DP[0][0] = 0
for i in range(n):
for j in range(sv+1):
if j < V[i]:
DP[i+1][j] = DP[i][j]
else:
DP[i+1][j] = min(DP[i][j], DP[i][j-V[i]]+W[i])
for j in range(sv, 0, -1):
if DP[n][j] <= m:
print(j)
exit()
print(0)
``` | instruction | 0 | 46,157 | 10 | 92,314 |
Yes | output | 1 | 46,157 | 10 | 92,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
import sys
input = sys.stdin.readline
M, W = map(int, input().split()) # M: 品物の種類 W: 重量制限
single = True # True = 重複なし
price_list = [0]
weight_list = [0]
for _ in range(M):
price, weight = map(int, input().split())
price_list.append(price)
weight_list.append(weight)
################################################################################################################################################
V = sum(price_list)
dp_max = W + 1 # 総和重量の最大値
dp = [[dp_max] * (V + 1) for _ in range(M + 1)] # 左端と上端の境界条件で W, M を一個ずつ多めに取る
""" dp[item <= M ][weight <= V] = 価値を固定した時の"最小"重量 """
for item in range(M+1): # 左端の境界条件
dp[item][0] = 0 # 価値 0 を実現するのに必要な品物の個数は 0 個
# for weight in range(W+1): # 上端の境界条件
# dp[0][weight] = dp_min # 0 種類の品物で実現できる価値総額は存在しない （dpの初期化で自動的に課せる場合が多い）
for item in range(1,M+1): # 境界条件を除いた M 回のループ（※ f[item] = g.f[item-i] の漸化式なので、list index out of range となる）
for price in range(1, V+1):
if price < price_list[item]: # price (総価値) < price_list[item] (品物の価値) の時は無視してよい
dp[item][price] = dp[item - 1][price]
else:
temp = dp[item - single][price - price_list[item]] + weight_list[item] # single = True: 重複なし
if temp < dp[item-1][price]: # min(dp[item-1][price], temp)
dp[item][price] = temp
else:
dp[item][price] = dp[item-1][price]
print(max(price for price in range(V+1) if dp[M][price] <= W))
``` | instruction | 0 | 46,158 | 10 | 92,316 |
Yes | output | 1 | 46,158 | 10 | 92,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
import sys
# import re
import math
import collections
# import decimal
import bisect
import itertools
import fractions
# import functools
import copy
# import heapq
import decimal
# import statistics
import queue
# import numpy as np
sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7
# MOD = 998244353
ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))
# ===CODE===
def main():
n, w = ns()
dp = [INF for _ in range(n * 100 + 1)]
dp[0] = 0
for _ in range(n):
vi, wi = ns()
for i in range(n * 100 - vi, -1, -1):
if dp[i] == INF:
continue
if dp[i] + wi <= w:
dp[i + vi] = min(dp[i + vi], dp[i] + wi)
for i in range(n * 100, -1, -1):
if dp[i] != INF:
print(i)
exit(0)
if __name__ == '__main__':
main()
``` | instruction | 0 | 46,159 | 10 | 92,318 |
Yes | output | 1 | 46,159 | 10 | 92,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
N, W = map(int, input().split())
dp = [0] + [W+1] * (N*100)
for i in range(N):
x, y = map(int, input().split())
for j in range(i*100, -1, -1):
if dp[j+y] > dp[j] + x:
dp[j+y] = dp[j] + x
for v, w in zip(range(N*100, -1, -1), dp[::-1]):
if w < W+1:
print(v)
break
``` | instruction | 0 | 46,160 | 10 | 92,320 |
No | output | 1 | 46,160 | 10 | 92,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
a,b = map(int,input())
dp = [0 for i in range(b+1)]
for i in range(a):
x,y = map(int,input())
for j in range(b,y+1,y):
pp = x + dp[j-y]
if pp > dp[j]:
dp[j] = pp
print(max(dp))
``` | instruction | 0 | 46,161 | 10 | 92,322 |
No | output | 1 | 46,161 | 10 | 92,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
a,b = map(int,input().split())
dp = [0 for i in range(b+1)]
for i in range(a):
x,y = map(int,input().split())
for j in range(b,y-1,-y):
pp = x + dp[j-y]
if pp > dp[j]:
dp[j] = pp
print(max(dp))
``` | instruction | 0 | 46,162 | 10 | 92,324 |
No | output | 1 | 46,162 | 10 | 92,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 100
* 1 ≤ wi ≤ 10,000,000
* 1 ≤ W ≤ 1,000,000,000
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
a,b = map(int,input().split())
dp = [0 for i in range(b+1)]
for i in range(a):
x,y = map(int,input().split())
for j in range(b,y-1,-1):
pp = x + dp[j-y]
if pp > dp[j]:
dp[j] = pp
print(max(dp))
``` | instruction | 0 | 46,163 | 10 | 92,326 |
No | output | 1 | 46,163 | 10 | 92,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,279 | 10 | 92,558 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
m,n=map(int,input().split())
l=[]
#print(m,n)
for i in range(m):
l+=[list(map(int,input().split()))]
l=[set(i[1:]) for i in l]
#k=0
#print(n,m)
for i in range(m):
for j in range(i+1,m):
#print(a,b,a.intersection(b))
if len(l[i].intersection(l[j]))==0:
print('impossible')
exit()
"""count=0
for j in range(m):
if i in l[j]:
count+=1
if count==m:
print('possible')
exit()"""
print('possible')
``` | output | 1 | 46,279 | 10 | 92,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,280 | 10 | 92,560 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve():
m, n = map(int, input().split())
c = [0]*(n+1)
a = [None]*m
for i in range(m):
s = map(int, input().split())
next(s)
a[i] = list(s)
for i in range(m):
for v in a[i]:
c[v] += 1
for j in range(i+1,m):
for v in a[j]:
if c[v] > 0:
break
else:
print('impossible')
return
for v in a[i]:
c[v] -= 1
print('possible')
solve()
``` | output | 1 | 46,280 | 10 | 92,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,281 | 10 | 92,562 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
m, n = map(int, input().split())
Dora = []
Swiper = []
all = set([i for i in range(1, n+1)])
for _ in range(m):
a = set(list(map(int, input().split()))[1:])
Dora.append(a)
Swiper.append(all.difference(a))
flag = 1
for i in range(m):
for j in range(m):
if Dora[i] | Swiper[j] == Swiper[j]:
flag = 0
break
if flag == 0:
break
if flag == 1:
print("possible")
else:
print("impossible")
``` | output | 1 | 46,281 | 10 | 92,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,282 | 10 | 92,564 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
m, n = map(int, input().split())
cnt = [0] * (n + 1)
p = 1
arr = []
for _ in range(m):
*a, = map(int, input().split())
s = {a[i] for i in range(1, a[0] + 1)}
for i in arr:
if len(i | s) == len(i) + a[0]:
print('impossible')
exit()
arr.append(s)
print('possible')
``` | output | 1 | 46,282 | 10 | 92,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,283 | 10 | 92,566 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
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")
#-------------------------------------------------------------------------------------------------------------------
m,n=map(int,input().split())
l=[]
for i in range(m):
e=list(map(int,input().split()))
l.append(set(e[1:]))
for i in range(m):
for j in range(i+1,m):
if len(l[i]&l[j])==0:
print("impossible")
sys.exit(0)
print("possible")
``` | output | 1 | 46,283 | 10 | 92,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,284 | 10 | 92,568 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
m,n = map(int,input().split())
t = []
for i in range(m):
temp = list(map(int,input().split()))[1:]
bit = 0
for i in temp:
bit ^= (1<<(i-1))
t.append(bit)
ans = True
for i in range(m-1):
for j in range(i+1,m):
if t[i] & t[j] == 0:
ans = False
print('possible' if ans else 'impossible')
``` | output | 1 | 46,284 | 10 | 92,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,285 | 10 | 92,570 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
m,n=map(int,input().split())
l=[]
l=[list(map(int,input().split())) for i in range(m)]
l=[set(i[1:]) for i in l]
for i in range(m):
for j in range(i+1,m):
if len(l[i].intersection(l[j]))==0:
print('impossible')
exit()
print('possible')
``` | output | 1 | 46,285 | 10 | 92,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12. | instruction | 0 | 46,286 | 10 | 92,572 |
Tags: bitmasks, brute force, constructive algorithms, math, number theory
Correct Solution:
```
M, N = map(int, input().split())
X = []
for i in range(M):
X.append(set(list(map(int, input().split()))[1:]))
flg = 1
for i in range(M):
for j in range(i):
if len(X[i] & X[j]) == 0:
flg = 0
break
if flg == 0:
break
if flg:
print("possible")
else:
print("impossible")
``` | output | 1 | 46,286 | 10 | 92,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dora the explorer has decided to use her money after several years of juicy royalties to go shopping. What better place to shop than Nlogonia?
There are n stores numbered from 1 to n in Nlogonia. The i-th of these stores offers a positive integer a_i.
Each day among the last m days Dora bought a single integer from some of the stores. The same day, Swiper the fox bought a single integer from all the stores that Dora did not buy an integer from on that day.
Dora considers Swiper to be her rival, and she considers that she beat Swiper on day i if and only if the least common multiple of the numbers she bought on day i is strictly greater than the least common multiple of the numbers that Swiper bought on day i.
The least common multiple (LCM) of a collection of integers is the smallest positive integer that is divisible by all the integers in the collection.
However, Dora forgot the values of a_i. Help Dora find out if there are positive integer values of a_i such that she beat Swiper on every day. You don't need to find what are the possible values of a_i though.
Note that it is possible for some values of a_i to coincide in a solution.
Input
The first line contains integers m and n (1≤ m ≤ 50, 1≤ n ≤ 10^4) — the number of days and the number of stores.
After this m lines follow, the i-th line starts with an integer s_i (1≤ s_i ≤ n-1), the number of integers Dora bought on day i, followed by s_i distinct integers, the indices of the stores where Dora bought an integer on the i-th day. The indices are between 1 and n.
Output
Output must consist of a single line containing "possible" if there exist positive integers a_i such that for each day the least common multiple of the integers bought by Dora is strictly greater than the least common multiple of the integers bought by Swiper on that day. Otherwise, print "impossible".
Note that you don't have to restore the integers themselves.
Examples
Input
2 5
3 1 2 3
3 3 4 5
Output
possible
Input
10 10
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
Output
impossible
Note
In the first sample, a possible choice for the values of the a_i is 3, 4, 3, 5, 2. On the first day, Dora buys the integers 3, 4 and 3, whose LCM is 12, while Swiper buys integers 5 and 2, whose LCM is 10. On the second day, Dora buys 3, 5 and 2, whose LCM is 30, and Swiper buys integers 3 and 4, whose LCM is 12.
Submitted Solution:
```
m,n=map(int,input().split())
s=set(range(1,n+1))
d=[[None,None]for i in range(m)]
for i in range(m):
x=list(map(int,input().split()))[1:]
x=set(x)
d[i][0]=x
d[i][1]=s-x
for i in range(m):
for j in range(i+1,m):
bi,si=d[i]
bj,sj=d[j]
if sj.issuperset(bi) and bj.issubset(si):
print('impossible')
quit()
print('possible')
``` | instruction | 0 | 46,287 | 10 | 92,574 |
Yes | output | 1 | 46,287 | 10 | 92,575 |
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