message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
def buscaEnergia(perna,qtd):
total = 0
if(qtd == 0):
return total
for i in range(len(d)-1,-1,-1):
for v in d[i]:
if(v <perna):
qtd-=1
total+=i
else:
break
if(qtd == 0):
break
if(qtd == 0):
break
return total
n = int(input())
tamanhoPerna = [int(i) for i in input().split()]
energiaPerna = [int(i) for i in input().split()]
d = [[] for i in range(201)]
cntPerna = [0]*100001
sumPerna = [0]*100001
corteTotal = 0
maxTam = 0
for i in range(len(tamanhoPerna)):
d[energiaPerna[i]].append(tamanhoPerna[i])
sumPerna[tamanhoPerna[i]]+=energiaPerna[i]
cntPerna[tamanhoPerna[i]]+=1
corteTotal+=energiaPerna[i]
maxTam = max(tamanhoPerna[i],maxTam)
for i in range(len(d)):
d[i].sort()
currentMin = float("inf")
for perna in tamanhoPerna:
somaMesa = sumPerna[perna]
somaMesa+=buscaEnergia(perna,cntPerna[perna]-1)
currentMin = min(currentMin,corteTotal-somaMesa)
print(currentMin)
``` | instruction | 0 | 24,258 | 8 | 48,516 |
Yes | output | 1 | 24,258 | 8 | 48,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
n = int(input())
leng = list(map(int, input().split()))
energy = list(map(int, input().split()))
sum = 0
lenSum = {}
lenCount = {}
custos = {}
for i in range(n):
length, custo = leng[i], energy[i]
sum += custo
lenSum[length] = lenSum.setdefault(length, 0) + custo
lenCount[length] = lenCount.setdefault(length, 0) + 1
custos.setdefault(custo, []).append(length)
conjuntoLen = set(leng)
for leng in custos.values():
leng.sort()
listaEnergia = list(reversed(sorted(custos.keys())))
ans = -1
for length in conjuntoLen:
total = sum - lenSum[length]
find = lenCount[length] - 1
if find != 0:
for custo in listaEnergia:
for x in custos[custo]:
if x >= length:
break
total -= custo
find -= 1
if find == 0:
break
if find == 0:
break
if ans == -1 or total < ans:
ans = total
print(ans)
``` | instruction | 0 | 24,259 | 8 | 48,518 |
Yes | output | 1 | 24,259 | 8 | 48,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
f = lambda: map(int, input().split())
n, p, m, s = input(), {}, 0, sorted(zip(f(), f()), key=lambda q: -q[1])
for L, d in s:
k, D = p.get(L, (-1, 0))
p[L] = (k + 1, D + d)
for L, (k, D) in p.items():
if k:
for l, d in s:
if l < L:
D += d
k -= 1
if k == 0: break
m = max(D, m)
print(sum(d for l, d in s) - m)
``` | instruction | 0 | 24,260 | 8 | 48,520 |
Yes | output | 1 | 24,260 | 8 | 48,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
d = list(map(int, input().split()))
from collections import defaultdict, Counter
ld = [(li, di) for li, di in zip(l, d)]
ld.sort(key=lambda x : x[0])
ls = sorted(list(set(l)))
surld = defaultdict(int)
for i in range(len(ld) - 1, -1, -1):
surld[ld[i][0]] += ld[i][1]
if len(ls) > 1:
for i in range(len(ls) - 2, -1, -1):
surld[ls[i]] += surld[ls[i+1]]
preld = defaultdict(list)
for li, di in ld:
for lj in ls:
if lj > li:
preld[lj].append(di)
for k in preld:
preld[k].sort()
lcnt = Counter(l)
llcnt = defaultdict(int)
for i in range(len(ls)):
if i > 0:
llcnt[ls[i]] = lcnt[ls[i]] + llcnt[ls[i-1]]
res = float('inf')
for i in range(len(ls)):
need = 0
if i != len(ls) - 1:
need = surld[ls[i+1]]
cut = max(0, (llcnt[ls[i]] - lcnt[ls[i]]) + 1)
need += sum(preld[ls[i]][:cut])
res = min(res, need)
print(res)
``` | instruction | 0 | 24,261 | 8 | 48,522 |
No | output | 1 | 24,261 | 8 | 48,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
def increment(d,key,energy):
if key in d.keys():
d[key][0] += 1
d[key][1] += energy
else:
d[key] = [1,energy]
legs = int(input())
lengths = list(map(int,input().split()))
energies = list(map(int,input().split()))
pairsEL = []
for x in range(legs):
pairsEL.append([energies[x],lengths[x]])
pairsEL.sort()
pairsEL.reverse()
lengthMap = {}
for x in range(legs):
increment(lengthMap,lengths[x],energies[x])
'''
minEnergy = 20000001
currCost = 0
currLegs = 0
for key in sorted(lengthMap.keys())[::-1]:
minLength = min(legs,2*lengthMap[key][0]-1)
legsToRemove = legs - minLength - currLegs
currEnergy = currCost
x=0
shift=0
while x < legsToRemove:
if pairsEL[x+shift][1] < key:
currEnergy += pairsEL[x+shift][0]
x+=1
else:
shift+=1
minEnergy = min(minEnergy, currEnergy)
if currCost > minEnergy:
break
currLegs += lengthMap[key][0]
currCost += lengthMap[key][1]
print(minEnergy)
'''
totalEnergy = sum(energies)
maxEnergy = 0
for key in reversed(sorted(lengthMap.keys())):
currEnergy = 0
legsToAdd = lengthMap[key][0]-1
for pair in pairsEL:
if pair[1] < key:
currEnergy += pair[0]
legsToAdd -= 1
if legsToAdd == 0:
break
maxEnergy = max(maxEnergy,currEnergy)
print(totalEnergy-maxEnergy)
``` | instruction | 0 | 24,262 | 8 | 48,524 |
No | output | 1 | 24,262 | 8 | 48,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
class Noha:
def __init__(self, dlzka, celkova_cena, pocet, vsetky_nohy):
self.dlzka = dlzka
self.celkova_cena = celkova_cena
self.pocet = pocet
self.vsetky_nohy = vsetky_nohy
def __repr__(self):
return 'D: {} - C: {} - P: {}\n'.format(self.dlzka, self.celkova_cena, self.pocet)
n = int(input())
legs = [(int(x), int(y)) for x, y in zip(input().split(), input().split())]
# Cena odstranenia noh...
cena_nohy = {}
pocet_noh = {}
rozne_nohy = {}
for noha in legs:
dlzka, cena = noha
cena_nohy[dlzka] = cena_nohy.get(dlzka, 0) + cena
pocet_noh[dlzka] = pocet_noh.get(dlzka, 0) + 1
try:
rozne_nohy[dlzka].append(cena)
except:
rozne_nohy[dlzka] = []
nohy = []
for key in cena_nohy.keys():
nohy.append(Noha(dlzka=key, celkova_cena=cena_nohy[key], pocet=pocet_noh[key], vsetky_nohy=rozne_nohy[key]))
nohy.sort(key=lambda x: x.dlzka)
pocet_lacnych_noh = [0] * 220
for noha in legs:
pocet_lacnych_noh[noha[1]] += 1
# Pome odstranovat nohy....
answer = float('inf')
pocet_noh_doteraz = 0
cena_noh_doteraz = 0
for i in range(len(nohy)-1, -1, -1):
dlzka = nohy[i].dlzka
cenaa = nohy[i].celkova_cena
pocett = nohy[i].pocet
vymaz_tieto_nohy = nohy[i].vsetky_nohy
nohy_na_vymazanie = n - ((pocett * 2)-1) - pocet_noh_doteraz
#print(pocet_lacnych_noh[:10])
for noha in vymaz_tieto_nohy:
pocet_lacnych_noh[noha] -= 1
tmp = cena_noh_doteraz
for cena, pocet in enumerate(pocet_lacnych_noh):
if pocet == 0: continue
#print('cena', cena)
#print('pocet', pocet)
if pocet <= nohy_na_vymazanie:
nohy_na_vymazanie -= pocet
tmp += pocet * cena
else:
tmp += nohy_na_vymazanie * cena
break
answer = min(tmp, answer)
pocet_noh_doteraz += pocett
cena_noh_doteraz += cenaa
print(answer)
``` | instruction | 0 | 24,263 | 8 | 48,526 |
No | output | 1 | 24,263 | 8 | 48,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.
In total the table Arthur bought has n legs, the length of the i-th leg is li.
Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.
A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.
Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.
Input
The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.
The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.
The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.
Output
Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.
Examples
Input
2
1 5
3 2
Output
2
Input
3
2 4 4
1 1 1
Output
0
Input
6
2 2 1 1 3 3
4 3 5 5 2 1
Output
8
Submitted Solution:
```
M = 100002
def main():
n = int(input())
length = [int(i) for i in input().split(" ")]
d = [int(i) for i in input().split(" ")]
a1 = [0]*M
a2 = [0]*M
a3 = [[length[i], d[i]] for i in range(n)]
a3 = sorted(a3, key = lambda x : x[1])
for i in range(n):
a1[length[i]] += 1
a2[length[i]] += d[i]
rleg, re = n, 0
for i in range(M-1, 0, -1):
if (a1[i] > rleg/2):
break
else:
cost, nleg, j = 0, 0, 0
while (a1[i] <= (rleg - nleg)/2):
if (j < len(a3) and a3[j][0] < a1[i]):
cost += a3[j][1]
nleg += 1
j += 1
elif (j < len(a3)):
j += 1
continue
else:
break
if (cost > a2[i]):
break
if (cost <= a2[i] and j < len(a3)):
re += cost
break
else:
re += a2[i]
rleg -= a1[i]
print(re)
if __name__ == "__main__":
main()
``` | instruction | 0 | 24,264 | 8 | 48,528 |
No | output | 1 | 24,264 | 8 | 48,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An infinitely long railway has a train consisting of n cars, numbered from 1 to n (the numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train?
Input
The first line of the input contains integer n (1 ≤ n ≤ 100 000) — the number of cars in the train.
The second line contains n integers pi (1 ≤ pi ≤ n, pi ≠ pj if i ≠ j) — the sequence of the numbers of the cars in the train.
Output
Print a single integer — the minimum number of actions needed to sort the railway cars.
Examples
Input
5
4 1 2 5 3
Output
2
Input
4
4 1 3 2
Output
2
Note
In the first sample you need first to teleport the 4-th car, and then the 5-th car to the end of the train.
Submitted Solution:
```
#!/usr/bin/env python3
# 606C_sort.py - Codeforces.com/problemset/problem/606/C by Sergey 2015
import unittest
import sys
###############################################################################
# Sort Class (Main Program)
###############################################################################
class Sort:
""" Sort representation """
def __init__(self, test_inputs=None):
""" Default constructor """
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
return next(it) if it else sys.stdin.readline().rstrip()
# Reading single elements
[self.n] = map(int, uinput().split())
# Reading a single line of multiple elements
self.nums = list(map(int, uinput().split()))
# Translate number to position
tr = [0] * self.n
for i in range(self.n):
tr[self.nums[i]-1] = i
# Longest increasing subarray
max_len = 1
for i in range(self.n):
if i > 0 and tr[i] > tr[i-1]:
cur_len += 1
max_len = max(max_len, cur_len)
else:
cur_len = 1
self.result = self.n - max_len
def calculate(self):
""" Main calcualtion function of the class """
return str(self.result)
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_single_test(self):
""" Sort class testing """
# Constructor test
test = "5\n4 1 2 5 3"
d = Sort(test)
self.assertEqual(d.n, 5)
self.assertEqual(d.nums, [4, 1, 2, 5, 3])
# Sample test
self.assertEqual(Sort(test).calculate(), "2")
# Sample test
test = "4\n4 1 3 2"
self.assertEqual(Sort(test).calculate(), "2")
# Sample test
test = "8\n6 2 1 8 5 7 3 4"
self.assertEqual(Sort(test).calculate(), "5")
# My tests
test = ""
# self.assertEqual(Sort(test).calculate(), "0")
# Time limit test
# self.time_limit_test(5000)
def time_limit_test(self, nmax):
""" Timelimit testing """
import random
import timeit
# Random inputs
test = str(nmax) + " " + str(nmax) + "\n"
numnums = [str(i) + " " + str(i+1) for i in range(nmax)]
test += "\n".join(numnums) + "\n"
nums = [random.randint(1, 10000) for i in range(nmax)]
test += " ".join(map(str, nums)) + "\n"
# Run the test
start = timeit.default_timer()
d = Sort(test)
calc = timeit.default_timer()
d.calculate()
stop = timeit.default_timer()
print("\nTimelimit Test: " +
"{0:.3f}s (init {1:.3f}s calc {2:.3f}s)".
format(stop-start, calc-start, stop-calc))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(Sort().calculate())
``` | instruction | 0 | 24,290 | 8 | 48,580 |
Yes | output | 1 | 24,290 | 8 | 48,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
n = int(input())
l = sorted(list(map(int,input().split())))
l = l[::-1]
MOD = 10**9+7
power = []
two = 1
for i in range(n+3):
power.append(two)
two *= 2
two%=MOD
ans = 0
for i in range(n):
ans = (ans + (l[i] * power[n-i-1])) % MOD
ans = (ans - (l[i] * power[i])) % MOD
print(ans)
``` | instruction | 0 | 24,404 | 8 | 48,808 |
Yes | output | 1 | 24,404 | 8 | 48,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
n=int(input())
li=list(map(int,input().split()))
modulo=1000000007
temp=0
res=0
counter=2
li.sort()
for i in range(1,n):
temp=2*temp%modulo+(li[i]-li[i-1])*((counter-1)%modulo)
res=(res+temp)%modulo
counter=(counter<<1)%modulo
print(res)
``` | instruction | 0 | 24,405 | 8 | 48,810 |
Yes | output | 1 | 24,405 | 8 | 48,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
n=int (input ())
A=list (map(int, input (). split ()))
A=sorted (A)
M=10**9+7
pw=[1]
sumF, sumL=0,0
for i in range (n-1):
pw. append ((2*pw[i])%M)
for i in range (n) :
sumF=(sumF+A[i]*pw[n-1-i])%M
for i in range (n) :
sumL=(sumL+A[i]*pw[i])%M
print ((sumL-sumF+M)%M)
``` | instruction | 0 | 24,406 | 8 | 48,812 |
Yes | output | 1 | 24,406 | 8 | 48,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
import sys
def solution():
MOD = int(1e9 + 7)
n = int( input() )
x = list( map(int, sys.stdin.readline().split()) )
x.sort(reverse = True)
ac1 = [ x[0] ]
for i in range( 1, len(x) ):
ac1.append( ( ac1[-1] + x[i] ) % MOD )
x = x[::-1]
ac2 = [ x[0] ]
for i in range( 1, len(x) ):
ac2.append( ( ac2[-1] + x[i] ) % MOD )
ans = 0
lim = n // 2
for i in range(lim):
ans = ( ans + ( pow(2, i, MOD) * ( ac1[i] - ac2[i] + MOD ) ) % MOD ) % MOD
if n % 2 == 0:
lim -= 1
for i in range(lim):
ans = ( ans + ( pow(2, n - 2 - i, MOD) * ( ac1[i] - ac2[i] ) + MOD ) % MOD ) % MOD
print(ans)
if __name__ == '__main__':
solution()
``` | instruction | 0 | 24,407 | 8 | 48,814 |
Yes | output | 1 | 24,407 | 8 | 48,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
def mp(): return map(int,input().split())
def lt(): return list(map(int,input().split()))
def pt(x): print(x)
def ip(): return input()
def it(): return int(input())
def sl(x): return [t for t in x]
def spl(x): return x.split()
def aj(liste, item): liste.append(item)
def bin(x): return "{0:b}".format(x)
def listring(l): return ' '.join([str(x) for x in l])
def ptlist(l): print(' '.join([str(x) for x in l]))
M = 1000000007
n = it()
w = lt()
def exp(k,t,w):
if t == 0:
return w
if t == 1:
return k*w
elif t%2 == 0:
return exp(k*k%M,t//2,w)
else:
return exp(k*k%M,t//2,k*w%M)
expp = {}
w.sort()
m = 0
for i in range(n):
for j in range(i+1,n):
if j-i-1 not in expp:
expp[j-i-1] = exp(2,j-i-1,1)
m += expp[j-i-1]*w[j]-w[i]
pt(m)
``` | instruction | 0 | 24,408 | 8 | 48,816 |
No | output | 1 | 24,408 | 8 | 48,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
import sys
read=lambda:sys.stdin.readline().rstrip()
readi=lambda:int(sys.stdin.readline())
writeln=lambda x:sys.stdout.write(str(x)+"\n")
write=lambda x:sys.stdout.write(x)
MOD = (10**9)+7
N = readi()
ns = list(map(int, read().split()))
ns.sort()
s = 0
for i in range(N):
val = (ns[i] - ns[-(1+i)])*(2**i)
if val > 0:
val %= MOD
s += val
writeln(s)
``` | instruction | 0 | 24,409 | 8 | 48,818 |
No | output | 1 | 24,409 | 8 | 48,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
def mp(): return map(int,input().split())
def lt(): return list(map(int,input().split()))
def pt(x): print(x)
def ip(): return input()
def it(): return int(input())
def sl(x): return [t for t in x]
def spl(x): return x.split()
def aj(liste, item): liste.append(item)
def bin(x): return "{0:b}".format(x)
def listring(l): return ' '.join([str(x) for x in l])
def ptlist(l): print(' '.join([str(x) for x in l]))
M = 1000000007
n = it()
x = lt()
def exp(k,t,w):
if t == 0:
return w
if t == 1:
return k*w
elif t%2 == 0:
return exp(k*k%M,t//2,w)
else:
return exp(k*k%M,t//2,k*w%M)
expp = {0:1}
expp[n] = exp(2,n,1)
x.sort()
m = 0
for i in range(n):
if i not in expp:
expp[i] = expp[i-1] * 2
expp[i] %= M
if n-i-1 not in expp:
expp[n-1-i] = expp[n-i] // 2
expp[n-i-1] %= M
m += (expp[i]-1)*x[i]
m -= (expp[n-i-1]-1)*x[i]
m += M
m %= M
pt(m)
``` | instruction | 0 | 24,410 | 8 | 48,820 |
No | output | 1 | 24,410 | 8 | 48,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression <image>. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, <image>. Since the required sum can be quite large Noora asks to find it modulo 109 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7
Output
3
Input
3
4 3 1
Output
9
Note
There are three non-empty subsets in the first sample test:<image>, <image> and <image>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <image>, <image>, <image>, <image>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
# from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n=Int()
a=sorted(array())
ans=0
high=1
low=pow(2,n-1,M)
for i in a:
ans=(ans+(high*i)%M)%M
ans=(M+ans-(low*i)%M)%M
high=high<<1
low=low>>1
print(ans)
``` | instruction | 0 | 24,411 | 8 | 48,822 |
No | output | 1 | 24,411 | 8 | 48,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
conv_map = [[3, 3, 0, 4, 4, 0, 3, 3], [2, 2, 0, 3, 3, 0, 2, 2], [1, 1, 0, 2, 2, 0, 1, 1]]
res, prev = 0, None
a = [list(input()) for _ in range(6)]
for i in range(6):
for j in range(8):
if a[i][j] == '.':
if conv_map[i // 2][j] > res:
a[i][j] = 'P'
if prev:
a[prev[0]][prev[1]] = '.'
prev = (i, j)
res = conv_map[i // 2][j]
print('\n'.join(''.join(ai) for ai in a))
``` | instruction | 0 | 24,439 | 8 | 48,878 |
Yes | output | 1 | 24,439 | 8 | 48,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
import sys
fin = sys.stdin
fout = sys.stdout
a = [0] * 6
for i in range(6):
a[i] = [0] * 6
for i in range(2):
a[i][0] = 3
a[i][1] = 3
a[i][2] = 4
a[i][3] = 4
a[i][4] = 3
a[i][5] = 3
for i in range(2, 4):
a[i][0] = 2
a[i][1] = 2
a[i][2] = 3
a[i][3] = 3
a[i][4] = 2
a[i][5] = 2
for i in range(4, 6):
a[i][0] = 1
a[i][1] = 1
a[i][2] = 2
a[i][3] = 2
a[i][4] = 1
a[i][5] = 1
ansI = -1
ansJ = -1
max = -1
ansL = []
for i in range(6):
s = fin.readline().strip()
ansL.append(s)
s = s.replace("-", "")
for j in range(6):
if s[j] == '.' and a[i][j] > max:
max = a[i][j]
ansI = i
ansJ = j
# print(ansI, ansJ)
for i in range(len(ansL)):
cur = ansL[i]
realJ = -1
for j in range(len(cur)):
if (cur[j] != '-'):
realJ += 1
if i == ansI and realJ == ansJ and cur[j] != '-':
fout.write('P')
else:
fout.write(cur[j])
fout.write("\n")
fin.close()
fout.close()
``` | instruction | 0 | 24,440 | 8 | 48,880 |
Yes | output | 1 | 24,440 | 8 | 48,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
a=[[3,3,0,4,4,0,3,3],[3,3,0,4,4,0,3,3],[2,2,0,3,3,0,2,2],[2,2,0,3,3,0,2,2],[1,1,0,2,2,0,1,1],[1,1,0,2,2,0,1,1]]
mx=0
k=[]
for i in range(6):
s=input()
k.append(s)
for j in range(len(s)):
if(s[j]=='.'):
if(a[i][j]>mx):
mx=a[i][j]
i1=i
i2=j
for i in range(6):
for j in range(8):
if(i==i1 and j==i2):
print("P",end='')
else:
print(k[i][j],end='')
print()
``` | instruction | 0 | 24,441 | 8 | 48,882 |
Yes | output | 1 | 24,441 | 8 | 48,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
ar = [
[3, 3, -1, 4, 4, -1, 3, 3],
[3, 3, -1, 4, 4, -1, 3, 3],
[2, 2, -1, 3, 3, -1, 2, 2],
[2, 2, -1, 3, 3, -1, 2, 2],
[1, 1, -1, 2, 2, -1, 1, 1],
[1, 1, -1, 2, 2, -1, 1, 1],
]
std = []
def main():
mx = -1
mq = -1
mi = -1
for q in range(6):
s = input()
std.append(s)
for i in range(8):
if( s[i] == '.' and ar[q][i] > mx):
mx = ar[q][i]
mq = q
mi = i
for q in range(6):
s = ""
for i in range(8):
if(q == mq and i == mi):
s += "P"
else:
s += std[q][i]
print(s)
main()
``` | instruction | 0 | 24,442 | 8 | 48,884 |
Yes | output | 1 | 24,442 | 8 | 48,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
a = [[''] * 8 for i in range(6)]
for i in range(6):
a[i] = input()
b = []
ok = 0
for i in range(6):
for j in range(8):
if a[i][j] == '.' and not ok:
if j != 7:
print('P', end='')
else:
print('P')
ok = 1
else:
if j != 7:
print(a[i][j], end='')
else:
print(a[i][j])
``` | instruction | 0 | 24,443 | 8 | 48,886 |
No | output | 1 | 24,443 | 8 | 48,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
places = []
isPlace = False
for i in range(6):
places += input()
step = 3
for i in range(3):
for j in range(6):
if not(isPlace):
if places[step] == ".":
places[step] = "P"
isPlace = True
if i == 0:
if step % 2 != 0:
step += 1
else:
step += 7
else:
if step % 2 == 0:
step += 1
else:
step += 7
if i == 0:
step = 0
else:
step = 6
for i in range(6):
s = ""
for j in range(8):
s += places[j+8*i]
print(s)
``` | instruction | 0 | 24,444 | 8 | 48,888 |
No | output | 1 | 24,444 | 8 | 48,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
def print_results(rows):
for item in rows:
print(item)
def main():
rows = []
for i in range(6):
rows.append(input())
new_row = ""
for i, row in (enumerate(rows)):
if row[3] == '.':
rows[i] = row[:3] + "P" + row[6:]
print_results(rows)
return
elif row[4] == '.':
rows[i] = row[:4] + "P" + row[5:]
print_results(rows)
return
if (new_row != ""):
rows[i - 1] = new_row
print_results(rows)
return
if row[0] == '.':
new_row = "P" + row[1:]
elif row[1] == '.':
new_row = row[0] + "P" + row[2:]
elif row[6] == '.':
new_row = row[:6] + "P" + row[7]
elif row[7] == '.':
new_row = row[:7] + "P"
rows[5] = new_row
print_results(rows)
return
main()
``` | instruction | 0 | 24,445 | 8 | 48,890 |
No | output | 1 | 24,445 | 8 | 48,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A classroom in a school has six rows with 3 desks in each row. Two people can use the same desk: one sitting on the left and one sitting on the right.
Some places are already occupied, and some places are vacant. Petya has just entered the class and wants to occupy the most convenient place. The conveniences of the places are shown on the picture:
<image>
Here, the desks in the top row are the closest to the blackboard, while the desks in the bottom row are the furthest from the blackboard.
You are given a plan of the class, where '*' denotes an occupied place, '.' denotes a vacant place, and the aisles are denoted by '-'.
Find any of the most convenient vacant places for Petya.
Input
The input consists of 6 lines. Each line describes one row of desks, starting from the closest to the blackboard. Each line is given in the following format: two characters, each is '*' or '.' — the description of the left desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the center desk in the current row; a character '-' — the aisle; two characters, each is '*' or '.' — the description of the right desk in the current row. So, the length of each of the six lines is 8.
It is guaranteed that there is at least one vacant place in the classroom.
Output
Print the plan of the classroom after Petya takes one of the most convenient for him places. Mark this place with the letter 'P'. There should be exactly one letter 'P' in the plan. Petya can only take a vacant place. In all other places the output should coincide with the input.
If there are multiple answers, print any.
Examples
Input
..-**-..
..-**-..
..-..-..
..-..-..
..-..-..
..-..-..
Output
..-**-..
..-**-..
..-..-..
..-P.-..
..-..-..
..-..-..
Input
**-**-**
**-**-**
..-**-.*
**-**-**
..-..-..
..-**-..
Output
**-**-**
**-**-**
..-**-.*
**-**-**
..-P.-..
..-**-..
Input
**-**-*.
*.-*.-**
**-**-**
**-**-**
..-..-..
..-**-..
Output
**-**-*.
*.-*P-**
**-**-**
**-**-**
..-..-..
..-**-..
Note
In the first example the maximum convenience is 3.
In the second example the maximum convenience is 2.
In the third example the maximum convenience is 4.
Submitted Solution:
```
a=[[3,3,0,4,4,0,3,3],[3,3,0,4,4,0,3,3],[2,2,0,3,3,0,2,2],[2,2,0,3,3,0,2,2],[1,1,0,2,2,0,1,1],[1,1,0,2,2,0,1,1]]
mx=0
k=[]
for i in range(6):
s=input()
k.append(s)
for j in range(len(s)):
if(s[j]=='.'):
if(a[i][j]>mx):
mx=a[i][j]
i1=i
i2=j
for i in range(6):
for j in range(8):
if(i==i1 and j==i2):
print("P",end='')
else:
print(k[i][j],end='')
print("\n")
``` | instruction | 0 | 24,446 | 8 | 48,892 |
No | output | 1 | 24,446 | 8 | 48,893 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,463 | 8 | 48,926 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
n,k=in_arr()
pre=[0]*(n+1)
l=in_arr()
for i in range(1,n+1):
pre[i]+=pre[i-1]
pre[i]+=l[i-1]
ans=0
for bit in range(60,-1,-1):
tar=ans|(1<<bit)
dp=[[0 for i in range(n+1)] for i in range(k+1)]
dp[0][0]=1
for j in range(1,k+1):
for k2 in range(1,n+1):
for k1 in range(k2):
if dp[j-1][k1] and (pre[k2]-pre[k1])&tar==tar:
dp[j][k2]=1
if dp[k][n]:
ans=tar
print ans
``` | output | 1 | 24,463 | 8 | 48,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,464 | 8 | 48,928 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
#!/usr/bin/env python3
[n, k] = map(int, input().strip().split())
ais = list(map(int, input().strip().split()))
iais = [0 for _ in range(n + 1)]
for i in range(n):
iais[i + 1] = iais[i] + ais[i]
def calc(k, split):
res = 0
for i in range(k):
res &= iais[split[i + 1]] - iais[split[i]]
return res
def check_mask(mask):
dp = [[False for j in range(n + 1)] for i in range(k + 1)]
for j in range(1, n - k + 1 + 1):
dp[1][j] = (iais[j] & mask == mask)
if not any(dp[1]):
return False
for i in range(2, k + 1):
for j in range(i, n - (k - i) + 1):
dp[i][j] = any(dp[i - 1][r] and ((iais[j] - iais[r]) & mask == mask) for r in range(i - 1, j - 1 + 1))
if not any(dp[i]):
return False
return dp[k][n]
mask = 0
for i in range(55, -1, -1):
if check_mask(mask | (1 << i)):
mask |= 1 << i
print (mask)
``` | output | 1 | 24,464 | 8 | 48,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,465 | 8 | 48,930 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
[n, k] = [int(x) for x in input().split()]
sum = [int(x) for x in input().split()]
for i in range(1, n):
sum[i] += sum[i - 1]
def check(mask, all):
dp = [[False for j in range(n)] for i in range(k)]
for i in range(n):
dp[0][i] = ((sum[i] & all & mask) == mask)
for i in range(1, k):
for j in range(n):
dp[i][j] = any(dp[i - 1][p] and ((sum[j] - sum[p]) & all & mask) == mask for p in range(0, j))
return dp[k - 1][n - 1]
ans = 0
for i in range(60, -1, -1):
if(check(ans | (1 << i), ~((1 << i) - 1))):
ans |= 1 << i
print(ans)
``` | output | 1 | 24,465 | 8 | 48,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,466 | 8 | 48,932 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# 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")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
# ------------------------------
def main():
n, k = RL()
arr = RLL()
def dpf(num):
dp = [[0]*(k+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(1, n+1):
for j in range(1, k+1):
now = 0
for l in range(i, j-2, -1):
now+=arr[l-1]
dp[i][j] |= dp[l-1][j-1] and (now & num == num)
# for i in dp: print(i)
# print(num)
return dp[-1][-1]
res = 0
inc = 0
for i in range(55, -1, -1):
if dpf(inc+(1<<i)):
inc+=(1<<i)
res = max(res, inc)
# break
print(res)
if __name__ == "__main__":
main()
``` | output | 1 | 24,466 | 8 | 48,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,467 | 8 | 48,934 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
from collections import Counter
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
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")
#-------------------game starts now-----------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10001)]
prime[0]=prime[1]=False
#pp=[0]*10000
def SieveOfEratosthenes(n=10000):
p = 2
c=0
while (p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
#pp[i]=1
prime[i] = False
p += 1
#-----------------------------------DSU--------------------------------------------------
class DSU:
def __init__(self, R, C):
#R * C is the source, and isn't a grid square
self.par = range(R*C + 1)
self.rnk = [0] * (R*C + 1)
self.sz = [1] * (R*C + 1)
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
if self.rnk[xr] < self.rnk[yr]:
xr, yr = yr, xr
if self.rnk[xr] == self.rnk[yr]:
self.rnk[xr] += 1
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
def size(self, x):
return self.sz[self.find(x)]
def top(self):
# Size of component at ephemeral "source" node at index R*C,
# minus 1 to not count the source itself in the size
return self.size(len(self.sz) - 1) - 1
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#---------------------------------Pollard rho--------------------------------------------
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return math.gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = math.gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = math.gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=n
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
res=mid
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=-1
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
right = mid-1
else:
res=mid
left = mid + 1
return res
#---------------------------------running code------------------------------------------
n,k=map(int,input().split())
l=list(map(int,input().split()))
s=[0]*(n+1)
for i in range(1,n+1):
s[i]=s[i-1]+l[i-1]
#s=SegmentTree(l)
def check(x):
dp=[[0 for i in range(k+1)]for j in range(n+1)]
dp[0][0]=1
for i in range(1,n+1):
for j in range(1,min(i,k)+1):
for h in range(i-1,-1,-1):
cur=s[i]-s[h]
if cur&x==x:
dp[i][j]=max(dp[i][j],dp[h][j-1])
return dp[n][k]
temp=0
cou=2**57
for i in range(57,-1,-1):
temp^=cou
if check(temp):
ans=temp
else:
temp^=cou
ans=temp
cou//=2
print(ans)
``` | output | 1 | 24,467 | 8 | 48,935 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,468 | 8 | 48,936 |
Tags: bitmasks, dp, greedy
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,k=map(int,input().split())
l=list(map(int,input().split()))
s=[0]*(n+1)
for i in range(1,n+1):
s[i]=s[i-1]+l[i-1]
#s=SegmentTree(l)
def check(x):
dp=[[0 for i in range(k+1)]for j in range(n+1)]
dp[0][0]=1
for i in range(1,n+1):
for j in range(1,min(i,k)+1):
for h in range(i-1,-1,-1):
cur=s[i]-s[h]
if cur&x==x:
dp[i][j]=max(dp[i][j],dp[h][j-1])
return dp[n][k]
temp=0
cou=2**57
for i in range(57,-1,-1):
temp^=cou
if check(temp)==1:
ans=temp
else:
temp^=cou
ans=temp
cou//=2
print(ans)
``` | output | 1 | 24,468 | 8 | 48,937 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,469 | 8 | 48,938 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
a_sum = [0]
for i in range(n):
a_sum.append(a_sum[-1] + a[i])
a = [0] + a
def get(i, j):
assert 0 <= i <= j <= n
if i > 0:
return a_sum[j] - a_sum[i - 1]
else:
return a_sum[j]
ans = 0
nowmax = 0
for bit in range(60, -1, -1):
tmpmax = nowmax + (1 << bit)
dp = [[0]*(n + 10) for i in range(n + 10)]
dp[0][0] = tmpmax
for pos in range(1, n + 1):
for div in range(1, k + 1):
for l in range(1, pos + 1):
dp[pos][div] = max(dp[pos][div], dp[pos - l]
[div - 1] & get(pos - l + 1, pos) & tmpmax)
if dp[n][k] == tmpmax:
nowmax += (1 << bit)
print(nowmax)
``` | output | 1 | 24,469 | 8 | 48,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,470 | 8 | 48,940 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def check(trans,k,n):
dp = [[0]*(k+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(n):
for y in range(k):
if not dp[i][y]:
continue
for x in trans[i]:
dp[x][y+1] = 1
return dp[n][k]
def main():
n,k = map(int,input().split())
a = list(map(int,input().split()))
ans,de,val = 0,[],1<<60
for i in range(59,-1,-1):
trans = [[] for _ in range(n)]
val >>= 1
for j in range(n):
su = 0
for l in range(j,n):
su += a[l]
fl = 1
for z in de:
if not su&z:
fl = 0
break
if not fl or not su&val:
continue
trans[j].append(l+1)
if check(trans,k,n):
ans += val
de.append(val)
print(ans)
#Fast IO Region
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")
if __name__ == '__main__':
main()
``` | output | 1 | 24,470 | 8 | 48,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$ | instruction | 0 | 24,471 | 8 | 48,942 |
Tags: bitmasks, dp, greedy
Correct Solution:
```
import time;
def getIntList():
return list(map(int, input().split()));
def getTransIntList(n):
first = getIntList();
m = len(first);
result = [[0] * n for _ in range(m)];
for i in range(m):
result[i][0] = first[i];
for j in range(1, n):
curr = getIntList();
for i in range(m):
result[i][j] = curr[i];
return result;
n, k = getIntList();
a = getIntList();
#sums[i][j] - сумма a по индексам от i до j не включая j
sums = [[0]*(n+1) for _ in range(n+1)];
for i in range(n):
for j in range(i+1, n+1):
sums[i][j]=sums[i][j-1]+a[j-1];
class SearchProblem:
def __init__(self, a, n, k, tiLim):
self.a=a;
self.n=n;
self.k=k;
self.maxResult=0;
self.tiLim=time.time()+tiLim;
def search(self, currResult, currIndex, currLines):
#Время вышло - заканчиваем.
if time.time()>self.tiLim:
return;
if currLines>0 and currResult<=self.maxResult:
return;
if currLines==self.k-1:
lastSum = sums[currIndex][self.n];
currResult = currResult & lastSum;
if currResult > self.maxResult:
self.maxResult = currResult;
for nextIndex in range(currIndex+1, self.n+1):
currSum=sums[currIndex][nextIndex];
if currLines==0:
nextResult=currSum
else:
nextResult=currResult & currSum;
self.search(nextResult, nextIndex, currLines+1);
flag=True;
if time.time() > self.tiLim:
flag=False;
return self.maxResult, flag;
#upLim[i][j] - оценка сверху на красоту разбиения книг с номерами с j до конца по i полкам
upLim=[[0]*(n+1) for _ in range(k+1)];
for i in range(1, k+1):
if i==1:
for j in range(0, n):
upLim[i][j]=sums[j][n];
else:
for j in range(n-i, -1, -1):
upLim[i][j]=0;
for j1 in range(j+1, n):
curr=min(sums[j][j1], upLim[i-1][j1]);
upLim[i][j]=max(upLim[i][j], curr);
def solve():
#Сначала ищем начальное решение deepfirst
problem=SearchProblem(a, n, k, 0.1);
if k==1:
return sum(a);
maxResult, solved = problem.search(0, 0, 0);
if solved:
#print("deep first succeed")
return maxResult;
results = [[set() for _ in range(n + 1)] for _ in range(k + 1)];
# А теперь ищем полное решение динамическим программированием
for i in range(1, n + 1):
for firstIndexSum in range(0, i):
# print(firstIndexSum, i);
currSum = sums[firstIndexSum][i];
if firstIndexSum == 0:
if currSum>maxResult:
results[1][i].add(currSum);
if k == 2:
lastSum = sums[i][n];
currResult = currSum & lastSum;
if currResult > maxResult:
maxResult = currResult;
for lines1 in range(k):
for j in range(n):
results[lines1][j] = {r for r in results[lines1][j] if r >= maxResult}
else:
for lines in range(k):
for prevSum in results[lines][firstIndexSum]:
# print(prevSum,"&",currSum,"=",prevSum&currSum)
fullSum = prevSum & currSum;
currLines = lines + 1;
if currLines == k - 1:
lastSum = sums[i][n];
currResult = fullSum & lastSum;
if currResult > maxResult:
maxResult = currResult;
for lines1 in range(k):
for j in range(n):
results[lines1][j] = {r for r in results[lines1][j] if r >= maxResult}
if upLim[k-currLines][i]<=maxResult:
continue;
if fullSum>maxResult:
results[currLines][i].add(fullSum);
# print(i, results[i])
return maxResult;
print(solve());
``` | output | 1 | 24,471 | 8 | 48,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$
Submitted Solution:
```
#!/usr/bin/python3.7
import sys
n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
cur = 0
def is_submask(n, mask):
return (n & mask) == mask
def f(mask):
s = [[False for j in range(n)] for i in range(n)]
for l in range(n):
cur = 0
for r in range(l, n):
cur += a[r]
s[l][r] = is_submask(cur, mask)
dp = [[False for j in range(n)] for i in range(k)]
dp[0] = s[0][:]
for k1 in range(1, k):
for r in range(n):
for l in range(1, r + 1):
dp[k1][r] |= dp[k1 - 1][l - 1] & s[l][r]
return dp[k - 1][n - 1]
cur = 0
for i in range(56, -1, -1):
if f(cur + 2 ** i):
cur += 2 ** i
print(cur)
``` | instruction | 0 | 24,472 | 8 | 48,944 |
Yes | output | 1 | 24,472 | 8 | 48,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$
Submitted Solution:
```
def solve(a,n,k,start,shelf,depth,sol):
if(depth==k-1):
shelf[-1] = sum(a[start:])
res = shelf[0]
for i in range(1,k):
res = res & shelf[i]
sol.append(res)
return
val = 0
# print("At depth "+str(depth))
for i in range(start,n-k+1):
val += a[i]
shelf[depth] = val
solve(a,n,k,i+1,shelf,depth+1,sol)
n,k = map(int,input().split())
a = [int(x) for x in input().split()]
shelf = [None for i in range(k)]
sol = []
solve(a,n,k,0,shelf,0,sol)
ans = max(sol)
print(ans)
``` | instruction | 0 | 24,473 | 8 | 48,946 |
No | output | 1 | 24,473 | 8 | 48,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$
Submitted Solution:
```
from functools import reduce
N=55
dp = [ [-1 for i in range(N)] for i in range(N) ]
N,K = list(map(int,input().strip().split()))
A = list(map(int, input().strip().split()))
# solve(i,k) or dp[i][k] => divide [i:N) into k segments
def solve(i,k):
if(dp[i][k]!=-1):
return dp[i][k]
if(N-i<k):
assert False
if(k==1):
m = sum(A[i:])
elif(N-i==k):
m = reduce(lambda x,y: x&y, A[i:], 1)
else:
m = 0
for j in range(i+1, N+1):
if N-j>=k-1:
t = solve(j,k-1)&sum(A[i:j])
m = max(m,t)
else:
break
print(m)
dp[i][k]=m
return m
print(solve(0,K))
``` | instruction | 0 | 24,474 | 8 | 48,948 |
No | output | 1 | 24,474 | 8 | 48,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$
Submitted Solution:
```
def solve(a,k,start,n,toGo):
if(toGo==0):
val = sum(a[start:])
return val
val = 0
ans = 0
for i in range(start,n-toGo):
val += a[i]
temp = solve(a,k,i+1,n,toGo-1)
ans = max(ans,val & temp)
return ans
n,k = map(int,input().split())
a = [int(x) for x in input().split()]
print(solve(a,k,0,n,k-1))
``` | instruction | 0 | 24,475 | 8 | 48,950 |
No | output | 1 | 24,475 | 8 | 48,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr Keks is a typical white-collar in Byteland.
He has a bookshelf in his office with some books on it, each book has an integer positive price.
Mr Keks defines the value of a shelf as the sum of books prices on it.
Miraculously, Mr Keks was promoted and now he is moving into a new office.
He learned that in the new office he will have not a single bookshelf, but exactly k bookshelves. He decided that the beauty of the k shelves is the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the values of all the shelves.
He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on k shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 50) — the number of books and the number of shelves in the new office.
The second line contains n integers a_1, a_2, … a_n, (0 < a_i < 2^{50}) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of k shelves in the new office.
Examples
Input
10 4
9 14 28 1 7 13 15 29 2 31
Output
24
Input
7 3
3 14 15 92 65 35 89
Output
64
Note
In the first example you can split the books as follows:
$$$(9 + 14 + 28 + 1 + 7) \& (13 + 15) \& (29 + 2) \& (31) = 24.$$$
In the second example you can split the books as follows:
$$$(3 + 14 + 15 + 92) \& (65) \& (35 + 89) = 64.$$$
Submitted Solution:
```
def getIntList():
return list(map(int, input().split()));
def getTransIntList(n):
first=getIntList();
m=len(first);
result=[[0]*n for _ in range(m)];
for i in range(m):
result[i][0]=first[i];
for j in range(1, n):
curr=getIntList();
for i in range(m):
result[i][j]=curr[i];
return result;
n, k=getIntList();
a= getIntList();
#Возможные варианты результата расстановки с учетом количество полок
#result[i] - расставили книги с номерами до i-1
results=[set() for _ in range(n+1)];
results[0].add((2**51-1, 0));
def solve():
#Последняя сумма от firstIndexSum до i не включая i
for i in range(1, n+1):
for firstIndexSum in range(0, i):
#print(firstIndexSum, i);
currSum=sum(a[firstIndexSum: i]);
for result in results[firstIndexSum]:
prevSum, prevLines=result;
#print(prevSum,"&",currSum,"=",prevSum&currSum)
fullSum=prevSum&currSum;
currLines=prevLines+1;
if currLines>k or currLines+n-i<k:
continue;
results[i].add((fullSum, currLines));
#print(i, results[i])
maxResult=0;
for result in results[i]:
if result[1]==k:
maxResult=max(maxResult, result[0])
return maxResult;
print(solve());
``` | instruction | 0 | 24,476 | 8 | 48,952 |
No | output | 1 | 24,476 | 8 | 48,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,919 | 8 | 49,838 |
Tags: dp, math, number theory
Correct Solution:
```
from sys import stdin
for _ in range(int(input())):
n = int(stdin.readline().rstrip())
l = list(map(int, stdin.readline().rstrip().split(" ")))
z = [0]*(n+1)
z[1]=1
for i in range(n,1,-1):
j = 2
c = 0
while i*j <= n:
if l[i - 1] < l[i*j -1 ]:
c = max(c, z[i*j])
j+=1
z[i]=c+1
m = 1
for i in range(2,n+1):
if l[0]<l[i-1]:
z[i]+=1
m = max(m, z[i])
# print(*z)
print(m)
``` | output | 1 | 24,919 | 8 | 49,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,920 | 8 | 49,840 |
Tags: dp, math, number theory
Correct Solution:
```
from sys import stdin
def main():
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
ar = list(map(int, input().split()))
dp = [1] * (n + 1)
for i in range(2, n):
if ar[i - 1] > ar[0]:
dp[i - 1] = max(dp[i - 1], dp[0] + 1)
for i in range(2, n + 1):
for j in range(i + i, n + 1, i):
if ar[j - 1] > ar[i - 1]:
dp[j - 1] = max(dp[i - 1] + 1, dp[j - 1])
print(max(dp))
if __name__ == "__main__":
main()
``` | output | 1 | 24,920 | 8 | 49,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,921 | 8 | 49,842 |
Tags: dp, math, number theory
Correct Solution:
```
for i in range(int(input())):
n=int(input())
s=list(map(int,input().split()))
ans=[1 for i in range(n)]
for i in range(n):
size=s[i]
for j in range(i+1,n+1,i+1):
index=j-1
if(s[index]>size):
ans[index]=max(ans[index],ans[i]+1)
print(max(ans))
``` | output | 1 | 24,921 | 8 | 49,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,922 | 8 | 49,844 |
Tags: dp, math, number theory
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
arre=list(map(int,input().split()))
arr=[1 for j in range(n+1)]
for j in range(1,n+1):
arr[j]=arre[j-1]
f=[1 for j in range(n+1)]
for j in range(1,n+1):
p=j*2
for k in range(p,n+1,j):
if(arr[j]<arr[k]):
f[k]=max(f[k],f[j]+1)
ans=0
for j in range(1,n+1):
ans=max(ans,f[j])
print(ans)
``` | output | 1 | 24,922 | 8 | 49,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,923 | 8 | 49,846 |
Tags: dp, math, number theory
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
s=list(map(int,input().split()))
beauty=[1]*n
for i in range(n):
for j in range(2*i+1,n,i+1):
if s[j]>s[i]:
beauty[j]=max(beauty[i]+1,beauty[j])
print(max(beauty))
``` | output | 1 | 24,923 | 8 | 49,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,924 | 8 | 49,848 |
Tags: dp, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
status = [1]*(n+1)
for i in range(n):
for j in range(i*2 + 1, n, i+1):
if(a[i] < a[j]):
status[j] = max(status[i]+1, status[j])
print(max(status))
``` | output | 1 | 24,924 | 8 | 49,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,925 | 8 | 49,850 |
Tags: dp, math, number theory
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
arr=[1]*(n)
for i in range(1,n):
if a[i]>a[0]:
arr[i]+=1
for i in range(2,n+1):
for j in range(2*i,n+1,i):
if a[j-1]>a[i-1]:
arr[j-1]=max(arr[j-1],arr[i-1]+1)
print(max(arr))
``` | output | 1 | 24,925 | 8 | 49,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,926 | 8 | 49,852 |
Tags: dp, math, number theory
Correct Solution:
```
import math
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
l=[0]+l
arr = [1] * (n + 1)
ans=1
for i in range(1,n+1):
for j in range(1,int(math.sqrt(i))+1):
if(i%j==0):
d1 = j
d2 = int(i / j)
if (l[i] > l[d1]):
arr[i] = max(arr[i], arr[d1]+1)
if (l[i] > l[d2]):
arr[i] = max(arr[i], arr[d2]+1)
ans=max(ans,arr[i])
print(ans)
``` | output | 1 | 24,926 | 8 | 49,853 |
Provide tags and a correct Python 2 solution for this coding contest problem.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model. | instruction | 0 | 24,927 | 8 | 49,854 |
Tags: dp, math, number theory
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pl(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(input()):
n=ni()
l=[0]+li()
dp=[1]*(n+1)
for i in range(1,n+1):
for j in range(2*i,n+1,i):
if l[i]<l[j]:
dp[j]=max(dp[j],dp[i]+1)
pn(max(dp))
``` | output | 1 | 24,927 | 8 | 49,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model.
Submitted Solution:
```
t=int(input())
for w in range(t):
n=int(input())
l=[int(i) for i in input().split()]
if n==1:
print (1)
else:
l1=[1]*(n+1)
for i in range(1,n+1):
j=2*i
while j<=n:
if l[j-1]>l[i-1]:
l1[j-1]=max(l1[j-1],l1[i-1]+1)
j+=i
k=0
for i in l1:
k=max(i,k)
print(k)
``` | instruction | 0 | 24,928 | 8 | 49,856 |
Yes | output | 1 | 24,928 | 8 | 49,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model.
Submitted Solution:
```
T = int(input())
for _ in range(T):
N = int(input())
S = list(map(int, input().split()))
# maxCount = 0
counts = [1]*N
for i in range(N-1, -1, -1):
last = i
j = 2*i + 1
for j in range(2*i+1, N, i+1):
if(S[i] < S[j]):
counts[i] = max(counts[i], counts[j]+1)
print(max(counts))
``` | instruction | 0 | 24,929 | 8 | 49,858 |
Yes | output | 1 | 24,929 | 8 | 49,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n models in the shop numbered from 1 to n, with sizes s_1, s_2, …, s_n.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices i_j and i_{j+1} (note that i_j < i_{j+1}, because Orac arranged them properly), i_{j+1} is divisible by i_j and s_{i_j} < s_{i_{j+1}}.
For example, for 6 models with sizes \{3, 6, 7, 7, 7, 7\}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t\ (1 ≤ t≤ 100): the number of queries.
Each query contains two lines. The first line contains one integer n\ (1≤ n≤ 100 000): the number of models in the shop, and the second line contains n integers s_1,...,s_n\ (1≤ s_i≤ 10^9): the sizes of models.
It is guaranteed that the total sum of n is at most 100 000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
Input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
Output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model.
Submitted Solution:
```
t = int(input())
for tc in range(t):
n = int(input())
s = [0] + list(map(int, input().split()))
dp = [1] * (n+1)
for i in range(n,0,-1):
for j in range(2,n+2):
if i * j > n:
break
else:
if s[i] < s[i*j]:
dp[i] = max(dp[i], dp[i*j] + 1)
print(max(dp))
``` | instruction | 0 | 24,930 | 8 | 49,860 |
Yes | output | 1 | 24,930 | 8 | 49,861 |
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