message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used. | instruction | 0 | 7,772 | 19 | 15,544 |
Tags: brute force, constructive algorithms, greedy, math, probabilities
Correct Solution:
```
t = int(input())
chisls = [2 ** (i + 1) - 2 for i in range(1, 72)]
# print(chisls)
for i in range(t):
n = int(input())
if n % 2 == 1:
print(-1)
else:
# print(1, end=' ')
data = [1]
n -= 2
p = len(chisls) - 1
while n > 0:
while chisls[p] > n:
p -= 1
# print(p)
while n >= chisls[p]:
data += [1] + [0 for j in range(p)]
n -= chisls[p]
if len(data) > 2000:
break
if len(data) > 2000:
print(-1)
else:
print(len(data))
print(*data)
``` | output | 1 | 7,772 | 19 | 15,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used. | instruction | 0 | 7,773 | 19 | 15,546 |
Tags: brute force, constructive algorithms, greedy, math, probabilities
Correct Solution:
```
import sys
from math import gcd,sqrt,ceil,log2
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
sys.setrecursionlimit(2*10**5+10)
import heapq
from itertools import permutations
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
aa='abcdefghijklmnopqrstuvwxyz'
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")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def primeFactors(n):
sa = []
# sa.add(n)
while n % 2 == 0:
sa.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.append(i)
n = n // i
# sa.add(n)
if n > 2:
sa.append(n)
return sa
def seive(n):
pri = [True]*(n+1)
p = 2
while p*p<=n:
if pri[p] == True:
for i in range(p*p,n+1,p):
pri[i] = False
p+=1
return pri
def check_prim(n):
if n<0:
return False
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
return False
return True
def getZarr(string, z):
n = len(string)
# [L,R] make a window which matches
# with prefix of s
l, r, k = 0, 0, 0
for i in range(1, n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
# Create concatenated string "P$T"
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
ha = []
for i in range(l):
if z[i] == len(pattern):
ha.append(i - len(pattern) - 1)
return ha
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# n = int(input())
# l = list(map(int,input().split()))
#
# hash = defaultdict(list)
# la = []
#
# for i in range(n):
# la.append([l[i],i+1])
#
# la.sort(key = lambda x: (x[0],-x[1]))
# ans = []
# r = n
# flag = 0
# lo = []
# ha = [i for i in range(n,0,-1)]
# yo = []
# for a,b in la:
#
# if a == 1:
# ans.append([r,b])
# # hash[(1,1)].append([b,r])
# lo.append((r,b))
# ha.pop(0)
# yo.append([r,b])
# r-=1
#
# elif a == 2:
# # print(yo,lo)
# # print(hash[1,1])
# if lo == []:
# flag = 1
# break
# c,d = lo.pop(0)
# yo.pop(0)
# if b>=d:
# flag = 1
# break
# ans.append([c,b])
# yo.append([c,b])
#
#
#
# elif a == 3:
#
# if yo == []:
# flag = 1
# break
# c,d = yo.pop(0)
# if b>=d:
# flag = 1
# break
# if ha == []:
# flag = 1
# break
#
# ka = ha.pop(0)
#
# ans.append([ka,b])
# ans.append([ka,d])
# yo.append([ka,b])
#
# if flag:
# print(-1)
# else:
# print(len(ans))
# for a,b in ans:
# print(a,b)
def mergeIntervals(arr):
# Sorting based on the increasing order
# of the start intervals
arr.sort(key = lambda x: x[0])
# array to hold the merged intervals
m = []
s = -10000
max = -100000
for i in range(len(arr)):
a = arr[i]
if a[0] > max:
if i != 0:
m.append([s,max])
max = a[1]
s = a[0]
else:
if a[1] >= max:
max = a[1]
#'max' value gives the last point of
# that particular interval
# 's' gives the starting point of that interval
# 'm' array contains the list of all merged intervals
if max != -100000 and [s, max] not in m:
m.append([s, max])
return m
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def sol(n):
seti = set()
for i in range(1,int(sqrt(n))+1):
if n%i == 0:
seti.add(n//i)
seti.add(i)
return seti
def lcm(a,b):
return (a*b)//gcd(a,b)
#
# n,p = map(int,input().split())
#
# s = input()
#
# if n <=2:
# if n == 1:
# pass
# if n == 2:
# pass
# i = n-1
# idx = -1
# while i>=0:
# z = ord(s[i])-96
# k = chr(z+1+96)
# flag = 1
# if i-1>=0:
# if s[i-1]!=k:
# flag+=1
# else:
# flag+=1
# if i-2>=0:
# if s[i-2]!=k:
# flag+=1
# else:
# flag+=1
# if flag == 2:
# idx = i
# s[i] = k
# break
# if idx == -1:
# print('NO')
# exit()
# for i in range(idx+1,n):
# if
#
def moore_voting(l):
count1 = 0
count2 = 0
first = 10**18
second = 10**18
n = len(l)
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
elif count1 == 0:
count1+=1
first = l[i]
elif count2 == 0:
count2+=1
second = l[i]
else:
count1-=1
count2-=1
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
if count1>n//3:
return first
if count2>n//3:
return second
return -1
def find_parent(u,parent):
if u!=parent[u]:
parent[u] = find_parent(parent[u],parent)
return parent[u]
def dis_union(n,e):
par = [i for i in range(n+1)]
rank = [1]*(n+1)
for a,b in e:
z1,z2 = find_parent(a,par),find_parent(b,par)
if rank[z1]>rank[z2]:
z1,z2 = z2,z1
if z1!=z2:
par[z1] = z2
rank[z2]+=rank[z1]
else:
return a,b
def dijkstra(n,tot,hash):
hea = [[0,n]]
dis = [10**18]*(tot+1)
dis[n] = 0
boo = defaultdict(bool)
check = defaultdict(int)
while hea:
a,b = heapq.heappop(hea)
if boo[b]:
continue
boo[b] = True
for i,w in hash[b]:
if b == 1:
c = 0
if (1,i,w) in nodes:
c = nodes[(1,i,w)]
del nodes[(1,i,w)]
if dis[b]+w<dis[i]:
dis[i] = dis[b]+w
check[i] = c
elif dis[b]+w == dis[i] and c == 0:
dis[i] = dis[b]+w
check[i] = c
else:
if dis[b]+w<=dis[i]:
dis[i] = dis[b]+w
check[i] = check[b]
heapq.heappush(hea,[dis[i],i])
return check
def power(x,y,p):
res = 1
x = x%p
if x == 0:
return 0
while y>0:
if (y&1) == 1:
res*=x
x = x*x
y = y>>1
return res
import sys
from math import ceil,log2
INT_MAX = sys.maxsize
def minVal(x, y) :
return x if (x < y) else y
def getMid(s, e) :
return s + (e - s) // 2
def RMQUtil( st, ss, se, qs, qe, index) :
if (qs <= ss and qe >= se) :
return st[index]
if (se < qs or ss > qe) :
return INT_MAX
mid = getMid(ss, se)
return minVal(RMQUtil(st, ss, mid, qs,
qe, 2 * index + 1),
RMQUtil(st, mid + 1, se,
qs, qe, 2 * index + 2))
def RMQ( st, n, qs, qe) :
if (qs < 0 or qe > n - 1 or qs > qe) :
print("Invalid Input")
return -1
return RMQUtil(st, 0, n - 1, qs, qe, 0)
def constructSTUtil(arr, ss, se, st, si) :
if (ss == se) :
st[si] = arr[ss]
return arr[ss]
mid = getMid(ss, se)
st[si] = minVal(constructSTUtil(arr, ss, mid,
st, si * 2 + 1),
constructSTUtil(arr, mid + 1, se,
st, si * 2 + 2))
return st[si]
def constructST( arr, n) :
x = (int)(ceil(log2(n)))
max_size = 2 * (int)(2**x) - 1
st = [0] * (max_size)
constructSTUtil(arr, 0, n - 1, st, 0)
return st
# t = int(input())
# for _ in range(t):
#
# n = int(input())
# l = list(map(int,input().split()))
# # x,y = 0,10
# st = constructST(l, n)
#
# pre = [0]
# suf = [0]
# for i in range(n):
# pre.append(max(pre[-1],l[i]))
# for i in range(n-1,-1,-1):
# suf.append(max(suf[-1],l[i]))
#
#
# i = 1
# # print(pre,suf)
# flag = 0
# x,y,z = -1,-1,-1
# # suf.reverse()
# print(suf)
# while i<len(pre):
#
# z = pre[i]
# j = bisect_left(suf,z)
# if suf[j] == z:
# while i<n and l[i]<=z:
# i+=1
# if pre[i]>z:
# break
# while j<n and l[n-j]<=z:
# j+=1
# if suf[j]>z:
# break
# # j-=1
# print(i,n-j)
# # break/
# if RMQ(st,n,i,j) == z:
# c = i+j-i+1
# x,y,z = i,j-i+1,n-c
# break
# else:
# i+=1
#
# else:
# i+=1
#
#
#
# if x!=-1:
# print('Yes')
# print(x,y,z)
# else:
# print('No')
# t = int(input())
#
# for _ in range(t):
#
# def debug(n):
# ans = []
# for i in range(1,n+1):
# for j in range(i+1,n+1):
# if (i*(j+1))%(j-i) == 0 :
# ans.append([i,j])
# return ans
#
#
# n = int(input())
# print(debug(n))
# import sys
# input = sys.stdin.readline
# import bisect
#
# t=int(input())
# for tests in range(t):
# n=int(input())
# A=list(map(int,input().split()))
#
# LEN = len(A)
# Sparse_table = [A]
#
# for i in range(LEN.bit_length()-1):
# j = 1<<i
# B = []
# for k in range(len(Sparse_table[-1])-j):
# B.append(min(Sparse_table[-1][k], Sparse_table[-1][k+j]))
# Sparse_table.append(B)
#
# def query(l,r): # [l,r)におけるminを求める.
# i=(r-l).bit_length()-1 # 何番目のSparse_tableを見るか.
#
# return min(Sparse_table[i][l],Sparse_table[i][r-(1<<i)]) # (1<<i)個あれば[l, r)が埋まるので, それを使ってminを求める.
#
# LMAX=[A[0]]
# for i in range(1,n):
# LMAX.append(max(LMAX[-1],A[i]))
#
# RMAX=A[-1]
#
# for i in range(n-1,-1,-1):
# RMAX=max(RMAX,A[i])
#
# x=bisect.bisect(LMAX,RMAX)
# #print(RMAX,x)
# print(RMAX,x,i)
# if x==0:
# continue
#
# v=min(x,i-1)
# if v<=0:
# continue
#
# if LMAX[v-1]==query(v,i)==RMAX:
# print("YES")
# print(v,i-v,n-i)
# break
#
# v-=1
# if v<=0:
# continue
# if LMAX[v-1]==query(v,i)==RMAX:
# print("YES")
# print(v,i-v,n-i)
# break
# else:
# print("NO")
#
#
#
#
#
#
#
#
#
# t = int(input())
#
# for _ in range(t):
#
# x = int(input())
# mini = 10**18
# n = ceil((-1 + sqrt(1+8*x))/2)
# for i in range(-100,1):
# z = x+-1*i
# z1 = (abs(i)*(abs(i)+1))//2
# z+=z1
# # print(z)
# n = ceil((-1 + sqrt(1+8*z))/2)
#
# y = (n*(n+1))//2
# # print(n,y,z,i)
# mini = min(n+y-z,mini)
# print(n+y-z,i)
#
#
# print(mini)
#
t = int(input())
for _ in range(t):
k = int(input())
ans = []
if k%2!=0:
print(-1)
continue
ans = [1]
su = 2
k1 = 2
while su-k!=0:
su+=2**k1
if su>k:
su-=2**k1
su+=2
ans.append(1)
k1 = 2
continue
ans.append(0)
k1+=1
k-=su
ans+=[1]*(k//2)
if len(ans)>2000:
print(-1)
continue
print(len(ans))
print(*ans)
``` | output | 1 | 7,773 | 19 | 15,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used. | instruction | 0 | 7,774 | 19 | 15,548 |
Tags: brute force, constructive algorithms, greedy, math, probabilities
Correct Solution:
```
two = []
cur = 1
for i in range(100):
two.append(cur)
cur *= 2
t = int(input())
for _ in range(t):
k = int(input())
if k % 2:
print(-1)
continue
res = []
def get(x):
a = 0
while 1:
if two[a+3]-2 > x:
break
a += 1
res.append('1')
res.extend(['0']*a)
return x-two[a+2]+2
while k:
k = get(k)
n = len(res)
if n > 2000:
print(-1)
continue
print(n)
print(' '.join(res))
``` | output | 1 | 7,774 | 19 | 15,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
#from random import randint
#
#
#arr=[1,1,1,1,1,1]
#m=len(arr)
#prevOne=[0 for _ in range(m)]
#for i in range(m):
# if arr[i]==1:
# prevOne[i]=i
# elif i>0:
# prevOne[i]=prevOne[i-1]
#
#t=0
#n=10000
#for _ in range(n):
# curr=0
# while curr<m:
# if randint(1,2)==2:
# curr+=1
# else:
# curr=prevOne[curr]
# t+=1
#
#print(t/n)
#each 1 contributes 2
#each 0 contributes 2**(1 + number of zeros in a row including itself)
#k=n1*2 + sum(4*(2**l - 1)) where l is the number of consequtive 0s in a row
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def calcTurnsForLZeroesInARow(L):
return 4*(2**L-1)
t=int(input())
for _ in range(t):
k=int(input())
if k%2==1:
print(-1)
else:
ans=[1]
k-=2
while k>0:
l=0
b=2000
while b>0:
while calcTurnsForLZeroesInARow(l+b)<=k:
l+=b
b//=2
ans=ans+[0]*l
k-=calcTurnsForLZeroesInARow(l)
if k==0:
break
ans.append(1)
k-=2
# print('here')
print(len(ans))
oneLineArrayPrint(ans)
``` | instruction | 0 | 7,775 | 19 | 15,550 |
Yes | output | 1 | 7,775 | 19 | 15,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
t = int(input())
add = 4
arr = [2]
for i in range(70):
arr += [arr[-1]+add]
add *= 2
# print(arr)
for _ in range(t):
k = int(input())
ans = [1]
flag = 1
while k != 0:
# print(k, ans)
flag = 0
for i in range(69, -1, -1):
if arr[i] <= k:
ans += [0]*i + [1]
flag = 1
k -= arr[i]
if flag == 0:
ans = -1
break
if ans != -1:
ans.pop()
print(len(ans))
print(*ans)
else:
print(ans)
``` | instruction | 0 | 7,776 | 19 | 15,552 |
Yes | output | 1 | 7,776 | 19 | 15,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
import sys
from collections import defaultdict as dd
from collections import Counter as cc
from queue import Queue
import math
import itertools
try:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
input = lambda: sys.stdin.readline().rstrip()
for _ in range(int(input())):
q=int(input())
if q%2==1:
print(-1)
else:
w=[1]
i=2
q-=2
while q>0:
if 2*i<q:
i*=2
q-=i
w.append(0)
else:
i=2
q-=2
w.append(1)
print(len(w))
print(*w)
``` | instruction | 0 | 7,777 | 19 | 15,554 |
Yes | output | 1 | 7,777 | 19 | 15,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
from collections import deque
import sys
serie = ""
d = deque()
n = 2
d.append(n)
while n<10**18:
n = 2*n +2
d.append(n)
r = 58
def binarySearch(l,r,x):
if r >= l:
mid = l + (r - l) // 2
if d[mid] == x:
return mid
elif d[mid] > x:
return binarySearch(l, mid-1, x)
else:
return binarySearch(mid + 1, r, x)
else:
return l-1
def p(x,s = ""):
if x == 0:
return s
i = binarySearch(0,r,x)
s+="1 "+("0 "*i)
x-=d[i]
return p(x,s)
t = int(sys.stdin.readline().strip())
while t>0:
k = int(sys.stdin.readline().strip())
serie = ""
if k%2 == 1:
print(-1)
else:
s = p(k)
print(len(s)//2)
print(s)
t-=1
``` | instruction | 0 | 7,778 | 19 | 15,556 |
Yes | output | 1 | 7,778 | 19 | 15,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
T = int(input())
for _ in range(T):
k = int(input())
if k % 2: print(-1); continue
res = []
for u in range(60, 1, -1):
p = 2**u + 2
if k // p: res.append((u, k//p))
k %= p
res.append((1, k//2))
ls = []
for u, cc in res:
ls += sum([ [0]*(u-1)+[1] ]*cc, [])
print(len(ls))
print(*ls)
``` | instruction | 0 | 7,779 | 19 | 15,558 |
No | output | 1 | 7,779 | 19 | 15,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
mod = 10**9 + 7
def solve():
n = int(input())
ans = []
if n == 1:
ans.append(-1)
else:
while n > 1 and len(ans) < 2000:
if n % 2 == 0:
ans.append(1)
n //= 2;
else:
ans.append(0)
n += 1
if n > 1:
ans = [-1]
print(len(ans))
print(*ans)
t = 1
t = int(input())
while t > 0:
solve()
t -= 1
``` | instruction | 0 | 7,780 | 19 | 15,560 |
No | output | 1 | 7,780 | 19 | 15,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
import sys
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
k = int(stdin.readline())
ans = []
if k % 2 == 1:
print (-1)
continue
tmp = 1
while k > 0:
if k & (2**tmp) > 0:
ans.append(1)
for i in range(tmp-1):
ans.append(0)
k ^= 2**tmp
tmp += 1
if len(ans) <= 2000:
print (len(ans))
print (*ans)
else:
print (-1)
``` | instruction | 0 | 7,781 | 19 | 15,562 |
No | output | 1 | 7,781 | 19 | 15,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.
There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.
For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.
For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.
Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).
Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.
Output
For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.
Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.
Example
Input
4
1
2
8
12
Output
-1
1
1
4
1 1 1 1
5
1 1 0 1 1
Note
In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.
In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used.
Submitted Solution:
```
import math
a = int(input())
for i in range(a):
k = int(input())
if k %2 == 1:
print("-1")
else:
ans = ""
time = 0
while (k>0):
n = int(math.log(k+2,2))-1
ans += "1 "
k -= pow(2,n)
n -= 1
time += 1
while(n>0):
ans += "0 "
k -= pow(2,n)
n -= 1
time += 1
print(time)
print(ans[:-1])
``` | instruction | 0 | 7,782 | 19 | 15,564 |
No | output | 1 | 7,782 | 19 | 15,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,919 | 19 | 15,838 |
Tags: greedy
Correct Solution:
```
from collections import defaultdict, deque
from heapq import heappush, heappop
from math import inf
ri = lambda : map(int, input().split())
def sm(x):
ans = 0
while x:
ans += (x % 10)
x //= 10
return ans
def solve():
n = int(input())
A = list(ri())
ans = 0
cnt = defaultdict(int)
for x in A:
d = sm(x)
if d % 3 == 0:
if cnt[0] > 0:
cnt[0] -= 1
ans += 1
else:
cnt[0] += 1
elif cnt[3-(d%3)] > 0:
cnt[3-(d%3)] -= 1
ans += 1
else:
cnt[d%3] += 1
print(ans)
t = 1
#t = int(input())
while t:
t -= 1
solve()
``` | output | 1 | 7,919 | 19 | 15,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,920 | 19 | 15,840 |
Tags: greedy
Correct Solution:
```
d=[0,0,0];input()
for i in map(int,input().split()):d[i%3]+=1
print(d[0]//2+min(d[1],d[2]))
``` | output | 1 | 7,920 | 19 | 15,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,921 | 19 | 15,842 |
Tags: greedy
Correct Solution:
```
a=input()
b=map(int,input().split())
c=0
d=0
e=0
for i in b:
if i%3==0:
c+=1
elif i%3==1:
d+=1
else:
e+=1
print(min(d,e)+c//2)
``` | output | 1 | 7,921 | 19 | 15,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,923 | 19 | 15,846 |
Tags: greedy
Correct Solution:
```
n=int(input())
ar=list(map(lambda x:int(x)%3,input().split(' ')))
print((ar.count(0)//2)+(min(ar.count(1),ar.count(2))))
``` | output | 1 | 7,923 | 19 | 15,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,924 | 19 | 15,848 |
Tags: greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
n=int(input())
a=list(map(int,input().split()))
one=0
zero=0
two=0
for i in range(0,len(a)):
if a[i]%3==0:
zero+=1
if a[i]%3==1:
one+=1
elif a[i]%3==2:
two+=1
print(zero//2+min(one,two))
``` | output | 1 | 7,924 | 19 | 15,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,925 | 19 | 15,850 |
Tags: greedy
Correct Solution:
```
import sys
from array import array # noqa: F401
from collections import Counter
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
cnt = Counter(x % 3 for x in map(int, input().split()))
print(cnt[0] // 2 + min(cnt[1], cnt[2]))
``` | output | 1 | 7,925 | 19 | 15,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1 | instruction | 0 | 7,926 | 19 | 15,852 |
Tags: greedy
Correct Solution:
```
s=int(input())
ar=[sum([int(x) for x in e])%3 for e in input().split()]
x,y,z=ar.count(0),ar.count(1),ar.count(2)
print(x//2+min(y,z))
``` | output | 1 | 7,926 | 19 | 15,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
n=int(input())
arr=list(map(int,(input().split())))
a,b,c=0,0,0
for i in arr:
if i%3==0:
a+=1
elif i%3==1:
b+=1
else:
c+=1
print((a//2)+min(b,c))
``` | instruction | 0 | 7,927 | 19 | 15,854 |
Yes | output | 1 | 7,927 | 19 | 15,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
#from dust i have come dust i will be
n=int(input())
a=list(map(int,input().split()))
x=0
y=0
z=0
for i in range(n):
if a[i]%3==0:
x+=1
elif a[i]%3==1:
y+=1
else:
z+=1
t=(x//2)+min(y,z)
print(t)
``` | instruction | 0 | 7,928 | 19 | 15,856 |
Yes | output | 1 | 7,928 | 19 | 15,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
n = int(input())
ost = [0,0,0]
for i in map(int, input().split()):
ost[i%3] += 1
# print(ost)
print(min(ost[1], ost[2])+(ost[0]//2))
``` | instruction | 0 | 7,929 | 19 | 15,858 |
Yes | output | 1 | 7,929 | 19 | 15,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
n = int(input())
l = [int(i) for i in input().split()]
c1 = 0
c2 = 0
c3 = 0
ans = 0
for i in range(n):
if (l[i]%3 == 0):
c1 = c1 + 1
elif (l[i]%3 == 1):
c2 = c2 + 1
else:
c3 = c3 + 1
ans += c1//2
min1 = min(c2,c3)
max1 = max(c2,c3)
ans += min1
print (ans)
``` | instruction | 0 | 7,930 | 19 | 15,860 |
Yes | output | 1 | 7,930 | 19 | 15,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
input()
t = list(sum(int(i) for i in j) % 3 for j in input().split())
p = [0] * 3
for i in t:
p[i] += 1
x = min(p[1], p[2])
y = p[1] + p[2] - x
print(p[0] // 2 + x + y // 2)
``` | instruction | 0 | 7,931 | 19 | 15,862 |
No | output | 1 | 7,931 | 19 | 15,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
a=input()
b=map(int,input().split())
c=0
d=0
e=0
for i in b:
print(i)
if i%3==0:
c+=1
elif i%3==1:
d+=1
else:
e+=1
print(min(d,e)+c//2)
``` | instruction | 0 | 7,932 | 19 | 15,864 |
No | output | 1 | 7,932 | 19 | 15,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky.
When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest.
Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123.
What maximum number of tickets could Vasya get after that?
Input
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Output
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
Examples
Input
3
123 123 99
Output
1
Input
6
1 1 1 23 10 3
Output
1
Submitted Solution:
```
from collections import defaultdict, deque
from heapq import heappush, heappop
from math import inf
ri = lambda : map(int, input().split())
def sm(x):
ans = 0
while x:
ans += (x % 10)
x //= 10
return ans
def solve():
n = int(input())
A = list(ri())
ans = 0
cnt = defaultdict(int)
for x in A:
d = sm(x)
if d % 3 == 0:
if cnt[0] > 1:
cnt[0] -= 1
ans += 1
else:
cnt[0] += 1
elif cnt[3-(d%3)] > 1:
cnt[3-(d%3)] -= 1
ans += 1
else:
cnt[d%3] += 1
print(ans)
t = 1
#t = int(input())
while t:
t -= 1
solve()
``` | instruction | 0 | 7,933 | 19 | 15,866 |
No | output | 1 | 7,933 | 19 | 15,867 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,200 | 19 | 16,400 |
"Correct Solution:
```
N,K = map(int,input().split())
R,S,P = map(int,input().split())
T = list(input())
d = {'r':P,'s':R,'p':S}
ans = 0
for i,c in enumerate(T):
if i < K or c != T[i-K]:
ans += d[c]
continue
else:
T[i] = '_'
print(ans)
``` | output | 1 | 8,200 | 19 | 16,401 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,201 | 19 | 16,402 |
"Correct Solution:
```
from itertools import groupby
N, K = map(int, input().split())
R, S, P = map(int, input().split())
T = input()
d = {"r": P, "s": R, "p": S}
print(sum(sum(map(lambda p: d[p[0]]*((len(list(p[1]))+1)//2), groupby(T[k::K]))) for k in range(K)))
``` | output | 1 | 8,201 | 19 | 16,403 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,202 | 19 | 16,404 |
"Correct Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = list(input())
dict = {"r":p, "s":r, "p":s}
val = 0
for i in range(n):
if i < k:
val += dict[t[i]]
elif t[i-k] != t[i]:
val += dict[t[i]]
else:
t[i] = " "
print(val)
``` | output | 1 | 8,202 | 19 | 16,405 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,203 | 19 | 16,406 |
"Correct Solution:
```
n,k = map(int,input().split())
r,s,p = map(int,input().split())
t = list(input())
for i in range(k,n):
if t[i]==t[i-k]:
t[i]=''
ans = t.count('r')*p+t.count('s')*r+t.count('p')*s
print(ans)
``` | output | 1 | 8,203 | 19 | 16,407 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,204 | 19 | 16,408 |
"Correct Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = list(input())
d = {'r': p, 's': r, 'p': s}
point = 0
for i in range(k):
point += d[t[i]]
for i in range(k, n):
if t[i] == t[i - k]:
t[i] = 'x'
continue
point += d[t[i]]
print(point)
``` | output | 1 | 8,204 | 19 | 16,409 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,205 | 19 | 16,410 |
"Correct Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
d = {"r":p, "s":r, "p":s}
t = list(input())
dp = [0]*n
for i in range(n):
if i>=k and t[i]==t[i-k]:
dp[i] = 0
t[i] = "x"
else:
dp[i] = d[t[i]]
print(sum(dp))
``` | output | 1 | 8,205 | 19 | 16,411 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,206 | 19 | 16,412 |
"Correct Solution:
```
N, K = map(int, input().split())
points = list(map(int, input().split()))
hand = {'r': 0, 's': 1, 'p': 2}
T = [hand[t] for t in input()]
win = [False] * N
ans = 0
for i in range(N):
if i < K or T[i] != T[i-K] or not win[i-K]:
win[i] = True
ans += points[(T[i]+2) % 3]
print(ans)
``` | output | 1 | 8,206 | 19 | 16,413 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996 | instruction | 0 | 8,207 | 19 | 16,414 |
"Correct Solution:
```
N,K=map(int,input().split())
R,S,P=map(int,input().split())
T=list(input())
for i in range(N-K):
if T[K+i]==T[i]:
T[K+i]=""
ans=T.count("r")*P+T.count("s")*R+T.count("p")*S
print(ans)
``` | output | 1 | 8,207 | 19 | 16,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n,k=map(int,input().split())
r,s,p=map(int,input().split())
t=list(input())
cnt={'r':0,'s':0,'p':0}
for i in range(n):
if i<=k-1:
cnt[t[i]]+=1
elif t[i]==t[i-k]:
t[i]='o'
continue
else:
cnt[t[i]]+=1
ans=p*cnt['r']+r*cnt['s']+s*cnt['p']
print(ans)
``` | instruction | 0 | 8,208 | 19 | 16,416 |
Yes | output | 1 | 8,208 | 19 | 16,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = list(input())
score = {"r":p, "s":r, "p":s}
ans = 0
for i in range(n):
if i<k:
ans += score[t[i]]
else :
if t[i] != t[i-k] :
ans += score[t[i]]
else :
t[i]="x"
print(ans)
``` | instruction | 0 | 8,209 | 19 | 16,418 |
Yes | output | 1 | 8,209 | 19 | 16,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n,k = map(int,input().split())
r,s,p = map(int,input().split())
t = list(input())
ans = 0
for i in range(k,n) :
if t[i] == t[i-k] :
t[i] = 'x'
for i in range(n) :
if t[i] == 'r' :
ans += p
elif t[i] == 's' :
ans += r
elif t[i] == 'p' :
ans += s
print(ans)
``` | instruction | 0 | 8,210 | 19 | 16,420 |
Yes | output | 1 | 8,210 | 19 | 16,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
N,K=map(int,input().split())
R,S,P=map(int,input().split())
T=input()
arr=[0]*N
d={'r':P,'s':R,'p':S}
for i,t in enumerate(T):
if i-K>=0:
if T[i-K]!=t:
arr[i]=d[t]
elif arr[i-K]==0:
arr[i]=d[t]
else:
arr[i]=d[t]
print(sum(arr))
``` | instruction | 0 | 8,211 | 19 | 16,422 |
Yes | output | 1 | 8,211 | 19 | 16,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
from collections import deque
N, K = [int(x) for x in input().split()]
r, s, p = [int(x) for x in input().split()]
win_point = {
'r': r,
's': s,
'p': p,
}
next_hands = {
'r': ['s', 'p'],
's': ['r', 'p'],
'p': ['r', 's'],
}
enemy_hands = input()
def can_win(enemy_hand, my_hands):
# 勝てる場合にはTrueと勝てる手を教えてくれる
if enemy_hand == 'r' and 'p' in my_hands:
return True, 'p'
if enemy_hand == 's' and 'r' in my_hands:
return True, 'r'
if enemy_hand == 'p' and 's' in my_hands:
return True, 's'
# 勝てる手がない場合
return False, None
def choose_hand(enemy_hand, next_enemy_hand, now_hands):
# 次の手に勝てるようにその手以外を選択する
if next_enemy_hand == enemy_hand:
# 次回も同じ手ならどっちでも良い
return now_hands[0]
win, hand = can_win(next_enemy_hand, enemy_hand)
# 次の手と今の相手の手を戦わせる
# 後の手が勝ったら今の手と同じものを出す
# 今の手が勝ったら後の手と同じものと出す
if win:
return now_hands[enemy_hand]
else:
return now_hands[next_enemy_hand]
point = 0
for index in range(K):
target = index
now_hands = ['r', 'p', 's']
for i in range(index, N, K):
win, hand = can_win(enemy_hands[i], now_hands)
if win:
point += win_point[hand]
now_hands = next_hands[hand]
else:
if i + K < N:
next_enemy_hand = enemy_hands[i+K]
now_hands = next_hands[choose_hand(
enemy_hands[i], next_enemy_hand, now_hands)]
print(point)
``` | instruction | 0 | 8,212 | 19 | 16,424 |
No | output | 1 | 8,212 | 19 | 16,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = str(input())
``` | instruction | 0 | 8,213 | 19 | 16,426 |
No | output | 1 | 8,213 | 19 | 16,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n, k = map(int, input().split())
R, S, P = map(int, input().split())
t = input()
ans = 0
hand = []
for i in t:
if i == "r":
hand.append("p")
if i == "s":
hand.append("r")
if i == "p":
hand.append("s")
for i, h in enumerate(t):
if i >= k:
if h == "r":
if hand[i-k] != "p":
ans += P
else:
hand[i] = "r"
if h == "s":
if hand[i-k] != "r":
ans += R
else:
hand[i] = "s"
if h == "p":
if hand[i-k] != "s":
ans += S
else:
hand[i] = "p"
else:
if h == "r":
ans += P
if h == "s":
ans += R
if h == "p":
ans += S
print(ans)
``` | instruction | 0 | 8,214 | 19 | 16,428 |
No | output | 1 | 8,214 | 19 | 16,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n,k=map(int,input().split())
p,r,s=map(int,input().split())
mc=input()
ans=0
for i in range(len(mc)):
if i>=k and mc[i-k]!=mc[i]:
if mc[i]=='r':
ans+=r
elif mc[i]=='p':
ans+=p
else:
ans+=s
``` | instruction | 0 | 8,215 | 19 | 16,430 |
No | output | 1 | 8,215 | 19 | 16,431 |
Provide a correct Python 3 solution for this coding contest problem.
Consider the following game. There are k pairs of n cards with numbers from 1 to n written one by one. Shuffle these kn cards well to make piles of k cards and arrange them in a horizontal row. The i-th (k-card) pile from the left of the n piles created in this way is called "mountain i".
<image>
The game starts at mountain 1. Draw the top card of the pile (the drawn card is not returned to the original pile), and if the number written on that card is i, draw the top card of the pile i Pull. In this way, draw the top card of the pile numbered by the number written on the drawn card repeatedly, and if all the piles have no cards, it is successful. If there are still piles of cards left, but there are no more piles to draw next, it is a failure.
If it fails in the middle, it ends with a failure, or the remaining card pile is left as it is (the pile number is also kept as it is) and the game is restarted. When restarting the game, the first card to draw is from the leftmost pile of the remaining piles (the top card of that pile is the first card to be drawn). After resuming, proceed with the game in the same way as before resuming, and if there are no cards in all the piles, it is a success. It is a failure.
<image>
Such a game shall be restarted up to m times. However, m is 0 or 1. In other words, it either does not restart once or restarts only once. The initial placement of cards differs depending on how you shuffle before the game starts. Of course, depending on the initial placement of the card, it may succeed without resuming, it may resume and succeed, or it may resume and fail. Since it is shuffled enough, we assume that all initial arrangements appear with the same probability, and we want to find the probability p that the restart will succeed within m times. Express this probability p as a decimal number, and create a program that finds and outputs to the decimal place r. However, output so that the following conditions are met.
* If p × 10K becomes an integer when a sufficiently large positive integer K is taken, 0 continues from the middle of the decimal part, but that 0 should also be output. For example, if p = 3/8 = 0.375, then 0.37500 is output if r = 5, and 0.37 is output if r = 2. Similarly, when p = 1.0, for example, if r = 3, output 1.000.
* For example, 0.150000 ... can be expressed as a recurring decimal 0.1499999 ..., but in such a case, the former expression is used.
On the first line of the input file, the integers n, k, m, and r are written in this order with a blank as the delimiter. 1 ≤ n ≤ 10000, 1 ≤ k ≤ 100, m = 0 or m = 1, 1 ≤ r ≤ 10000.
Input example 1 | Input example 2 | Input example 3
--- | --- | ---
|
2 1 0 5 | 3 1 1 3 | 2 2 1 3 |
Output example 1 | Output example 2 | Output example 3
0.50000 | 0.833 | 1.000 |
input
The input consists of multiple datasets. Input ends when n, k, m, and r are all 0. The number of datasets does not exceed 5.
output
For each dataset, p is output on one line as specified.
Example
Input
2 1 0 5
3 1 1 3
2 2 1 3
0 0 0 0
Output
0.50000
0.833
1.000 | instruction | 0 | 8,342 | 19 | 16,684 |
"Correct Solution:
```
# AOJ 0504: Card Game II
# Python3 2018.7.1 bal4u
from decimal import *
while True:
n, k, m, r = map(int, input().split())
if n == 0: break
setcontext(Context(prec=r, rounding=ROUND_HALF_UP))
one = Decimal(1)
ans = one/Decimal(n)
if m == 1:
s = 0
for i in range(1, n): s += one/Decimal(i)
ans *= 1+s
ans = str(ans)[:r+2]
if len(ans) < r+2: ans += '0'*(r+2-len(ans))
print(ans)
``` | output | 1 | 8,342 | 19 | 16,685 |
Provide a correct Python 3 solution for this coding contest problem.
Consider the following game. There are k pairs of n cards with numbers from 1 to n written one by one. Shuffle these kn cards well to make piles of k cards and arrange them in a horizontal row. The i-th (k-card) pile from the left of the n piles created in this way is called "mountain i".
<image>
The game starts at mountain 1. Draw the top card of the pile (the drawn card is not returned to the original pile), and if the number written on that card is i, draw the top card of the pile i Pull. In this way, draw the top card of the pile numbered by the number written on the drawn card repeatedly, and if all the piles have no cards, it is successful. If there are still piles of cards left, but there are no more piles to draw next, it is a failure.
If it fails in the middle, it ends with a failure, or the remaining card pile is left as it is (the pile number is also kept as it is) and the game is restarted. When restarting the game, the first card to draw is from the leftmost pile of the remaining piles (the top card of that pile is the first card to be drawn). After resuming, proceed with the game in the same way as before resuming, and if there are no cards in all the piles, it is a success. It is a failure.
<image>
Such a game shall be restarted up to m times. However, m is 0 or 1. In other words, it either does not restart once or restarts only once. The initial placement of cards differs depending on how you shuffle before the game starts. Of course, depending on the initial placement of the card, it may succeed without resuming, it may resume and succeed, or it may resume and fail. Since it is shuffled enough, we assume that all initial arrangements appear with the same probability, and we want to find the probability p that the restart will succeed within m times. Express this probability p as a decimal number, and create a program that finds and outputs to the decimal place r. However, output so that the following conditions are met.
* If p × 10K becomes an integer when a sufficiently large positive integer K is taken, 0 continues from the middle of the decimal part, but that 0 should also be output. For example, if p = 3/8 = 0.375, then 0.37500 is output if r = 5, and 0.37 is output if r = 2. Similarly, when p = 1.0, for example, if r = 3, output 1.000.
* For example, 0.150000 ... can be expressed as a recurring decimal 0.1499999 ..., but in such a case, the former expression is used.
On the first line of the input file, the integers n, k, m, and r are written in this order with a blank as the delimiter. 1 ≤ n ≤ 10000, 1 ≤ k ≤ 100, m = 0 or m = 1, 1 ≤ r ≤ 10000.
Input example 1 | Input example 2 | Input example 3
--- | --- | ---
|
2 1 0 5 | 3 1 1 3 | 2 2 1 3 |
Output example 1 | Output example 2 | Output example 3
0.50000 | 0.833 | 1.000 |
input
The input consists of multiple datasets. Input ends when n, k, m, and r are all 0. The number of datasets does not exceed 5.
output
For each dataset, p is output on one line as specified.
Example
Input
2 1 0 5
3 1 1 3
2 2 1 3
0 0 0 0
Output
0.50000
0.833
1.000 | instruction | 0 | 8,343 | 19 | 16,686 |
"Correct Solution:
```
from decimal import Decimal, getcontext
d1 = Decimal(1)
while True:
n, k, m, r = map(int, input().split())
if not n:
break
getcontext().prec = r + 1
ans = d1 / Decimal(n)
if m:
ans *= 1 + sum(d1 / Decimal(i) for i in range(1, n))
print('{{:.{}f}}'.format(r + 1).format(ans)[:-1])
``` | output | 1 | 8,343 | 19 | 16,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following game. There are k pairs of n cards with numbers from 1 to n written one by one. Shuffle these kn cards well to make piles of k cards and arrange them in a horizontal row. The i-th (k-card) pile from the left of the n piles created in this way is called "mountain i".
<image>
The game starts at mountain 1. Draw the top card of the pile (the drawn card is not returned to the original pile), and if the number written on that card is i, draw the top card of the pile i Pull. In this way, draw the top card of the pile numbered by the number written on the drawn card repeatedly, and if all the piles have no cards, it is successful. If there are still piles of cards left, but there are no more piles to draw next, it is a failure.
If it fails in the middle, it ends with a failure, or the remaining card pile is left as it is (the pile number is also kept as it is) and the game is restarted. When restarting the game, the first card to draw is from the leftmost pile of the remaining piles (the top card of that pile is the first card to be drawn). After resuming, proceed with the game in the same way as before resuming, and if there are no cards in all the piles, it is a success. It is a failure.
<image>
Such a game shall be restarted up to m times. However, m is 0 or 1. In other words, it either does not restart once or restarts only once. The initial placement of cards differs depending on how you shuffle before the game starts. Of course, depending on the initial placement of the card, it may succeed without resuming, it may resume and succeed, or it may resume and fail. Since it is shuffled enough, we assume that all initial arrangements appear with the same probability, and we want to find the probability p that the restart will succeed within m times. Express this probability p as a decimal number, and create a program that finds and outputs to the decimal place r. However, output so that the following conditions are met.
* If p × 10K becomes an integer when a sufficiently large positive integer K is taken, 0 continues from the middle of the decimal part, but that 0 should also be output. For example, if p = 3/8 = 0.375, then 0.37500 is output if r = 5, and 0.37 is output if r = 2. Similarly, when p = 1.0, for example, if r = 3, output 1.000.
* For example, 0.150000 ... can be expressed as a recurring decimal 0.1499999 ..., but in such a case, the former expression is used.
On the first line of the input file, the integers n, k, m, and r are written in this order with a blank as the delimiter. 1 ≤ n ≤ 10000, 1 ≤ k ≤ 100, m = 0 or m = 1, 1 ≤ r ≤ 10000.
Input example 1 | Input example 2 | Input example 3
--- | --- | ---
|
2 1 0 5 | 3 1 1 3 | 2 2 1 3 |
Output example 1 | Output example 2 | Output example 3
0.50000 | 0.833 | 1.000 |
input
The input consists of multiple datasets. Input ends when n, k, m, and r are all 0. The number of datasets does not exceed 5.
output
For each dataset, p is output on one line as specified.
Example
Input
2 1 0 5
3 1 1 3
2 2 1 3
0 0 0 0
Output
0.50000
0.833
1.000
Submitted Solution:
```
from decimal import Decimal, getcontext
d1 = Decimal(1)
while True:
n, k, m, r = map(int, input().split())
if not n:
break
getcontext().prec = r + 1
ans = d1 / Decimal(n)
if m:
ans *= 1 + sum(d1 / Decimal(i) for i in range(1, n))
print('{{:.{}f}}'.format(r).format(ans))
``` | instruction | 0 | 8,344 | 19 | 16,688 |
No | output | 1 | 8,344 | 19 | 16,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the following game. There are k pairs of n cards with numbers from 1 to n written one by one. Shuffle these kn cards well to make piles of k cards and arrange them in a horizontal row. The i-th (k-card) pile from the left of the n piles created in this way is called "mountain i".
<image>
The game starts at mountain 1. Draw the top card of the pile (the drawn card is not returned to the original pile), and if the number written on that card is i, draw the top card of the pile i Pull. In this way, draw the top card of the pile numbered by the number written on the drawn card repeatedly, and if all the piles have no cards, it is successful. If there are still piles of cards left, but there are no more piles to draw next, it is a failure.
If it fails in the middle, it ends with a failure, or the remaining card pile is left as it is (the pile number is also kept as it is) and the game is restarted. When restarting the game, the first card to draw is from the leftmost pile of the remaining piles (the top card of that pile is the first card to be drawn). After resuming, proceed with the game in the same way as before resuming, and if there are no cards in all the piles, it is a success. It is a failure.
<image>
Such a game shall be restarted up to m times. However, m is 0 or 1. In other words, it either does not restart once or restarts only once. The initial placement of cards differs depending on how you shuffle before the game starts. Of course, depending on the initial placement of the card, it may succeed without resuming, it may resume and succeed, or it may resume and fail. Since it is shuffled enough, we assume that all initial arrangements appear with the same probability, and we want to find the probability p that the restart will succeed within m times. Express this probability p as a decimal number, and create a program that finds and outputs to the decimal place r. However, output so that the following conditions are met.
* If p × 10K becomes an integer when a sufficiently large positive integer K is taken, 0 continues from the middle of the decimal part, but that 0 should also be output. For example, if p = 3/8 = 0.375, then 0.37500 is output if r = 5, and 0.37 is output if r = 2. Similarly, when p = 1.0, for example, if r = 3, output 1.000.
* For example, 0.150000 ... can be expressed as a recurring decimal 0.1499999 ..., but in such a case, the former expression is used.
On the first line of the input file, the integers n, k, m, and r are written in this order with a blank as the delimiter. 1 ≤ n ≤ 10000, 1 ≤ k ≤ 100, m = 0 or m = 1, 1 ≤ r ≤ 10000.
Input example 1 | Input example 2 | Input example 3
--- | --- | ---
|
2 1 0 5 | 3 1 1 3 | 2 2 1 3 |
Output example 1 | Output example 2 | Output example 3
0.50000 | 0.833 | 1.000 |
input
The input consists of multiple datasets. Input ends when n, k, m, and r are all 0. The number of datasets does not exceed 5.
output
For each dataset, p is output on one line as specified.
Example
Input
2 1 0 5
3 1 1 3
2 2 1 3
0 0 0 0
Output
0.50000
0.833
1.000
Submitted Solution:
```
# AOJ 0504: Card Game II
# Python3 2018.7.1 bal4u
def calc(b, n):
a = 1
for i in range(n+1):
if a >= b:
d[i] += a//b
a %= b
if a == 0: break
a *= 10
ans = ['0']*10005
while True:
n, k, m, r = map(int, input().split())
if n == 0: break
r += 6
d = [0]*10005
calc(n, r)
if m == 1:
for i in range(1, n): calc(n*i, r)
k = 0
for i in range(r, -1, -1):
k += d[i]
if k >= 10:
ans[i] = str(k%10)
k //= 10
else:
ans[i] = str(k)
k = 0
print(ans[0], '.', ''.join(ans[1:r-5]), sep='')
``` | instruction | 0 | 8,345 | 19 | 16,690 |
No | output | 1 | 8,345 | 19 | 16,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of n numbers: p1, p2, ..., pn. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes i and i + 1, for which an inequality pi > pi + 1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes i and i + 1, for which the inequality pi < pi + 1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
Input
The first line contains integer n (1 ≤ n ≤ 3000). The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation p. The numbers are separated by spaces.
Output
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
2
1 2
Output
0.000000
Input
5
3 5 2 4 1
Output
13.000000
Note
In the first test the sequence is already sorted, so the answer is 0. | instruction | 0 | 8,629 | 19 | 17,258 |
Tags: combinatorics, dp, probabilities
Correct Solution:
```
from sys import *
n=int(stdin.readline().strip())
s1=stdin.readline().strip()
a=list(map(int,s1.split()))
if n==1:
print('0')
else:
x=0
n0=1
while n0<n:
n0=n0*2
b=[0]*(2*n0+10)
'''for i in range (n-1):
for j in range (i+1,n):
if a[i]>a[j]:
x+=1'''
for i in range (n):
t=a[i]
u=0
k=1
j=1
while t>0:
if (t>>j)<<j!=t:
u=u+b[(n0+t-1)>>(j-1)]
t=t-k
k=k<<1
j=j+1
x=x+u
j=n0+a[i]-1
while j>0:
b[j]+=1
j=j>>1
x=((n*(n-1))//2)-x
'''n=x//2
print(x,n,' !!!')
r=x
i=1
bi=n
eps=0.0000001
if x>0:
while (x+2*i)*bi*((0.5)**i)>eps:
r=r+(x+2*i)*bi*((0.5)**i)
#print(r)
bi=(bi*(n+i))//(i+1)
i=i+1
#print(bi,i)
else:
r=0
r=r*((0.5)**n)
print("%.7f"%r)'''
if x%2 ==1:
print(2*x-1)
else:
print(2*x)
``` | output | 1 | 8,629 | 19 | 17,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of n numbers: p1, p2, ..., pn. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes i and i + 1, for which an inequality pi > pi + 1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes i and i + 1, for which the inequality pi < pi + 1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
Input
The first line contains integer n (1 ≤ n ≤ 3000). The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation p. The numbers are separated by spaces.
Output
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
2
1 2
Output
0.000000
Input
5
3 5 2 4 1
Output
13.000000
Note
In the first test the sequence is already sorted, so the answer is 0. | instruction | 0 | 8,630 | 19 | 17,260 |
Tags: combinatorics, dp, probabilities
Correct Solution:
```
n=int(input().strip())
p=[0]+list(map(int,input().split()))
c=[0]*(n+1)
def lowbit(x):
return x&(-x)
def add(x,v):
while x<=n:
c[x]+=v
x+=lowbit(x)
def get(x):
ans=0
while x:
ans+=c[x]
x-=lowbit(x)
return ans
ans=0
for i in range(n,0,-1):
ans+=get(p[i])
add(p[i],1)
if ans%2:
print(2*ans-1)
else:
print(2*ans)
``` | output | 1 | 8,630 | 19 | 17,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of n numbers: p1, p2, ..., pn. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes i and i + 1, for which an inequality pi > pi + 1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes i and i + 1, for which the inequality pi < pi + 1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
Input
The first line contains integer n (1 ≤ n ≤ 3000). The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation p. The numbers are separated by spaces.
Output
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
2
1 2
Output
0.000000
Input
5
3 5 2 4 1
Output
13.000000
Note
In the first test the sequence is already sorted, so the answer is 0. | instruction | 0 | 8,631 | 19 | 17,262 |
Tags: combinatorics, dp, probabilities
Correct Solution:
```
n=int(input())
b=list(map(int,input().split()))
cnt=0
for i in range(n):
j=i-1
while(j>=0):
if b[j]>b[i]:
cnt+=1
j+=-1
i+=-1
print(2*cnt-cnt%2)
``` | output | 1 | 8,631 | 19 | 17,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of n numbers: p1, p2, ..., pn. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes i and i + 1, for which an inequality pi > pi + 1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes i and i + 1, for which the inequality pi < pi + 1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
Input
The first line contains integer n (1 ≤ n ≤ 3000). The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation p. The numbers are separated by spaces.
Output
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
2
1 2
Output
0.000000
Input
5
3 5 2 4 1
Output
13.000000
Note
In the first test the sequence is already sorted, so the answer is 0. | instruction | 0 | 8,632 | 19 | 17,264 |
Tags: combinatorics, dp, probabilities
Correct Solution:
```
#!/usr/bin/python3
import sys
class CumTree:
def __init__(self, a, b):
self.a = a
self.b = b
self.count = 0
if a == b:
return
mid = (a + b) // 2
self.levo = CumTree(a, mid)
self.desno = CumTree(mid+1, b)
def manjsi(self, t):
if self.a >= t:
return 0
if self.b < t:
return self.count
return self.levo.manjsi(t) + self.desno.manjsi(t)
def vstavi(self, t):
if self.a <= t <= self.b:
self.count += 1
if self.a == self.b:
return
self.levo.vstavi(t)
self.desno.vstavi(t)
n = int(sys.stdin.readline())
p = [int(x) for x in sys.stdin.readline().strip().split()]
ct = CumTree(1, 4096)
vsota = 0
while len(p) > 0:
x = p.pop()
vsota += ct.manjsi(x)
ct.vstavi(x)
k, d = vsota // 2, vsota % 2
print("%f" % (4*k + d))
``` | output | 1 | 8,632 | 19 | 17,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jeff has become friends with Furik. Now these two are going to play one quite amusing game.
At the beginning of the game Jeff takes a piece of paper and writes down a permutation consisting of n numbers: p1, p2, ..., pn. Then the guys take turns to make moves, Jeff moves first. During his move, Jeff chooses two adjacent permutation elements and then the boy swaps them. During his move, Furic tosses a coin and if the coin shows "heads" he chooses a random pair of adjacent elements with indexes i and i + 1, for which an inequality pi > pi + 1 holds, and swaps them. But if the coin shows "tails", Furik chooses a random pair of adjacent elements with indexes i and i + 1, for which the inequality pi < pi + 1 holds, and swaps them. If the coin shows "heads" or "tails" and Furik has multiple ways of adjacent pairs to take, then he uniformly takes one of the pairs. If Furik doesn't have any pair to take, he tosses a coin one more time. The game ends when the permutation is sorted in the increasing order.
Jeff wants the game to finish as quickly as possible (that is, he wants both players to make as few moves as possible). Help Jeff find the minimum mathematical expectation of the number of moves in the game if he moves optimally well.
You can consider that the coin shows the heads (or tails) with the probability of 50 percent.
Input
The first line contains integer n (1 ≤ n ≤ 3000). The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation p. The numbers are separated by spaces.
Output
In a single line print a single real value — the answer to the problem. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
2
1 2
Output
0.000000
Input
5
3 5 2 4 1
Output
13.000000
Note
In the first test the sequence is already sorted, so the answer is 0. | instruction | 0 | 8,633 | 19 | 17,266 |
Tags: combinatorics, dp, probabilities
Correct Solution:
```
def count(a):
n = len(a)
cnt = 0
for i in range(n):
for j in range(i+1, n):
if a[i] > a[j]:
cnt+=1
return cnt
n = int(input())
p = list(map(int, input().split()))
num = count(p)
print(num*2 - num%2)
``` | output | 1 | 8,633 | 19 | 17,267 |
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