message stringlengths 2 433k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6 | instruction | 0 | 73,534 | 12 | 147,068 |
Tags: brute force, math, two pointers
Correct Solution:
```
import io,os
from math import *
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
manual = 20
n,Q = map(int,input().split())
after = min(manual, n)
before = n -after
ind = 0
def unrankperm(i):
unused = [i+1 for i in range(after)]
r = []
for j in range(after)[::-1]:
use = i // factorial(j)
r.append(unused[use])
del unused[use]
i -= use * factorial(j)
return r
p = unrankperm(ind)
#for i in range(100):
# print(unrankperm(i))
for _ in range(Q):
q = list(map(int,input().split()))
if q[0] == 1:
l,r = q[1:]
amt = 0
if l <= before:
if r <= before:
amt += r * (r + 1) // 2
else:
amt += before * (before + 1) // 2
amt -= l * (l - 1) // 2
l = before + 1
if r > before:
amt += sum(p[l-1-before:r-before]) + (r - l + 1) * (before)
print(amt)
else:
ind += q[1]
p = unrankperm(ind)
``` | output | 1 | 73,534 | 12 | 147,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6 | instruction | 0 | 73,535 | 12 | 147,070 |
Tags: brute force, math, two pointers
Correct Solution:
```
# No need to use segment Tree because apart from the last 14 numbers, nothing will change since 14! > 2*pow(10, 10)
def modify(rest):
for i in range(len(rest)):
prefix[-len(rest)+i] = prefix[-len(rest)+(i-1)] + rest[i]
# print("prefix", prefix)
def next_perm_direct(permc): # Gets the next permc lexographically bigger permutation
rest = [x for x in range(max(1, n-14), n+1)]
temp = 1
for i in range(1, len(rest)):
temp = temp*i
for i in range(len(rest)-1):
if permc >= temp:
change = permc//temp
rest[i], rest[i+change] = rest[i+change], rest[i]
rest = rest[:i+1] + list(sorted(rest[i+1:]))
permc = permc%temp
temp = temp//(len(rest)-1-i)
# print("rest", rest)
modify(rest)
n, q = map(int, input().split())
prefix = [0 for i in range(n+1)]
for i in range(1, n+1):
prefix[i] = prefix[i-1]+i
# print(prefix)
permc = 0
for _ in range(q):
quer = list(map(int, input().split()))
if len(quer) == 2:
permc += quer[1]
next_perm_direct(permc)
else:
l, r = quer[1], quer[2]
print(prefix[r]-prefix[l-1])
``` | output | 1 | 73,535 | 12 | 147,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class smt:
def __init__(self,l,r,arr):
self.l=l
self.r=r
self.value=(1<<31)-1 if l<r else arr[l]
mid=(l+r)//2
if(l<r):
self.left=smt(l,mid,arr)
self.right=smt(mid+1,r,arr)
self.value&=self.left.value&self.right.value
#print(l,r,self.value)
def setvalue(self,x,val):
if(self.l==self.r):
self.value=val
return
mid=(self.l+self.r)//2
if(x<=mid):
self.left.setvalue(x,val)
else:
self.right.setvalue(x,val)
self.value=self.left.value&self.right.value
def ask(self,l,r):
if(l<=self.l and r>=self.r):
return self.value
val=(1<<31)-1
mid=(self.l+self.r)//2
if(l<=mid):
val&=self.left.ask(l,r)
if(r>mid):
val&=self.right.ask(l,r)
return val
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it):
res=[]
for ch,g in groupby(it):
res.append((ch,len(list(g))))
return res
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def cal(x,k):
ans=[]
used=(1<<k)-1
for i in range(k-1,-1,-1):
cur,x=divmod(x,fact(i))
c=0
for j in range(k):
if used&1<<j:
c+=1
if c==cur+1:
ans.append(j+1)
used^=1<<j
break
return ans
t=1
for i in range(t):
n,q=RL()
cur=0
k=min(n,14)
for i in range(q):
res=RLL()
if len(res)==2:
cur+=res[1]
else:
l,r=res[1:]
if r<=n-k:
ans=(l+r)*(r-l+1)//2
else:
ans=0
if l<=n-k:
ans=(l+n-k)*(n-k-l+1)//2
l=n-k+1
res=cal(cur,k)
#print(res)
ans+=sum(res[l-n+k-1:r-n+k])+(r-l+1)*(n-k)
print(ans)
``` | instruction | 0 | 73,536 | 12 | 147,072 |
Yes | output | 1 | 73,536 | 12 | 147,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
import sys
import itertools as it
import math as mt
import bisect as bi
import collections as cc
import itertools as it
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
n,q=I()
ar=list(range(1,n+1))
pref=[0]
for i in range(n):
pref.append(pref[-1]+ar[i])
arr=ar[-15::]
clen=len(arr)
nxt=0
fact=[1,1]
for i in range(2,20):
fact.append(fact[-1]*i)
def ch(n):
temp=[i for i in arr]
re=[]
for i in range(clen,0,-1):
cur=n//fact[i-1]
n-=(cur*(fact[i-1]))
re.append(temp.pop(cur))
return re
for tc in range(q):
now=I()
if now[0]==2:
nxt+=now[1]
continue
left,right=now[1],now[2]
if right<=n-clen:
print(pref[right]-pref[left-1])
continue
elif left<=n-clen:
ans=pref[-clen-1]-pref[left-1]
left=0
right-=n-clen+1
else:
ans=0
left-=n-clen+1
right-=n-clen+1
cur=ch(nxt)
print(ans+sum(cur[left:right+1]))
``` | instruction | 0 | 73,537 | 12 | 147,074 |
Yes | output | 1 | 73,537 | 12 | 147,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
import bisect
factor = [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200]
n,q = map(int,input().split())
def getpermute(x,n):
m = bisect.bisect(factor,x)+1
output = [0]*m
have = [i for i in range(m)]
for i in range(m-1):
add = x//factor[m-i-2]
x -= add*factor[m-i-2]
output[i] = have[add]+(n-m+1)
have.pop(add)
output[-1] = have[0]+(n-m+1)
return output
t = 1
x = 0
arr = [n]
while t<=q:
inp = list(map(int,input().split()))
if inp[0]==2:
x += inp[1]
arr = getpermute(x,n)
t += 1
continue
l,r = inp[1],inp[2]
base = n - len(arr)
if r<=base:
ans = (l+r)*(r-l+1)//2
print(ans)
elif l<=base:
ans = (l+base)*(base-l+1)//2
ans += sum(arr[:r-base])
print(ans)
else:
ans = sum(arr[l-base-1:r-base])
print(ans)
t += 1
``` | instruction | 0 | 73,538 | 12 | 147,076 |
Yes | output | 1 | 73,538 | 12 | 147,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
n, Q = map(int, input().split())
cnt = 1
F = [1]
for i in range(1, 17):
F.append(F[-1] * i)
st = max(0, n - 15)
a = [0] * (n + 1)
vs = [0] * (n + 1)
for i in range(1, n + 1):
a[i] = a[i - 1] + i
for _ in range(Q):
q = list(map(int, input().split()))
if q[0] == 1:
l, r = q[1], q[2]
print(a[r] - a[l - 1])
continue
x = q[1]
cnt += x
hav = cnt
for i in range(st, n + 1):
vs[i] = 0
for i in range(st, n):
tmp = 0
for j in range(st, n + 1):
if not vs[j]:
tmp += F[n - i]
if tmp >= hav:
hav -= tmp - F[n - i]
a[i] = a[i - 1] + j
vs[j] = 1
break
for i in range(st, n + 1):
if not vs[i]:
a[n] = a[n - 1] + i
``` | instruction | 0 | 73,539 | 12 | 147,078 |
Yes | output | 1 | 73,539 | 12 | 147,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
n, q = list(map(int, input().split()))
def calc(left, right):
return right * (right+1) // 2 - left * (left-1) // 2
def calc2(left, right, shifts, nn, facs):
res = 0
if nn > 15:
count = right - left + 1
diff = nn - 15
res += count * diff
nn = 15
arr = []
a = shifts
b = facs[nn-1]
z = nn - 1
unused = [i+1 for i in range(nn)]
for i in range(nn):
cur = (a // b)
arr.append(unused[cur])
del unused[cur]
a = a % facs[z]
z = z - 1
b = facs[z]
for i in range(left-1, right):
res += arr[i]
return res
fac = [1]
s = 1
for i in range(1, 15):
s = s * i
fac.append(s)
total = 0
for i in range(q):
line = list(map(int, input().split()))
if len(line) == 3:
# type 1
l = line[1]
r = line[2]
res = 0
if l <= n - 15:
r2 = min(r, n-15)
res += calc(l, r2)
if r > n - 15:
l2 = max(l, n - 13)
r2 = r
if n > 15:
l2 = l2 - (n-15)
r2 = r2 - (n-15)
res += calc2(l2, r2, total, n, fac)
print(res)
else:
# type 2
x = line[1]
total += x
if n < 16:
total = total % fac[n]
``` | instruction | 0 | 73,540 | 12 | 147,080 |
No | output | 1 | 73,540 | 12 | 147,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
from math import factorial
def per(i, s1):
r = []
for j in range(len(s1))[::-1]: use = i // factorial(j);r.append(s1[use]);del s1[use];i -= use * factorial(j)
return r
n = list(map(int, input().split()))
s = []
s1 = []
s12 = []
s0 = []
for i in range(n[0]):
s.append(i + 1)
if n[0] - i < 16:
s1.append(i + 1)
else:
s0.append(i + 1)
f = 0
s12 = s1
for i in range(n[1]):
k = list(map(int, input().split()))
if k[0] == 1:
print(sum(s[k[1] - 1: k[2]]))
else:
f += k[1]
s1 = per(f, s12)
s = s0 + s1
``` | instruction | 0 | 73,541 | 12 | 147,082 |
No | output | 1 | 73,541 | 12 | 147,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
n, q = list(map(int, input().split()))
def calc(left, right):
return right * (right+1) / 2 - left * (left-1) / 2
def calc2(left, right, shifts, nn, facs):
res = 0
if nn > 14:
count = right - left + 1
diff = nn - 14
res += count * diff
nn = 14
arr = []
a = shifts
b = facs[nn-1]
z = nn - 1
for i in range(nn):
cur = (a // b) + 1
for j in arr:
if j <= cur:
cur += 1
arr.append(cur)
a = a % facs[z]
z = z - 1
b = facs[z]
for i in range(left-1, right):
res += arr[i]
return res
fac = [1]
s = 1
for i in range(1, 15):
s = s * i
fac.append(s)
total = 0
for i in range(q):
line = list(map(int, input().split()))
if len(line) == 3:
# type 1
l = line[1]
r = line[2]
res = 0
if l <= n - 14:
r2 = min(r, n-14)
res += calc(l, r2)
if r > n - 14:
l2 = max(l, n - 13)
r2 = r
if n > 14:
l2 = l2 - (n-14)
r2 = r2 - (n-14)
res += calc2(l2, r2, total, n, fac)
print(int(res))
else:
# type 2
x = line[1]
total += x
if n < 15:
total = total % fac[n]
``` | instruction | 0 | 73,542 | 12 | 147,084 |
No | output | 1 | 73,542 | 12 | 147,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.
Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.
The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example:
* for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4];
* for permutation [1, 2, 3] the next permutation is [1, 3, 2];
* for permutation [2, 1] next permutation does not exist.
You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).
You need to process q queries of two types:
* 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r].
* 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2].
For each query of the 1-st type output the required sum.
Input
The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.
The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).
It is guaranteed that all requests of the 2-nd type are possible to process.
Output
For each query of the 1-st type, output on a separate line one integer — the required sum.
Example
Input
4 4
1 2 4
2 3
1 1 2
1 3 4
Output
9
4
6
Note
Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows:
1. 2 + 3 + 4 = 9;
2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2];
3. 1 + 3 = 4;
4. 4 + 2 = 6
Submitted Solution:
```
from math import factorial
def short_perm_to_num(perm, n): # O(n**2), but n <= 70
result = 0
is_used = {}
for i in range(n):
for j in range(perm[i]):
if not (j in is_used):
result += factorial(n-1-i)
is_used[perm[i]] = True
return result
def perm_to_num(perm, n):
max_len = 15
if n <= max_len:
return short_perm_to_num(perm, n)
else:
last_part = [x - (n-max_len) for x in perm[n-max_len:]]
return short_perm_to_num(last_part, max_len)
def short_num_to_perm(num, n):
is_used = {}
perm = []
for i in range(n):
deg = num // factorial(n-1-i)
num %= factorial(n-1-i)
j = 0
while j in is_used:
j += 1
# now j is 1st unused item
deg_of_j = 0
while (deg_of_j < deg) or (j in is_used):
if not (j in is_used):
deg_of_j += 1
j += 1
perm.append(j)
is_used[j] = True
return perm
def num_to_perm(num, n): # RETURNS ONLY LAST 15 SYMBOLS
max_len = 15
if n <= max_len:
return short_num_to_perm(num, n)
else:
last_part_low = short_num_to_perm(num, max_len)
last_part = [x + (n-max_len) for x in last_part_low]
return last_part
[n, q] = [int(x) for x in input().split()]
full_perm = [i for i in range(0, n)] # full permutation as an array, but don't forget to add 1 when printing
num_of_perm = 0
# 70! is just above 10**100, so elements before the last 70 are still 1, 2, 3, ... n-7
for quarry_num in range(q):
quarry = [int(x) for x in input().split()]
if quarry[0] == 1:
perm = num_to_perm(num_of_perm, n)
l = quarry[1]
r = quarry[2]
if n <= 15:
print(sum(perm[l-1:r]) + r-l+1)
else: # n-15, n-14, ... n-1 are in the perm; we must compute sum(perm[l-1:r])
if l-1 >= n-15:
print(sum(perm[l-1-n+15:r-n+15]) + r-l+1)
else: # l-1 < n-15
if r < n-15:
s = ((l-1) + r) / 2 * (r - l + 2)
print(s + r-l+1)
else:
s = ((l-1) + n-16) / 2 * (n-16 - l + 2)
rest = sum(perm[:r-n+15])
print(s + rest + r-l+1)
if quarry[0] == 2:
num_of_perm += quarry[1]
``` | instruction | 0 | 73,543 | 12 | 147,086 |
No | output | 1 | 73,543 | 12 | 147,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.
For permutation p = p0, p1, ..., pn, Polo has defined its beauty — number <image>.
Expression <image> means applying the operation of bitwise excluding "OR" to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".
Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input
The single line contains a positive integer n (1 ≤ n ≤ 106).
Output
In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.
If there are several suitable permutations, you are allowed to print any of them.
Examples
Input
4
Output
20
0 2 1 4 3 | instruction | 0 | 73,615 | 12 | 147,230 |
Tags: implementation, math
Correct Solution:
```
import math
def solve():
n, = map(int, input().split())
top = 1
while top <= n:
top *= 2
top -= 1
ans = [0] * (n + 1)
i = n
cur = n
while i > 0:
if top - i > cur:
cur = i
top = 1
while top <= cur:
top *= 2
top -= 1
ans[i] = top - i
ans[top - i] = i
i -= 1
score = 0
if n % 2 == 1:
ans[top] = 0
ans[0] = top
for i in range(n + 1):
score += ans[i] ^ i
print(score)
print(*ans)
if __name__ == "__main__":
t = 1
# t = int(input())
for _ in range(t):
solve()
``` | output | 1 | 73,615 | 12 | 147,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.
For permutation p = p0, p1, ..., pn, Polo has defined its beauty — number <image>.
Expression <image> means applying the operation of bitwise excluding "OR" to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".
Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input
The single line contains a positive integer n (1 ≤ n ≤ 106).
Output
In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.
If there are several suitable permutations, you are allowed to print any of them.
Examples
Input
4
Output
20
0 2 1 4 3 | instruction | 0 | 73,616 | 12 | 147,232 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
p = [i for i in range(n + 1)]
k = 1
while(2 * k <= n):
k *= 2
m = n + 1
while m > 0:
while k >= m:
k //= 2
for i in range(m - k):
if k - i - 1 >= 0:
p[k + i], p[k - i - 1] = p[k - i - 1], p[k + i]
m = k - i - 1
print(n * (n + 1))
print(' '.join(map(str, p)))
``` | output | 1 | 73,616 | 12 | 147,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.
For permutation p = p0, p1, ..., pn, Polo has defined its beauty — number <image>.
Expression <image> means applying the operation of bitwise excluding "OR" to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".
Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input
The single line contains a positive integer n (1 ≤ n ≤ 106).
Output
In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.
If there are several suitable permutations, you are allowed to print any of them.
Examples
Input
4
Output
20
0 2 1 4 3 | instruction | 0 | 73,617 | 12 | 147,234 |
Tags: implementation, math
Correct Solution:
```
import math
def ones(n):
number_of_bits = (int)(math.floor(math.log(n) /
math.log(2))) + 1;
return [((1 << number_of_bits) - 1) ^ n,(1 << number_of_bits) - 1];
n = int(input())
has=[0]*(n+1)
ans=[0]*(n+1)
fin=0
for i in range(n,0,-1):
if has[i]==0:
com,fi = ones(i)
# print(com,fi,i)
has[com]=1
fin+=2*fi
ans[com]=i
ans[i]=com
print(fin)
print(*ans)
``` | output | 1 | 73,617 | 12 | 147,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.
For permutation p = p0, p1, ..., pn, Polo has defined its beauty — number <image>.
Expression <image> means applying the operation of bitwise excluding "OR" to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".
Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input
The single line contains a positive integer n (1 ≤ n ≤ 106).
Output
In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.
If there are several suitable permutations, you are allowed to print any of them.
Examples
Input
4
Output
20
0 2 1 4 3 | instruction | 0 | 73,618 | 12 | 147,236 |
Tags: implementation, math
Correct Solution:
```
from bisect import insort,bisect_right,bisect_left
from sys import stdout, stdin, setrecursionlimit
from math import sqrt,ceil,floor,factorial,gcd,log2,log10
from io import BytesIO, IOBase
from collections import *
from itertools import *
from random import *
from string import *
from queue import *
from heapq import heappush, heappop, heapify
from re import *
from os import *
####################################---fast-input-output----#########################################
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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
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:
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")
stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
graph, mod, szzz = {}, 10**9 + 7, lambda: sorted(zzz())
def getStr(): return input()
def getInt(): return int(input())
def listStr(): return list(input())
def getStrs(): return input().split()
def isInt(s): return '0' <= s[0] <= '9'
def input(): return stdin.readline().strip()
def zzz(): return [int(i) for i in input().split()]
def output(answer, end='\n'): stdout.write(str(answer) + end)
def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def getPrimes(N = 10**5):
SN = int(sqrt(N))
sieve = [i for i in range(N+1)]
sieve[1] = 0
for i in sieve:
if i > SN:
break
if i == 0:
continue
for j in range(2*i, N+1, i):
sieve[j] = 0
prime = [i for i in range(N+1) if sieve[i] != 0]
return prime
def primeFactor(n,prime=getPrimes()):
lst = []
mx=int(sqrt(n))+1
for i in prime:
if i>mx:break
while n%i==0:
lst.append(i)
n//=i
if n>1:
lst.append(n)
return lst
dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]
daysInMounth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
#################################################---Some Rule For Me To Follow---#################################
"""
--instants of Reading problem continuously try to understand them.
--If you Know some-one , Then you probably don't know him !
--Try & again try, maybe you're just one statement away!
"""
##################################################---START-CODING---###############################################
s = n = getInt()+1
lst,new=[],list(range(n))[::-1]
k = 2**20
while s:
while k>=2*s:k//=2
lst=new[n-s:n+s-k]+lst
s=k-s
print(n*(n-1))
for i in lst:
output(i)
``` | output | 1 | 73,618 | 12 | 147,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.
For permutation p = p0, p1, ..., pn, Polo has defined its beauty — number <image>.
Expression <image> means applying the operation of bitwise excluding "OR" to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".
Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input
The single line contains a positive integer n (1 ≤ n ≤ 106).
Output
In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.
If there are several suitable permutations, you are allowed to print any of them.
Examples
Input
4
Output
20
0 2 1 4 3 | instruction | 0 | 73,619 | 12 | 147,238 |
Tags: implementation, math
Correct Solution:
```
import sys
from math import gcd,sqrt,ceil
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
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
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 = set()
sa.add(n)
while n % 2 == 0:
sa.add(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.add(i)
n = n // i
# sa.add(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 debug(n):
l = [i for i in range(n+1)]
z = permutations(l)
maxi = 0
for i in z:
fin = 0
for j in range(n+1):
fin+=j^i[j]
maxi = max(maxi,fin)
return maxi
n = int(input())
hash = defaultdict(int)
ans = [0]*(n+1)
seti = set()
for i in range(n,0,-1):
if i not in seti:
z1 = i^int('1'*len((bin(i)[2:])),2)
seti.add(z1)
ans[z1] = i
ans[i] = z1
# print(ans)
fin = 0
for i in range(n+1):
fin+=i^ans[i]
print(fin)
print(*ans)
``` | output | 1 | 73,619 | 12 | 147,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little penguin Polo likes permutations. But most of all he likes permutations of integers from 0 to n, inclusive.
For permutation p = p0, p1, ..., pn, Polo has defined its beauty — number <image>.
Expression <image> means applying the operation of bitwise excluding "OR" to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is represented as "^" and in Pascal — as "xor".
Help him find among all permutations of integers from 0 to n the permutation with the maximum beauty.
Input
The single line contains a positive integer n (1 ≤ n ≤ 106).
Output
In the first line print integer m the maximum possible beauty. In the second line print any permutation of integers from 0 to n with the beauty equal to m.
If there are several suitable permutations, you are allowed to print any of them.
Examples
Input
4
Output
20
0 2 1 4 3
Submitted Solution:
```
from bisect import insort,bisect_right,bisect_left
from sys import stdout, stdin, setrecursionlimit
from math import sqrt,ceil,floor,factorial,gcd,log2,log10
from io import BytesIO, IOBase
from collections import *
from itertools import *
from random import *
from string import *
from queue import *
from heapq import heappush, heappop, heapify
from re import *
from os import *
####################################---fast-input-output----#########################################
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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
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:
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")
stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
graph, mod, szzz = {}, 10**9 + 7, lambda: sorted(zzz())
def getStr(): return input()
def getInt(): return int(input())
def listStr(): return list(input())
def getStrs(): return input().split()
def isInt(s): return '0' <= s[0] <= '9'
def input(): return stdin.readline().strip()
def zzz(): return [int(i) for i in input().split()]
def output(answer, end='\n'): stdout.write(str(answer) + end)
def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def getPrimes(N = 10**5):
SN = int(sqrt(N))
sieve = [i for i in range(N+1)]
sieve[1] = 0
for i in sieve:
if i > SN:
break
if i == 0:
continue
for j in range(2*i, N+1, i):
sieve[j] = 0
prime = [i for i in range(N+1) if sieve[i] != 0]
return prime
def primeFactor(n,prime=getPrimes()):
lst = []
mx=int(sqrt(n))+1
for i in prime:
if i>mx:break
while n%i==0:
lst.append(i)
n//=i
if n>1:
lst.append(n)
return lst
dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]
daysInMounth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
#################################################---Some Rule For Me To Follow---#################################
"""
--instants of Reading problem continuously try to understand them.
--If you Know some-one , Then you probably don't know him !
--Try & again try, maybe you're just one statement away!
"""
##################################################---START-CODING---###############################################
s = n = getInt()+1
lst,new=[],list(range(n))[::-1]
k = 1<<21
while s:
while k>=2*s:k//=2
lst+=new[n-s:n+s-k]
s=k-s
print(n*(n-1))
print(*lst[::-1])
``` | instruction | 0 | 73,620 | 12 | 147,240 |
No | output | 1 | 73,620 | 12 | 147,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,621 | 12 | 147,242 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n=int(input())
a=list(map(float, input().split()))
a.sort(reverse=True)
su=0
while(a):
su += sum(a)
a=a[:len(a)//4]
print(int(su))
``` | output | 1 | 73,621 | 12 | 147,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,622 | 12 | 147,244 |
Tags: constructive algorithms, greedy, implementation, sortings
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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
################################
n=int(input())
from math import log
l=list(map(int,input().split()))
l.sort(reverse=True)
a=int(log(n,4))
if n==1:
print(l[0])
else:
ans=l[0]*(a+1)
i=1
c=0
while i<n:
if a>1:
for j in range((4**c)*3):
ans+=l[i+j]*a
a-=1
i+=(4**c)*3
c+=1
else:
ans+=l[i]
i+=1
print(ans)
``` | output | 1 | 73,622 | 12 | 147,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,623 | 12 | 147,246 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
def arr_inp():
return [float(x) for x in input().split()]
def solve(n):
global s
if n == 1:
s += a[0]
return
else:
solve(n // 4)
s += sum(a[:n])
n, a, s = int(input()), sorted(arr_inp(), reverse=True), 0
solve(n)
print(int(s))
``` | output | 1 | 73,623 | 12 | 147,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,624 | 12 | 147,248 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n = int(input())
arr = list(map(float, input().split()))
arr.sort(reverse = True)
res = 0
while arr:
res += sum(arr)
arr = arr[:len(arr)//4]
print(int(res))
``` | output | 1 | 73,624 | 12 | 147,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,625 | 12 | 147,250 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
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 gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
n=int(input())
a=list(map(int, input().split()))
a.sort()
a.reverse()
ans=0
i=1
while(i<=n):
ans+=sum(a[0:i])
i*=4
print(ans)
return
if __name__ == "__main__":
main()
``` | output | 1 | 73,625 | 12 | 147,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,626 | 12 | 147,252 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
n=Int()
a=sorted(array(),reverse=True)
k=int(log(n,4))+1
ans=0
end=1
c=0
for i in range(n):
if(i==end):
k-=1
c+=1
end=4**(c)
# print(i,k,end)
ans+=k*a[i]
print(ans)
``` | output | 1 | 73,626 | 12 | 147,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,627 | 12 | 147,254 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
#!/usr/bin/env python
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")
def main():
n = int(input())
numbers = [int(c) for c in input().split()]
numbers.sort(reverse=True)
acc = [0]
for e in numbers:
acc.append(acc[-1] + e)
i = 1
res = 0
while i <= n:
cnt = n // i
res += acc[cnt]
i *= 4
print(res)
if __name__ == "__main__":
main()
``` | output | 1 | 73,627 | 12 | 147,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14. | instruction | 0 | 73,628 | 12 | 147,256 |
Tags: constructive algorithms, greedy, implementation, sortings
Correct Solution:
```
n=int(input())
lis=list(map(float,input().split()))
ans=0
lis.sort(reverse=1)
while lis:
ans+=sum(lis)
lis=lis[:len(lis)//4]
print(int(ans))
``` | output | 1 | 73,628 | 12 | 147,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14.
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
l.sort()
temp = l[::]
for i in range(1, n):
temp[i] = temp[i] + temp[i-1]
su = 0
j = n-1
i = 4
while j-1 >= 0:
su = su + temp[n-1] - temp[j-1]
j = n - i
i = i*2
su = su + temp[n-1]
print(su)
``` | instruction | 0 | 73,633 | 12 | 147,266 |
No | output | 1 | 73,633 | 12 | 147,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14.
Submitted Solution:
```
def arr_inp():
return [int(x) for x in input().split()]
n, a = int(input()), sorted(arr_inp(), reverse=True)
if n == 1:
print(a[0])
else:
print(a[0] + sum(a[:4]))
``` | instruction | 0 | 73,634 | 12 | 147,268 |
No | output | 1 | 73,634 | 12 | 147,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.
He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.
The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:
1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.
As you can see, the algorithm is recursive.
Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.
Input
The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.
Output
On a single line print the maximum value of the beauty of the described matrix.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
1
13
Output
13
Input
4
1 2 3 4
Output
14
Note
Consider the second sample. You need to arrange the numbers in the matrix as follows:
1 2
3 4
Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14.
Submitted Solution:
```
def arr_inp():
return [int(x) for x in input().split()]
def arr_sum(arr):
return list(accumulate(arr, lambda x, y: x + y))
def solve(n):
global s
if n == 1:
s += a[0]
return
else:
solve(n // 4)
s += a[n-1]
from itertools import *
n, a, s = int(input()), arr_sum(sorted(arr_inp(), reverse=True)), 0
print(a)
solve(n)
print(s)
``` | instruction | 0 | 73,636 | 12 | 147,272 |
No | output | 1 | 73,636 | 12 | 147,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,780 | 12 | 147,560 |
Tags: dp, sortings
Correct Solution:
```
N=int(input())
s=list(map(int,input().split()))
for i in range(N):s[i]-=i
X=sorted(s)
dp=[0]*N
for i in s:
mi = 7e77
for j in range(N):
mi = min(mi, dp[j])
dp[j] = mi + abs(i-X[j])
print(min(dp))
``` | output | 1 | 73,780 | 12 | 147,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,781 | 12 | 147,562 |
Tags: dp, sortings
Correct Solution:
```
# Idea:
# convert the problem to making array a non decreasing by decreasing each ai by i
# Now we can argue that the optimal final array must have
# each element equal to some element of the array a.
# Proof is to assume optimal solution does not satisfy this then you can
# always increase or decrease one element while maintaining the
# non decreasing property and decreasing or not changing the number of
# operations used. The point at which we cannot keep increasing or decreasing
# is when the element i we are changing reaches some a[j] since then we may
# go from removing to increasing operations (kind of like slope/events trick)
n = int(input())
a = list(map(int,input().split()))
for i in range(n):
a[i] -= i
sorted_a = sorted(a)
dp = [0.0]*n # dp[j] = min operations such a[i] = sorted_a[j]
dp2 = [0.0]*n
for i in range(n):
# we are changing a[i]
mn_prev_state = 1e13
for j in range(n):
mn_prev_state = min(mn_prev_state, dp[j])
# we change a[i] to sorted_a[j]
# ofc this means the changed value of a[i-1] <= sorted_a[j]
# so the changed value of a[i-1] = sorted_a[0...j]
# hence mn_prev_state is all min(dp[0...j])
dp2[j] = mn_prev_state + abs(a[i] - sorted_a[j])
for j in range(n):
dp[j] = dp2[j]
print(int(min(dp)))
``` | output | 1 | 73,781 | 12 | 147,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,782 | 12 | 147,564 |
Tags: dp, sortings
Correct Solution:
```
from heapq import *
n = int(input())
turn = [99**9]
opty = 0
for i,a in enumerate(map(int,input().split())):
a-= i
optx = -turn[0]
if optx <= a:
heappush(turn, -a)
else:
heappush(turn, -a)
heappop(turn)
heappush(turn, -a)
opty+= optx-a
print(opty)
``` | output | 1 | 73,782 | 12 | 147,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,783 | 12 | 147,566 |
Tags: dp, sortings
Correct Solution:
```
import heapq
n = int(input())
d = list(map(int,input().split()))
pq = [-d[0]]
heapq.heapify(pq)
ans = 0
for i in range(1,n):
temp = i - d[i]
heapq.heappush(pq,temp)
if heapq.nsmallest(1,pq)[0] < temp:
ans += temp - heapq.nsmallest(1,pq)[0]
heapq.heappushpop(pq,temp)
print(ans)
``` | output | 1 | 73,783 | 12 | 147,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,784 | 12 | 147,568 |
Tags: dp, sortings
Correct Solution:
```
from heapq import *
class Maxheap:
def __init__(_): _.h = []
def add(_, v): heappush(_.h, -v)
def top(_): return -_.h[0]
def pop(_): return -heappop(_.h)
class Graph:
def __init__(_):
_.change = Maxheap() # increment slope at ...
_.change.add(-10**18)
_.a = _.y = 0 # last line has slope a, starts from y
_.dx = 0 # the whole graph is shifted right by ...
def __repr__(_): return f"<{[x+_.dx for x in _.change]}; {_.a} {_.y}>"
def shiftx(_, v): _.dx+= v
def shifty(_, v): _.y+= v
def addleft(_, v):
if _.change.top() < v-_.dx:
dx = v-_.dx - _.change.top()
_.y+= _.a*dx
_.change.add(v-_.dx)
def addright(_, v):
if _.change.top() < v-_.dx:
dx = v-_.dx - _.change.top()
_.y+= _.a*dx; _.a+= 1
_.change.add(v-_.dx)
return
_.change.add(v-_.dx)
_.a+= 1; _.y+= _.change.top()-(v-_.dx)
def cutright(_):
dx = _.change.pop()-_.change.top()
_.a-= 1; _.y-= _.a*dx
n = int(input())
G = Graph()
for x in map(int,input().split()):
G.shiftx(1)
G.addleft(x)
G.addright(x)
while G.a > 0: G.cutright()
print(G.y)
``` | output | 1 | 73,784 | 12 | 147,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,785 | 12 | 147,570 |
Tags: dp, sortings
Correct Solution:
```
from bisect import bisect_left as BL
n = int(input())
graph = [(-99**9, 0)] # x, slope
ans = 0
for i,a in enumerate(map(int,input().split())):
a-= i
new = []
turnj = BL(graph, (a,99**9)) - 1
if turnj != len(graph)-1:
ans+= graph[-1][0] - a
# add |x-a|
for j in range(turnj):
x, sl = graph[j]
new.append((x, sl-1))
for j in range(turnj, len(graph)):
x, sl = graph[j]
if j == turnj:
new.append((x, sl-1))
new.append((a, sl+1))
else: new.append((x, sl+1))
# remove positive slopes
graph = new
while graph[-1][1] > 0: x, sl = graph.pop()
if graph[-1][1] != 0: graph.append((x, 0))
print(ans)
``` | output | 1 | 73,785 | 12 | 147,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,786 | 12 | 147,572 |
Tags: dp, sortings
Correct Solution:
```
from bisect import bisect_left as BL
n = int(input())
graph = [(-99**9, 0, 0)] # x, y, slope
for i,a in enumerate(map(int,input().split())):
a-= i
new = []
turnj = BL(graph, (a,99**9)) - 1
# add |x-a|
for j in range(turnj):
x, y, sl = graph[j]
new.append((x, y+(a-x), sl-1))
for j in range(turnj, len(graph)):
x, y, sl = graph[j]
if j == turnj:
new.append((x, y+(a-x), sl-1))
new.append((a, y+(a-x)+(sl-1)*(a-x), sl+1))
else: new.append((x, y+(x-a), sl+1))
# remove positive slopes
graph = new
while graph[-1][2] > 0: x, y, sl = graph.pop()
if graph[-1][2] != 0: graph.append((x, y, 0))
print(graph[-1][1])
``` | output | 1 | 73,786 | 12 | 147,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12 | instruction | 0 | 73,787 | 12 | 147,574 |
Tags: dp, sortings
Correct Solution:
```
from bisect import insort
class Graph:
def __init__(_):
_.change = [-10**27] # increment slope at ...
_.a = _.y = 0 # last line has slope a, starts from y
_.dx = 0 # the whole graph is shifted right by ...
def __repr__(_): return f"<{[x+_.dx for x in _.change]}; {_.a} {_.y}>"
def shiftx(_, v): _.dx+= v
def shifty(_, v): _.y+= v
def addleft(_, v):
if _.change[-1] < v-_.dx:
dx = v-_.dx - _.change[-1]
_.y+= _.a*dx
insort(_.change, v-_.dx)
def addright(_, v):
if _.change[-1] < v-_.dx:
dx = v-_.dx - _.change[-1]
_.y+= _.a*dx; _.a+= 1
insort(_.change, v-_.dx)
return
insort(_.change, v-_.dx)
_.a+= 1; _.y+= _.change[-1]-(v-_.dx)
#def remleft(_, v): change.remove(v-_.dx)
def cutright(_):
dx = _.change.pop()-_.change[-1]
_.a-= 1; _.y-= _.a*dx
n = int(input())
G = Graph()
for x in map(int,input().split()):
G.shiftx(1)
G.addleft(x)
G.addright(x)
while G.a > 0: G.cutright()
print(G.y)
``` | output | 1 | 73,787 | 12 | 147,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12
Submitted Solution:
```
def mySol(arr, n):
mod = 0
if arr[0] > arr[1]:
mod = arr[0]-arr[1] + 1
arr[0] = arr[1]-1
for i in range(1, n-1):
am = arr[i]
a0 = arr[i-1]
a1 = arr[i+1]
if a0 < am < a1:
continue
if a1 < am :
possDed = am - a0 - 1
if possDed > (am-a1) + 1:
mod += am - a1
am = a1
else:
mod += possDed
am -= possDed
mod += am - a1 + 1
a1 = am + 1
arr[i],arr[i-1],arr[i+1] = am, a0, a1
# print(arr)
return mod
n = int(input())
arr = list(map(int, input().split()))
print(mySol(arr,n))
``` | instruction | 0 | 73,788 | 12 | 147,576 |
No | output | 1 | 73,788 | 12 | 147,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12
Submitted Solution:
```
import math
import random
def IsRW (River):
t=1
for i in range(len(River)-1):
if River[i+1]<=River[i]:
t=0
return t
def SwitchRivers (River,L,R,k):
for i in range (len(River)-2):
j=i+1
if (River[j]<=River[j-1]) and (River[j]<=River[j+1]) and (River[j+1]>River[j-1]):
River[j]=River[j-1]+1
else:
if (River[j]>=River[j-1]) and (River[j]>=River[j+1]) and (River[j+1]>River[j-1]):
River[j]=River[j+1]-1
'''
print (River)
for i in range (len(River)-1):
j=i+1
if (River[j]<=River[j-1]):
River[j]=River[j-1]+1
'''
a=[]
k=0
for i in range (R-L+1):
k=k+River[L+i]
n=int(k/(R-L+1))
# M=River[L]
# for i in range ((R-L+1)):
# if abs(River[i]-n)<abs(M-n):
# M=River[i]
l=L
r=R
M=n
m = l+int((r-l+(r-l)%2)/2)
k=k+abs(River[m]-M)
if (IsRW(River)==0):
River[m]=M
for i in range(m-l):
if (River[l+i]>=M) and (IsRW(River)==0):
c=River[l+i]-(M-(m-l-i))
River[l+i]=River[l+i]-c
k=k+c
for i in range(r-m):
if (River[m+i+1]<=M) and (IsRW(River)==0):
c=(M+i+1)-River[m+i+1]
River[m+i+1]=River[m+i+1]+c
k=k+c
if (m>L) and (m<=R) and (IsRW(River)==0):
River=SwitchRivers (River,L,m-1,k)
if (m>=L) and (m<R) and (IsRW(River)==0):
River=SwitchRivers (River,m+1,R,k)
'''
if (IsRW(River)==0):
River=SwitchRivers (River,L,R,k)
'''
return (River)
Rivers = []
River = []
#River = [2,1,5,11,5,9,11]
#River = [5,4,3,2,1]
#River = [6,7,0,99,23,53]
n=input()
n=int(n)
s=input()
b=''
for i in range (len(s)):
a=s[i]
if a!=' ':
b=b+a
else:
c = int(b)
b=''
River.append(c)
c = int(b)
River.append(c)
for i in range(len(River)):
Rivers.append(River[i])
SwitchRivers (River,0,len(River)-1,0)
k=0
for i in range (len(River)):
k=k+abs(River[i]-Rivers[i])
print(k)
``` | instruction | 0 | 73,789 | 12 | 147,578 |
No | output | 1 | 73,789 | 12 | 147,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12
Submitted Solution:
```
import math
import random
def IsRW (River):
t=1
for i in range(len(River)-1):
if River[i+1]<=River[i]:
t=0
return t
def SwitchRivers (River,L,R,k):
print (River)
for i in range (len(River)-2):
j=i+1
if (River[j]<=River[j-1]) and (River[j]<=River[j+1]) and (River[j+1]>River[j-1]):
River[j]=River[j-1]+1
else:
if (River[j]>=River[j-1]) and (River[j]>=River[j+1]) and (River[j+1]>River[j-1]):
River[j]=River[j+1]-1
'''
print (River)
for i in range (len(River)-1):
j=i+1
if (River[j]<=River[j-1]):
River[j]=River[j-1]+1
'''
a=[]
k=0
for i in range (R-L+1):
k=k+River[L+i]
n=int(k/(R-L+1))
# M=River[L]
# for i in range ((R-L+1)):
# if abs(River[i]-n)<abs(M-n):
# M=River[i]
l=L
r=R
M=n
m = l+int((r-l+(r-l)%2)/2)
k=k+abs(River[m]-M)
if (IsRW(River)==0):
River[m]=M
for i in range(m-l):
if (River[l+i]>=M) and (IsRW(River)==0):
c=River[l+i]-(M-(m-l-i))
River[l+i]=River[l+i]-c
k=k+c
for i in range(r-m):
if (River[m+i+1]<=M) and (IsRW(River)==0):
c=(M+i+1)-River[m+i+1]
River[m+i+1]=River[m+i+1]+c
k=k+c
if (m>L) and (m<=R) and (IsRW(River)==0):
River=SwitchRivers (River,L,m-1,k)
if (m>=L) and (m<R) and (IsRW(River)==0):
River=SwitchRivers (River,m+1,R,k)
'''
if (IsRW(River)==0):
River=SwitchRivers (River,L,R,k)
'''
return (River)
Rivers = []
River = []
#River = [2,1,5,11,5,9,11]
#River = [5,4,3,2,1]
#River = [6,7,0,99,23,53]
n=input()
n=int(n)
s=input()
b=''
for i in range (len(s)):
a=s[i]
if a!=' ':
b=b+a
else:
c = int(b)
b=''
River.append(c)
c = int(b)
River.append(c)
for i in range(len(River)):
Rivers.append(River[i])
SwitchRivers (River,0,len(River)-1,0)
k=0
for i in range (len(River)):
k=k+abs(River[i]-Rivers[i])
print (River)
print(k)
``` | instruction | 0 | 73,790 | 12 | 147,580 |
No | output | 1 | 73,790 | 12 | 147,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
Output
Print the minimum number of operation required to make the array strictly increasing.
Examples
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 4 3 2 1
Output
12
Note
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12
Submitted Solution:
```
import math
import random
def IsRW (River):
t=1
for i in range(len(River)-1):
if River[i+1]<=River[i]:
t=0
return t
def SwitchRivers (River,L,R,k):
for i in range (len(River)-2):
j=i+1
if (River[j]<=River[j-1]) and (River[j]<=River[j+1]) and (River[j+1]>River[j-1]):
River[j]=River[j-1]+1
else:
if (River[j]>=River[j-1]) and (River[j]>=River[j+1]) and (River[j+1]>River[j-1]):
River[j]=River[j+1]-1
a=[]
k=0
for i in range (R-L+1):
k=k+River[L+i]
n=int(k/(R-L+1))
# M=River[L]
# for i in range ((R-L+1)):
# if abs(River[i]-n)<abs(M-n):
# M=River[i]
l=L
r=R
M=n
m = l+int((r-l+(r-l)%2)/2)
k=k+abs(River[m]-M)
if (IsRW(River)==0):
River[m]=M
for i in range(m-l):
if (River[l+i]>=M) and (IsRW(River)==0):
c=River[l+i]-(M-(m-l-i))
River[l+i]=River[l+i]-c
k=k+c
for i in range(r-m):
if (River[m+i+1]<=M) and (IsRW(River)==0):
c=(M+i+1)-River[m+i+1]
River[m+i+1]=River[m+i+1]+c
k=k+c
if (m>L) and (m<=R) and (IsRW(River)==0):
River=SwitchRivers (River,L,m-1,k)
if (m>=L) and (m<R) and (IsRW(River)==0):
River=SwitchRivers (River,m+1,R,k)
return (River)
Rivers = []
River = []
#River = [2,1,5,11,5,9,11]
#River = [5,4,3,2,1]
#River = [6,7,0,99,23,53]
n=input()
n=int(n)
s=input()
b=''
for i in range (len(s)):
a=s[i]
if a!=' ':
b=b+a
else:
c = int(b)
b=''
River.append(c)
c = int(b)
River.append(c)
for i in range(len(River)):
Rivers.append(River[i])
SwitchRivers (River,0,len(River)-1,0)
k=0
for i in range (len(River)):
k=k+abs(River[i]-Rivers[i])
print(k)
``` | instruction | 0 | 73,791 | 12 | 147,582 |
No | output | 1 | 73,791 | 12 | 147,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The marmots have prepared a very easy problem for this year's HC2 – this one. It involves numbers n, k and a sequence of n positive integers a1, a2, ..., an. They also came up with a beautiful and riveting story for the problem statement. It explains what the input means, what the program should output, and it also reads like a good criminal.
However I, Heidi, will have none of that. As my joke for today, I am removing the story from the statement and replacing it with these two unhelpful paragraphs. Now solve the problem, fools!
Input
The first line of the input contains two space-separated integers n and k (1 ≤ k ≤ n ≤ 2200). The second line contains n space-separated integers a1, ..., an (1 ≤ ai ≤ 104).
Output
Output one number.
Examples
Input
8 5
1 1 1 1 1 1 1 1
Output
5
Input
10 3
16 8 2 4 512 256 32 128 64 1
Output
7
Input
5 1
20 10 50 30 46
Output
10
Input
6 6
6 6 6 6 6 6
Output
36
Input
1 1
100
Output
100 | instruction | 0 | 73,828 | 12 | 147,656 |
Tags: greedy, sortings
Correct Solution:
```
import math
import queue
from itertools import permutations
n,k=map(int,input().split())
a=[int(cost) for cost in input().split()]
a.sort()
sum=0
for i in range(0,k):
sum+=a[i]
print(sum)
``` | output | 1 | 73,828 | 12 | 147,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,900 | 12 | 147,800 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from heapq import *
n, c = list(map(int, input().split()))
a = list(map(int, input().split()))
if c > n:
print(sum(a))
exit()
b = [0] * n
s = 0
h = []
for i in range(n):
s = s + a[i] - a[i-c] if i + 1 > c else s + a[i]
heappush(h, (a[i], i))
if i + 1 < c:
b[i] = s
else:
while h[0][1] <= i - c:
heappop(h)
v1 = b[i-1] + a[i]
v2 = b[i-c] + s - h[0][0]
b[i] = min(v1, v2)
print(b[-1])
``` | output | 1 | 73,900 | 12 | 147,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,901 | 12 | 147,802 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from heapq import *
n, c = map(int, input().split())
a = list(map(int, input().split()))
if c > n:
print(sum(a))
exit()
b = [0] * n
s = 0
h = []
for i in range(n):
s = s + a[i] - a[i-c] if i + 1 > c else s + a[i]
heappush(h, (a[i],i))
if i + 1 < c:
b[i] = s
else:
while h[0][1] <= i-c:
heappop(h)
v1 = b[i-1] + a[i]
v2 = b[i-c] + s - h[0][0]
b[i] = min(v1, v2)
print(b[-1])
``` | output | 1 | 73,901 | 12 | 147,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,902 | 12 | 147,804 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from functools import *
from time import time
from heapq import *
def measure_time(fun):
def wrapper(*args, **kwargs):
begin = time()
rv = fun(*args, **kwargs)
end = time()
print(end - begin)
return rv
return wrapper
def our_cache(func):
memo = [-1]*(max+1)
@wraps(func)
def wrapper(arg):
if memo[arg] == -1:
memo[arg] = func(arg)
return memo[arg]
return wrapper
def main():
n, c = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
if c > n:
print(sum(a))
exit(0)
b = n * [0]
s = 0
h = []
for i in range(n):
s = s + a[i] if i <= c-1 else s + a[i] - a[i-c]
heappush(h, (a[i], i))
if i <= c-2:
b[i] = s
else:
while i-h[0][1] >= c:
heappop(h)
v1 = b[i-c] + s - h[0][0]
v2 = a[i] + b[i-1]
b[i] = min(v1, v2)
print(b[-1])
if __name__ == "__main__":
main()
``` | output | 1 | 73,902 | 12 | 147,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,903 | 12 | 147,806 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
str=input().split()
n=int(str[0])
len=int(str[1])
a=[]
Q=[]
F=[]
for i in range(0,n+1):
a.append(0)
Q.append(0)
F.append(0)
sum=0
h=1
t=0
str=input().split()
for i in range(1,n+1):
a[i]=int(str[i-1])
sum+=a[i]
#print (sum)
while h<=t and Q[h]<=i-len:
h=h+1
while h<=t and a[i]<=a[Q[t]]:
t=t-1
t=t+1;Q[t]=i;
if (i<len) :
F[i]=0
else :
F[i]=F[i-len]+a[Q[h]]
F[i]=max(F[i],F[i-1])
print (sum-F[n])
``` | output | 1 | 73,903 | 12 | 147,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,904 | 12 | 147,808 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
class SegTree:
def __init__(self, init_val, ide_ele, segfunc):
self.n = len(init_val)
self.num = 2**(self.n-1).bit_length()
self.ide_ele = ide_ele
self.segfunc = segfunc
self.seg = [ide_ele]*2*self.num
# set_val
for i in range(self.n):
self.seg[i+self.num] = init_val[i]
# built
for i in range(self.num-1, 0, -1):
self.seg[i] = self.segfunc(self.seg[2*i], self.seg[2*i+1])
def add(self, k, x):
k += self.num
self.seg[k] += x
while k:
k = k >> 1
self.seg[k] = self.segfunc(self.seg[2*k], self.seg[2*k+1])
def query(self, l, r):
if r <= l:
return self.ide_ele
l += self.num
r += self.num
lres = self.ide_ele
rres = self.ide_ele
while l < r:
if r & 1:
r -= 1
rres = self.segfunc(self.seg[r], rres)
if l & 1:
lres = self.segfunc(lres, self.seg[l])
l += 1
l = l >> 1
r = r >> 1
res = self.segfunc(lres, rres)
return res
def __str__(self): # for debug
arr = [self.query(i,i+1) for i in range(self.n)]
return str(arr)
INF = 10**18
def main():
n, k = map(int, input().split())
A = list(map(int, input().split()))
if k > n:
print(sum(A))
exit()
if k == n:
print(sum(A)-min(A))
exit()
res = []
from collections import deque
d = deque([])
for i in range(n):
while d:
if d[0] <= i-k:
d.popleft()
else:
break
while d:
if A[d[-1]] > A[i]:
d.pop()
else:
break
d.append(i)
if i-k+1 >= 0:
res.append(A[d[0]])
#print(res)
seg = SegTree(res, -INF, max)
for i in range(len(res)):
t = max(0, seg.query(0, i-k+1))
seg.add(i, t)
#print(seg)
print(sum(A)-seg.query(0, seg.n))
if __name__ == '__main__':
main()
``` | output | 1 | 73,904 | 12 | 147,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,905 | 12 | 147,810 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
def queries(l,r):
mini =float("inf")
while l<=r:
if l %2 ==1:
mini =min(mini,sgn_tree[l])
l+=1
if r % 2==0:
mini =min(mini,sgn_tree[r])
r-=1
l =l>>1
r =r>>1
return mini
n,c=map(int,input().split())
arr=list(map(int,input().split()))
sgn_tree=[0 for i in range(2*n)]
sgn_tree[n:] =arr
for i in range(n-1,0,-1):
sgn_tree[i] =min(sgn_tree[2*i],sgn_tree[2*i+1])
dp=[0 for i in range(n)]
for i in range(n):
if i-c>=-1:
dp[i] =max(dp[i-c] +queries(n+i-c+1,n+i),dp[i-1])
s=sum(arr)
print(s -dp[n-1])
``` | output | 1 | 73,905 | 12 | 147,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,906 | 12 | 147,812 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from functools import *
from time import time
from heapq import *
#
# def measure_time(fun):
# def wrapper(*args, **kwargs):
# begin = time()
# rv = fun(*args, **kwargs)
# end = time()
# print(end - begin)
# return rv
# return wrapper
#
#
# def our_cache(func):
# memo = [-1]*(max+1)
#
# @wraps(func)
# def wrapper(arg):
# if memo[arg] == -1:
# memo[arg] = func(arg)
# return memo[arg]
# return wrapper
def main():
n, c = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
if c > n:
print(sum(a))
exit(0)
b = n * [0]
s = 0
h = []
for i in range(n):
s = s + a[i] if i <= c-1 else s + a[i] - a[i-c]
heappush(h, (a[i], i))
if i <= c-2:
b[i] = s
else:
while i-h[0][1] >= c:
heappop(h)
v1 = b[i-c] + s - h[0][0]
v2 = a[i] + b[i-1]
b[i] = min(v1, v2)
print(b[-1])
if __name__ == "__main__":
main()
``` | output | 1 | 73,906 | 12 | 147,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively. | instruction | 0 | 73,907 | 12 | 147,814 |
Tags: data structures, dp, greedy, math
Correct Solution:
```
from heapq import *
f = lambda: map(int, input().split())
n, c = f()
a = list(f()) + [0] * c
b = [0] * (n + c)
s = 0
h = [(0, -1)]
for i in range(n):
s += a[i] - a[i - c]
heappush(h, (a[i], i))
while h[0][1] <= i - c: heappop(h)
b[i] = min(b[i - 1] + a[i], b[i - c] + s - h[0][0])
print(b[n - 1])
``` | output | 1 | 73,907 | 12 | 147,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively.
Submitted Solution:
```
n, c = map(int, input().split())
a = sorted(list(map(int, input().split())))
print(sum(a[n//c:]))
``` | instruction | 0 | 73,908 | 12 | 147,816 |
No | output | 1 | 73,908 | 12 | 147,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively.
Submitted Solution:
```
def iter(ar, l, r, c, result, min_):
sum_ = ar[l]
pos = -1
for i in range(l + 1, r):
if (r - i) < c:
break
elif (sum_ < ar[i]) or ((i -l) > c):
pos = i
sum_+=ar[i]
if (pos > 0 and (pos -l) >= min_):
# print(str(l)+" " + str(pos))
# print(str(pos)+" " + str(r))
iter(ar, l, pos, c, result, -1)
iter(ar, pos, r, c, result, 2)
else:
delta = (r - l) // c
result[0] += sum(ar[l + delta:r])
pairs= [int(s) for s in input().split(" ")]
n= pairs[0]
c= pairs[1]
a= [int(s) for s in input().split(" ")]
a.sort()
result=[0]
iter(a,0,n,c,result,-1)
print(result[0])
``` | instruction | 0 | 73,909 | 12 | 147,818 |
No | output | 1 | 73,909 | 12 | 147,819 |
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