qyliang/CodeNanoFix huggingface dataset:
Takahashi, a merchant, travels from town 1 to town N, buying and selling apples. He starts with no apples and can move between towns and buy/sell apples at distinct prices. The total number of apples bought and sold must be at most T. Takahashi aims to maximize profit, while Aoki, a rival, can change apple prices to decrease Takahashi's profit by at least 1 yen. The task is to find the minimum cost for Aoki to achieve this. Constraints: 1 ≤ N ≤ 10^5, 1 ≤ Ai ≤ 10^9, Ai are distinct, 2 ≤ T ≤ 10^9, and Takahashi's initial profit is at least 1 yen. Input format: N T, followed by N apple prices. Output: Minimum total cost to decrease Takahashi's profit by at least 1 yen.

# Buggy version:
def slove():
    import sys
    import collections
    input = sys.stdin.readline
    n, t = list(map(int, input().rstrip('\n').split()))
    a = list(map(int, input().rstrip('\n').split()))
    b = collections.defaultdict(list)
    m_c = 10 ** 9
    for i, v in enumerate(a):
        if i != 0 and m_c < v:
            b[v-m_c] += [m_c]
        m_c = min(m_c, v)
    b = sorted(b.items(), reverse=True)[0]
    print(b)
    print(len(collections.Counter(b[1])))


if __name__ == '__main__':
    slove()


# Fixed version:
def slove():
    import sys
    import collections
    input = sys.stdin.readline
    n, t = list(map(int, input().rstrip('\n').split()))
    a = list(map(int, input().rstrip('\n').split()))
    b = collections.defaultdict(list)
    m_c = 10 ** 9
    for i, v in enumerate(a):
        if i != 0 and m_c < v:
            b[v-m_c] += [m_c]
        m_c = min(m_c, v)
    b = sorted(b.items(), reverse=True)[0]
    print(len(collections.Counter(b[1])))


if __name__ == '__main__':
    slove()
Score: 900 points
Problem Statement
In 2937, DISCO creates a new universe called DISCOSMOS to celebrate its 1000-th anniversary.
DISCOSMOS can be described as an H x W grid. Let (i, j) (1 ≤ i ≤ H, 1 ≤ j ≤ W) denote the square at the i-th row from the top and the j-th column from the left.
At time 0, one robot will be placed onto each square. Each robot is one of the following three types:
- Type-H: Does not move at all.
- Type-R: If a robot of this type is in (i, j) at time t, it will be in (i, j+1) at time t+1. If it is in (i, W) at time t, however, it will be instead in (i, 1) at time t+1. (The robots do not collide with each other.)
- Type-D: If a robot of this type is in (i, j) at time t, it will be in (i+1, j) at time t+1. If it is in (H, j) at time t, however, it will be instead in (1, j) at time t+1.
There are 3^(H x W) possible ways to place these robots. In how many of them will every square be occupied by one robot at times 0, T, 2T, 3T, 4T, and all subsequent multiples of T?
Since the count can be enormous, compute it modulo (10^9 + 7).
Constraints
- 1 ≤ H ≤ 10^9
- 1 ≤ W ≤ 10^9
- 1 ≤ T ≤ 10^9
- H, W, T are all integers.
Input
Input is given from Standard Input in the following format:
H W T
Output
Print the number of ways to place the robots that satisfy the condition, modulo (10^9 + 7).
Sample Input 1
2 2 1
Sample Output 1
9
Shown below are some of the ways to place the robots that satisfy the condition, where ., >, and v stand for Type-H, Type-R, and Type-D, respectively:
>>  ..  vv
..  ..  vv
Sample Input 2
869 120 1001
Sample Output 2
672919729

# Buggy version:
def gcd(x,y):
    if x<y:
        x,y=y,x
    if y==0:
        return x
    else:
        return gcd(y,x%y)
    
H,W,T=map(int,input().split())
mod=10**9+7

h=gcd(H,T)
w=gcd(W,T)
H//=h
W//=w

print(h,w,H,W)
ans=pow(2,H,mod)+pow(2,W,mod)+pow(2,gcd(H,W),mod)-3
print(pow(ans,h*w,mod))

# Fixed version:
def gcd(x,y):
    if x<y:
        x,y=y,x
    if y==0:
        return x
    else:
        return gcd(y,x%y)
    
H,W,T=map(int,input().split())
mod=10**9+7

h=gcd(H,T)
w=gcd(W,T)
H//=h
W//=w

ans=pow(2,H,mod)+pow(2,W,mod)+pow(2,gcd(H,W),mod)-3
print(pow(ans,h*w,mod))
You are given N points arranged in a line. Takahashi adds edges between these points using M operations. Each operation connects points within a given range with a specific length. Find the shortest path from point 1 to point N in the final graph. If no path exists, output -1. Constraints: 2 ≤ N ≤ 10^5, 1 ≤ M ≤ 10^5, 1 ≤ Li < Ri ≤ N, 1 ≤ Ci ≤ 10^9. Input format: N M, followed by M lines of Li, Ri, and Ci. Output the length of the shortest path or -1.

# Buggy version:
import sys
input=sys.stdin.readline
import heapq
n,m=map(int,input().split())

D=[0]
P=[]
ans=float("inf")

for i in range(m):
  l,r,c=map(int,input().split())
  P.append((l,r,c))
P.sort(key=lambda x:x[0])
heapq.heappush(D,1*10**15+0)
for l,r,c in P:
  while D:
    ld=heapq.heappop(D)
    ll=ld//(10**15)
    dd=ld%(10**15)
    if ll<l:
      continue
    else:
      break
  if ll>=l:
    heapq.heappush(D,r*10**15+dd+c)
  heapq.heappush(D,ll*10**15+dd)
print(D)
while D:
  ld=heapq.heappop(D)
  ll=ld//(10**15)
  dd=ld%(10**15)
  if ll==n:
    ans=min(ans,dd)
print(-1 if ans==float("inf") else ans)


  

# Fixed version:
import sys
input=sys.stdin.readline
import heapq
n,m=map(int,input().split())

D=[0]
P=[]
ans=float("inf")

for i in range(m):
  l,r,c=map(int,input().split())
  P.append((l,r,c))
P.sort(key=lambda x:x[0])
heapq.heappush(D,0*10**15+1)
for l,r,c in P:
  while D:
    ld=heapq.heappop(D)
    dd=ld//(10**15)
    ll=ld%(10**15)
    if ll<l:
      continue
    else:
      break
  if ll>=l:
    heapq.heappush(D,r+(dd+c)*10**15)
  heapq.heappush(D,ll+dd*10**15)
while D:
  ld=heapq.heappop(D)
  dd=ld//(10**15)
  ll=ld%(10**15)
  if ll==n:
    ans=min(ans,dd)
print(-1 if ans==float("inf") else ans)

# Buggy version:
# -*- coding: utf-8 -*-
import sys
from collections import defaultdict
from operator import itemgetter
from heapq import heappop, heappush
input = sys.stdin.readline
def inpl(): return list(map(int, input().split()))

N, M = inpl()
D = [0] + [10**18]*(N-1)
g0 = itemgetter(0)
Q = sorted([inpl() for _ in range(M)], key=g0, reverse=True)
#print(*Q, sep="\n")

HQ = [(0, 0)]

for i in range(N):
    while HQ[0][1] < i:
        heappop(HQ)
    v = HQ[0][0]
    while len(Q) and (i == (Q[-1][0]-1)):
        l, r, c = Q.pop()
        heappush(HQ, (v+c, r-1))


if v == 10**18:
    print(-1)
else:
    print(v)

# Fixed version:
# -*- coding: utf-8 -*-
import sys
from collections import defaultdict
from operator import itemgetter
from heapq import heappop, heappush
#input = sys.stdin.readline
def inpl(): return list(map(int, input().split()))

N, M = inpl()
D = [0] + [10**18]*(N-1)
g0 = itemgetter(0)
Q = sorted([inpl() for _ in range(M)], key=g0, reverse=True)
#print(*Q, sep="\n")

HQ = [(0, 0), (10**18, N)]

for i in range(N):
    while HQ[0][1] < i:
        heappop(HQ)
    v = HQ[0][0]
    while len(Q) and (i == (Q[-1][0]-1)):
        l, r, c = Q.pop()
        heappush(HQ, (v+c, r-1))


if v == 10**18:
    print(-1)
else:
    print(v)
You are given a positive even number N. Find the number of strings 's' of length N consisting of 'A', 'B', and 'C' that satisfy the condition: 's' can be converted to the empty string by erasing two consecutive characters, but 'AB' or 'BA' cannot be erased. Compute the count modulo 998244353. Constraints: 2 ≤ N ≤ 10^7. Input is given as N, and the output should be the number of strings that satisfy the conditions, modulo 998244353.

# Buggy version:

n=int(input())
pre=[1]*(n//2+200)
pp=[1]*(n+1)
mod=998244353
for i in range(1,n//2+100):
    pre[i]=((pre[i-1]*2)%mod)
    
p=[1]*(n+1)
for i in range(n):
    p[i+1]=((p[i]*(i+1))%mod)
    pp[i+1]=pow(p[i+1],mod-2,mod)
tot=1
for i in range(n):
    tot*=3
    tot%=mod
#print(1)
#print(tot)
cc=0
for i in range(n//2+1,n+1):
    c=(((((p[n]*pp[i])%mod)*pp[n-i])%mod)*pre[n-i])%mod
    c%=mod
    cc+=c
tot-=cc
tot-=cc
tot%=mod
print(tot)


# Fixed version:

n=int(input())
pre=[1]*(n//2+200)
pp=[1]*(n+1)
mod=998244353
for i in range(1,n//2+100):
    pre[i]=((pre[i-1]*2)%mod)
    
p=[1]*(n+1)
for i in range(n):
    p[i+1]=((p[i]*(i+1))%mod)
pp[-1] = pow(p[-1], mod - 2, mod)
 
for i in range(2, n + 1):
	pp[-i] = int((pp[-i + 1] * (n + 2 - i)) % mod)
tot=1
for i in range(n):
    tot*=3
    tot%=mod
#print(1)
#print(tot)
cc=0
for i in range(n//2+1,n+1):
    c=(((((p[n]*pp[i])%mod)*pp[n-i])%mod)*pre[n-i])%mod
    c%=mod
    cc+=c
tot-=cc
tot-=cc
tot%=mod
print(tot)
Title: Gossip

Problem:
There are n idols numbered from 1 to n lined up in a row. Each idol can pass information to the idol on their left or right in one unit of time. Idol 1 can only pass information to idol 2, and idol n can only pass information to idol n-1. At time 0, m idols hold secret information with numbers a1, a2, ..., am. Find the minimum time for all idols to receive the secret information.

Input:
The input consists of:
- Two integers n and m on the first line separated by a space.
- m integers a1, a2, ..., am on the second line separated by spaces.

Constraints:
- 2 ≤ n ≤ 10^5
- 1 ≤ m ≤ n
- 1 ≤ ai ≤ n
- All ai values are distinct and given in ascending order.

Output:
Output the minimum time for all idols to receive the information in a single line.

Sample Input 1:
3 2
1 3

Sample Output 1:
1

Sample Input 2:
10 3
2 5 7

Sample Output 2:
3

Sample Input 3:
10 5
2 5 6 8 10

Sample Output 3:
1

Sample Input 4:
100000 1
1

Sample Output 4:
99999

# Buggy version:
n, m = map(int, input().split())
a =[0] + list(map(int, input().split())) + [n]
t = max(a[0] - 1, n - a[-1])
for i in range(1, m):t = max(t, -((a[i - 1] - a[i]) // 2))
print(t)


# Fixed version:
n,m = map(int,input().split())
a = list(map(int,input().split()))
c = [a[0]-1, n-a[m-1]]
for i in range(m-1):
    c.append((a[i+1]-a[i])//2)
print(max(c))
ICPC Calculator

In mathematics, the order of operations is usually specified using parentheses. However, the International Counter of Parentheses Council (ICPC) aims to establish a notation without parentheses as the world standard. Dr. Tsukuba, a member of ICPC, has invented a new parenthesis-free notation. In this notation, a single expression is represented by multiple lines, each containing an addition operator (+), a multiplication operator (*), or an integer. Expressions may be arbitrarily nested, and each expression is given a nesting level. The top-level expression has a nesting level of 0. The task is to write a program that computes the value of expressions written in Dr. Tsukuba's notation. The input consists of multiple datasets, each starting with a line containing a positive integer n, followed by n lines denoting a single expression in Dr. Tsukuba's notation. The output should contain a single line for each dataset, with an integer representing the value of the given expression.

# Buggy version:
while True:
    n = int(input())
    if n == 0:break
    a = [input() for _ in range(n)]
    while True:
        if len(a) == 1:break
        c = 0
        s = 0
        for i in range(len(a)):
            if a[i].count('.') > c:
                c = a[i].count('.')
                s = i
        e = s
        while e < len(a) and a[e].count('.') == a[s].count('.'):
            e += 1
        k = a[s - 1].replace('.', '')
        b = [a[i] for i in range(s, e)]
        for i in range(len(b)):
            b[i] = int(b[i].replace('.', ''))
        if k == '+':
            a[s] = '.'*a[s].count('.')+str(sum(b))
            del a[s + 1:e]
        else:
            m = 1
            for i in b:
                m *= i
            a[s] = '.'*a[s].count('.')+str(m)
            del a[s + 1:e]
    print(a[0])
    


# Fixed version:
while True:
    n = int(input())
    if n == 0:break
    a = [input() for _ in range(n)]
    while True:
        if len(a) == 1:break
        c = 0
        s = 0
        for i in range(len(a)):
            if a[i].count('.') > c:
                c = a[i].count('.')
                s = i
        e = s
        while e < len(a) and a[e].count('.') == a[s].count('.'):
            e += 1
        k = a[s - 1].replace('.', '')
        b = [a[i] for i in range(s, e)]
        for i in range(len(b)):
            b[i] = int(b[i].replace('.', ''))
        if k == '+':
            a[s - 1] = '.'*a[s - 1].count('.')+str(sum(b))
            del a[s:e]
        else:
            m = 1
            for i in b:
                m *= i
            a[s - 1] = '.'*a[s - 1].count('.')+str(m)
            del a[s:e]
    print(a[0])
Dr. Tsukuba wants to compare the effectiveness of his new programming training method with the conventional one. He plans to conduct an experiment with two students, each trained under a different method. To ensure a fair comparison, he needs two students with the closest scores from a list of student scores. The task is to write a program that selects two students with the smallest score difference. The input consists of multiple datasets, each containing the number of students and their respective scores. The program should output the absolute difference in scores for each dataset.

# Buggy version:
while True:
    n = int(input())
    a = list(map(int, input().split()))
    l = list()

    for i in range(1, n-1):
        for j in range(i+1,n):
            l.append(abs(a[i]-a[j]))
    if l = []:
        break
    if n == 0:
        break
    print(min(l))



# Fixed version:
while True:
    n = int(input())
    if n==0: break
    l = list(map(int,input().split()))
    r = []
    for i in range(n):
        for j in range(i+1,n):
            r += [abs(l[i]-l[j])]
    print(min(r))
In TKB City, citizens are eager to know the winner of an election as soon as possible. The task is to create a program that, as votes are counted one by one, identifies the winner and determines when the winner is guaranteed. The input consists of datasets, each containing the number of votes and the votes themselves. The output should indicate the winner and the number of votes at which the winner is identified, unless there is a tie, in which case "TIE" should be output.

# Buggy version:
def secMax(a):
    mx = 0
    smx = 0
    for i in range(0,len(a)):
        if a[i] > mx:
            smx = mx
            mx = a[i]
        elif a[i] > smx:
            smx = a[i]
    else:
        return smx

while True:
    n = int(input())
    if n == 0:
        break
    c = list(map(str,input().split()))
    ##alphabet = char('a')##char型ねぇよって言われた(´・ω・`)
    alph = ["A","B","C","D","E","F","G","H",
            "I","J","K","L","M","N","O","P","Q",
            "R","S","T","U","V","W","X","Y","Z"]
    abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    for i in range(0, n):
        for j in range(0, 26):
            if c[i] == alph[j]:
                abcCount[j] = abcCount[j] + 1
                break
        print(abcCount)
        if max(abcCount) > n-1-i+secMax(abcCount):
            print(alph[abcCount.index(max(abcCount))], sum(abcCount))
            break
    else:
        print("TIE")



# Fixed version:
def secMax(a):
    mx = 0
    smx = 0
    for i in range(0,len(a)):
        if a[i] > mx:
            smx = mx
            mx = a[i]
        elif a[i] > smx:
            smx = a[i]
    else:
        return smx

while True:
    n = int(input())
    if n == 0:
        break
    c = list(map(str,input().split()))
    ##alphabet = char('a')##char型ねぇよって言われた(´・ω・`)
    alph = ["A","B","C","D","E","F","G","H",
            "I","J","K","L","M","N","O","P","Q",
            "R","S","T","U","V","W","X","Y","Z"]
    abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    for i in range(0, n):
        for j in range(0, 26):
            if c[i] == alph[j]:
                abcCount[j] = abcCount[j] + 1
                break
        #print(abcCount)
        if max(abcCount) > n-1-i+secMax(abcCount):
            print(alph[abcCount.index(max(abcCount))], sum(abcCount))
            break
    else:
        print("TIE")
Mr. Gardiner, a modern garden designer, plans ponds by first deciding their locations and then designing them with the terrain features intact. His ponds are rectangular with simple aspect ratios. He divides the land into cells of unit square and annotates each cell with its elevation. A pond occupies a rectangular area where the outermost cells are higher than the inner cells. The width and depth of the pond are at least three. The capacity of a pond is the maximum amount of water it can keep. Given a grid map describing the elevation of each unit square cell, the task is to calculate the largest possible capacity of a pond built in the site. The input consists of datasets, each containing the depth and width of the garden site, followed by the elevation values. The output should provide the largest possible capacity of a pond for each dataset, or zero if no ponds can be built.

# Buggy version:
while True:
    d, w = map(int, input().split())
    if d==w==0:
        break
    E=[list(map(int, input().split())) for i in range(d)]
    a=0
    for x2 in range(d):
        for y2 in range(w):
            for x1 in range(x2-1):
                for y1 in range(y2-1):
                    co=10**18
                    for x in range(x1, x2+1):
                        co=min(co, E[x][y1], E[x][y2])
                    for y in range(y1, y2+1):
                        co=min(co, E[x1][y], E[x2][y])
                    ci=cnt=0
                    for x in range(x1+1, x2):
                        for y in range(y1+1, y2):
                            ci=max(ci, E[x][y])
                            cnt+=E[x][y]
                    if ci < co:
                        a=max(ans, (x2-x1-1)*(y2-y1-1)*co - cnt)
    print(a)



# Fixed version:
while 1:
    d, w = map(int, input().split())
    if d == w == 0:
        break
    E = [list(map(int, input().split())) for i in range(d)]
    ans = 0
    for x2 in range(d):
        for y2 in range(w):
            for x1 in range(x2-1):
                for y1 in range(y2-1):
                    co = 10**18
                    for x in range(x1, x2+1):
                        co = min(co, E[x][y1], E[x][y2])
                    for y in range(y1, y2+1):
                        co = min(co, E[x1][y], E[x2][y])
                    ci = cnt = 0
                    for x in range(x1+1, x2):
                        for y in range(y1+1, y2):
                            ci = max(ci, E[x][y])
                            cnt += E[x][y]
                    if ci < co:
                        ans = max(ans, (x2-x1-1)*(y2-y1-1)*co - cnt)
    print(ans)
Shell Sort is a variation of Insertion Sort used to arrange a list of n elements A. The algorithm involves two functions: insertionSort(A, n, g) and shellSort(A, n). The latter calls the former, considering every g-th element. The task is to complete the program by filling in the missing values. The program should read an integer n and a sequence A, then print m, G_i (i = 0, 1, ..., m − 1), and the sorted sequence A. Constraints include 1 ≤ m ≤ 100, 0 ≤ G_i ≤ n, and cnt not exceeding ⌈n^1.5⌉. The input consists of an integer n followed by n lines of A_i values. The output includes m, G_i values, cnt, and the sorted A_i values. Constraints are 1 ≤ n ≤ 1,000,000 and 0 ≤ A_i ≤ 10^9.

# Buggy version:
def insertionSort(A,n,g,cnt):
	for i in range(g,n):
		v = A[i]
		j = i - g
		while j >= 0 and A[j] > v:
			A[j+g] = A[j]
			j = j - g
			cnt += 1
		A[j+g] = v
	return A,cnt

def shellSort(A,n):
	cnt = 0
	G = [1]
	while G[-1] <= n:
		G.append(G[-1]*3 + 1)
	G = G[::-1]
	m = len(G)
	for g in G:
		A,cnt = insertionSort(A,n,g,cnt)
	return A,m,G,cnt

if __name__ == '__main__':
	
	n = int(input())
	A = []
	[A.append(int(input())) for i in range(n)]
	
	'''
	data = [int(d) for d in data.split('\n')]
	n = data.pop(0)
	A = data[:]
	'''
	
	A,m,G,cnt = shellSort(A,n)
	print(m)
	print(' '.join([str(i) for i in G]))
	print(cnt)
	[print(a) for a in A]

# Fixed version:
def insertionSort(A,n,g,cnt):
	for i in range(g,n):
		v = A[i]
		j = i - g
		while j >= 0 and A[j] > v:
			A[j+g] = A[j]
			j = j - g
			cnt += 1
		A[j+g] = v
	return A,cnt

def shellSort(A,n):
	cnt = 0
	G = [1]
	while G[-1]*3+1 <= n:
		G.append(G[-1]*3 + 1)
	G = G[::-1]
	m = len(G)
	for g in G:
		A,cnt = insertionSort(A,n,g,cnt)
	return A,m,G,cnt

if __name__ == '__main__':
	
	n = int(input())
	A = []
	[A.append(int(input())) for i in range(n)]
	
	'''
	data = [int(d) for d in data.split('\n')]
	n = data.pop(0)
	A = data[:]
	'''
	
	A,m,G,cnt = shellSort(A,n)
	print(m)
	print(' '.join([str(i) for i in G]))
	print(cnt)
	[print(a) for a in A]
Title: Merge Sort

Write a program for the Merge Sort algorithm using the provided pseudocode. Report the number of comparisons in the Merge function.

Merge(A, left, mid, right):
  n1 = mid - left
  n2 = right - mid
  create arrays L[0...n1], R[0...n2]
  for i = 0 to n1-1:
    L[i] = A[left + i]
  for i = 0 to n2-1:
    R[i] = A[mid + i]
  L[n1] = SENTINEL
  R[n2] = SENTINEL
  i = 0
  j = 0
  for k = left to right-1:
    if L[i] <= R[j]:
      A[k] = L[i]
      i = i + 1
    else:
      A[k] = R[j]
      j = j + 1

Merge-Sort(A, left, right):
  if left+1 < right:
    mid = (left + right)/2
    call Merge-Sort(A, left, mid)
    call Merge-Sort(A, mid, right)
    call Merge(A, left, mid, right)

Input:
- The first line contains n.
- The second line contains n integers.

Output:
- Print the sequence S on the first line, with consecutive elements separated by a space.
- Print the number of comparisons on the second line.

Constraints:
- n ≤ 500000
- 0 ≤ an element in S ≤ 10^9

Sample Input 1:
10
8 5 9 2 6 3 7 1 10 4

Sample Output 1:
1 2 3 4 5 6 7 8 9 10
34

Notes:

# Buggy version:
def g(A,l,m,r):
 global c
 L=A[l:m]+[1e10]
 R=A[m:r]+[1e10]
 i=j=0
 for k in range(l,r):
  if L[i]<R[j]:A[k]=L[i];i+=1
  else:A[k]=R[j];j+=1
  c+=1
def s(A,l,r):
 if l+1<r:
  m=(l+r)//2
  s(A,l,m)
  s(A,m,r)
  g(A,l,m,r)
c=0
n=int(input())
A=list(map(int,input().split()))
s(A,0,n)
print(*A)


# Fixed version:
def g(A,l,m,r):
 global c
 L,R=A[l:m]+[1e10],A[m:r]+[1e10]
 i=j=0
 for k in range(l,r):
  if L[i]<R[j]:A[k]=L[i];i+=1
  else:A[k]=R[j];j+=1
  c+=1
def s(A,l,r):
 if l+1<r:
  m=(l+r)//2
  s(A,l,m);s(A,m,r)
  g(A,l,m,r)
c=0
n=int(input())
A=list(map(int,input().split()))
s(A,0,n)
print(*A)
print(c)

# Buggy version:
n = int(input())
S = list(map(int, input().split()))
SENTINEL = 10 ** 9 + 1
count = 0

def merge(A, left, mid, right):
    n1 = mid - left
    n2 = right - mid
    L = A[left:mid]
    R = A[mid:right]
    for i in range(n1):
        L[i] = A[left + i]
    for i in range(n2):
        R[i] = A[mid + i]
    L[n1] = r[n2] = SENTINEL
    i = j = 0
    for k in range(left, right):
        count += 1
        if(L[i] <= R[i]):
            A[k] = L[i]
            i += 1
        else:
            A[k] = R[j]
            j += 1

mergeSort(A, left, right):
    if(left + 1 < right):
        mid = (left + right) // 2
        mergeSort(A, left, mid)
        mergeSort(A, mid, right)
        merge(A, left, mid, right)

mergeSort(A, 0, n)
print(*A)
print(count)

# Fixed version:
INFTY = 10 ** 9
count = 0

def merge(A, left, mid, right):
    global count

    n1 = mid - left
    n2 = right - mid

    L = A[left:mid] + [INFTY]
    R = A[mid:right] + [INFTY]

    i = 0
    j = 0
    for k in range(left, right):
        if L[i] <= R[j]:
            A[k] = L[i]
            i += 1
        else:
            A[k] = R[j]
            j += 1
        
        count += 1


def mergeSort(A, left, right):
    if left + 1 < right:
        mid = (left + right) // 2
        mergeSort(A, left, mid)
        mergeSort(A, mid, right)
        merge(A, left, mid, right)



n = int(input())
S = list(map(int, input().split()))

mergeSort(S, 0, len(S))
print(*S)
print(count)