TnT/CodeNet4Repair huggingface dataset:
Problem:
You are given an integer N. Determine if there exists a tree with 2N vertices numbered 1 to 2N satisfying the following condition, and show one such tree if the answer is yes.    * Assume that, for each integer i between 1 and N (inclusive), Vertex i and N+i have the weight i. Then, for each integer i between 1 and N, the bitwise XOR of the weights of the vertices on the path between Vertex i and N+i (including themselves) is i.

Incorrect submission (verdict: Wrong Answer):
n = int(input())

if bin(n).count('1') == 1:
    print('No')
else:
    print('1 2')
    print('2 3')
    print('3 {0}'.format(n + 1))
    print('{0} {1}'.format(n + 1, n + 2))
    print('{0} {1}'.format(n + 2, n + 3))

    if n >= 5:
        if n % 2 == 1:
            for i in range(2, n + 1):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
        else:
            for i in range(2, n):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
            print('{0} {1}'.format(n - 1, n))
            print('{0} {1}'.format((n - 1) ^ 1 ^ n, n * 2))


Fixed, accepted submission:
n = int(input())

if bin(n).count('1') == 1:
    print('No')
else:
    print('Yes')
    print('1 2')
    print('2 3')
    print('3 {0}'.format(n + 1))
    print('{0} {1}'.format(n + 1, n + 2))
    print('{0} {1}'.format(n + 2, n + 3))

    if n >= 5:
        if n % 2 == 1:
            for i in range(4, n + 1):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
        else:
            for i in range(4, n):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
            print('{0} {1}'.format(n - 1, n))
            print('{0} {1}'.format((n - 1) ^ 1 ^ n, n * 2))
Problem:
You have an integer variable x. Initially, x=0.  Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`.  Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).

Incorrect submission (verdict: Wrong Answer):
n = int(input())
s = input()
x = 0
for i in range(n):
    if s[i] == "I":
        x += 1
    elif s[i] == "D":
        x -= 1
print(x)


Fixed, accepted submission:
n = int(input())
S = input()

x = 0
ans = x
for s in S:
    if s == 'I':
        x += 1
    else:
        x -= 1
    ans = max(ans, x)
print(ans)
Problem:
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.  You are given an integer N. Find the N-th Lucas number.  Here, the i-th Lucas number L_i is defined as follows:    * L_0=2   * L_1=1   * L_i=L_{i-1}+L_{i-2} (i≥2)

Incorrect submission (verdict: Wrong Answer):
n = int(input())
def lucas(n):
    a,b = 2,1
    if n == 1:
        return a
    elif n == 2:
        return b
    for _ in range(n-2):
        a,b = b,a+b
    return b
for i in range(1,10):
    print(lucas(i))

Fixed, accepted submission:
n = int(input())
def lucas(n):
    a,b = 2,1
    if n == 0:
        return a
    elif n == 1:
        return b
    for _ in range(n-1):
        a,b = b,a+b
    return b
print(lucas(n))
Problem:
There is a grid with H rows and W columns.  The square at the i-th row and j-th column contains a string S_{i,j} of length 5.  The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet.    Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location.  For example, the square at the 6-th row and 8-th column should be reported as `H6`.

Incorrect submission (verdict: Wrong Answer):
#Where's Snuke?
H, W = map(int, input().split())
ab = [chr(ord('A') + i) for i in range(26)]
for i in range(H):
    S = input().split()
    for j in range(W):
        if S[j] == "snuke":
            print(ab[i]+str(i+1)) 
            break
        break
        

Fixed, accepted submission:
#Where's Snuke?
H, W = map(int, input().split())
ab = [chr(ord('A') + i) for i in range(26)]
for i in range(H):
    S = input().split()
    for j in range(W):
        if S[j] == "snuke":
            print(ab[j]+str(i+1)) 
            break
Problem:
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.

Incorrect submission (verdict: Wrong Answer):
a,b,c = map(int,input().split(" "))
if (a < c and c < b) or (a > c and c > b):
    print("yes")
else:
    print("no")


Fixed, accepted submission:
a,b,c = map(int,input().split(" "))
if (a < c and c < b) or (a > c and c > b):
    print("Yes")
else:
    print("No")
Problem:
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b.  The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

Incorrect submission (verdict: Wrong Answer):
a, op, b = input().split()
a, b = int(a), int(b)
if op == '?':
  quit()
else:
  if op == '+':
    result = a+b
  elif op == '-':
    result = a-b
  elif op == '*':
    result = a*b
  elif op == '/':
    result = a/b
  print(result)

Fixed, accepted submission:
while True:
  a, op, b = input().split()
  a, b = int(a), int(b)
  if op == '?':
    quit()
  else:
    if op == '+':
      result = a+b
    elif op == '-':
      result = a-b
    elif op == '*':
      result = a*b
    elif op == '/':
      result = int(a/b)
    print(result)
Problem:
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once.  Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

Incorrect submission (verdict: Wrong Answer):
S = input()
for i in range(len(S)-7):
  S_cut = S[:i]+S[i+len(S)-7:]
  print(S_cut)
  if S_cut=='keyence':
    print('YES')
    exit()
print('NO')

Fixed, accepted submission:
S = input()
for i in range(len(S)-5):
  S_cut = S[:i]+S[i+len(S)-7:]
  #print(S_cut)
  if S_cut=='keyence':
    print('YES')
    exit()
print('NO')
Problem:
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i.  Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

Incorrect submission (verdict: Wrong Answer):
S=input()
S_half_front=int(S[:2])
S_half_back=int(S[2:])
S_bool=S_half_front is not S_half_back
if S_bool:
    print("No")
else:
    print("Yes")

Fixed, accepted submission:
S=input()
print(S.count("1"))
Problem:
Write a program which reads two integers x and y, and prints them in ascending order.

Incorrect submission (verdict: Wrong Answer):
x,y = input().split()
if x > y:
  print(y,x)
print(x,y)

Fixed, accepted submission:
while True:
    x,y = map(int,input().split())
    if x == y == 0:
        break
    if x > y:
        print(y,x)
        continue
    print(x,y)
Problem:
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode:                 for i = 1 to A.length-1     key = A[i]     /* insert A[i] into the sorted sequence A[0,...,j-1] */     j = i - 1     while j >= 0 and A[j] > key         A[j+1] = A[j]         j--     A[j+1] = key   Note that, indices for array elements are based on 0-origin.  To illustrate the algorithms, your program should trace intermediate result for each step.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
l=list(map(int, input().split()))
for i in range(n):
    l[0:i]=sorted(l[0:i])
    print(l)

Fixed, accepted submission:
n=int(input())
l=list(map(int, input().split()))
for i in range(1,n+1):
    l[0:i]=sorted(l[0:i])
    print(*l)
Problem:
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.  The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.  One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.  If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.    * X < Z \leq Y   * x_1, x_2, ..., x_N < Z   * y_1, y_2, ..., y_M \geq Z  Determine if war will break out.

Incorrect submission (verdict: Wrong Answer):
N, M, X, Y = map(int, input().split())
xlst = list(map(int, input().split()))
ylst = list(map(int, input().split()))

xmax = max(xlst)
ymin = min(ylst)
cand = []
for i in range(xmax+1, ymin+1):
    cand.append(i)
print(cand)
    
warflag = True
for c in cand:
    if X < c <= Y:
        warflag = False
if warflag:
    print('War')
else:
    print('No War')

Fixed, accepted submission:
N, M, X, Y = map(int, input().split())
xlst = list(map(int, input().split()))
ylst = list(map(int, input().split()))

xmax = max(xlst)
ymin = min(ylst)
cand = []
for i in range(xmax+1, ymin+1):
    cand.append(i)
    
warflag = True
for c in cand:
    if X < c <= Y:
        warflag = False
if warflag:
    print('War')
else:
    print('No War')
Problem:
Write a program which calculates the area and perimeter of a given rectangle.

Incorrect submission (verdict: Wrong Answer):
x,y=[int(i) for i in input().split()]
print(x*y)

Fixed, accepted submission:
x,y = [int(i) for i in input().split()]
area = x*y
length = (x*2)+(y*2)
print(area,length)
Problem:
地勢を示す縦 12, 横 12 のマスからなる平面図があります。おのおののマスは白か黒に塗られています。白は海を、黒は陸地を表します。二つの黒いマスが上下、あるいは左右に接しているとき、これらは地続きであるといいます。この平面図では、黒いマス一つのみ、あるいは地続きの黒いマスが作る領域を「島」といいます。例えば下図には、5 つの島があります。                 ■■■■□□□□■■■■ ■■■□□□□□■■■■ ■■□□□□□□■■■■ ■□□□□□□□■■■■ □□□■□□□■□□□□ □□□□□□■■■□□□ □□□□□■■■■■□□ ■□□□■■■■■■■□ ■■□□□■■■■■□□ ■■■□□□■■■□□□ ■■■■□□□■□□□□ □□□□□□□□□□□□   マスのデータを読み込んで、島の数を出力するプログラムを作成してください。

Incorrect submission (verdict: Wrong Answer):
def f1():
    t = 0
    for y, line in enumerate(M):
        for x, cell in enumerate(line):
            if M[y][x] == '1':
                f2(x, y)
                t += 1
    return t


def f2(x, y):
    if x < 0 or len(M[0]) == x or y < 0 or len(M) == y:
        return
    if M[y][x] == '1':
        M[y][x] = '0'
        f2(x, y-1)
        f2(x, y+1)
        f2(x+1, y)
        f2(x-1, y)


M = []
while True:
    try:
        line = input()
        if line == '':
            print(f1())
            M = []
        else:
            M.append(list(input()))
    except EOFError:
        print(f1())
        break

Fixed, accepted submission:
def f1():
    t = 0
    for y in range(len(M)):
        for x in range(len(M[0])):
            if M[y][x] == '1':
                f2(x, y)
                t += 1
    return t


def f2(x, y):
    if x < 0 or len(M[0]) == x or y < 0 or len(M) == y:
        return
    if M[y][x] == '1':
        M[y][x] = '0'
        f2(x, y-1)
        f2(x, y+1)
        f2(x+1, y)
        f2(x-1, y)


M = []
while True:
    try:
        line = input()
        if line == '':
            print(f1())
            M = []
        else:
            M.append(list(line))
    except EOFError:
        print(f1())
        break
Problem:
E869120 found a chest which is likely to contain treasure.   However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.   He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.  One more thing he found is a sheet of paper with the following facts written on it:    * Condition 1: The string S contains a string T as a contiguous substring.   * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.  Print the string S.   If such a string does not exist, print `UNRESTORABLE`.

Incorrect submission (verdict: Wrong Answer):
# time complexity: O(|S - T|*|T|)
# space complexity: O(|S - T|)

S_d = input()
T = input()
len_T = len(T)


def replacable(string, T):
    for char_s, char_t in zip(string, T):
        if char_s == char_t or char_s == '?':
            continue
        else:
            return False
    else:
        return True


# see if S_d can contain T
t_start_index_list = []
# O(|S - T|)
for start_idx in range(len(S_d) - len_T + 1):
    string = S_d[start_idx:start_idx + len_T]
    # O(|T|)
    if replacable(string, T):
        t_start_index_list.append(start_idx)

if len(t_start_index_list) == 0:
    print("UNRESTORABLE")
else:
    S_candidate_list = []
    for t_start_index in t_start_index_list:
        print(t_start_index)
        S = S_d[:t_start_index] + T + S_d[t_start_index + len_T:]
        S = S.replace('?', 'a')
        S_candidate_list.append(S)
    S_candidate_list.sort()
    print(S_candidate_list[0])


Fixed, accepted submission:
# time complexity: O(|S - T|*|T|)
# space complexity: O(|S - T|)

S_d = input()
T = input()
len_T = len(T)


def replacable(string, T):
    for char_s, char_t in zip(string, T):
        if char_s == char_t or char_s == '?':
            continue
        else:
            return False
    else:
        return True


# see if S_d can contain T
t_start_index_list = []
# O(|S - T|)
for start_idx in range(len(S_d) - len_T + 1):
    string = S_d[start_idx:start_idx + len_T]
    # O(|T|)
    if replacable(string, T):
        t_start_index_list.append(start_idx)

if len(t_start_index_list) == 0:
    print("UNRESTORABLE")
else:
    S_candidate_list = []
    for t_start_index in t_start_index_list:
        S = S_d[:t_start_index] + T + S_d[t_start_index + len_T:]
        S = S.replace('?', 'a')
        S_candidate_list.append(S)
    S_candidate_list.sort()
    print(S_candidate_list[0])
Problem:
Write a program which calculates the cube of a given integer x.

Incorrect submission (verdict: Wrong Answer):
def cubic(x):
    x = x^3
    return x

Fixed, accepted submission:
x = int(input())
y = x **3
print(y, end = "\n")
Problem:
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

Incorrect submission (verdict: Wrong Answer):
A, B, C = map(int, input().split())
print("YES" if A <= C and C <= B else "NO")

Fixed, accepted submission:
A, B, C = map(int, input().split())
print("Yes" if A <= C and C <= B else "No")
Problem:
We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square.  Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors.  Find the number of such ways to paint the dominoes, modulo 1000000007.  The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner:    * Each domino is represented by a different English letter (lowercase or uppercase).   * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

Incorrect submission (verdict: Wrong Answer):
import sys
n = int(input())
s = input()
#s = [input() for _ in range(2)]
print(len(s))
t = []

if (len(s)==1):
    print(3)
    sys.exit()
if (len(s)==2):
    print(6)
    sys.exit()
for i in range(len(s)-1):
    if (i>0):
        if(s[i-1]==s[i]):
            if (i == len(s)-2):
                t.append(1)
                continue
            else:
                continue
    
    if (s[i]==s[i+1]):
        t.append(2)
    else:
        t.append(1)
print(t)
if (t[0]==1):
    ans = 3
else:
    ans = 6
for i in range(len(t)-1):
    print(ans)
    if (t[i]==1):
        if (t[i+1]==1):
            ans *= 2
        else:
            ans *= 2
    else:
        if (t[i+1]==1):
            ans *= 1
        else:
            ans *= 3
print(ans % 1000000007)

Fixed, accepted submission:
import sys
n = int(input())
s = input()
t = []

if (len(s)==1):
    print(3)
    sys.exit()
if (len(s)==2):
    print(6)
    sys.exit()

for i in range(len(s)-1):
    if (i>0):
        if(s[i-1]==s[i]):
            if (i == len(s)-2):
                t.append(1)
                continue
            else:
                continue
        elif (i == len(s)-2 and s[i]!=s[i+1]):
            t.append(1)
            
    
    if (s[i]==s[i+1]):
        t.append(2)
    else:
        t.append(1)

if (t[0]==1):
    ans = 3
else:
    ans = 6

for i in range(len(t)-1):
    
    if (t[i]==1):
        if (t[i+1]==1):
            ans *= 2
        else:
            ans *= 2
    else:
        if (t[i+1]==1):
            ans *= 1
        else:
            ans *= 3
print(ans % 1000000007)
Problem:
Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7.  Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation:    * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward.  For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back.    Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.

Incorrect submission (verdict: Wrong Answer):
x=int(input())

ans=x/11

if x%11==0:
    print(ans)
elif x%11<=6:
    print(ans+1)
else:
    print(ans+2)

Fixed, accepted submission:
x=int(input())

ans=int(x/11)*2

if x%11==0:
    print(ans)
elif x%11<=6:
    print(ans+1)
else:
    print(ans+2)
Problem:
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`.  Consider doing the following operation:    * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.  You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Incorrect submission (verdict: Wrong Answer):
h, w, k = map(int, input().split())

c = [list(input()) for _ in range(h)]

ans = 0

#make binary numbers ranging from 0 to 2**h -1 or 2**w -1
for row in range(2**h):
    for column in range(2**w):
        count = 0
        # Neither ith digit of row nor jth digit of column is 1 and c[i][j] is black.
        for i in range(h):
            for j in range(w):
                if ((row << i)&1) == 0 and ((column << j)&1) == 0 and c[i][j] == '#':
                    count += 1
        if count == k:
            ans += 1

print(ans)

Fixed, accepted submission:
h, w, k = map(int, input().split())

c = [list(input()) for _ in range(h)]

ans = 0

#make binary numbers ranging from 0 to 2**h -1 or 2**w -1
for row in range(2**h):
    for column in range(2**w):
        count = 0
        # Neither ith digit of row nor jth digit of column is 1 and c[i][j] is black.
        for i in range(h):
            for j in range(w):
                if ((row >> i)&1) == 0 and ((column >> j)&1) == 0 and c[i][j] == '#':
                    count += 1
        if count == k:
            ans += 1

print(ans)
Problem:
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?  Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

Incorrect submission (verdict: Wrong Answer):
a = int(input())
b = int(input())
c = int(input())
x = int(input())

total = 0
ans = 0
for i in range(a):
    for j in range(b):
        for k in range(c):
            total = 500*a+100*b+50*c
            if total==x:
                ans += 1
print(ans)


Fixed, accepted submission:
a = int(input())
b = int(input())
c = int(input())
x = int(input())

total = 0
ans = 0
for i in range(a+1):
    for j in range(b+1):
        for k in range(c+1):
            total = 500*i+100*j+50*k
            if total==x:
                ans += 1
print(ans)
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Incorrect submission (verdict: Wrong Answer):
a,b,c = [int(i) for i in input().split()]

if (a+b==c) or (b+c==a) or (c+a==b):
  print("YES")
else:
  print("NO")

Fixed, accepted submission:
a,b,c = [int(i) for i in input().split()]

if (a+b==c) or (b+c==a) or (c+a==b):
  print("Yes")
else:
  print("No")
Problem:
Joisino the magical girl has decided to turn every single digit that exists on this world into 1.  Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points).  She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive).  You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows:    * If A_{i,j}≠-1, the square contains a digit A_{i,j}.   * If A_{i,j}=-1, the square does not contain a digit.  Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.

Incorrect submission (verdict: Wrong Answer):
h,w = map(int,input().split())
C = [list(map(int,input().split())) for i in range(10)]
A = [list(map(int,input().split())) for i in range(h)]
INF = float('inf')
cost = 0
D = [[INF]*10 for i in range(10)]
for k in range(10):
    for i in range(10):
        for j in range(10):
            D[i][j] = min(C[i][j],C[i][k]+C[k][j])

for y in range(h):
    for x in range(w):
        if A[y][x] == -1 or A[y][x] == 1:
            continue
        else:
            cost += D[A[y][x]][1]
print(cost)


Fixed, accepted submission:
from scipy.sparse.csgraph import floyd_warshall as wf
h,w = map(int,input().split())
C = [list(map(int,input().split())) for i in range(10)]
A = [list(map(int,input().split())) for i in range(h)]
C = wf(C)
cost = 0
for y in range(h):
    for x in range(w):
        if A[y][x] != -1:
            cost += C[A[y][x]][1]
print(int(cost))