TnT/CodeNet4Repair huggingface dataset:
Problem:
We ask you to select some number of positive integers, and calculate the sum of them.  It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.  Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.  If the objective is achievable, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
A,B,C = map(int,input().split())
for i in range(B):
  if (i+1)*A%B== C:
    print('Yes')
    exit()
print('No')


Fixed, accepted submission:
A,B,C = map(int,input().split())
for i in range(B):
  if (i+1)*A%B== C:
    print('YES')
    exit()
print('NO')
Problem:
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room.  For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left.  Assume that initially no person lives in the building.

Incorrect submission (verdict: Wrong Answer):
nums = [[[0 for x in range(0, 10)] for x in range(0, 3)] for x in range(0, 4)]

input1='''1 1 3 8
3 2 2 7
4 3 8 1'''.split("\n")
print(input1)

for e in input1:
    efbv =  e.split(" ")
    e_b, e_f, e_r, e_v = map(int, efbv)
    nums[e_b - 1][e_f - 1][e_r -1] += e_v

for b in range(4):
    for f in range (3):
        print(" " + " ".join(map(str, nums[b][f])))
    if b != 3: print("#" * 20)

Fixed, accepted submission:
buildings =  [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for _ in range(n):
    b, f, r, v = map(int, input().split())

    buildings[b-1][f-1][r-1] += v
sep_count = len(buildings) - 1
for b in buildings:
    for f in b:
        print(" ",end="")
        print(*f)
    if sep_count != 0:
        print("#" * 20)
        sep_count -= 1
Problem:
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.

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

from itertools import product
tot_comb = [el for el in product(list(range(-120, 121)), list(range(-120, 121))) if el[0]!=el[1]]
tot_comb += [(0,0)]
num_dict = {el[0]**5 - el[1]**5: el for el in tot_comb}
print(num_dict[x])

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

from itertools import product
tot_comb = [el for el in product(list(range(-120, 121)), list(range(-120, 121))) if el[0]!=el[1]]
tot_comb += [(0,0)]
num_dict = {el[0]**5 - el[1]**5: el for el in tot_comb}
pair = num_dict[x]
print(pair[0], pair[1])
Problem:
There are N positive integers written on a blackboard: A_1, ..., A_N.  Snuke can perform the following operation when all integers on the blackboard are even:    * Replace each integer X on the blackboard by X divided by 2.  Find the maximum possible number of operations that Snuke can perform.

Incorrect submission (verdict: Wrong Answer):
N=int(input())
suji=input()
I1=suji.split(" ")
I=list(map(int,I1))
wari=2
n=0
kazu=0
X=1
while n<3:
    if I[n]%wari==0:
        n+=1
        if I[N-1]%2==0:
            wari=wari*2
            kazu+=1
            n=0
    else:
        X=0
        break
print(kazu)

Fixed, accepted submission:
import math
import numpy as np

N = int(input())
A_list = list(map(int, input().split()))
np_A = np.array(A_list)
tmp_cou = 0
x = 1
while x==1:
    for i in np_A:
        if i % 2 == 0:
            tmp_cou += 1
        else:
            x = 2
    np_A = np_A/2

cou = math.floor(tmp_cou/N)
print(cou)
Problem:
Write a program which calculates the cube of a given integer x.

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

print(x ^ 3)

Fixed, accepted submission:
x = input()
x = int(x)
print(x ** 3)
Problem:
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives.  You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative.  For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins.  If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

Incorrect submission (verdict: Wrong Answer):
n=int(input())
c=0
for i in range(n):
    x,u=input().split()
    if u=='JPY':
        c+=int(x)
    elif u=='BTC':
        c+=int(380000 * float(x))
print(c)

Fixed, accepted submission:
n=int(input())
c=0
for i in range(n):
    x,u=input().split()
    if u=='JPY':
        c+=int(x)
    elif u=='BTC':
        c+=380000 * float(x)
print(c)
Problem:
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:    * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

Incorrect submission (verdict: Wrong Answer):
s=input()

s=s.replace("eraser","")
s=s.replace("erase","")
s=s.replace("dreamer","")
s=s.replace("dream","")

if s=="":
    print("Yes")
else:
    print("No")


Fixed, accepted submission:
s=input()

s=s.replace("eraser","")
s=s.replace("erase","")
s=s.replace("dreamer","")
s=s.replace("dream","")

if s=="":
    print("YES")
else:
    print("NO")
Problem:
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi:    * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person.  Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.

Incorrect submission (verdict: Wrong Answer):
a, b, k = map(int, input().split())
for i in range(k):
    if i%2 == 0:
        if a%2 == 0:
            a /= 2
            b += 2
        elif a%2 == 1:
            a -= 1
            a /= 2
            b += a
    elif i%2 == 1:
        if b%2 == 0:
            b /= 2
            a += b
        elif a%2 == 1:
            b -= 1
            b /= 2
            a += 2
    print(int(a),int(b))
print(int(a),int(b))


Fixed, accepted submission:
a, b, k = map(int, input().split())
for i in range(k):
    if i%2 == 0:
        if a%2 == 0:
            a /= 2
            b += a
        elif a%2 == 1:
            a -= 1
            a /= 2
            b += a
    elif i%2 == 1:
        if b%2 == 0:
            b /= 2
            a += b
        elif b%2 == 1:
            b -= 1
            b /= 2
            a += b
print(int(a),int(b))
Problem:
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

Incorrect submission (verdict: Wrong Answer):
n,m=map(int,input().split())
roads=[]
for i in range(m):
  roads.append(input().split())
count=0
for j in range(1,n+1):
  for k in range(m):
    if roads[k][0]==j or roads[k][1]==j:
      count+=1
  print(count)
  count=0

Fixed, accepted submission:
n,m=map(int,input().split())
roads=[]
for i in range(m):
  roads.append(input().split())
count=0
for j in range(1,n+1):
  for k in range(m):
    if int(roads[k][0])==j or int(roads[k][1])==j:
      count+=1
      #print('road')
  print(count)
  count=0
Problem:
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.   The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.  An edge whose removal disconnects the graph is called a _bridge_.   Find the number of the edges that are bridges among the M edges.

Incorrect submission (verdict: Wrong Answer):
n, m = map(int, input().split(" "))
a = [list(map(int, input().split(" "))) for i in range(m)]

a_counts = [0]*n
for a_i in sum(a, []):
	a_counts[a_i-1] += 1

bridge_count = 0
for i, count in enumerate(a_counts):
	if count == 1:
		p = i+1
		while a_counts[p-1] <= 2: 
			for j, a_i in enumerate(a):
				if a_i and p in a_i and a_counts[p-1] <= 2:
					print(a_i)
					a_i.remove(p)
					print(a_i)
					p = a_i[0]
					a[j] = None
					bridge_count += 1
			break
			
print(bridge_count)

Fixed, accepted submission:
n, m = map(int, input().split(" "))
a = [list(map(int, input().split(" "))) for i in range(m)]

a_connects = [[] for i in range(n)]
for x, y in a:
	a_connects[x-1].append(y-1)
	a_connects[y-1].append(x-1)

a_connect_sums = [len(connect) for connect in a_connects]
bridge_count = 0
while 1 in a_connect_sums:
	x = a_connect_sums.index(1)
	y = a_connects[x][0]

	a_connects[x].remove(y)
	a_connects[y].remove(x)

	a_connect_sums[x] -= 1
	a_connect_sums[y] -= 1

	bridge_count += 1

print(bridge_count)
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())
Xpoints=list(map(int, input().split()))
Ypoints=list(map(int, input().split()))

for z in range(-100, 101):
    if X < z and z <= Y:
        if max(Xpoints) < z:
            if min(Ypoints) >= z:
                print('No War')
else:
    print('War')


Fixed, accepted submission:
N,M,X,Y=map(int, input().split())
Xpoints=list(map(int, input().split()))
Ypoints=list(map(int, input().split()))
isWar='War'
for z in range(-99, 101):
    if X < z and z <= Y:
        if max(Xpoints) < z:
            if min(Ypoints) >= z:
                isWar = 'No War'
                break

print(isWar)
Problem:
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:    * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

Incorrect submission (verdict: Wrong Answer):
def main(s):
    t = ""
    e = ["dream","erase","eraser","dreamer"]
    while len(t) < len(s) :
        val = ""
        c = t
        
        for i in e:
            val = t + i
            if s[:len(val)] != val:
                continue
            
            else:
                t = val
                break
        
        if s == t:
            return True
        
        if t == c:
            return False


s = input()
print("Yes" if main(s) else "No")

Fixed, accepted submission:
def main(s):
    e = ["dreamer","eraser","dream","erase"]
    
    t = [""]
    
    while len(t) > 0:
        val = t.pop(0)
        for i in e:
            tar = val + i
            
            if s[:len(tar)] == tar:
                if s == tar:
                    return True
                else:
                    t.append(tar)
    return False

s = input()
print("YES" if main(s) else "NO")
Problem:
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.  To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Incorrect submission (verdict: Wrong Answer):
a = input().split()
five = a.count(5)
seven = a.count(7)

if five == 2 and seven == 1:
  print('YES')
else:
  print('NO')

Fixed, accepted submission:
a = input().split()
five = a.count('5')
seven = a.count('7')

if five == 2 and seven == 1:
  print('YES')
else:
  print('NO')
Problem:
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N.  There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i.  Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.  Help him.

Incorrect submission (verdict: Wrong Answer):
import math

def solve(n, m, a, b):
    islands = []

    for i in range(m):
        if(a[i] == 1):
            islands.append(b[i])
        if(b[i] == 1):
            islands.append(a[i])

    for i in range(m):
        if(a[i] == m and b[i] in islands):
            return "POSSIBLE"
        if(b[i] == m and a[i] in islands):
            return "POSSIBLE"
    return "IMPOSSIBLE"

(n, m) = map(int, input().split())
a = []
b = []

for i in range(m):
    (ai, bi) = map(int, input().split())
    a.append(ai)
    b.append(bi)

print(solve(n, m, a, b))


Fixed, accepted submission:
def solve(n, m, a, b):
    islands = set()

    for i in range(m):
        if(a[i] == 1):
            islands.add(b[i])
        if(b[i] == 1):
            islands.add(a[i])

    for i in range(m):
        if(a[i] == n and b[i] in islands):
            return "POSSIBLE"
        if(b[i] == n and a[i] in islands):
            return "POSSIBLE"
    return "IMPOSSIBLE"

(n, m) = map(int, input().split())
a = []
b = []

for i in range(m):
    (ai, bi) = map(int, input().split())
    a.append(ai)
    b.append(bi)

print(solve(n, m, a, b))
Problem:
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes:    * Hit his pocket, which magically increases the number of biscuits by one.   * Exchange A biscuits to 1 yen.   * Exchange 1 yen to B biscuits.  Find the maximum possible number of biscuits in Snuke's pocket after K operations.

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

if b-a < 2 or a > k - 2:
    print(k)
else:
    
    num = a 
    c = k - a
    
    if c % 2 != 0:
        c -= 1
        num += 1
    
    while c > 0:
        c -= 2
        num += b - a
        
    print(num)

Fixed, accepted submission:
k, a, b = map(int, input().split())

if b - a < 2 or a > k - 1:
    print(k + 1)
else:
    
    num = a
    c = k - a + 1

    if c % 2 != 0:
        c -= 1
        num += 1

    num += c//2*(b-a)

    print(num)
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())
if a <= b <= c:
  print("Yes")
else:
  print("No")
  

Fixed, accepted submission:
a,b,c = map(int,input().split())
if a<=c<=b:
  print("Yes")
else:
  print("No")
Problem:
Write a program which reads two integers a and b, and calculates the following values:    * a ÷ b: d (in integer)   * remainder of a ÷ b: r (in integer)   * a ÷ b: f (in real number)

Incorrect submission (verdict: Wrong Answer):
a, b = input().split()
a = int(a)
b = int(b)

print(a//b, a%b, a/b)

Fixed, accepted submission:
a, b = input().split()
a = int(a)
b = int(b)
 
print(a//b, a%b, "{0:.5f}".format(a/b))
Problem:
An elementary school student Takahashi has come to a variety store.  He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?  Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

Incorrect submission (verdict: Wrong Answer):
a,b,c = map(int,input().split())
if a + b >= c:
    print("YES")
else:
    print("No")

Fixed, accepted submission:
a,b,c = map(int,input().split())
if a + b >= c:
    print("Yes")
else:
    print("No")
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<B and B<C:
  print("Yes")
elif C<B and B<A:
  print("Yes")
else:
  print("No")


Fixed, accepted submission:
A,C,B=map(int,input().split())
if A<B and B<C:
  print("Yes")
elif C<B and B<A:
  print("Yes")
else:
  print("No")
Problem:
平面上の異なる 4 点 $A (x_a, y_a)$, $B (x_b, y_b)$, $C (x_c, y_c)$, $D(x_d, y_d)$ の座標を読み込んで、それら 4 点を頂点とした四角形 $ABCD$ に凹みがなければ YES、凹みがあれば NO と出力するプログラムを作成してください。  凹みのある四角形とは図 1 のような四角形です。  

Incorrect submission (verdict: Wrong Answer):
def judge(p1,p2,p3,p4):
    t1 = (p1[0] - p3[0]) * (p2[1] - p1[1]) + (p1[1] - p3[1]) * (p1[0] - p2[0])
    t2 = (p1[0] - p3[0]) * (p4[1] - p1[1]) + (p1[1] - p3[1]) * (p1[0] - p4[0])
    t3 = (p2[0] - p4[0]) * (p1[1] - p2[1]) + (p2[1] - p4[1]) * (p2[0] - p1[0])
    t4 = (p2[0] - p4[0]) * (p3[1] - p2[1]) + (p2[1] - p4[1]) * (p2[0] - p3[0])
    return t1*t2<0 and t3*t4<0
while True:
	try:
		x1,y1,x2,y2,x3,y3,x4,y4=map(float, input().split(","))
		A=[x1,y1]
		B=[x2,y2]
		C=[x3,y3]
		D=[x4,y4]
		print(A,B,C,D)
		if judge(A,B,C,D)==True or judge(A,D,B,C)==True :
			print("YES")
		else:
			print("NO")
	except EOFError:
		break


Fixed, accepted submission:
def judge(p1,p2,p3,p4):
    t1 = (p3[0] - p4[0]) * (p1[1] - p3[1]) + (p3[1] - p4[1]) * (p3[0] - p1[0])
    t2 = (p3[0] - p4[0]) * (p2[1] - p3[1]) + (p3[1] - p4[1]) * (p3[0] - p2[0])
    t3 = (p1[0] - p2[0]) * (p3[1] - p1[1]) + (p1[1] - p2[1]) * (p1[0] - p3[0])
    t4 = (p1[0] - p2[0]) * (p4[1] - p1[1]) + (p1[1] - p2[1]) * (p1[0] - p4[0])
    return t1*t2>0 and t3*t4>0 
while True:
	try:
		x1,y1,x2,y2,x3,y3,x4,y4=map(float, input().split(","))
		A=[x1,y1]
		B=[x2,y2]
		C=[x3,y3]
		D=[x4,y4]
		if judge(A,B,C,D)==False :
			print("NO")
		else:
			print("YES")
	except EOFError:
		break
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,b,c,x = open(0)
d=0
x = int(x)

for i in range(int(a)+1):
  for j in range(int(b)+1):
    for k in range(int(c)+1):
      500*i + 100*j + 50*k == x
      d += 1
      
print(d)

      

Fixed, accepted submission:
a,b,c,x = open(0)
d=0
x = int(x)
 
for i in range(int(a)+1):
  for j in range(int(b)+1):
    for k in range(int(c)+1):
      if  500*i + 100*j + 50*k == x:
        d += 1
     
      
print(d)
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

Incorrect submission (verdict: Wrong Answer):
a, b = map(str, input().split())
n = int(a+b)
m = n**0.5

if int(m)-m == 0:
    print('Yes')
else:
    print('No')

Fixed, accepted submission:
a, b = map(int, input().split())
if(a*b%2 == 0 ):
  print('Even')
else:
  print('Odd')
Problem:
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank.  The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.)  Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

Incorrect submission (verdict: Wrong Answer):
import math

X = int(input())
money = 100
year = 0

while(money <= X):
    money += math.floor(money * 0.01)
    year += 1

print(year)

Fixed, accepted submission:
import math

X = int(input())
money = 100
year = 0

while(money < X):
    money += math.floor(money * 0.01)
    year += 1

print(year)
Problem:
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

Incorrect submission (verdict: Wrong Answer):
S = list(input())
N = len(S)
T = list(input())
M = len(T)
S.sort()
T.sort()
mmm = min(N, M)

for i in range(mmm):
    if S[i] > T[i]:
        print("No")
        exit()
    elif S[i] < T[i]:
        print("Yes")
        exit()
    else:
        continue

if N < M:
    print("Yes")
else:
    print("No")


Fixed, accepted submission:
S = list(input())
N = len(S)
T = list(input())
M = len(T)
S.sort()
T.sort(reverse = True)
mmm = min(N, M)

for i in range(mmm):
    if S[i] > T[i]:
        print("No")
        exit()
    elif S[i] < T[i]:
        print("Yes")
        exit()
    else:
        continue

if N < M:
    print("Yes")
else:
    print("No")
Problem:
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.  Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Incorrect submission (verdict: Wrong Answer):
n = int(input())
mochi = []
for i in range(n):
    mochi.append(int(input()))

print (mochi)

mochi_use = sorted(mochi,reverse = True)

ans = 1

for i in range(n-1):
    if mochi_use[i] > mochi_use[i+1]:
        ans += 1

print(ans)


Fixed, accepted submission:
n = int(input())
mochi = []
for i in range(n):
    mochi.append(int(input()))

mochi_use = sorted(mochi,reverse = True)

ans = 1

for i in range(n-1):
    if mochi_use[i] > mochi_use[i+1]:
        ans += 1

print(ans)
Problem:
E869120 has A 1-yen coins and infinitely many 500-yen coins.   Determine if he can pay exactly N yen using only these coins.

Incorrect submission (verdict: Wrong Answer):
#88

N = int(input())
A = int(input())

if N%500 >= A:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
#88

N = int(input())
A = int(input())

if N%500 <= A:
    print("Yes")
else:
    print("No")
Problem:
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.

Incorrect submission (verdict: Wrong Answer):
print(set(map(chr,range(97,123)))-set(input())or"None")

Fixed, accepted submission:
letters="abcdefghijklmnopqrstuvwxyz"
ans=sorted(set(letters)^set(input()))
print("None" if len(ans) == 0 else ans[0])
Problem:
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

Incorrect submission (verdict: Wrong Answer):
n,m = map(int, input().split())
ab = []
for _ in range(m):
    ab.append(list(map(int, input().split())))
ans = {}
for i in range(1, m+1):
    ans.update({i : 0})
for i in range(m):
    for j in range(2):
        if ab[i][j] in ans:
            ans[ab[i][j]] += 1
for v in ans.values():
    print(v)


Fixed, accepted submission:
n,m = map(int, input().split())
ab = []
for _ in range(m):
    ab.append(list(map(int, input().split())))
ans = {}
for i in range(1, n+1):
    ans.update({i : 0})
for i in range(m):
    for j in range(2):
        if ab[i][j] in ans:
            ans[ab[i][j]] += 1
for v in ans.values():
    print(v)