Problem:


Score : 600 points

Problem Statement

There is a game that involves three variables, denoted A, B, and C.

As the game progresses, there will be N events where you are asked to make a choice.
Each of these choices is represented by a string s_i. If s_i is AB, you must add 1 to A or B then subtract 1 from the other; if s_i is AC, you must add 1 to A or C then subtract 1 from the other; if s_i is BC, you must add 1 to B or C then subtract 1 from the other.

After each choice, none of A, B, and C should be negative.

Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Constraints

1 \leq N \leq 10^5

0 \leq A,B,C \leq 10^9

N, A, B, C are integers.

s_i is AB, AC, or BC.

Input

Input is given from Standard Input in the following format:

N A B C
s_1
s_2
:
s_N

Output

If it is possible to make N choices under the condition, print Yes; otherwise, print No.

Also, in the former case, show one such way to make the choices in the subsequent N lines. The (i+1)-th line should contain the name of the variable (A, B, or C) to which you add 1 in the i-th choice.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A



Sample input:
2 1 3 0
AB
AC


Sample output:
Yes
A
C

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	N,A,B,C = inputMap()
	stat = [0,0,0]
	Q = []
	for i in range(N):
		s = inputStr()
		Q.append(s)
		if s == "AB":
			stat[0] += 1
			stat[1] += 1
		elif s == "AC":
			stat[0] += 1
			stat[2] += 1
		elif s == "BC":
			stat[1] += 1
			stat[2] += 1

	#print(stat)

	hon = "Yes"
	ans = []
	for i in Q:
		if s == "AB":
			if A <= B:
				B -= 1
				A += 1
				ans.append("A")
			else:
				B += 1
				A -= 1
				ans.append("B")
			if A < 0 or B < 0:
				hon = "No"
				break
		elif s == "AC":
			if A <= C:
				C -= 1
				A += 1
				ans.append("A")
			else:
				C += 1
				A -= 1
				ans.append("C")
			if A < 0 or C < 0:
				hon = "No"
				break
		elif s == "BC":
			if B <= C:
				C -= 1
				B += 1
				ans.append("B")
			else:
				C += 1
				B -= 1
				ans.append("C")
			if C < 0 or B < 0:
				hon = "No"
				break

	if hon == "No":
		print(hon)
	else:
		print(hon)
		for i in ans:
			print(i)


#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
import sys,bisect
input = sys.stdin.readline

n,a,b,c = map(int,input().split())

ab = []
ac = []
bc = []


s = [0]*n
for i in range(n):
    inp = input()
    #print(inp)
    if inp == "AB\n":
        ab.append(i)
    elif inp == "AC\n":
        ac.append(i)
        s[i] = 1
    else:
        bc.append(i)
        s[i] = 2

res = []
for i in range(n):
    if s[i] == 0:
        if a == b:
            if a == 0:
                print("NO")
                exit()
            p = bisect.bisect_left(ac,i)
            q = bisect.bisect_left(bc,i)
            if p >= len(ac):
                a -= 1
                b += 1
                res.append("B")
            elif q >= len(bc):
                a += 1
                b -= 1
                res.append("A")
            else:
                if ac[p] > bc[q]:
                    a -= 1
                    b += 1
                    res.append("B")
                else:
                    a += 1
                    b -= 1
                    res.append("A")
        elif a > b:
            a -= 1
            b += 1
            res.append("B")
        else:
            a += 1
            b -= 1
            res.append("A")
    elif s[i] == 1:
        if a == c:
            if a == 0:
                print("NO")
                exit()
            p = bisect.bisect_left(ab,i)
            q = bisect.bisect_left(bc,i)
            if p >= len(ab):
                a -= 1
                c += 1
                res.append("C")
            elif q >= len(bc):
                a += 1
                c -= 1
                res.append("A")
            else:
                if ab[p] > bc[q]:
                    a -= 1
                    c += 1
                    res.append("C")
                else:
                    a += 1
                    c -= 1
                    res.append("A")
        elif a > c:
            a -= 1
            c += 1
            res.append("C")
        else:
            a += 1
            c -= 1
            res.append("A")
    elif s[i] == 2:
        if b == c:
            if b == 0:
                print("NO")
                exit()
            p = bisect.bisect_left(ab,i)
            q = bisect.bisect_left(ac,i)
            if p >= len(ab):
                b -= 1
                c += 1
                res.append("C")
            elif q >= len(ac):
                b += 1
                c -= 1
                res.append("B")
            else:
                if ab[p] > ac[q]:
                    b -= 1
                    c += 1
                    res.append("C")
                else:
                    b += 1
                    c -= 1
                    res.append("B")
        elif b > c:
            b -= 1
            c += 1
            res.append("C")
        else:
            b += 1
            c -= 1
            res.append("B")   

print("YES")
print(*res,sep='\n')

Submission status: Wrong Answer

Solution code:
import sys
from math import ceil, floor, sqrt, sin, cos, pi
from itertools import accumulate, permutations, combinations
from fractions import gcd # 最大公約数
from collections import deque, Counter
from operator import itemgetter
from heapq import heappop,heappush
sys.setrecursionlimit(10**7)
def lcm(x, y): return ((x * y) // gcd(x, y)) # 最小公倍数
# list(map(int, input().split()))
n, a, b, c = map(int, input().split())
ans = []
cnt_ab = 0
cnt_ac = 0
cnt_bc = 0
s = []
for _ in range(n):
    x = input()
    if x == 'AB':
        cnt_ab += 1
        s.append(0)
    elif x == 'AC':
        cnt_ac += 1
        s.append(1)
    else:
        cnt_bc += 1
        s.append(2)
for x in s:
    if x == 0:
        if a > b:
            a -= 1
            b += 1
            ans.append('B')
        elif a < b:
            a += 1
            b -= 1
            ans.append('A')
        else:
            if cnt_ac > cnt_bc:
                b -= 1
                a += 1
                ans.append('A')
            else:
                a -= 1
                b += 1
                ans.append('B')
        cnt_ab -= 1
    elif x == 1:
        if a > c:
            a -= 1
            c += 1
            ans.append('C')
        elif a < c:
            a += 1
            c -= 1
            ans.append('A')
        else:
            if cnt_ab > cnt_bc:
                c -= 1
                a += 1
                ans.append('A')
            else:
                a -= 1
                c += 1
                ans.append('C')
        cnt_ac -= 1
    else:
        if b > c:
            b -= 1
            c += 1
            ans.append('C')
        elif b < c:
            b += 1
            c -= 1
            ans.append('B')
        else:
            if cnt_ac > cnt_ab:
                c -= 1
                b += 1
                ans.append('B')
            else:
                b -= 1
                c += 1
                ans.append('C')
        cnt_bc -= 1
    if a < 0 or b < 0 or c < 0:
        print('No')
        sys.exit()
print('Yes')
for x in ans:
    print(x)

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	N,A,B,C = inputMap()
	stat = [0,0,0]
	Q = []
	for i in range(N):
		s = inputStr()
		Q.append(s)
		if s == "AB":
			stat[0] += 1
			stat[1] += 1
		elif s == "AC":
			stat[0] += 1
			stat[2] += 1
		elif s == "BC":
			stat[1] += 1
			stat[2] += 1

	#print(stat)

	hon = "Yes"
	ans = []
	for i in Q:
		if s == "AB":
			if A < B:
				B -= 1
				A += 1
				ans.append("A")
			else:
				B += 1
				A -= 1
				ans.append("B")
			if A < 0 or B < 0:
				hon = "No"
				break
		elif s == "AC":
			if A <= C:
				C -= 1
				A += 1
				ans.append("A")
			else:
				C += 1
				A -= 1
				ans.append("C")
			if A < 0 or C < 0:
				hon = "No"
				break
		elif s == "BC":
			if B <= C:
				C -= 1
				B += 1
				ans.append("B")
			else:
				C += 1
				B -= 1
				ans.append("C")
			if C < 0 or B < 0:
				hon = "No"
				break

	if hon == "No":
		print(hon)
	else:
		print(hon)
		for i in ans:
			print(i)


#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
import sys
input = lambda: sys.stdin.readline().rstrip()
D = {"AB": 2, "AC": 1, "BC": 0}
N, a, b, c = map(int, input().split())
X = [a, b, c]
S = [D[input()] for _ in range(N)]
ANS = []
if max(X) == 0:
    print("No")
    exit()

if sum(X) == 1:
    t = 0 if a else 1 if b else 2
    for s in S:
        if X[s]:
            print("No")
            exit()
        t ^= s ^ 3
        ANS.append(t)
        X[s-1] ^= 1
        X[s-2] ^= 1

s = S[0]
if X[s-1] + X[s-2] == 0:
    print("No")
    exit()

for i, s in enumerate(S):
    if X[s-1] > X[s-2]:
        ANS.append((s-2) % 3)
        X[s-1] -= 1
        X[s-2] += 1
    elif X[s-1] < X[s-2]:
        ANS.append((s-1) % 3)
        X[s-1] += 1
        X[s-2] -= 1
    elif i == N - 1 or S[i+1] == s-1:
        ANS.append((s-2) % 3)
        X[s-1] -= 1
        X[s-2] += 1
    else:
        ANS.append((s-1) % 3)
        X[s-1] += 1
        X[s-2] -= 1

print("Yes")
print("\n".join([chr(t+65) for t in ANS]))

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	N,A,B,C = inputMap()
	stat = [0,0,0]
	Q = []
	for i in range(N):
		s = inputStr()
		Q.append(s)
		if s == "AB":
			stat[0] += 1
			stat[1] += 1
		elif s == "AC":
			stat[0] += 1
			stat[2] += 1
		elif s == "BC":
			stat[1] += 1
			stat[2] += 1

	#print(stat)

	hon = "Yes"
	ans = []
	for i in Q:
		if s == "AB":
			if A <= B:
				B -= 1
				A += 1
				ans.append("A")
			else:
				B += 1
				A -= 1
				ans.append("B")
			if A < 0 or B < 0:
				hon = "No"
				break
		elif s == "AC":
			if A < C:
				C -= 1
				A += 1
				ans.append("A")
			else:
				C += 1
				A -= 1
				ans.append("C")
			if A < 0 or C < 0:
				hon = "No"
				break
		elif s == "BC":
			if B <= C:
				C -= 1
				B += 1
				ans.append("B")
			else:
				C += 1
				B -= 1
				ans.append("C")
			if C < 0 or B < 0:
				hon = "No"
				break

	if hon == "No":
		print(hon)
	else:
		print(hon)
		for i in ans:
			print(i)


#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import itertools
input = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush

from decimal import *

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

##### 累乗 #####
# pow(x, y, z) -> x**y % z

##### 割り算の切り上げ #####
# tmp = -(-4 // 3)

##### dict の for文 #####
# for k, v in d.items():
#     print(k, v)

def inputInt(): return int(input())
def inputMap(): return map(int, input().split())
def inputList(): return list(map(int, input().split()))
def inputStr(): return input()[:-1]

inf = float('inf')
mod = 1000000007

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-

def main():
	N,A,B,C = inputMap()
	stat = [0,0,0]
	Q = []
	for i in range(N):
		s = inputStr()
		Q.append(s)
		if s == "AB":
			stat[0] += 1
			stat[1] += 1
		elif s == "AC":
			stat[0] += 1
			stat[2] += 1
		elif s == "BC":
			stat[1] += 1
			stat[2] += 1

	#print(stat)

	hon = "Yes"
	ans = []
	for i in Q:
		if s == "AB":
			if A <= B:
				B -= 1
				A += 1
				ans.append("A")
			else:
				B += 1
				A -= 1
				ans.append("B")
			if A < 0 or B < 0:
				hon = "No"
				break
		elif s == "AC":
			if A <= C:
				C -= 1
				A += 1
				ans.append("A")
			else:
				C += 1
				A -= 1
				ans.append("C")
			if A < 0 or C < 0:
				hon = "No"
				break
		elif s == "BC":
			if B < C:
				C -= 1
				B += 1
				ans.append("B")
			else:
				C += 1
				B -= 1
				ans.append("C")
			if C < 0 or B < 0:
				hon = "No"
				break

	if hon == "No":
		print(hon)
	else:
		print(hon)
		for i in ans:
			print(i)


#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# nCr mod m
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10**9+7):
    r = min(r, n-r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i+1, mod) % mod
    return res

# mを法とするaの乗法的逆元
def modinv(a, mod=10**9+7):
    return pow(a, mod-2, mod)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# nHr mod m
# 問題によって、combination()を切り替えること
def H(n, r, mod=10**9+7):
	# comb = Combination(n+r-1, mod)
	# return comb(n+r-1, r)
    return combination(n+r-1, r, mod)


class Combination:
    """
    O(n)の前計算を1回行うことで，O(1)でnCr mod mを求められる
    n_max = 10**6のとき前処理は約950ms (PyPyなら約340ms, 10**7で約1800ms)
    使用例：
    comb = Combination(1000000)
    print(comb(5, 3))  # 10
    """
    def __init__(self, n_max, mod=10**9+7):
        self.mod = mod
        self.modinv = self.make_modinv_list(n_max)
        self.fac, self.facinv = self.make_factorial_list(n_max)

    def __call__(self, n, r):
        return self.fac[n] * self.facinv[r] % self.mod * self.facinv[n-r] % self.mod

    def make_factorial_list(self, n):
        # 階乗のリストと階乗のmod逆元のリストを返す O(n)
        # self.make_modinv_list()が先に実行されている必要がある
        fac = [1]
        facinv = [1]
        for i in range(1, n+1):
            fac.append(fac[i-1] * i % self.mod)
            facinv.append(facinv[i-1] * self.modinv[i] % self.mod)
        return fac, facinv

    def make_modinv_list(self, n):
        # 0からnまでのmod逆元のリストを返す O(n)
        modinv = [0] * (n+1)
        modinv[1] = 1
        for i in range(2, n+1):
            modinv[i] = self.mod - self.mod//i * modinv[self.mod%i] % self.mod
        return modinv

#-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-
# dfs のサンプル
def dfs(graph,parent,counter,edge):
    stk = []
    stk.append(edge)
    while len(stk) > 0:
        p = stk.pop()
        for e in graph[p]:
            if parent[p] == e:
                continue
            else:
                parent[e] = p
                counter[e] += counter[p]
                stk.append(e)

if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
import sys
input = sys.stdin.readline

N,A,B,C=map(int,input().split())
S=[input().strip() for i in range(N)]

ANS=[]
for i in range(N):
    s=S[i]
    if s=="AB":
        if A==B==1 and i<N-1:
            if S[i+1]=="AC":
                A+=1
                B-=1
                ANS.append("A")
            else:
                B+=1
                A-=1
                ANS.append("B")

        else:
            if min(A,B)==A:
                A+=1
                B-=1
                ANS.append("A")
            else:
                B+=1
                A-=1
                ANS.append("B")

    elif s=="BC":
        if A==B==1 and i<N-1:
            if S[i+1]=="AC":
                C+=1
                B-=1
                ANS.append("C")
            else:
                B+=1
                C-=1
                ANS.append("B")

        else:
            if min(B,C)==B:
                B+=1
                C-=1
                ANS.append("B")
            else:
                C+=1
                B-=1
                ANS.append("C")

    else:
        if A==C==1 and i<N-1:
            if S[i+1]=="AB":
                A+=1
                C-=1
                ANS.append("A")
            else:
                C+=1
                A-=1
                ANS.append("C")

        else:
            if min(A,C)==A:
                A+=1
                C-=1
                ANS.append("A")
            else:
                C+=1
                A-=1
                ANS.append("C")

    if min(A,B,C)<0:
        print("No")
        sys.exit()

print("Yes")
print("\n".join(ANS))

Submission status: Wrong Answer

Solution code:
n, *a = [int(i) for i in input().split()]
abc = "ABC"
xy = []
for _ in range(n):
    xy.append([abc.index(i) for i in input()])
i = 0
ans = []
while i < n:
    if a[xy[i][0]] == 1 and a[xy[i][1]] == 1:
        j = list(set([0, 1, 2])-set(xy[i]))[0]
        if i == n - 1 or a[j] != 0 or xy[i] == xy[i+1]:
            if a[xy[i][0]] > a[xy[i][1]]:
                ans.append(abc[xy[i][1]])
                a[xy[i][0]] -= 1
                a[xy[i][1]] += 1
            else:
                ans.append(abc[xy[i][0]])
                a[xy[i][1]] -= 1
                a[xy[i][0]] += 1
            i += 1
        else:
            k = list(set(xy[i + 1])-set([j]))[0]
            ans.append(abc[k])
            a[k] += 1
            a[list(set(xy[i]) - set([k]))[0]] -= 1
            i += 1

    else:
        if a[xy[i][0]] > a[xy[i][1]]:
            ans.append(abc[xy[i][1]])
            a[xy[i][0]] -= 1
            a[xy[i][1]] += 1
        else:
            ans.append(abc[xy[i][0]])
            a[xy[i][1]] -= 1
            a[xy[i][0]] += 1
        i += 1
    if min(a) < 0:
        print("No")
        exit()

for i in ans:
    print(i)