結果

問題 No.2437 Fragile Multiples of 11
ユーザー tassei903
提出日時 2024-06-20 14:40:26
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 77 ms / 2,000 ms
コード長 3,365 bytes
コンパイル時間 472 ms
コンパイル使用メモリ 82,096 KB
実行使用メモリ 75,152 KB
最終ジャッジ日時 2024-06-20 14:40:30
合計ジャッジ時間 4,426 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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ファイルパターン 結果
other AC * 34
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ソースコード

diff # 

import sys
input = lambda :sys.stdin.readline()[:-1]
ni = lambda :int(input())
na = lambda :list(map(int,input().split()))
yes = lambda :print("yes");Yes = lambda :print("Yes");YES = lambda : print("YES")
no = lambda :print("no");No = lambda :print("No");NO = lambda : print("NO")
#######################################################################

n = ni()
X = [int(x) for x in input()]

m = 11
mod = 998244353

def solve(n, X):
    dp = [[0 for i in range(4)] for i in range(m)]
    dp[0][0] = 1

    for i in range(n):
        ndp = [[0 for i in range(4)] for i in range(m)]
        ndp, dp = dp, ndp
        for j in range(m):
            for k in range(4):
                p, q = k//2, k%2
                if ndp[j][k] == 0:continue
                for x in range(10):
                    np = p
                    nq = q
                    xj = x if (n-i-1) % 2 else (-x)%m
                    nj = (j + xj)%m
                    if (xj+j * 2)%m == 0 and p:continue
                    if x != 0:
                        np = 1
                    if q == 0 and x > X[i]:
                        continue
                    if x < X[i]:
                        nq = 1
                    nk = np*2 + nq
                    dp[nj][nk] += ndp[j][k]
                    dp[nj][nk] %= mod

    return (dp[0][2]+dp[0][3])%mod

e0 = set()
e1 = {0}
def f(s, x):
    return set([y * 10 + x for y in s])
def merge(s1, s2):
    return s1 | s2

def solve_greedy(n, X):
    dp = [[e0 for i in range(4)] for i in range(m)]
    dp[0][0] = e1

    for i in range(n):
        ndp = [[set() for i in range(4)] for i in range(m)]
        ndp, dp = dp, ndp
        for j in range(m):
            for k in range(4):
                p, q = k//2, k%2
                if len(ndp[j][k]) == 0:continue
                for x in range(10):
                    np = p
                    nq = q
                    xj = x if (n-i-1) % 2 else (-x)%m
                    nj = (j + xj)%m
                    if (xj+j * 2)%m == 0 and p:continue
                    if x != 0:
                        np = 1
                    if q == 0 and x > X[i]:
                        continue
                    if x < X[i]:
                        nq = 1
                    nk = np*2 + nq
                    dp[nj][nk] = merge(dp[nj][nk], f(ndp[j][k], x))
        #print(dp)
    return dp[0]

def solve2(n, X):
    dp = [[0 for i in range(4)] for i in range(m)]
    dp[0][0] = 1

    for i in range(n):
        ndp = [[0 for i in range(4)] for i in range(m)]
        ndp, dp = dp, ndp
        #print(ndp)
        for j in range(m):
            for k in range(4):
                p, q = k//2, k%2
                if ndp[j][k] == 0:continue
                for x in range(10):
                    np = p
                    nq = q
                    xj = x if (n-i) % 2 else (-x)%m
                    nj = (j + xj)%m
                    if x != 0:
                        np = 1
                    if q == 0 and x > X[i]:
                        continue
                    if x < X[i]:
                        nq = 1
                    nk = np*2 + nq
                    #print((j,p,q),(nj, np,nq))
                    dp[nj][nk] += ndp[j][k]
                    dp[nj][nk] %= mod

    return (dp[0][2]+dp[0][3])%mod

print(solve(n, X))
#print(solve_greedy(n, X))
#print(solve2(n, X))
0