Submission #18525716

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Source Code Expand

import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")

### 素数の逆元とCombination


import numpy as np
from numba import njit, b1, i4, i8, f8
@njit((i8[:],i8), cache=True)
def main(vs, n):
    M = 10**9+7 # 出力の制限
    N = 300 # 必要なテーブルサイズ
    g1 = [0] * (N+1) # 元テーブル
    g2 = [0] * (N+1) #逆元テーブル
    inverse = [0] * (N+1) #逆元テーブル計算用テーブル
    g1[0] = g1[1] = g2[0] = g2[1] = 1
    inverse[0], inverse[1] = [0, 1] 
    for i in range( 2, N + 1 ):
        g1[i] = ( g1[i-1] * i ) % M 
        inverse[i] = ( -inverse[M % i] * (M//i) ) % M # ai+b==0 mod M <=> i==-b*a^(-1) <=> i^(-1)==-b^(-1)*aより
        g2[i] = (g2[i-1] * inverse[i]) % M 
    def cmb(n, r, M):
        if ( r<0 or r>n ):
            return 0
        r = min(r, n-r)
        return (g1[n] * g2[r] * g2[n-r]) % M
    def perm(n, r, M):
        if (r<0 or r>n):
            return 0
        return (g1[n] * g2[n-r]) % M

    dp = np.zeros(n+1, dtype=np.int64)
    dp[0] = 1
    s = 0
    for v in vs:
        if v==0:
            continue
        ndp = np.zeros(n+1, dtype=np.int64)
        for j in range(n+1):
            if dp[j]==0:
                continue
            for a in range(s+1-j+1):
                for b in range(j+1):
                    ind = j+v-a-b-b
                    val = dp[j] * cmb(s+1-j,a,M) % M
                    val *= cmb(j,b,M)
                    val %= M
                    val *= cmb(v-1,a+b-1,M)
                    val %= M
                    ndp[ind] += val
                    ndp[ind] %= M
    #                 print(a,b,dp[j] * cmb(s+1-j,a,M) * cmb(j,b,M) * cmb(v-1,a+b-1,M))
        s += v
        dp = ndp
    #     print(dp)
    ans = dp[0]
    print(ans%M)
vs = np.array(list(map(int, input().split())), dtype=np.int64)
n = sum(vs)
main(vs,n)

Submission Info

Judge Result