結果

問題 No.1532 Different Products
コンテスト
ユーザー chineristAC
提出日時 2021-05-02 03:47:49
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,415 ms / 4,000 ms
コード長 1,967 bytes
コンパイル時間 171 ms
コンパイル使用メモリ 82,352 KB
実行使用メモリ 258,048 KB
最終ジャッジ日時 2024-10-10 04:49:41
合計ジャッジ時間 52,343 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 62
権限があれば一括ダウンロードができます

ソースコード

diff # 

"""
Dirichlet Product
"""
def mul(N,K,L,data_as,data_ab,data_bs,data_bb):
    data_rs = [0 for i in range(K+1)]
    data_rb = [0 for i in range(L+1)]

    for i in range(1,K+1):
        for j in range(1,K//i+1):
            data_rs[i*j] += data_as[i] * data_bs[j]

    data_as_cum = [data_as[i] for i in range(K+1)]
    data_bs_cum = [data_bs[i] for i in range(K+1)]
    for i in range(1,K+1):
        data_as_cum[i] += data_as_cum[i-1]
        data_bs_cum[i] += data_bs_cum[i-1]

    def cumA(i):
        if 1<=i<=L:
            return data_ab[i]
        else:
            return data_as_cum[(N//i)]

    def cumB(i):
        if 1<=i<=L:
            return data_bb[i]
        else:
            return data_bs_cum[(N//i)]

    for i in range(1,L+1):
        M = int((N//i)**.5)
        for j in range(1,M+1):
            data_rb[i] += data_as[j] * cumB(i*j)
        for j in range(1,M+1):
            data_rb[i] += data_bs[j] * (cumA(i*j)-cumA(N//M))

    return data_rs,data_rb

def single_mul(N,K,L,a,A,val):
    res = [a[i] for i in range(K+1)]
    Res = [A[i] for i in range(L+1)]

    for i in range(1,K+1):
        if i*val<=K:
            res[i*val] += a[i]

    cum = [a[i] for i in range(K+1)]
    for i in range(1,K+1):
        cum[i] += cum[i-1]

    def cumA(i):
        if 1<=i<=L:
            return A[i]
        else:
            return cum[N//i]

    for i in range(1,L+1):
        Res[i] += cumA(i*val)

    return res,Res

import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

_N,_K = mi()

N = _K
K = min(N,int(N**.5)+1)
L = K

res = [0 for i in range(K+1)]
res[1] = 1
Res = [0 for i in range(L+1)]
for i in range(1,L+1):
    Res[i] = 1

for i in range(1,_N+1):
    res,Res = single_mul(N,K,L,res,Res,i)

print((Res[1]-1))
0