思路是將大整數(A*B)通過分治法拆分成(A1+A2)B,如果A1數位長度仍然大於某個閾值(以下代碼爲8位數),則繼續拆分。B也一樣。
辛辛苦苦寫完後,發現題友直接使用python自帶運算AB也能通過,查證得知Python整數大小沒有限制。如果用java或者c++實現的話他們的方法肯定是不合格的了。人生苦短,python牛逼。
import sys
def add(n1, n2):
n1 = n1[::-1]
n2 = n2[::-1]
# 補齊到和的最大位數
if len(n1)<len(n2):
n2 += '0'
n1 += '0'*(len(n2)-len(n1))
else:
n1 += '0'
n2 += '0'*(len(n1)-len(n2))
carry,sum = 0,''
for i in range(len(n1)):
cur_sum = int(n1[i]) + int(n2[i]) + carry
carry = cur_sum // 10
sum += str(cur_sum % 10)
sum = sum[::-1]
# 去掉前面的零
while len(sum)>1 and sum[0]=='0':
sum = sum[1:]
return sum
def muti(x1, x2):
# xi = ['123435',10]分別表示分解後乘數,數位
len1,len2 = len(x1[0]),len(x2[0])
if len1 > 8:
cut_point = len1//2
x1_0 = [x1[0][:cut_point],x1[1]+len1-cut_point]
x1_1 = [x1[0][cut_point:],x1[1]]
return add(muti(x1_0, x2),muti(x1_1,x2))
if len2 > 8:
cut_point = len2//2
x2_0 = [x2[0][:cut_point],x2[1]+len2-cut_point]
x2_1 = [x2[0][cut_point:],x2[1]]
return add(muti(x1, x2_0),muti(x1,x2_1))
val = str(int(x1[0])*int(x2[0]))
bit = x1[1]+x2[1]
val += '0'*bit
return val
if __name__ == "__main__":
for line in sys.stdin:
n1,n2 = line.strip().split()
print(muti([n1,0], [n2,0]))