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[Other] 乘式还原 DL : 0: A代表数字0到9中的前五个数字，Z代表后五个数字，请还原下列乘式。 A Z A × A A Z ------------ A A A A A A Z Z Z A A ------------ Z A Z A A *问题分析与算法设计问题本身并不复杂，可以对乘式中的每一位使用穷举法，最终可以得到结果。本题的关键在于怎样有效的判断每个部分积的每一位是否满足题意，这一问题处理不好，编写的程序会很长。程序实现中采用了一个判断函数，通过传入函数的标志字符串对所有的数进行统一的判断处理。-0-9 in the first five digits, Z, after five figures, would reinstate the following multiplier formula. A Z-A A A Z --- --- A A A A Z Z --- --- A A A Z * Analysis design problems with the algorithm itself is not complicated and can take every type of use of an exhaustive law, the final results can be. This is the key to judging how effective each part of the plot of each one intended to satisfy that, the problem is not handled well, the preparation process will be very long. Program adopted a judgment function, function through the introduction of a symbol for all string a few unified handling of judgment.
Date : 2008-10-13 Size : 1.6kb User : 姚紫欣
[Other] 乘式还原(2) DL : 0: 有乘法算式如下： ○○○ × ○○ ------------ ○○○○ ○○○○ ------------ ○○○○○ 18个○的位置上全部是素数(1、3、5或7)，请还原此算式。 *问题分析与算法设计问题中虽然有18数位，但只要确定乘数和被乘数后经过计算就可确定其它的数位。-multiplication formula is as follows : 2000 2000 --- --- 2000 2000 --- --- 2000 18 hours position all prime numbers (1, 3, 5 or 7), this reduction formula. * Analysis and design of algorithms Although the number of 18-bit, but as long as the multiplicand and multiplier determined after a calculation can be determined by several other.
Date : 2008-10-13 Size : 2.27kb User : 姚紫欣
[Other] 2060 DL : 0: 判断另一种斐波那契系列是否是3 的倍数问题该题如果用递归求出F(n),然后判断将会超时。注意到F(n) = F(n-1) + F(n-2)根据和的模等于模的和。即有F(n)％3=(F(n-1)％3+F(n-2)％3)％3 F(0)％3=1,F(1)％3=2 计算F(n)％3 : 1,2,0,2,2,1,0,1,1,2,0,2,2,1…………… 可以看出F(n)％3 以1,2,0,2,2,1,0,1, 重复出现 F(n)％3=0 即F(n)是3 的倍数此时的n 等于8k+2 或8k+6-Fibonacci Series 3 is whether the multiplier issue that if the calculated using recursive F (n), The judgment will then overtime. Notes F (n) = F (n-1) F (n-2) and the modulus and the same model. That is, F (n) = 3% (F (n-1)% F 3 (n-2)% 3)% 3 F (0%) 3 = 1, F (1)% 2 3 = calculated F (n)% 3 : 1,2,0,2,2,1,0,1,1,2,0. 2,2,1 ... ... ... can be seen F (n) 3% to 1,2,0,2,2,1,0,1, repeated F (n) = 0% 3 F (n) 3 is a multiple of n time to 8k or 8k 6 2
Date : 2008-10-13 Size : 471.45kb User : rcponder