The code performs a number (ITERS) of iterations of the
Bailey s 6-step FFT algorithm (following the ideas in the
CMU Task parallel suite).
1.- Generates an input signal vector (dgen) with size
n=n1xn2 stored in row major order
In this code the size of the input signal
is NN=NxN (n=NN, n1=n2=N)
2.- Transpose (tpose) A to have it stored in column
major order
3.- Perform independent FFTs on the rows (cffts)
4.- Scale each element of the resulting array by a
factor of w[n]**(p*q)
5.- Transpose (tpose) to prepair it for the next step
6.- Perform independent FFTs on the rows (cffts)
7.- Transpose the resulting matrix
The code requires nested Parallelism.
替代加密:
A B C D E F G H I J K L M N O P Q R S T U V W 密文
Y Z D M R N H X J L I O Q U W A C B E G F K P 明文
X Y Z
T S V
I HAVE A DREAM！#
密文？？
用ARM編程實(shí)現(xiàn)替代加密。
RSA算法 :首先, 找出三個(gè)數(shù), p, q, r, 其中 p, q 是兩個(gè)相異的質(zhì)數(shù), r 是與 (p-1)(q-1) 互質(zhì)的數(shù)...... p, q, r 這三個(gè)數(shù)便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個(gè) m 一定存在, 因?yàn)?r 與 (p-1)(q-1) 互質(zhì), 用輾轉(zhuǎn)相除法就可以得到了..... 再來, 計(jì)算 n = pq....... m, n 這兩個(gè)數(shù)便是 public_key ，編碼過程是, 若資料為 a, 將其看成是一個(gè)大整數(shù), 假設(shè) a < n.... 如果 a >= n 的話, 就將 a 表成 s 進(jìn)位 (s
