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(lái), 計(jì)算 n = pq....... m, n 這兩個(gè)數(shù)便是 public_key ,編碼過(guò)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
標(biāo)簽:
person_key
RSA
算法
上傳時(shí)間:
2013-12-14
上傳用戶:zhuyibin
數(shù)字運(yùn)算,判斷一個(gè)數(shù)是否接近素?cái)?shù)
A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.
Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.
Input
Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.
Output
For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.
Sample Input
10 111
2 110
10 123
6 1000
8 2314
0
Sample Output
yes
yes
no
yes
no
標(biāo)簽:
數(shù)字
運(yùn)算
上傳時(shí)間:
2015-05-21
上傳用戶:daguda