Re: [問題] 關於(96)計概第十四題...
續前文...
formula = g^(xy) mod n
所以題目 = 3^(8*10) mod 47
解答：
3^(80) mod 47
= 3^(5*16) mod 47 ...3^5 = 243 , 243 = 47*5 + 8
= 8^(2*8) mod 47 ...8^2 = 64 , 64 = 47*1 + 17
= 17^(2*4) mod 47 ...17^2 = 289 , 289 = 47*6 + 7
= 7^(2*2) mod 47 ...7^2 = 49 , 49 = 47*1 + 2
= 2^2 mod 47 ...2^2 = 4 , 4 = 47*0 + 4
= 4
mod就是剝洋蔥 很簡單...
所有的題目都可以這樣解!! 找到最接近 mod 對象的值就對了!!
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