我一直在研究从大量模的维基百科链接，下面是伪code。
I have been studying this link from wikipedia of modulo of a large number, Here is the pseudocode.
function modular_pow(base, exponent, modulus)
result := 1
while exponent > 0
if (exponent mod 2 == 1):
result := (result * base) mod modulus
exponent := exponent >> 1
base = (base * base) mod modulus
return result
我不明白wiki.Why给出的解释，我要检查是否EXP％2为偶数或奇数。也就是为什么我做了三个操作？
I don't understand the explanation given in wiki.Why I have to check if exp%2 is even or odd. also why I am doing the three operations?
推荐答案
该算法的组合 幂的平方的算法和模运算。
This algorithm is a combination of the Exponentiation by Squaring algorithm and modulo arithmetic.
要理解这是怎么回事，首先要考虑一个情况，当指数是 2 的力量。然后，假设指数= 2 ^氏/ code>，其结果可能是由于平方的结果计算 K 倍，即
To understand what's going on, first consider a situation when exponent is a power of 2. Then, assuming that exponent = 2 ^ k, the result could be computed by squaring the result k times, i.e.
res = (...((base ^ 2) ^2 ) ... ) ^2))
---------------------
k times
在指数不是 2 的力量，我们需要做更多的乘法。事实证明，如果我们可以把指数 2无余，我们可以广场基地，并划分的指数。但是，如果有一个部分，我们必须另外由电流基础值乘以中间结果。
When exponent is not a power of 2, we need to make additional multiplications. It turns out that if we can divide exponent by 2 without remainder, we can square the base, and divide the exponent. If, however, there is a remainder, we must additionally multiply the intermediate result by the value of the current base.
您看到的是相同幂的平方应用于模乘法。 ＆gt;使用指数＆GT的算法表示整数除以2; 1 运行，这等同于地板（指数/ 2）。
What you see is the same exponentiation by squaring applied to modulo multiplication. The algorithm denotes integer division by two using the exponent >> 1 operation, which is identical to floor(exponent / 2).
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