问题描述
我正在用C语言编写RSA加密算法。我不打算在任何地方生产它,主要是为了扩大我对加密的理解。
I am writing an RSA encryption algorithm in C. Im not planning to put it in production anywhere, it is mainly just so I can broaden my understanding of encryption.
如何处理RSA产生的大量数字?即使使用相对较小的私钥(例如103)执行解密时,我仍然遇到以下问题:
How do I deal with the massive numbers that RSA generates? Even when performing decryption with a relatively small private key like 103, I still have the issue of dealing with things like this:
存储这种大小的数字的最佳方法是什么?使用标准库有什么办法吗?
。
What is the best way to store a number of that size? Is there any way to do it using standard libraries?.
推荐答案
方法错误。 67 ^ 103 mod 143
不需要先计算 67 ^ 103
。
Wrong approach. 67^103 mod 143
does not need to first calculate 67^103
.
循环计算模
,一次计算1位指数。
Calculate the modulo
in a loop, 1 bit of exponent at a time.
uint32_t powmod(uint32_t base, uint32_t expo, uint32_t mod) {
// % mod need only for the cases expo==0, mod<=1
uint32_t y = 1u % mod;
while (expo) {
if (expo & 1u) {
y = ((uint64_t) base * y) % mod;
}
expo >>= 1u;
base = ((uint64_t) base * base) % mod;
}
return y;
}
int main(void) {
printf("%lu",(unsigned long) powmod(67, 103, 143));
}
输出
89
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