remainder solution for large numbers

Big numbers out of bounds

When a increases, finally Han Hui's $3^{The\; size\; of\; a}$ grows exponentially, which may exceed the value range of int32 or even int64, resulting in an error in the return value.

Large number remainder problem: under the premise of only using int32 type storage, correctly calculate $x^{}$Remainder of${}^a$$x^a$ ⊙ p) problem.

Solution: Cyclic remainder, fast power remainder, the latter has lower time complexity, both methods are introduced based on the following remainder operation rules

$$(xy)\odot p=[(x\odot p\left)\right(and\odot p)]\odot p$$

1. Cyclic remainder:

• According to the nature of the remainder operation (∵ in this question $x<p$,∴$x\odot p=x$ ）:

$$x^a\odot p=\left[\right(x^{a-1}\odot p)(x\odot p)]\odot p=[(x^{a-1}\odot p)x]\odot p$$

• Analysis: Using this formula, you can find $x^1,x^2,x^3,...x^{}$For the remainder of${}^a$${}^{-}$${}^{1\; to\; p,\; the\; intermediate\; value\; of\; each\; round\; is\; guaranteed\; to\; be\; within\; the\; range\; of\; int32\; values.\; The\; code\; of\; the\; encapsulation\; method\; is\; as\; follows.}$
• Time complexity $O\left(N\right)$ : where N=a, which is the linear complexity of the loop.

// (x^a) % p —— 循环求余法
int remainder(int x,int a,int p):
{
    
    
	int rem = 1;
	for(int i=0:i < a;i++){
    
    
		rem = (rem * x) % p;
	}
	return rem;
}

2. Fast power remainder:

• According to the nature of the remainder operation, it can be deduced that:
• $$x^a\odot p=(x^2{)}^{a/2}\odot p=(x^2\odot p{)}^{a/2}\odot p$$
• When a is an odd number, a/2 is not an integer, so it is divided into the following two cases, ("//" represents division rounded down):
  $$x^a\odot p=\{\begin{array}{ll}(x^2\odot {)}^{a//2}\odot p,& \text{a is an even number}\\ [(x\odot p)(x^{a-1}\odot p)]\odot p=[x(x^2\odot p{)}^{a//2}]\odot p,& \text{a is}\end{array}$$
• Analysis : Using the above formula, the index a problem can be reduced to the index a//2 problem each time through the loop operation, only need to loop $log2\left(N\right)$ times, so the complexity can be reduced to logarithmic level. The encapsulation method code is as follows:

int reminder(int x,int a,int p)
{
    
    
	int rem = 1;
	while(a > 0){
    
    
		if (a % 0){
    
    
			rem = (rem * x) % p;
		}
		x = x ^ 2 % p;
		a //=2;
	}
	return rem;
}

• Help to understand: According to the following table, the initial state $rem=1,x=3,a=19,p=1000000007,$ finally$rem\ast （x^a\odot p)$ into$rem\ast (x^a\odot p)=rem\ast 1$ in the form of$rem$ is the remainder answer.
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