372. Super Pow_your task is to calculate ab mod 1337 where a is a-优快云博客

本文介绍了一种计算a^b mod c的快速幂取模算法，特别针对a为正整数，b为极大正整数并以数组形式给出的情况。通过实例展示了如何高效地解决此类问题，并提供了一个具体实现该算法的示例代码。

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.

Example1:

a = 2
b = [3]

Result: 8

Example2:

a = 2
b = [1,0]

Result: 1024

Credits:

Special thanks to @Stomach_ache for adding this problem and creating all test cases.

快速幂取模

求a^b % c ，得到以下函数 ,复杂度为O（logb）

public int PowerMod(int a, int b, int c)
	{
		int ans = 1;
		a = a % c;
		while (b > 0)
		{
			if (b % 2 == 1)
				ans = (ans * a) % c;
			b = b / 2;
			a = (a * a) % c;
		}
		return ans;
	}

这里的b表示为数组的形式，每次b/2比较麻烦，主要慢在这里，复杂度 O(length_of(b)*log b)

 public int superPow(int a, int[] b)
	{
		return PowerMod1(a, b, 1337);
	}
	
	public int PowerMod1(int a, int[] b, int c)
	{
		int ans = 1;
		a = a % c;
		int bheightestbit=0;
		int len=b.length;
		
		while (b[bheightestbit] > 0)
		{
			if (b[len-1] % 2 == 1)
				ans = (ans * a) % c;
			
			boolean setheightest=false;
			for(int i=bheightestbit;i<len;i++)
			{
				int bitnum=b[i];
				int mod=bitnum%2;
				b[i]=bitnum/2;
				if(mod==1&&i<len-1)
					b[i+1]+=10;
				if(!setheightest&&b[i]>0)
				{
					setheightest=true;
					bheightestbit=i;
				}
			}
				
			a = (a * a) % c;
		}
		return ans;
	}