karatsuba乘法

procedurekaratsuba(num1,num2)

if(num1<10)or(num2<10)

returnnum1*num2

/*calculatesthesizeofthenumbers*/

m=max(size(num1),size(num2))

m2=m/2

high1,low1=split_at(num1,m2)

high2,low2=split_at(num2,m2)

/*3callsmadetonumbersapproximatelyhalfthesize*/

z0=karatsuba(low1,low2)

z1=karatsuba((low1+high1),(low2+high2))

z2=karatsuba(high1,high2)

return(z2*10^(m))+(（z1-z2-z0）*10^(m/2))+(z0)

　　

#include <stdio.h>
#include <math.h>
 
//找到x的位数
int size(long x){
	int count=0;
	do{
		count++;
		x=x/10;
	}while(x);
	return count;
}
 
//找到x和y的最大值
int max(int x, int y){
	return x>y?x:y;
}
 
int getHigh(int x, int m){
	return x / (int)pow(10,m);
}
 
int getLow(int x,int m){
	return x - getHigh(x,m)*(int)pow(10,m); 
}
 
//大数相乘算法:比如1234,5678
//拆分为12,34   56,78
//x=x1*10^m+x0
//y=y1*10^m+y0
//满足：m<n且x0,y0<10^m
//
long karatsuba(long x, long y){
	int m;
	int x1,x0;
	int y1,y0;
 
	int z0;
	int z1;
	int z2;
 
	//结束递归
	if(x<10||y<10)
		return x*y;
	
	//获得拆分的位数
	//printf("%d
",max(size(x),size(y)));
	m = max(size(x),size(y)) / 2;
	printf("%d
",m);
 
	x0 = getLow(x,m);
	x1 = getHigh(x,m);
	y0 = getLow(y,m);
	y1 = getHigh(y,m);
 
	printf("分拆%d==%d
",x,y);	
	printf("分拆x0===%d
",x0);	
	printf("分拆x1===%d
",x1);
	printf("分拆y0===%d
",y0);
	printf("分拆y1===%d
",y1);
	
	z2 = karatsuba(x1,y1);
	printf("z2===%d
",z2);
	z0 = karatsuba(x0,y0);
	printf("z0===%d
",z0);
	z1 = karatsuba((x1+x0),(y1+y0)) - z2 - z0;
	printf("z1===%d
",z1);
	//return 0;
	return z2*(int)pow(10,2*m)+z1*(int)pow(10,m)+z0;
}
 
void main(){
	printf("%ld
",karatsuba(1234,56789));
	return;
}

　　

伪代码：

// 输入：n位长度的正整数x和y
// 输出：x*y的结果
函数 karatsuba (x, y) {
     a和b为x的前半部分和后半部分；
     c和d是y的前半部分和后半部分；
    递归调用函数karatsuba解决q=a*c， 递归karatsuba解决p=b*d，递归karatsuba解决k=(a+b)(c+d)
    返回10的n次方*q + 10的n/2次方 * (k - q - p) + p

}

https://zhuanlan.zhihu.com/p/144813558?from_voters_page=true