#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\n",max(size(x),size(y)));
m = max(size(x),size(y)) / 2;
printf("%d\n",m);
x0 = getLow(x,m);
x1 = getHigh(x,m);
y0 = getLow(y,m);
y1 = getHigh(y,m);
printf("分拆%d==%d\n",x,y);
printf("分拆x0===%d\n",x0);
printf("分拆x1===%d\n",x1);
printf("分拆y0===%d\n",y0);
printf("分拆y1===%d\n",y1);
z2 = karatsuba(x1,y1);
printf("z2===%d\n",z2);
z0 = karatsuba(x0,y0);
printf("z0===%d\n",z0);
z1 = karatsuba((x1+x0),(y1+y0)) - z2 - z0;
printf("z1===%d\n",z1);
//return 0;
return z2*(int)pow(10,2*m)+z1*(int)pow(10,m)+z0;
}
void main(){
printf("%ld\n",karatsuba(1234,56789));
return;
}
乘法——Karatsuba乘法
原文作者:大整数乘法问题
原文地址: https://blog.csdn.net/maoxunxing/article/details/41308247
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
原文地址: https://blog.csdn.net/maoxunxing/article/details/41308247
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。