大数运算之乘法

最新推荐文章于 2017-12-03 23:29:07 发布

原创最新推荐文章于 2017-12-03 23:29:07 发布 · 823 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#date

算法学习专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一种用于计算两个大整数相乘的简单算术单元方法。通过字符串逆序、逐位相乘和进位处理等步骤，实现了长整数的乘法运算，并提供了一个具体的C++代码示例。

/* * Author: Leng_que * Date: 2009/11/09 * E-mail: leng_que@yahoo.com.cn * Description: a simple arithmetic unit for compute product of two big positive integer. */ #include <iostream> using namespace std; void reverse(char* data) { char tmp=0; int len = strlen(data); int n_end = int(len/2); for ( int n=0; n<n_end; ++n ) { tmp = data[n]; data[n] = data[len-n-1]; data[len-n-1] = tmp; } } #define IToA(x) (x+48) #define AToI(x) (x-48) #define M 512 //乘积结果的位数最大值。注：该值应不小于len_a+len_b+1。 void muli(char* a, char* b) { int len_a = strlen(a); int len_b = strlen(b); int len = len_a+len_b; if ( len+1 > M ) { cout<<"保存结果的缓冲区不足……"<<endl; return ; } reverse(a); reverse(b); char ans[M]={0}; memset((void*)ans, '0', len); int i,j; int tens,unit; int n,mul; for ( i=0; i<len_a; ++i ) { for ( j=0; j<len_b; ++j ) { n = i+j; mul = AToI(a[i]) * AToI(b[j]); tens = mul/10; unit = mul%10; //这里考虑到可能有进位 tens += ( AToI(ans[n])+unit )/10; ans[n++] = IToA( (AToI(ans[n])+unit)%10 ); if ( ( AToI(ans[n])+tens )/10 == 1 ) { ans[n++] = IToA(( AToI(ans[n])+tens )%10); //循环检测进位 while (true) { if ( AToI(ans[n]) < 9 ) { ans[n] += 1; break; } else { ans[n++] = '0'; } } } else { ans[n] += tens; } } } if ( ans[len-1] == '0' ) { ans[len-1] = 0; } reverse(ans); cout<<ans<<endl; } #undef IToA #undef AToI #undef M //主函数 int main() { char a[]="999999999999999999999999999999999999999999999999999999999999999999123"; char b[]="999999999999999999999999999999999999999999999999999999999999999999777"; muli(a, b); return 0; }

程序运行结果：

999999999999999999999999999999999999999999999999999999999999999998900000000000000000000000000000000000000000000000000000000000000000195571