学步园

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

程序运行结果：

999999999999999999999999999999999999999999999999999999999999999998900000000000000000000000000000000000000000000000000000000000000000195571