任意长度的两个大数的乘法

来源：互联网发布：乐斗门派心法数据编辑：程序博客网时间：2024/05/07 06:06

方法（一）：

关于大数乘法，可以使用数组来模拟小学三年级的乘法竖式计算过程，代码如下：

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#include "iostream"
#include "string"
using namespace std;
int main(void)
{
char str1[1000],str2[1000];
int i,j,len1,len2,len;
bool flag=false;
cout<<"任意两个大数的乘法（用数组来模拟小学三年级的乘法竖式计算过程）："<<endl;
cout<<"请输入被乘数：";
gets(str1);
cout<<"请输入乘数：";
gets(str2);
len1=strlen(str1);
len2=strlen(str2);
int *a=new int[len1];
int *b=new int[len2];
len=len1*len2+1;
int *result= new int[len];
for(i=0;i<len;i++)
result[i]=0;
for(i=0;i<len1;i++) //将字符串转换为整数，并倒置过来
a[i]=str1[len1-1-i]-'0';
for(i=0;i<len2;i++)
b[i]=str2[len2-1-i]-'0';
for(i=0;i<len1;i++) //用数组来模拟小学三年级的乘法竖式计算过程
{
for(j=0;j<len2;j++)
result[i+j]+=a[i]*b[j];
}
for(i=0;i<len;i++) //处理进位的情况
{
if(result[i]>9)
{
result[i+1]+=result[i]/10;
result[i]%=10;
}
}
cout<<"两个大整数相乘的结果为：";
for(i=len-1;i>=0;i--)
{
if(flag)
cout<<result[i];
else if(result[i]!=0)
{
cout<<result[i];
flag=true;
}
}
cout<<endl;
delete []a;
delete []b;
delete []result;
system("pause");
return 0;
}

方法（二）：关于大数乘法，可以使用大整数乘法的分治方法：

设X和Y都是n位的整数，现在要计算它们的乘积XY。如果
**利用小学所学的方法，将每两个一位数都进行相乘，最后
**再相加，效率比较低下，乘法需要n^2次。分治的方法可以
**减少乘法的次数，设X被分成2部分A和B，即X=A*10^(n/2)+B
**Y也同样处理，即Y=C*10^(n/2)+D.
**那么，XY=(A*10^(n/2)+B)*(C*10^(n/2)+D)
=AC*10^n+(AD+BC)*10^(n/2)+BD ------>(1)
**AD和BC可以利用AC和BD来表示，AD+BC=(A-B)*(D-C)+AC+BD --->(2)
**这样(1)的乘法次数由4次减少到3次。

**最后的运算效率会有所提高。

***以上出自计算机算法设计与分析(王晓东) *******/

代码如下：

#include "iostream"
#include "list"
#include "string"
using namespace std;
/*******大整数减法*********/
list<char> long_sub(list<char> clist1, list<char> clist2);
/*******大整数加法*********/
list<char> long_add(list<char> clist1, list<char> clist2)
{
list<char> rs; //存放最后的结果
//如果一个正，一个负，做两数相减
if ( *(clist1.begin()) == '-' && *(clist2.begin()) != '-' )
{
clist1.erase(clist1.begin()); //去掉符号位
rs = long_sub(clist2, clist1);
return rs;
}
//如果一个负，一个正，做两数相减
if ( *(clist1.begin()) != '-' && *(clist2.begin()) == '-' )
{
clist2.erase(clist2.begin()); //去掉符号位
rs = long_sub(clist1, clist2);
return rs;
}
if ( *(clist1.begin()) == '-' && *(clist2.begin()) == '-' )
{
clist1.erase(clist1.begin());
clist2.erase(clist2.begin());
rs = long_add(clist1,clist2);
rs.push_front('-');
return rs;
}
if ( *(clist1.begin()) != '-' && *(clist2.begin()) != '-' )
{
//首先保证两数位数相同(填充0)
int length1 = clist1.size();
int length2 = clist2.size();
if( length1 < length2 )
{
for(int i=0; i<(length2 -length1); ++i)
{
clist1.push_front('0');
}
}
else if ( length1 > length2 )
{
for(int i=0; i<(length1 -length2); ++i)
{
clist2.push_front('0');
}
}
//整数加法，从低位加起，最低位的进位初始为0
int c=0; //低位借位初始为0
int low; //减完后本位的数值
list<char>::iterator iter1 = clist1.end();
--iter1;
list<char>::iterator iter2 = clist2.end();
--iter2;
for(; iter1!=clist1.begin() && iter2!=clist2.begin();--iter1,--iter2)
{
int num1 = *iter1 -'0';
int num2 = *iter2 -'0';
low = (num1+num2+c)%10;
c = (num1+num2+c)/10;
rs.push_front(low+'0');
}
//双方最高位相加的处理
int num1 = *iter1 -'0';
int num2 = *iter2 -'0';
low = (num1+num2+c)%10;
c = (num1+num2+c)/10;
rs.push_front(low+'0');
if ( c != 0 )
{
rs.push_front(c+'0');
}
return rs;
}
return rs;
}
/*******大整数减法*********/
list<char> long_sub(list<char> clist1, list<char> clist2)
{
list<char> rs; //存放最后的结果
//如果一正一负相减，做相加
if (*(clist1.begin()) != '-' && *(clist2.begin()) == '-' )
{
clist2.erase(clist2.begin()); //去掉符号位
rs = long_add(clist1, clist2);
return rs;
}
//如果一负一正相减，做相加(添符号)
if ( *(clist1.begin()) == '-' && *(clist2.begin()) != '-' )
{
clist1.erase(clist1.begin()); //去掉符号位
rs = long_add(clist1, clist2);
rs.push_front('-');
return rs;
}
//如果两负相减，作相减
if ( *(clist1.begin()) == '-' && *(clist2.begin()) == '-' )
{
clist1.erase(clist1.begin());
clist2.erase(clist2.begin()); //去掉符号位
rs = long_sub(clist2, clist1);
return rs;
}
//如果两正相减，做相减
if ( *(clist1.begin()) != '-' && *(clist2.begin()) != '-' )
{
int sign = -1; //代表两数相减结果的正负，如果最高位为'-'(ascii码为45)表示负，否则表示正
//首先保证加数位数相同(填充0)
int length1 = clist1.size();
int length2 = clist2.size();
if( length1 < length2 )
{
sign = '-';
for(int i=0; i<(length2 -length1); ++i)
{
clist1.push_front('0');
}
}
else if ( length1 > length2 )
{
for(int i=0; i<(length1 -length2); ++i)
{
clist2.push_front('0');
}
}
else if ( *(clist1.begin()) < *(clist2.begin()) )
{
sign = '-';
}
//整数减法，从低位减起，最低位的借位初始为0
int c = 0; //低位借位初始为0
int low; //减完后本位的数值
list<char>::iterator iter1 = clist1.end();
--iter1;
list<char>::iterator iter2 = clist2.end();
--iter2;
if (sign != '-' )
{
for(; iter1!=clist1.begin() && iter2!=clist2.begin();--iter1,--iter2)
{
int num1 = *iter1 -'0';
int num2 = *iter2 -'0';
int c_new = 0; //向高位的借位
if ( num1 < num2+c )
{
c_new = 1;
num1 = num1+10;
}
low = (num1-num2-c)%10;
c = c_new;
rs.push_front(low+'0');
}
//双方最高位相减的处理
int num1 = *iter1 -'0';
int num2 = *iter2 -'0';
low = (num1-num2-c)%10;
if ( low != 0 )
{
rs.push_front(low+'0');
}
}
else if ( sign == '-' )
{
for(; iter1!=clist1.begin() && iter2!=clist2.begin();--iter1,--iter2)
{
int num1 = *iter1 -'0';
int num2 = *iter2 -'0';
int c_new = 0; //向高位的借位
if ( num2 < num1+c )
{
c_new = 1;
num2 = num2+10;
}
low = (num2-num1-c)%10;
c = c_new;
rs.push_front(low+'0');
}
//双方最高位相减的处理
int num1 = *iter1 -'0';
int num2 = *iter2 -'0';
low = (num2-num1-c)%10;
if ( low != 0 )
{
rs.push_front(low+'0');
}
rs.push_front('-'); //最高位的'-'作为负数的标志
}
return rs;
}
return rs;
}
/*******大整数乘法*********/
list<char> long_mul(list<char> clist1, list<char> clist2)
{
list<char> rs; //保存最后的结果
//递归出口
if(clist1.size() == 1 && clist2.size() == 1) //两个数都成1位了
{
int num1 = *(clist1.begin())-'0';
int num2 = *(clist2.begin())-'0';
int high = (num1*num2)/10; //积的高位
int low = (num1*num2)%10; //积的低位
if ( high != 0 )
{
rs.push_back(high+'0');
}
rs.push_back(low+'0');
return rs;
}
if (clist1.size() == 1 && clist2.size() > 1)
{
int sign = -1; //最后结果的正负标志，'-'(ascii码为45)表示负，其他表示正
char high_bit2 = *(clist2.begin()); //clist2的最高位
if ( high_bit2 == '-' )
{
sign = '-'; //相乘结果为负
clist2.erase(clist2.begin()); //去掉高位的'-'
}
int length2 = clist2.size(); //clist2的有效长度(去除'-'号后)
if ( length2 > 1 )
{
//对clist2进行拆分
list<char> clist2_1; //高位部分
list<char> clist2_2; //低位部分
int i;
list<char>::iterator iter;
for (i=0,iter=clist2.begin(); i<length2/2; ++i,++iter)
{
clist2_1.push_back(*iter);
}
for(; iter != clist2.end(); ++iter)
{
clist2_2.push_back(*iter);
}
list<char> rs1; //高位部分和clist1的积
list<char> rs2; //低位部分和clist1的积
rs1 = long_mul(clist1, clist2_1);
rs2 = long_mul(clist1, clist2_2);
//对高位积进行移位(末尾添0)
for( i=0; i<clist2_2.size(); ++i )
{
rs1.push_back('0');
}
rs = long_add(rs1,rs2); //两部分积相加
if( sign == '-' )
{
rs.push_front('-');
}
}
else
{
rs = long_mul(clist1, clist2);
if ( sign == '-' )
{
rs.push_front('-'); //结果添上'-'号
}
}
return rs;
}
if (clist1.size() >1 && clist2.size() == 1)
{
int sign = -1; //最后结果的正负标志，'-'表示负，其他表示正
char high_bit1 = *(clist1.begin()); //clist1的最高位
if ( high_bit1 == '-' )
{
sign = '-'; //相乘结果为负
clist1.erase(clist1.begin()); //去掉高位的'-'
}
int length1 = clist1.size(); //clist1的有效长度(去除'-'号后)
if ( length1 > 1 )
{
//对clist1进行拆分
list<char> clist1_1; //高位部分
list<char> clist1_2; //低位部分
int i;
list<char>::iterator iter;
for (i=0,iter=clist1.begin(); i<length1/2; ++i,++iter)
{
clist1_1.push_back(*iter);
}
for(; iter != clist1.end(); ++iter)
{
clist1_2.push_back(*iter);
}
list<char> rs1; //高位部分和clist2的积
list<char> rs2; //低位部分和clist2的积
rs1 = long_mul(clist1_1, clist2);
rs2 = long_mul(clist1_2, clist2);
//对高位积进行移位(末尾添0)
for( i=0; i<clist1_2.size(); ++i )
{
rs1.push_back('0');
}
rs = long_add(rs1,rs2); //两部分积相加
if( sign == '-' )
{
rs.push_front('-');
}
}
else
{
rs = long_mul(clist1, clist2);
if ( sign == '-' )
{
rs.push_front('-'); //结果添上'-'号
}
}
return rs;
}
if (clist1.size() >1 && clist2.size() > 1 )
{
int sign = -1; //最后结果的正负标志，'-'表示负，其他表示正
char high_bit1 = *(clist1.begin()); //clist1的最高位
char high_bit2 = *(clist2.begin()); //clist2的最高位
if ( high_bit1 == '-' && high_bit2 != '-' )
{
sign = '-'; //相乘结果为负
clist1.erase(clist1.begin()); //去掉高位的'-'
}
if ( high_bit1 != '-' && high_bit2 == '-' )
{
sign = '-'; //相乘结果为负
clist2.erase(clist2.begin()); //去掉高位的'-'
}
if ( high_bit1 =='-' && high_bit2 == '-' )
{
clist1.erase(clist1.begin());
clist2.erase(clist2.begin()); //去掉高位的'-'
}
int length1 = clist1.size(); //clist1的有效长度
int length2 = clist2.size(); //clist2的有效长度
if ( length1 == 1 || length2 == 1 )
{
rs = long_mul(clist1, clist2);
if ( sign == '-' )
{
rs.push_front('-');
}
}
else if ( length1 > 1 && length2 > 1 )
{
//对clist1和clist2分别进行划分
list<char> clist1_1; //clist1的高位部分
list<char> clist1_2; //clist1的低位部分
list<char> clist2_1; //clist2的高位部分
list<char> clist2_2; //clist2的低位部分
int i;
list<char>::iterator iter;
for(i=0,iter=clist1.begin(); i<length1/2; ++i,++iter)
{
clist1_1.push_back(*iter);
}
for(; iter!=clist1.end(); ++iter)
{
clist1_2.push_back(*iter);
}
for(i=0,iter=clist2.begin(); i<length2/2; ++i,++iter)
{
clist2_1.push_back(*iter);
}
for(; iter!=clist2.end(); ++iter)
{
clist2_2.push_back(*iter);
}
list<char> rs_hh; //两个高位相乘的结果
list<char> rs_ll; //两个低位相乘的结果
rs_hh = long_mul(clist1_1, clist2_1);
//高位相乘结果移位(末尾添0)
for(i=0; i<clist1_2.size()+clist2_2.size(); ++i)
{
rs_hh.push_back('0');
}
rs_ll = long_mul(clist1_2, clist2_2);
list<char> sub1_hl; //clist1的高位和低位部分的差
list<char> sub2_lh; //clist2的低位和高位部分的差
//两个高位分别移位
for(i=0; i<clist1_2.size(); ++i)
{
clist1_1.push_back('0');
}
for(i=0; i<clist2_2.size(); ++i)
{
clist2_1.push_back('0');
}
sub1_hl = long_sub(clist1_1, clist1_2);
sub2_lh = long_sub(clist2_2, clist2_1);
list<char> rs_sub1_sub2; //两个差的乘积
rs_sub1_sub2 = long_mul(sub1_hl, sub2_lh);
//把几个乘积的结果加起来
list<char> tmp1 = long_add(rs_hh, rs_ll);
list<char> tmp2 = long_add(tmp1, rs_sub1_sub2);
list<char> tmp3 = long_add(tmp2, rs_hh);
rs = long_add(tmp3, rs_ll);
if ( sign == '-' )
{
rs.push_front('-');
}
}
return rs;
}
return rs;
}
int main(void)
{
list<char> clist1;
list<char> clist2;
cout <<"请您输入2个乘数: "<<endl;
cout <<"请您输入被乘数: ";
int i;
string str1;
cin >> str1;
for(i=0; i<str1.size(); ++i)
{
if (str1[i] >= '0' && str1[i] <= '9' )
{
clist1.push_back(str1[i]);
}
else
{
cout <<"被乘数中的数字只能为0~9" <<endl;
exit(1);
}
}
cout <<"请您输入乘数: ";
string str2;
cin >> str2;
for(i=0; i<str2.size(); ++i)
{
if ( str2[i] >= '0' && str2[i] <= '9' )
{
clist2.push_back(str2[i]);
}
else
{
cout <<"乘数中的数字只能为0~9" <<endl;
exit(1);
}
}
list<char> rs = long_mul(clist1,clist2);
cout << "两个大整数相乘的积为: ";
for(list<char>::iterator iter=rs.begin(); iter!=rs.end(); ++iter)
{
cout << *iter;
}
cout <<endl;
system("pause");
return 0;
}