方法（一）：

關于大數乘法，可以使用數組來模擬小學三年級的乘法豎式計算過程，代碼如下：

#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<<"任意兩個大數的乘法（用數組來模擬小學三年級的乘法豎式計算過程）："<<endl;cout<<"請輸入被乘數：";gets(str1);cout<<"請輸入乘數：";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++) //將字符串轉換為整數，并倒置過來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++) //用數組來模擬小學三年級的乘法豎式計算過程{for(j=0;j<len2;j++)result[i+j]+=a[i]*b[j];}for(i=0;i<len;i++) //處理進位的情況{if(result[i]>9){result[i+1]+=result[i]/10;result[i]%=10;}}cout<<"兩個大整數相乘的結果為：";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）：關于大數乘法，可以使用大整數乘法的分治方法：

設X和Y都是n位的整數，現在要計算它們的乘積XY。如果
**利用小學所學的方法，將每兩個一位數都進行相乘，最后
**再相加，效率比較低下，乘法需要n^2次。分治的方法可以
**減少乘法的次數，設X被分成2部分A和B，即X=A*10^(n/2)+B
**Y也同樣處理，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來表示，AD+BC=(A-B)*(D-C)+AC+BD --->(2)
**這樣(1)的乘法次數由4次減少到3次。

**最后的運算效率會有所提高。

***以上出自?? 計算機算法設計與分析(王曉東) *******/

代碼如下：

#include "iostream" #include "list" #include "string" using namespace std;/*******大整數減法*********/ list<char> long_sub(list<char> clist1, list<char> clist2); /*******大整數加法*********/ list<char> long_add(list<char> clist1, list<char> clist2) {list<char> rs; //存放最后的結果//如果一個正，一個負，做兩數相減if ( *(clist1.begin()) == '-' && *(clist2.begin()) != '-' ){clist1.erase(clist1.begin()); //去掉符號位rs = long_sub(clist2, clist1);return rs;}//如果一個負，一個正，做兩數相減if ( *(clist1.begin()) != '-' && *(clist2.begin()) == '-' ){clist2.erase(clist2.begin()); //去掉符號位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()) != '-' ){//首先保證兩數位數相同(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');}}//整數加法，從低位加起，最低位的進位初始為0int c=0; //低位借位初始為0int low; //減完后本位的數值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');}//雙方最高位相加的處理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; }/*******大整數減法*********/ list<char> long_sub(list<char> clist1, list<char> clist2) {list<char> rs; //存放最后的結果//如果一正一負相減，做相加if (*(clist1.begin()) != '-' && *(clist2.begin()) == '-' ){clist2.erase(clist2.begin()); //去掉符號位rs = long_add(clist1, clist2);return rs;}//如果一負一正相減，做相加(添符號)if ( *(clist1.begin()) == '-' && *(clist2.begin()) != '-' ){clist1.erase(clist1.begin()); //去掉符號位rs = long_add(clist1, clist2);rs.push_front('-');return rs;}//如果兩負相減，作相減if ( *(clist1.begin()) == '-' && *(clist2.begin()) == '-' ){clist1.erase(clist1.begin());clist2.erase(clist2.begin()); //去掉符號位rs = long_sub(clist2, clist1);return rs;}//如果兩正相減，做相減if ( *(clist1.begin()) != '-' && *(clist2.begin()) != '-' ){int sign = -1; //代表兩數相減結果的正負，如果最高位為'-'(ascii碼為45)表示負，否則表示正//首先保證加數位數相同(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 = '-';}//整數減法，從低位減起，最低位的借位初始為0int c = 0; //低位借位初始為0int low; //減完后本位的數值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');}//雙方最高位相減的處理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');}//雙方最高位相減的處理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'-'作為負數的標志}return rs;}return rs; } /*******大整數乘法*********/ list<char> long_mul(list<char> clist1, list<char> clist2) {list<char> rs; //保存最后的結果//遞歸出口if(clist1.size() == 1 && clist2.size() == 1) //兩個數都成1位了{int num1 = *(clist1.begin())-'0';int num2 = *(clist2.begin())-'0';int high = (num1*num2)/10; //積的高位int low = (num1*num2)%10; //積的低位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; //最后結果的正負標志，'-'(ascii碼為45)表示負，其他表示正char high_bit2 = *(clist2.begin()); //clist2的最高位if ( high_bit2 == '-' ){sign = '-'; //相乘結果為負clist2.erase(clist2.begin()); //去掉高位的'-'}int length2 = clist2.size(); //clist2的有效長度(去除'-'號后)if ( length2 > 1 ){//對clist2進行拆分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的積list<char> rs2; //低位部分和clist1的積rs1 = long_mul(clist1, clist2_1);rs2 = long_mul(clist1, clist2_2);//對高位積進行移位(末尾添0)for( i=0; i<clist2_2.size(); ++i ){rs1.push_back('0');}rs = long_add(rs1,rs2); //兩部分積相加if( sign == '-' ){rs.push_front('-');}}else{rs = long_mul(clist1, clist2);if ( sign == '-' ){rs.push_front('-'); //結果添上'-'號}}return rs;}if (clist1.size() >1 && clist2.size() == 1){int sign = -1; //最后結果的正負標志，'-'表示負，其他表示正char high_bit1 = *(clist1.begin()); //clist1的最高位if ( high_bit1 == '-' ){sign = '-'; //相乘結果為負clist1.erase(clist1.begin()); //去掉高位的'-'}int length1 = clist1.size(); //clist1的有效長度(去除'-'號后)if ( length1 > 1 ){//對clist1進行拆分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的積list<char> rs2; //低位部分和clist2的積rs1 = long_mul(clist1_1, clist2);rs2 = long_mul(clist1_2, clist2);//對高位積進行移位(末尾添0)for( i=0; i<clist1_2.size(); ++i ){rs1.push_back('0');}rs = long_add(rs1,rs2); //兩部分積相加if( sign == '-' ){rs.push_front('-');}}else{rs = long_mul(clist1, clist2);if ( sign == '-' ){rs.push_front('-'); //結果添上'-'號}}return rs;}if (clist1.size() >1 && clist2.size() > 1 ){int sign = -1; //最后結果的正負標志，'-'表示負，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 = '-'; //相乘結果為負clist1.erase(clist1.begin()); //去掉高位的'-'}if ( high_bit1 != '-' && high_bit2 == '-' ){sign = '-'; //相乘結果為負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的有效長度int length2 = clist2.size(); //clist2的有效長度if ( length1 == 1 || length2 == 1 ){rs = long_mul(clist1, clist2);if ( sign == '-' ){rs.push_front('-');}}else if ( length1 > 1 && length2 > 1 ){//對clist1和clist2分別進行劃分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; //兩個高位相乘的結果list<char> rs_ll; //兩個低位相乘的結果rs_hh = long_mul(clist1_1, clist2_1);//高位相乘結果移位(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//兩個高位分別移位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; //兩個差的乘積rs_sub1_sub2 = long_mul(sub1_hl, sub2_lh);//把幾個乘積的結果加起來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 <<"請您輸入2個乘數: "<<endl;cout <<"請您輸入被乘數: ";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 <<"被乘數中的數字只能為0~9" <<endl;exit(1);}}cout <<"請您輸入乘數: ";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 <<"乘數中的數字只能為0~9" <<endl;exit(1);}}list<char> rs = long_mul(clist1,clist2);cout << "兩個大整數相乘的積為: ";for(list<char>::iterator iter=rs.begin(); iter!=rs.end(); ++iter){cout << *iter;}cout <<endl;system("pause");return 0; }

總結

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