我正在准备面试问题，而不是作业。有一个关于如何将非常长的整数倍增的问题。任何人都可以提供任何 C++ 源代码来学习吗？我试图通过学习他人的解决方案来改善自己，从而缩小自己与他人之间的差距。
非常感谢！
如果您认为这里不适合提出此类问题，我们深表歉意。
【问题讨论】：
标签： c++
您可以将 GNU Multiple Precision Arithmetic Library 用于 C++。
如果您只是想要一种简单的方法来乘以大数（整数），那么您就是：
#include<iostream>
#include<string>
#include<sstream>
#define SIZE 700
using namespace std;
class Bignum{
int no[SIZE];
public:
Bignum operator *(Bignum& x){ // overload the * operator
/*
34 x 46
-------
204 // these values are stored in the
136 // two dimensional array mat[][];
-------
1564 // this the value stored in "Bignum ret"
*/
Bignum ret;
int carry=0;
int mat[2*SIZE+1][2*SIZE]={0};
for(int i=SIZE-1;i>=0;i--){
for(int j=SIZE-1;j>=0;j--){
carry += no[i]*x.no[j];
if(carry < 10){
mat[i][j-(SIZE-1-i)]=carry;
carry=0;
}
else{
mat[i][j-(SIZE-1-i)]=carry%10;
carry=carry/10;
}
}
}
for(int i=1;i<SIZE+1;i++){
for(int j=SIZE-1;j>=0;j--){
carry += mat[i][j]+mat[i-1][j];
if(carry < 10){
mat[i][j]=carry;
carry=0;
}
else{
mat[i][j]=carry%10;
carry=carry/10;
}
}
}
for(int i=0;i<SIZE;i++)
ret.no[i]=mat[SIZE][i];
return ret;
}
Bignum (){
for(int i=0;i<SIZE;i++)
no[i]=0;
}
Bignum (string _no){
for(int i=0;i<SIZE;i++)
no[i]=0;
int index=SIZE-1;
for(int i=_no.length()-1;i>=0;i--,index--){
no[index]=_no[i]-'0';
}
}
void print(){
int start=0;
for(int i=0;i<SIZE;i++)
if(no[i]!=0){
start=i;
break; // find the first non zero digit. store the index in start.
}
for(int i=start;i<SIZE;i++) // print the number starting from start till the end of array.
cout<<no[i];
cout<<endl;
return;
}
};
int main(){
Bignum n1("100122354123451234516326245372363523632123458913760187501287519875019671647109857108740138475018937460298374610938765410938457109384571039846");
Bignum n2("92759375839475239085472390845783940752398636109570251809571085701287505712857018570198713984570329867103986475103984765109384675109386713984751098570932847510938247510398475130984571093846571394675137846510874510847513049875610384750183274501978365109387460374651873496710394867103984761098347609138746297561762234873519257610");
Bignum n3 = n1*n2;
n3.print();
return 0;
}
如您所见，它乘以 2 个大整数 :) ...（最多 700 位）
【讨论】：
试试这个：
//------------DEVELOPED BY:Ighit F4YSAL-------------
#include<iostream>
#include<string>
#include<sstream>
#define BIG 250 //MAX length input
using namespace std;
int main(){
int DUC[BIG][BIG*2+1]={0},n0[BIG],n1[BIG],i,t,h,carry=0,res;
string _n0,_n1;
while(1){
//-----------------------------------get data------------------------------------------
cout<<"n0=";
cin>>_n0;
cout<<"n1=";
cin>>_n1;
//--------------------string to int[]----------------------------------------
for(i=_n0.length()-1,t=0;i>=0,t<=_n0.length()-1;i--,t++){
n0[i]=_n0[t]-'0';
}
i=0;
for(i=_n1.length()-1,t=0;i>=0,t<=_n1.length()-1;i--,t++){
n1[i]=_n1[t]-'0';
}
i=0;t=0;
//--------------------------produce lines of multiplication----------------
for(i=0;i<=_n1.length()-1;i++){
for(t=0;t<=_n0.length()-1;t++){
res=((n1[i]*n0[t])+carry);
carry=(res/10);
DUC[i][t+i]=res%10;
}
DUC[i][t+i]=carry;
carry=0;
}
i=0;t=0;res=0;carry=0;
//-----------------------------add the lines-------------------------------
for(i=0;i<=_n0.length()*2-1;i++){
for(t=0;t<=_n1.length()-1;t++){
DUC[BIG-1][i]+=DUC[t][i];
//cout<<DUC[t][i]<<"-";
}
res=((DUC[BIG-1][i])+carry);
carry=res/10;
DUC[BIG-1][i]=res%10;
//cout<<" ="<<DUC[BIG-1][i]<<endl;
}
i=0;t=0;
//------------------------print the result------------------------------------
cout<<"n1*n0=";
for(i=_n0.length()*2-1;i>=0;i--){
if((DUC[BIG-1][i]==0) and (t==0)){}else{cout<<DUC[BIG-1][i];t++;}
//cout<<DUC[BIG-1][i];
}
//-------------------------clear all-------------------------------------
for(i=0;i<=BIG-1;i++){
for(t=0;t<=BIG*2;t++){
DUC[i][t]=0;
}
n0[i]=0;n1[i]=0;
}
//--------------------------do it again-------------------------------------
cout<<"\n------------------------------------------------\n\n";
}
return 0;
}
【讨论】：
此解决方案适用于非常大的数字，但对于非常大的数字的阶乘计算效果不佳。希望它会对某人有所帮助。
#include <iostream>
#include <string>
using namespace std;
string addition(string a, string b) {
string ans = "";
int i, j, temp = 0;
i = a.length() - 1;
j = b.length() - 1;
while (i >= 0 || j >= 0) {
if (i >= 0 && a[i])
temp += a[i] - '0';
if (j >= 0 && b[j])
temp += b[j] - '0';
int t = (temp % 10);
char c = t + '0';
ans = ans + c;
temp = temp / 10;
i--;
j--;
}
while (temp > 0) {
int t = (temp % 10);
char c = t + '0';
ans = ans + c;
temp = temp / 10;
}
string fnal = "";
for (int i = ans.length() - 1;i >= 0;i--) {
fnal = fnal + ans[i];
}
return fnal;
}
string multiplication(string a, string b) {
string a1, b1 = "0";
int i, j, temp = 0, zero = 0;
i = a.length() - 1;
int m1, m2;
while (i >= 0) {
a1 = "";
m1 = a[i] - '0';
j = b.length() - 1;
while (j >= 0) {
m2 = b[j] - '0';
temp = temp + m1*m2;
int t = temp % 10;
char c = t + '0';
a1 = a1 + c;
temp = temp / 10;
j--;
}
while (temp > 0) {
int t = (temp % 10);
char c = t + '0';
a1 = a1 + c;
temp = temp / 10;
}
string fnal = "";
// reverse string
for (int i = a1.length() - 1;i >= 0;i--) {
fnal = fnal + a1[i];
}
a1 = fnal;
//add zero
for (int p = 0;p < zero;p++)
a1 = a1 + '0';
b1 = addition(a1, b1);
i--;
zero++;
}
return b1;
}
// upto 50 is ok
int factorial(int n) {
string a = "1";
for (int i = 2;i <= n;i++) {
string b = to_string(i);
a = multiplication(a, b);
}
cout << a << endl;
return a.length();
}
int main() {
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
int n;
cin >> n;
//cout << factorial(n) << endl;
string a, b;
a = "1281264836465376528195645386412541764536452813416724654125432754276451246124362456354236454857858653";
b = "3767523857619651386274519192362375426426534237624548235729562582916259723586347852943763548355248625";
//addition(a, b);
cout << multiplication(a, b);
return 0;
}
【讨论】：