How can I multiply really big numbers c++

Question

I have the following code

      int i, a, z;
i = 2343243443;
a = 5464354324324324;
z = i * a;
cout << z << endl;

When these are

Answer 1

// its may heplfull for you
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define MAX 1000
void reverse(char *from, char *to ){
    int len=strlen(from);
    int l;
    for(l=0;l<len;l++)to[l]=from[len-l-1];
    to[len]='\0';
}
void call_mult(char *first,char *sec,char *result){
    char F[MAX],S[MAX],temp[MAX];
    int f_len,s_len,f,s,r,t_len,hold,res;
    f_len=strlen(first);
    s_len=strlen(sec);
    reverse(first,F);
    reverse(sec,S);
    t_len=f_len+s_len;
    r=-1;
    for(f=0;f<=t_len;f++)temp[f]='0';
    temp[f]='\0';
    for(s=0;s<s_len;s++){
        hold=0;
        for(f=0;f<f_len;f++){
            res=(F[f]-'0')*(S[s]-'0') + hold+(temp[f+s]-'0');
            temp[f+s]=res%10+'0';
            hold=res/10;
            if(f+s>r) r=f+s;
        }
        while(hold!=0){
            res=hold+temp[f+s]-'0';
            hold=res/10;
            temp[f+s]=res%10+'0';
            if(r<f+s) r=f+s;
            f++;
        }
    }
    for(;r>0 && temp[r]=='0';r--);
    temp[r+1]='\0';
    reverse(temp,result);
}
int main(){
    char fir[MAX],sec[MAX],res[MAX];
    while(scanf("%s%s",&fir,&sec)==2){
        call_mult(fir,sec,res);
        int len=strlen(res);
        for(int i=0;i<len;i++)printf("%c",res[i]);
        printf("\n");
    }
    return 0;
}

Answer 2

First, you have to make your value long longs Second, you would want to add a (long long) in front of the multiplication. This is because when you have (ab) it returns an int therefore if you want it to return a long long you would want (long long)(ab).

Answer 3

As Jarod42 suggested is perfectly okay, but i am not sure whether overflow will take place or not ?

Try to store each and every digit of number in an array and after that multiply. You will definitely get the correct answer.

For more detail how to multiply using array follow this post http://discuss.codechef.com/questions/7349/computing-factorials-of-a-huge-number-in-cc-a-tutorial

Answer 4

Use pan paper approach as we used in 2nd standard. Store two numbers in two different array in reverse order. And take ans array as size of (arr1.size + arr2.size).And also initilize ans array to zero.

In your case arr1[10]={3,4,4,3,4,2,3,4,3,2}, arr2[15]={4,2,3,4,2,3,,4,5,3,4,5,3,4,6,4,5};

for(int i=0;i<arr1_length;i++)
{
    for(int j=0;j<arr2_length;j++)
    {

        ans[i+j]+=arr1[i]*arr2[j];
        ans[i+j+1]=ans[i+j+1]+ans[i+j]/10;
        ans[i+j]%=10;
    }
}

Then ans array contain the result.Please print carefully ans array. it may contain leading zero.

Answer 5

You can use queue data structure to find the product of two really big numbers with O(n*m). n and m are the number of digits in a number.

Answer 6

ints only hold 32 bits. When the result of a multiplication is larger than 2^31 - 1, the result rolls over to a large negative value. Instead of using the int data type, use long long int, which holds 64 bits.