write a code for large nos multilication (upto 200 digits)

Question Posted / atul kabra

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write a code for large nos multilication (upto 200 digits)..

Answer / atul kabra

#include<stdio.h>
#include <conio.h>
#include<string.h>
#include<alloc.h>
char * mul(char *, char);
char * add(char *, char *);
void main()
{

char *no1=(char * )malloc(100);
char *no2=(char *)malloc(100);
char *q="0";
char *p;
int i=0,j=0,t=0,l=0;
clrscr();
printf("\n Enter the large no ");
gets(no1);
printf("\n Enter the second no ");
gets(no2);

while(l<strlen(no2))
{
p=mul(no1,no2[l]);
for(j=1;j<=strlen(no2)-l-1;j++)
strcat(p,"0");
q=add(q,p);
l++;
}
printf("\n Multiplication is %s",q);
free(no1);
free(no2);
}

char * mul(char *x, char ch)
{
int i=0,j=0,t=0;
char *p=(char *)malloc(300);
char *a =(char *)malloc(300);
strcpy(p,x);
strrev(p);
while(p[i]!='\0')
{
a[j]=(p[i]-48)*(ch-48)+t;
t=a[j]/10;
a[j]=(a[j]%10)+48;
i++;
j++;
}
if(t!=0)
a[j]=t+48;
else
j--;
a[j+1]='\0';
strrev(a);
free(p);
return(a);
}
char * add(char *p, char *q)
{
char *t=(char *)malloc(300);
int i=0,j=0,x=0,a;
strrev(p);
strrev(q);
while(p[i]!='\0' && q[i]!='\0')
{
a=(p[i]-48)+(q[i]-48)+x;
x=a/10;
a=a%10;
t[i]=a+48;
i++;
}
while(i<strlen(p))
{
a=(p[i]-48)+x;
x=a/10;
a=a%10;
t[i]=a+48;
i++;
}
while(i<strlen(q))
{
a=(q[i]-48)+x;
x=a/10;
a=a%10;
t[i]=a+48;
i++;
}

if(x!=0)
t[i++]=x+48;
t[i]='\0';
strrev(t);
return(t);
}

Is This Answer Correct ?

5 Yes

2 No

write a code for large nos multilication (upto 200 digits)..

Answer / ajeet kumar

#include<iostream>
#include<string>
#include<sstream>
#include<vector>

using namespace std;

string toString(int a){
string s="";
stringstream buf;
buf<<a;
s=buf.str();
return s;
}

string ADD( string n1, string n2 ) {
int l1 = n1.size();
int l2 = n2.size();
int d = l1-l2;
d=(d>0)?d:-d;

string s1="";

if( d > 0 ){
for(int i=0;i<d;i++)
s1=s1+'0';
}
if(l1 > l2)
n2=s1+n2;
if(l2>l1)
n1=s1+n1;
int l = n1.size();
int carry = 0;
int x,y,z;
string ans="";

for(int j=l-1;j>=0;j--){
x=(int)n1[j]-48;
y=(int)n2[j]-48;
z=x+y+carry;

if(z>9){
carry=z/10;
string ad = toString(z%10);
ans=ad+ans;
}
else{
carry=0;
string ad = toString(z);
ans=ad+ans;
}
}
if(carry!=0)
ans='1'+ans;
return ans;
}

string remove_leading_zeroes(string tmp) {
string s="0";
for(int i=0;i<tmp.size();i++){
if(tmp[i]=='0')
continue;
tmp=tmp.substr(i,tmp.size()-i+1);
return tmp;
}
return s;
}

vector <string> equate_with_leading_zeros( string a, string
b ) {
int l1=a.size();
int l2=b.size();
int d=(l1-l2>0)?(l1-l2):(l2-l1);
string s="";
vector<string>v;

for(int i=0;i<d;i++)
s=s+'0';
if(l1>l2)
b=s+b;
if(l2>l1)
a=s+a;

v.push_back(a);
v.push_back(b);
return v;
}

string MULTIPLY( string s1, string s2 ) {
vector<string> v = equate_with_leading_zeros(s1,s2);
string a=v[0];
string b=v[1];
string op1,prev="";
int x,y,z;
int l=a.size();
int counter = 0;
int carry=0;

for(int i=l-1;i>=0;i--){
x = (int)a[i]-48;
op1="";
carry = 0;
for(int j=l-1;j>=0;j--){
y = (int)b[j]-48;
z = x*y+carry;
if(z>9){
op1=toString(z%10)+op1;
carry=z/10;
}
else{
op1=toString(z)+op1;
carry=0;
}
}
if(carry != 0){
op1=toString(carry)+op1;
}
carry=0;
string appString="";
for(int k=0;k<counter;k++)
appString=appString+'0';
op1=op1+appString;
prev = ADD(op1,prev);
counter++;
}
if(carry!=0){
prev = toString(carry)+prev;
}
string ans= remove_leading_zeroes(prev);
return ans;
}

int main()
{
string a,b;
cout<<"Enter first no.: ";
cin>>a;
cout<<"Enter second no.: ";
cin>>b;
string ans = MULTIPLY(a,b);
cout<<ans<<"\n";
return 0;
}

Is This Answer Correct ?

2 Yes

0 No

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