Integer high precision operation--addition

High-precision operation is an important algorithm of information science. This algorithm is calculated using multiple storage units, so it is more computationally scoped than a typical storage unit. It is also a frequently-tested topic in some informatics competitions.

The high-precision operation mainly has the following several steps:

1, read the string, converted to a number in reverse chronological storage into an array of integers;

2, arithmetic, pay attention to carry and borrow;

3, the reverse output integer array, the addition of attention to the highest bit carry, subtraction note high in the useless 0 do not output;

High-precision Addition code:

#include <stdio.h>
#include <string.h>
Char s[1000]; When the array is large, it should be defined as a global variable.
int a[1000]={0},b[1000]={0};
int main ()
{
int l1,l2,l,i;
scanf ("%s", s); Read the first integer
L1=strlen (s); To find the length of an integer (string)
for (i=0;i<=l1-1;i++)//To store an integer in reverse in the integer array a
{
a[l1-1-i]=s[i]-' 0 '; S[i] is a character type that gets an integer type by subtracting 0 of the ASCII code, such as: Converts the character 1 to the number 1
}
scanf ("%s", s); Read the second integer
L2=strlen (s);
for (i=0;i<=l2-1;i++)
{
b[l2-1-i]=s[i]-' 0 ';
}
if (L1>L2)
L=L1;
Else
L=L2;
for (i=0;i<=l-1;i++)//addition operation
{
A[i]=a[i]+b[i];
if (a[i]>=10)//greater than 10 go to a high
{
a[i+1]=a[i+1]+1;
a[i]=a[i]-10;
}
}
if (a[l]!=0)//Determine if the highest bit has a rounding
l++;
for (i=l-1;i>=0;i--)//output
printf ("%d", a[i]);
}

High-precision Subtraction code:

#include <stdio.h>
#include <string.h>
Char s[1000];
int a[1000]={0},b[1000]={0},c[1000]={0};
int main ()
{
int i,j,z,l1,l2,l,o;
Freopen ("Hp-.in", "R", stdin);
Freopen ("Hp-.out", "w", stdout);
scanf ("%s", s); A
L1=strlen (s);
for (i=0;i<=l1-1;i++)//convert S (meiosis) to numbers and in reverse to a array
a[l1-1-i]=s[i]-' 0 ';
scanf ("%s", s);
L2=strlen (s);
for (i=0;i<=l2-1;i++)//convert S (meiosis) to numbers, and in reverse to B array
b[l2-1-i]=s[i]-' 0 ';

if (L1==L2) {
for (j=l1;j>=0;j--)
{
if (A[j]>b[j])//a Large
{
o=0;
Break
}
if (B[j]>a[j])//b Large
{
O=1;
Break
}
}
}
if (l1>l2| | o==0)//a Large, A-b
{
L=L1;
for (i=0;i<l;i++)
{
if (A[i]<b[i])
{
a[i]=a[i]+10;
A[i+1]=a[i+1]-1;
}
C[i]=a[i]-b[i];
}
while (l>1&&c[l-1]==0) l--; To give up the useless 0 in the high
for (i=l-1;i>=0;i--)//output
{
printf ("%d", c[i]);
}
}
Else
{
printf ("-");
L=L2;
for (i=0;i<l;i++)
{
if (A[i]>b[i])
{
b[i]=b[i]+10;
B[i+1]-=1;
}
C[i]=b[i]-a[i];
}
while (l>1&&c[l-1]==0) l--; To give up the useless 0 in the high
for (i=l-1;i>=0;i--)//output
{
printf ("%d", c[i]);
}
}

return 0;
}

Integer high precision operation--addition