1.1 How it works#

First is addition between large numbers, which can be simulated by following the normal steps of addition. For example, see the figure below:

You can see that addition is performed from right to left, so we can reverse the obtained digits to invert their order, and after computation output in forward order.

1.2 Examples#

• AcWing 791. High-Precision Addition

Reference code：

1
void solve(){
2
  //...
3
  for(int i=0,c=0;i<max(a.length(),b.length()) || c;i++){
4
        int t1=0,t2=0;
5
        if(i<a.length()) t1 = a[i]-'0';
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        if(i<b.length()) t2 = b[i]-'0';
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        int t = t1 + t2 + c;
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        c = t / 10;t %= 10;
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        ans.push_back(t+'0');
10
    }
11
    //...
12
}

2.1 How it works#

Next is subtraction between large numbers. The method is still similar to high-precision addition; the difference is that due to the relative sizes of A and B, the result may be negative. In that case we can write A-B = -(B-A) to keep the subtraction result non-negative.

P.S. Note that because this is subtraction, leading zeros may appear; remember to remove them before output.

2.2 Examples#

• AcWing 792. High-precision Subtraction

Reference code：

1
bool comp(){
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    if(a.length()!=b.length()) return a.length()>b.length();
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    for(int i=a.length()-1;~i;i--)
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        if(a[i]!=b[i]) return a[i]>b[i];
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    return true;
6
}
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void solve(){
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    //...
9
    if(!comp()) p = 1,swap(a,b);
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    for(int i=0,c=0;i<a.length();i++){
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        int t1=0,t2=0;
12
        if(i<a.length()) t1 = a[i]-'0';
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        if(i<b.length()) t2 = b[i]-'0';
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        int t = t1 - t2 + c;
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        if(t < 0) c=-1,t+=10;
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        else c = 0;
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        ans.push_back(t+'0');
18
    }reverse(ans.begin(),ans.end());
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    if(p) cout<<"-";p=-1;
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    while(++p<ans.length() && ans[p]=='0');
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    ans = ans.substr(min(p,(int)(ans.length()-1)),max((int)(ans.length()-p),1));
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    //...
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}

3.1 How it works#

Multiplication between a large number and a small number is relatively straightforward: use a variable to record the carry of the products of each digit of A with b, and obtain the result for each digit in sequence.

P.S. Note that when the large number is 0, the product may have leading zeros; be sure to remove them.

3.2 Examples#

• AcWing 793. High-precision Multiplication

Reference code：

1
void solve(){
2
    for(int i=0,c=0;i<a.length() || c;i++){
3
        int t1 = 0;
4
        if(i<a.length()) t1 = a[i]-'0';
5
        int t = t1 * b + c;
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        ans.push_back(t%10+'0');c = t / 10;
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    }
8
}

4.1 How it works#

Similar to the multiplication of large numbers by small numbers, the division of a large number by a small number is also straightforward. You can divide the leading part of the large number from left to right by b, recording the quotient and remainder of each prefix with c/b.

P.S. Note that in earlier stages the dividend may be too small, causing leading zeros; remember to remove them before output.

4.2 Examples#

• AcWing 794. High-precision Division

Reference code：

1
void solve(){
2
    for(int i=0;i<a.length();i++){
3
        int t1 = a[i]-'0';
4
        int t = t1 + 10 * c;
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        ans.push_back(t/b + '0');
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        c = t%b;
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    }int p = -1;while(++p<ans.length() && ans[p]=='0');
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    ans = ans.substr(min(p,(int)(ans.length()-1)),max(1,(int)(ans.length()-p)));
9
}