Post

高精度

高精加

1
2
3
4
5
6
7
8
9
10
11
12
13
vector<int> add(vector<int> &A,vector<int>&B){
    if(A.size()<B.size())return add(B,A);
    vector<int> C;
    int t=0;
    for(int i=0;i<A.size();i++){
        t+=A[i];
        if(i<B.size())t+=B[i];
        C.push_back(t%10);
        t/=10;
    }
    if(t)C.push_back(t);
    return C;
}

高精减

1
2
3
4
5
6
7
8
9
10
11
12
13
vector<int> sub(vector<int>&A,vector<int> &B){
    vector<int> C;
    int t=0;
    for(int i=0;i<A.size();i++){
        t=A[i]-t;
        if(i<B.size())t-=B[i];
        C.push_back((t+10)%10);
        if(t<0)t=1;
        else t=0;
    }
    while(C.size()>1&&C.back()==0)C.pop_back();
    return C;
}

高精乘低精

1
2
3
4
5
6
7
8
9
10
11
vector<int> mul(vector<int>&A,int b){
    vector<int> C;
    int t=0;
    for(int i=0;i<A.size()||t;i++){
        if(i<A.size())t+=A[i]*b;
        C.push_back(t%10);
        t/=10;
    }
    while(C.size()>1&&C.back()==0)C.pop_back();
    return C;
}

高精除低精

1
2
3
4
5
6
7
8
9
10
11
12
vector<int> div(vector<int> &A,int b){
    vector<int> C;
    int r=0;
    for(int i=A.size()-1;i>=0;i--){
        r=r*10+A[i];
        C.push_back(r/b);
        r%=b;
    }
    reverse(C.begin(),C.end());
    while(C.size()>1&&C.back()==0)C.pop_back();
    return C;
}
This post is licensed under CC BY 4.0 by the author.