A / B

通过使用sub减操作来完成两个大整数的除法

样例

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

bool cmp(vector<int> &A, vector<int> &B){
    if(A.size()!=B.size()) return A.size()>B.size();
    for(int i=A.size()-1;i>=0;i--){
        if(A[i]!=B[i]) return A[i]>B[i]; 
    }
    return true;
}

vector<int> sub(vector<int> &A,vector<int> &B){
    vector<int> C;
    int t=0;
    for(int i=0;i<A.size()||t;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;
}

vector<int> div(vector<int> &A, vector<int> &B, vector<int> &r){
    vector<int> C;
    int j = B.size();
    r.assign(A.end()-j,A.end());
    while(j<=A.size()){
        int k=0;
        while(cmp(r,B)){
            vector<int> s = sub(r,B);
            r.clear();
            r.assign(s.begin(),s.end());
            k++;
        }
        C.push_back(k);
        if(j<A.size()) r.insert(r.begin(),A[A.size()-j-1]);
        if(r.size()>1&&r.back()==0) r.pop_back();
        j++;
    }
    reverse(C.begin(),C.end());
    while(C.size()>1&&C.back()==0) C.pop_back();
    return C;
}

int main(){
    string a,b;
    cin>>a>>b;
    vector<int> A,B,r;
    for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');
    for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');

    auto C = div(A,B,r);
    for(int i=C.size()-1;i>=0;i--) printf("%d",C[i]);
    printf("\n");
    for(int i=r.size()-1;i>=0;i--) printf("%d",r[i]);
    return 0;
}