$AC=Answer Coarse$=粗劣的答案

$WA=Wonderful Answer$=好答案

$TLE=Time Limit Enough$=时间充裕

$MLE=Memory Limit Enough$=内存充裕

$CE=Compile Easily$=轻松通过编译

$RE=Run Excellently$=完美运行

$UKE=Unbelievably Keep Enough Score$=难以置信地保持足够的分数

摆烂++++++++++！

高精度算法：

#include<iostream>
#include<vector>
using namespace std;
 
vector<int> CF(vector<int>&A,vector<int>&B){
    vector<int> C(A.size() + B.size(), 0);
    for (int i = 0; i < A.size(); i ++) 
        for (int j = 0; j < B.size(); j ++)
          C[i + j] += A[i] * B[j];
    for (int i = 0; i + 1 < C.size(); i ++){
        C[i + 1] += C[i] / 10;
        C[i] %= 10;
    }
    while(C.size()>1&&C.back()==0) C.pop_back();
    return C;
} 
int main(){
    string a,b;
    cin>>a>>b;
    vector<int> A,B;
    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');
    vector<int> C=CF(A,B);
    for(int i=C.size()-1;i>=0;i--) cout<<C[i];
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
vector<int> A,B,C;
string a,b;
vector<int> fan(vector<int> A){
	vector<int> B;
	for(int i=A.size()-1;i>=0;i--){
		B.push_back(A[i]);
	}
	return B;
}
int check(vector<int> A,vector<int> B){//if(a>b) return 1; else if(a==b) return 0; else return -1;
	if(A.size()>B.size()) return 1;
	else if(A.size()<B.size()) return -1;
	else{
		for(int i=A.size()-1;i>=0;i--){
			if(A[i]>B[i]) return 1; 
			else if(A[i]<B[i]) return -1;
		}
		return 0;
	}
}
vector<int> sub(vector<int> A,vector<int> B){
	vector<int> ans;
	int idx=0,x=0,sub_temp=0;
	for(;idx<A.size();idx++){
		sub_temp=0;
		if(idx<B.size()) sub_temp-=B[idx];
		sub_temp += (A[idx]-x);x=0;
		if(sub_temp<0){
			sub_temp+=10;
			x=1;
		}
		ans.push_back(sub_temp);
	}
	
	for(int i=ans.size()-1;ans[i]==0;i--){
		ans.pop_back();
	}
	return ans;
}
vector<int> suball(vector<int> A,vector<int> B){
	if(!check(A,B)){
		return sub(B,A);
	}else{		
		return sub(A,B);
	} 
}
vector<int> divall(vector<int> A,vector<int> B){
	vector<int> D,E,ans;//A/B
	for(int i=A.size()-1;i>=0;i--){
		D.push_back(A[i]);
		if(check(B,D)>0){
			ans.push_back(0);  //补零之后再下落 
			continue;
			cout<<"除数不够\n当前遍历到的下标："<<i<<"\n当前除数为：";
			for(int i=D.size()-1;i>=0;i--) cout<<D[i];
			cout<<endl;
			D.push_back(A[i]);
			cout<<"将"<<A[i]<<"压入被除数中。\n压入后被除数变为：";
			for(int i=0;i<D.size();i++) cout<<D[i];
			cout<<endl;
			
		} 
		else{
			cout<<"\n准备试除,第一个被除数\n"; 
			E = fan(D);
			for(int i=E.size()-1;i>=0;i--) cout<<E[i];
			cout<<endl;
			int idx=0;
			while(check(B,E)<=0){
				E = suball(E,B);
				idx++;
				cout<<"减完第"<<idx<<"次后，当前除数：";
				for(int i=E.size()-1;i>=0;i--) cout<<E[i];
				cout<<endl;
			}
			ans.push_back(idx);
			D = fan(E); // 上面把D反过来了，所以这里要再次反一下，否则结果有问题 
			cout<<"试除后的余数:";
			for(int i=D.size()-1;i>=0;i--) cout<<D[i];
			cout<<endl;
			cout<<"当时答案：";
			for(int i=0;i<ans.size();i++) cout<<ans[i];
			cout<<endl;
		}
	}
	ans = fan(ans);
	for(int i=ans.size()-1;ans[i]==0;i--){   //返回前过滤前导零 ，注意顺序为从前到后过滤 
		ans.pop_back();
	}
	return ans; 
}
int main(){
	cin>>a>>b;
	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');
	C = divall(A,B);
	for(int i=C.size()-1;i>=0;i--) cout<<C[i];
	return 0;
} 
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