# 11014: 【原1014】高精度加法

### 题目描述

author: 韩立 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/1014

## Input Format

$$0 \leq a , b \leq 10^{100000}$$。

## Sample Input

1234
1111111111


## Sample Output

1111112345


## FineArtz's solution

/* 高精度加法 */
#include <iostream>
#include <string>
using namespace std;

class BigInt{
friend BigInt operator +(const BigInt&, const BigInt&);
friend ostream& operator <<(ostream &, const BigInt&);
public:
//constructor
BigInt(string);

int getl() const { return len; }
private:
short data[100005];
int len;
};

BigInt::BigInt(string s){
len = s.size();
for (int i = 1; i <= len; ++i)
data[i] = s[len - i] - '0';
}

BigInt operator +(const BigInt &a, const BigInt &b){
BigInt ans("");
int l = (a.getl() > b.getl() ? a.getl() : b.getl());
ans.len = l;
for (int i = 1; i <= l; ++i)
ans.data[i] = a.data[i] + b.data[i];
int i = 1;
while (i <= l + 1){
if (ans.data[i] >= 10){
ans.data[i++] -= 10;
++ans.data[i];
}
else ++i;
}
ans.len = i;
while (ans.data[ans.len] == 0) --ans.len;
return ans;
}
ostream& operator <<(ostream &os, const BigInt &a){
for (int i = a.len; i >= 1; --i)
os << a.data[i];
return os;
}

int main(){
string s1, s2;
cin >> s1 >> s2;
BigInt a(s1), b(s2);
cout << a + b << endl;
return 0;
}


## yyong119's solution

#include <iostream>
#include <cstring>
int a[100001],b[100001],c[100001];
char p[100000],q[100000];
int i,length;
int main(){
using namespace std;
cin>>p>>q;
for (i=0; i<=strlen(p)-1; i++) a[strlen(p)-i]=int(p[i]-'0');
for (i=0; i<=strlen(q)-1; i++) b[strlen(q)-i]=int(q[i]-'0');
if (strlen(p)>=strlen(q)) length=strlen(p);
else length=strlen(q);
for (i=1; i<=length; i++){
c[i]+=a[i]+b[i];
c[i+1]+=c[i]/10; c[i]=c[i]%10;
}
if (c[length+1]) length++;
for (i=length; i>=1; i--) cout<<c[i];
return 0;
}