曾彪彪的个人博客 -- 高精加法优化

# P1601 A+B Problem（高精）

## 题目描述

高精度加法，相当于 a+b problem，**不用考虑负数**。

## 输入格式

分两行输入。$a,b \leq 10^{500}$。

## 输出格式

输出只有一行，代表 $a+b$ 的值。

## 输入输出样例 #1

### 输入 #1

```
1
1
```

### 输出 #1

```
2
```

## 输入输出样例 #2

### 输入 #2

```
1001
9099
```

### 输出 #2

```
10100
```

## 说明/提示

$20\%$ 的测试数据，$0\le a,b \le10^9$；

$40\%$ 的测试数据，$0\le a,b \le10^{18}$。

第一次自己做的思路：
```c++
/**
start time: 8:45
End time:
Type: simulator
Solution:
- use string to simulate a+b.
- just loop the short number
- handle the symbol
- if the two number have the same length, handle the symbol
-
test case:
-
99999
9999
-
9999
9999
-
99999
9
-
1
1
-
0
0
-
789
98789

first submit score: 100
cause:

**/

#include <bits/stdc++.h>
using namespace std;

string add(string a, string b)
{
    string result = "";
    if (a.size() < b.size())
    {
        swap(a, b);
    }
    reverse(a.begin(), a.end());
    reverse(b.begin(), b.end());
    int symbol = 0;
    for (int i = 0; i < b.size(); i++)
    {
        int v = (a[i] - '0') + (b[i] - '0') + symbol;
        symbol = v / 10;
        char data = (v % 10) + '0';
        result.push_back(data);
    }
    int index = b.size();
    while (symbol > 0)
    {
        if (a.size() == index)
        {
            result.push_back('0' + symbol);
            symbol = 0;
        }
        else
        {
            int v = (a[index] - '0') + symbol;
            result.push_back(v % 10 + '0');
            symbol = v / 10;
            index++;
        }
    }
    while (index < a.size())
    {
        result.push_back(a[index]);
        index++;
    }

reverse(result.begin(), result.end());
    return result;
}

int main()
{

// freopen("C:/Users/zengsam/Downloads/P1042_2.in", "r", stdin);
    string a, b;
    cin >> a >> b;
    string ans = add(a, b);
    cout << ans << endl;
    return 0;
}

```

发现有点复杂，看别人题解，优化如下：
```c++
/**
start time: 8:45
End time:
Type: simulator
Solution:
- use string to simulate a+b.
- just loop the short number
- handle the symbol
- if the two number have the same length, handle the symbol
-
test case:
-
99999
9999
-
9999
9999
-
99999
9
-
1
1
-
0
0
-
789
98789

first submit score: 100
cause:

**/

#include <bits/stdc++.h>
using namespace std;

string add(string a, string b)
{
    string result = "";
    if (a.size() < b.size())
    {
        swap(a, b);
    }
    reverse(a.begin(), a.end());
    reverse(b.begin(), b.end());
    int n = a.size() - b.size();
    for (int i = 0; i < n; i++)
    {
        b.push_back('0');
    }
    int symbol = 0;
    for (int i = 0; i < a.size(); i++)
    {
        int v = (a[i] - '0') + (b[i] - '0') + symbol;
        symbol = v / 10;
        char data = (v % 10) + '0';
        result.push_back(data);
    }
    if (symbol > 0)
    {
        result.push_back('0' + symbol);
    }

reverse(result.begin(), result.end());
    return result;
}

int main()
{

// freopen("C:/Users/zengsam/Downloads/P1042_2.in", "r", stdin);
    string a, b;
    cin >> a >> b;
    string ans = add(a, b);
    cout << ans << endl;
    return 0;
}

```

高精加法优化

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