高精度算法—加法篇

1 if (s1[0] ==‘0‘ && s2[0] == ‘0‘)
2         return ss += "0";

int tmp = 0;//与高精度减法不同，这里tmp不再为1，而是为0 解决例如999+1=1000，位数增多且其他位数都为0的情况
    while (ans[tmp] == 0)
        tmp++;

 1 #include<iostream>
 2 #include<string>
 3 #include<algorithm>
 4 using namespace std;
 5 int arr1[200000], arr2[200000], ans[200000];
 6 string jiafa(string s1, string s2)
 7 {
 8     string ss;
 9     if (s1[0] ==‘0‘ && s2[0] == ‘0‘)
10         return ss += "0";
11     if (s1.length() < s2.length() || (s1.length() == s2.length() && s1 < s2))
12     {
13         return jiafa(s2, s1);
14     }
15     for (int i = 1; i <= s1.length(); i++)
16         arr1[i] = s1[i - 1] - ‘0‘;
17     for (int i = 1; i <= s2.length(); i++)
18         arr2[i + s1.length() - s2.length()] = s2[i - 1] - ‘0‘;//这里是为了解决长度大的数减去长度小的数 例如89999-9
19     for (int i = s1.length(); i > 0; i--)
20         ans[i] = arr1[i] + arr2[i];
21     for (int i = s1.length(); i > 0; i--)
22     {
23         if (ans[i]>=10)
24         {
25             ans[i]-=10;
26             ans[i - 1]++;
27         }
28     }
29     int tmp = 0;
30     while (ans[tmp] == 0)
31         tmp++;
32     for (int i = tmp; i <= max(s1.length(), s2.length()); i++)
33         ss += ans[i] + ‘0‘;
34     return ss;
35 }
36 int main()
37 {
38     string s1, s2;
39     cin >> s1 >> s2;
40     cout << jiafa(s1, s2);
41 }