1.高精度加法
https://leetcode-cn.com/problems/add-strings/

https://leetcode-cn.com/problems/additive-number/

https://leetcode-cn.com/problems/plus-one/

https://leetcode-cn.com/problems/add-binary/submissions/

法一：板子

class Solution {
public:
    string addBinary(string a, string b) {
      vector<int>A,B,C;
      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');
      int t=0;
      for(int i=0;i<A.size()||i<B.size();i++)
      {
          if(i<A.size())  t+=A[i];
          if(i<B.size())  t+=B[i];
          C.push_back(t%2);
          t/=2;
      }
      if(t)  C.push_back(1);
      string c;
      for(int i=C.size()-1;i>=0;i--)  c+=to_string(C[i]);
      return c;
    }
};

法二：简化板子

class Solution {
public:
    string addBinary(string a, string b) {
        reverse(a.begin(),a.end());
        reverse(b.begin(),b.end());
        int t=0;
        string c;

        for(int i=0;i<a.size()||i<b.size()||t;i++)
        {
            if(i<a.size())  t+=a[i]-'0';
            if(i<b.size())  t+=b[i]-'0';
            c+=to_string(t%2);
            t/=2;
        }
        reverse(c.begin(),c.end());
        return c; 
    }
};

2.高进度乘法
https://leetcode-cn.com/problems/multiply-strings/

code
class Solution {
public:
    string multiply(string num1, string num2) {
        vector<int> A, B;
        int n = num1.size(), m = num2.size();
        for (int i = n - 1; i >= 0; i -- ) A.push_back(num1[i] - '0');
        for (int i = m - 1; i >= 0; i -- ) B.push_back(num2[i] - '0');

        vector<int> C(n + m);
        for (int i = 0; i < n; i ++ )
            for (int j = 0; j < m; j ++ )
                C[i + j] += A[i] * B[j];

        for (int i = 0, t = 0; i < C.size(); i ++ ) {
            t += C[i];
            C[i] = t % 10;
            t /= 10;
        }

        int k = C.size() - 1;
        while (k > 0 && !C[k]) k -- ;

        string res;
        while (k >= 0) res += C[k -- ] + '0';

        return res;
    }
};

3.差分
https://leetcode-cn.com/problems/car-pooling/