Cattle off network - Trend Micro -2020 session on campus recruitment examination on -1

Description Input:
Input of two non-negative number, each a number satisfying the following requirements:
(1) contains only 0-9, 0 or a decimal point ".";
(2) length is not greater than 200;
(3) without wrong ruling.
Description Output:
output and two numbers

1. Thoughts

• Beginning the problem too simple, direct string is converted int / double, then a simple digital addition. : Reflection two errors occurred during
  '' (1) that is a decimal decimal fraction, the integer part is 0, so long as the judgment s [1] == can know whether it is a decimal, so for granted;
  (2) did not notice length is less than 200, i.e., the number string can not speak into int / double, will overflow;
• Later look title problem before and think:
  (1) Since the length is less than 200, the character string can only add and subtract;
  (2) Since the number of inputs may also have decimal integers, it should be divided discussion, ( '.') determined by circumstances s.find: + decimal decimal decimal integer +, + a decimal integer, an integer + integer;
  (3) Consideration difficult situation, i.e., "+ decimal decimal", has three main parts calculated : left part decimal integer addition AddInte (), adding the fractional part of the decimal point right AddFrac (), and adding the carry portion after the decimal integer addition; in some circumstances other portions of the three-part calculation, i.e., the case covers the addition of all possible conditions;
  (4) which align the two numbers are integers and decimal integer string is right-aligned, the decimal string is left-aligned, the same length taken It is added. Integer carry occurs, and then adding the carry to the higher position; decimal carry occurs, and then adding the carry to the integer part.
• Relates to the role of function as follows:
  (. 1) the Add (): two strings are added, the current character string is determined by the find ( '.') Will be classified into four cases: + decimal decimal decimal integer + integer + decimal integer + integer; and all the fractional are divided into an integer part and the fractional part, for adding separately;
  (2) AddInte (): adding an integer, a string right alignment, alignment section () by adding AddNum; If there is a carry, then the carry bit in the higher performing AddInte () operation; Finally, the two longitudinal stitching together;
  (. 3) AddFrac (): decimal addition, the string left aligned, the alignment section by AddNum () with was added; then, the latter half of the splicing together; wherein the integer part to carry reused AddInte () for processing;
  (. 4) AddNum (): adding two integers of the same length.
  
  

2. Implement

#include <iostream>
#include <string>
using namespace std;

#define MAX(a,b) ((a>b)?a:b)

string AddNum(string s1, string s2){//相同长度的两数相加
    string res, s;
    int len = s1.size();
    int n = 0, c = 0;
    for (int i = len - 1; i >= 0; i--){
        n = (s1[i] - '0') + (s2[i] - '0') + c;
        if (n > 9){
            c = n / 10;
            s = '0' + (n % 10);
        }
        else{
            c = 0;
            s = '0' + n;
        }
        res = s + res;
    }
    if (c > 0){
        char cr = '0' + c;
        res = cr + res;
    }
    return res;
}

string AddFrac(string s1, string s2){//两个小数部分相加
    int l1 = s1.size(), l2 = s2.size();
    int len, maxlen, minlen;
    string f1, f2, res;

    //先确定两个小数的长度
    if (l1 > l2){
        len = l1;
        maxlen = l1;
        minlen = l2;
        f1 = s1;
        f2 = s2;
    }
    else{
        len = l2;
        maxlen = l2;
        minlen = l1;
        f1 = s2;
        f2 = s1;
    }

    //将位数更长的小数分成两部分
    string f11 = f1.substr(0, minlen);//与位数更短相加的前半部分
    string f12 = f1.substr(minlen, maxlen - minlen);//剩余后半部分

    string r1 = AddNum(f11, f2);
    int lr = r1.size();
    res = r1 + f12;
    
    return res;
}

string AddInte(string s1, string s2){//两个整数部分相加
    int l1 = s1.size(), l2 = s2.size();
    int len, maxlen, minlen;
    string i1, i2, res, n1, n2;
    
    //先确定两个整数的长度
    if (l1 > l2){
        len = l1;
        maxlen = l1;
        minlen = l2;
        i1 = s1;
        i2 = s2;
    }
    else{
        len = l2;
        maxlen = l2;
        minlen = l1;
        i1 = s2;
        i2 = s1;
    }

    //将位数更长的小数分成两部分
    string i11 = i1.substr(maxlen - minlen, minlen);//与位数更短相加的后半部分
    string i12 = i1.substr(0, maxlen - minlen);//剩余前半部分

    string r1 = AddNum(i11, i2);
    int lr = r1.size();

    if (lr > minlen){//小数部分存在进位
        //将小数进位与整数和进行两个整数相加操作
        n1 = AddInte(i12, r1.substr(0, lr - minlen));
        res = n1 + r1.substr(lr - minlen);
    }
    else{
        res = i12 + r1;
    }

    return res;
}

string Add(string s1, string s2){
    int d1 = s1.find('.');
    int d2 = s2.find('.');
    string i1, i2;
    string f1, f2;
    string n1, n2;
    string res;

    if (d1 > 0 && d2 > 0){//两个都是小数
        i1 = s1.substr(0, d1);
        f1 = s1.substr(d1+1);
        i2 = s2.substr(0, d2);
        f2 = s2.substr(d2 + 1);
        n1 = AddInte(i1, i2);
        n2 = AddFrac(f1, f2);
        int ml = MAX(f1.size(), f2.size());
        int ln2 = n2.size();
        if ( ln2 > ml ){//小数部分存在进位
            //将小数进位与整数和进行两个整数相加操作
            n1 = AddInte(n1, n2.substr(0, ln2 - ml));
            n2 = n2.substr(ln2 - ml);
        }
        res = n1 + '.' + n2;
    }
    else if (d1 > 0 && d2 < 0){//第一个是小数，第二个是整数
        i1 = s1.substr(0, d1);
        f1 = s1.substr(d1 + 1);
        i2 = s2;
        n1 = AddInte(i1, i2);
        n2 = f1;
        res = n1 + '.' + n2;
    }
    else if (d1 < 0 && d2 > 0){//第一个是整数，第二个是小数
        i1 = s1;
        i2 = s2.substr(0, d2);
        f2 = s2.substr(d2 + 1);
        n1 = AddInte(i1, i2);
        n2 = f2;
        res = n1 + '.' + n2;
    }
    else{//两个都是整数
        i1 = s1;
        i2 = s2;
        n1 = AddInte(i1, i2);
        res = n1;
    }

    return res; 
}

int main(){
    string s1, s2;
    while (cin >> s1 >> s2){
        string res;
        res = Add(s1, s2);
        cout << res << endl;
    }
    return 0;
}