大数相加和大数相乘以及打印从1到最大的n位数

string add(string a, string b){
    int nlength;
    int diff;
    if (a.size() > b.size()){
        nlength = a.size();
        diff = a.size() - b.size();
        b.insert(b.begin(), diff, '0');
        //cout << b << endl;
    }
    else{
        nlength = b.size();
        diff = b.size() - a.size();
        a.insert(a.begin(), diff, '0');
        //cout << a << endl;
    }
    //cout << a << endl;
    //cout << b << endl;
    //cout << c << endl;
    int takeover = 0;
    for (int i = nlength - 1; i >= 0; i--){
        int temp = a[i]-'0' + b[i] - '0' + takeover;
        cout << a[i] << " " << b[i] << endl;
        cout << temp << endl;
        if (temp >= 10){
            takeover = 1;
            b[i] = temp + '0' - 10;
            
            //cout << c[j] << endl;
        }
        else{
            b[i] = temp + '0';
            takeover = 0;
            
        }
    }
    //cout << takeover<<" " << j << endl;
    if (takeover == 1)b = '1' + b;
    return b;
}

void print(string str){
    int count = 0;
    for (int i = 0; i < str.size(); i++){
        if (str[i] == '0')count++;
        else
            break;
    }
    cout << str.substr(count, str.size() - count) << endl;;
}

 注：char a='9'；

        int b=a-'0';

        a的范围只能是0到9

better：

一位一位相加，然后算和算进位，最后根据进位情况看需不需要补一个高位

string addStrings(string num1, string num2) {
       string res = "";
        int m = num1.size(), n = num2.size(), i = m - 1, j = n - 1, carry = 0;
        while (i >= 0 || j >= 0) {
            int a = i >= 0 ? num1[i--] - '0' : 0;
            int b = j >= 0 ? num2[j--] - '0' : 0;
            int sum = a + b + carry;
            res.insert(res.begin(), sum % 10 + '0');
            carry = sum / 10;
        }
        return carry ? "1" + res : res;
    }

---

大数相乘

class Solution {
public:
    string multiply(string num1, string num2) {
        int n=num1.size();
        int m=num2.size();
        vector<int>temp(n+m,0);//一个m位和n位数相乘得到的数最多是n+m位
        for(int i=n-1;i>=0;i--){
            
            for(int j=m-1;j>=0;j--){
                 temp[i+j+1]+=(num1[i]-'0')*(num2[j]-'0');
            }
        }
        int bi=0;
        for(int i=temp.size()-1;i>=0;i--){
            temp[i]=temp[i]+bi;
            bi=temp[i]/10;
            temp[i]=temp[i]%10;
        }
        int i = 0;
        while (temp[i] == 0)i++;//跳过前面多余的0
        if(i>=temp.size())return "0";
        string res="";
        for(;i<temp.size();i++){
            res+=temp[i]+'0';
        }
        
        return res;
    }
};

---

打印从1到最大n位数

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


bool stop(string &str){
    bool stop = false;
    int i = str.size() - 1;
    int carry = 0;
    while (i >= 0){
        int tmp = str[i] - '0' + carry;
        if (i == str.size() - 1)tmp++;
        if (tmp >=10){
            if (i == 0){
                stop = true;
            }
            else{
                str[i] = tmp % 10 + '0';
                carry = tmp / 10;
                i--;
            }
            
        }
        else{
            str[i] = tmp + '0';
            break;
        }
    }
    return stop;
}

void print(string &str){
    bool flag = false;
    string res;
    for (int i = 0; i < str.size(); i++){
        if (str[i] == '0'&&flag == false)continue;
        else if (str[i] != '0'&&flag == false){
            flag = true;
            res.push_back(str[i]);
        }
        else{
            res.push_back(str[i]);
        }
    }
    cout << res << endl;
}
void PrintNumber(int n){
    string res(n,'0');
    while (!stop(res)){
        print(res);
    }
}
int main(){
    int n = 3;
    PrintNumber(n);
    system("pause");
    return 0;
}