高精度减法

题意：给你两个整数a,b，计算a+b 的和是多少？(a>0,b>0)

题解：模拟加法

没压位

/***
    高精度加法：模拟计算
    1.大整数存储:    A3A2A1A0
    A < 10 A:0~9
    len(A) < 10 A:0~999999999
***/
#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;
const int N = 1e6 + 7;

// C = A + B;
vector<int>add(vector<int> &A, vector<int>&B)
{
    vector<int> C;
    int t = 0;  //进位
    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 % 10);
        t /= 10;
    }
    if(t) C.push_back(1);  //是否进位
    return C;
}
int main()
{
    string a,b;
    vector<int> A,B;

    cin>>a>>b;
    for(int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0'); // A = [6 , 5, 4, 3, 2, 1]
    for(int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0'); // B = [6 , 5, .........1]

    vector<int> C = add(A, B);
    for(int i = C.size() - 1; i >= 0; i--) printf("%d",C[i]);
    return 0;
}

压位

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

const int base = 1e9;   //压9位

vector<int> add(vector<int> &A, vector<int> &B)
{
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        if(i < A.size()) t += A[i];
        if(i < B.size()) t += B[i];
        C.push_back(t % base);
        t /= base;
    }

    if (t) C.push_back(t);
    return C;
}

int main()
{
    string a, b;
    vector<int> A, B;
    cin >> a >> b;

    for (int i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i -- )
    {
        s += (a[i] - '0') * t;
        j ++, t *= 10;
        if (j == 9 || i == 0)
        {
            A.push_back(s);
            s = j = 0;
            t = 1;
        }
    }
    for (int i = b.size() - 1, s = 0, j = 0, t = 1; i >= 0; i -- )
    {
        s += (b[i] - '0') * t;
        j ++, t *= 10;
        if (j == 9 || i == 0)
        {
            B.push_back(s);
            s = j = 0;
            t = 1;
        }
    }

    vector<int> C = add(A, B);

    cout << C.back();
    for (int i = C.size() - 2; i >= 0; i -- ) printf("%09d", C[i]);   //前导 0 不输出
    cout << endl;

    return 0;
}