题解 | 高精度整数加法

#include <iostream>
#include <string>

using namespace std;

#define N 10010
int a[N];
int b[N];
int res[N];

void transfer(string& s1, string& s2){
    // 将s2 和 s2 转换成一个个数字并存储在a , b 数组中
    int len1 = s1.size();
    int len2 = s2.size();
    for(int i = 0; i < len1; ++i)   a[len1 - i - 1] = s1[i] - '0';
    for(int i = 0; i < len2; ++i)   b[len2 - i - 1] = s2[i] - '0';
}

void Add(){
    int carry = 0;  // 进位位
    for(int i = 0; i < N; ++i){
        res[i] = (a[i] + b[i] + carry) % 10;
        carry = (a[i] + b[i] + carry) / 10;
    }
}

int main()
{
    string str1;
    string str2;
    cin >> str1;
    cin >> str2;
    if(str1 == "0" && str2 == "0"){
        cout << 0 << endl;
        return 0;
    }
    transfer(str1,str2);
    Add();
    int i = N - 5;
    while(res[i] == 0)  --i;
    while(i >= 0)   printf("%d",res[i--]);
    cout << endl;
    return 0;
}