简单易懂，高精度加法

简单描述：给两个很大很大的非负整数，求它的和
我们用正常的加运算符，在数值肯定是超过了整型long long的数据范围，因此下面我们来用程序模拟加法

#include <bits/stdc++.h> 
using namespace std;
int main(){
    
    
	string a, b;
	getline(cin, a);
	getline(cin, b);
	int res[10001];
	int k = 0, len1 = a.length(), len2 = b.length();
	int len = min(len1, len2);
	if(len1 != len2){
    
    
		if(len1 > len2){
    
    
			for(int i = 0; i < abs(len2-len1); i++){
    
    
				res[i] = (a[i] - '0');
			}
			for(int i = abs(len2-len1); i < len1; i++){
    
    
				res[i] = (a[i] - '0') + (b[i - abs(len2-len1)] - '0');
			}
			k = len1;
		}else{
    
    
			for(int i = 0; i < abs(len2-len1); i++){
    
    
				res[i] = (b[i] - '0');
			}
			for(int i = abs(len2-len1); i < len2; i++){
    
    
				res[i] = (a[i - abs(len2-len1)] - '0') + (b[i] - '0');
			}
			k = len2;
		}
	}else{
    
    
		for(int i = 0; i < len1; i++){
    
    
			res[i] = (a[i] - '0') + (b[i] - '0');
		}
		k = len1;
		len = 0;
	}
	if(res[0] >= 10) cout<<res[0]/10;
 	for(int i = k; i > len; i--){
    
    
		if(res[i] >= 10){
    
    
			res[i-1] += res[i]/10;
			res[i] %= 10;
		}
	}
	for(int i = 0; i < k; i++){
    
    
		cout<<res[i];
	}
	return 0;
}
/*
输入：
11111111111111111111111111111111
985896848641681651551546
输出：
1111111110969107959752792762662657
*/