PAT B: A+B in 1022 D base (20 points)

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1. Topic

Enter two non-negative decimal integers A and B $(\le 2^{.\; 30}-.\; 1)$, the output A + B of the D (1 <D≤10) binary number.

Input format:

The input gives 3 integers A, B, and D in sequence in one line.

Output format:

Output the D number of A+B.

Input example 1:
123 456 8

Output sample 1:
3.2.1

Input example 2:
14.1.28 10.16.27

Output sample 2:
1103

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2. Ideas and precautions

analysis:

1. The conversion process of hexadecimal is to continuously divide the hexadecimal and take the modulus, and store it in the array
2. Possible reasons that are partially correct: No special judgment on sum=0
3. You can also use a container such as string or vector to store, and finally reverse, that is, reverse the order of all the elements.
   
   
   

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Three, AC code

(1) Array storage

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm> 
using namespace std;
int main()
{
    
    
	int M, N, p;
	int sum = 0, i = 0;
	int ans[100] = {
    
     0 };
	cin >> M >> N >> p;
	sum = M + N;
	if(sum==0)
    {
    
    
        printf("0\n");
        return 0;
    }
	while (sum != 0)
	{
    
    
		ans[i] = sum % p;
		sum /= p;
		i++;
	}
	
	for (int k = i-1; k >= 0; k--)
	{
    
    
		cout << ans[k];
	}
}

(2) Vector container

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm> 
using namespace std;
int convert(int k, int p);
int main()
{
    
    
	vector<int>ans;
	vector<int>::iterator it;

	int M, N, p, sum = 0;
	cin >> M >> N >> p;
	sum = M + N;
	
	if (sum == 0) 								//特判
	{
    
    
		cout << "0" << endl;
	}
	else
	{
    
    
			while (sum != 0)
		{
    
    
			ans.push_back(sum % p);
			sum /= p;
		}
		
			reverse(ans.begin(), ans.end());  //最后逆序输出
		
		for (it = ans.begin(); it != ans.end();++it)
		{
    
    
			cout << *it;
		}
	}
}

(3) stirng to store the hexadecimal number

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm> 
using namespace std;
int convert(int k, int p);
int main()
{
    
    
	int M, N, p, sum = 0;
	cin >> M >> N >> p;
	sum = M + N;
	if (sum == 0)
	{
    
    
		cout << "0" << endl;
		return 0;
	}
	else
	{
    
    
		string str ="";
		while (sum != 0)
		{
    
    
			to_string(sum % p);
			str=str+ to_string(sum % p);;
			sum /= p;
		}
		reverse(str.begin(), str.end());
		cout << str << endl;
	}
}

The above three are acceptable.