Informatics Olympiad all in one YBT 1175: divide by 13 | openjudge noi 1.13 27: divide by 13

【 Topic link 】

ybt 1175： Divide 13
OpenJudge NOI 1.13 27: Divide 13

【 Topic test site 】

1. High precision

Investigate ： High precision except low precision High precision die low precision
Explanation of high-precision calculation

【 Solution code 】

solution 1： Use functions and arrays

#include <bits/stdc++.h>
using namespace std;
#define N 105 
void toNum(char s[], int a[])
{
    
    a[0] = strlen(s);
    for(int i = 1; i <= a[0]; ++i)
        a[i] = s[a[0] - i] - '0';
}
void showNum(int a[])
{
    
    for(int i = a[0]; i >= 1; --i)
        cout << a[i];
}
void setLen(int a[], int i)
{
    
    while(a[i] == 0 && i > 1)
        i--;
    a[0] = i;
}
// High precision number divided by low precision number  a For divisor  b Divisor  r For business , Return value is remainder  
int Divide(int a[], int b, int r[])
{
    
    int x = 0;//x: Is the remainder of the last division , And as the next minuend 
    for(int i = a[0]; i >= 1; --i)
    {
    
        r[i] = (x * 10 + a[i]) / b;
        x = (x * 10 + a[i]) % b;
    }
    setLen(r, a[0]); 
    return x;
}
int main()
{
    
	int a[N] = {
    }, r[N] = {
    }, m;
    char s[N];
    cin >> s;
    toNum(s, a);
    m = Divide(a, 13, r);
    showNum(r);
    cout << endl << m;
	return 0;
}

solution 2： Overloaded operator in class

#include<bits/stdc++.h>
using namespace std;
#define N 105
struct HPN
{
    
	int a[N];
	HPN()
	{
    
		memset(a, 0, sizeof(a));
	}
	HPN(char s[])
	{
    
		memset(a, 0, sizeof(a));
		a[0] = strlen(s);
		for(int i = 1; i <= a[0]; ++i)
	        a[i] = s[a[0] - i] - '0';
	}
	int& operator [] (int i)
	{
    
		return a[i];
	}
	void setLen(int i)// Determine number of digits 
	{
    
		while(a[i] == 0 && i > 1)
	        i--;
	    a[0] = i;
	}
	void show()
	{
    
		for(int i = a[0]; i >= 1; --i)
        	cout << a[i];
	}
	HPN operator / (int b) // High precision except low precision 
    {
    
        int x = 0;
        HPN r;
        for(int i = a[0]; i >= 1; --i)
        {
    
            x = x * 10 + a[i];
            r[i] = x / b;
            x %= b;
        }
        r.setLen(a[0]);
        return r;
    }
    int operator % (int b) // High precision die low precision 
    {
    
        int x = 0;
        for(int i = a[0]; i >= 1; --i)
            x = (x * 10 + a[i]) % b;
        return x;
    }
};
int main()
{
    
	char s[N];
    cin >> s;
	HPN a(s), r;
	int m;
    r = a / 13;// High precision except low precision  
	m = a % 13;// High precision die low precision 
    r.show();
    cout << endl << m; 
    return 0;
}