UVa 256 quirksome Squares: enumeration | | Two times congruence

256-quirksome Squares

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=192

The number 3025 has a remarkable quirk:if you split it decimal representation in two strings of equal length (25) and square The sum of the numbers so obtained, you obtain the original number:

The problem is to determine all numbers and has a given even number of digits.

For example, 4-digit numbers run from 0000 to 9999. Note that leading zeroes should is taken into account. This means the 0001 which is equal to

is a quirksome number of 4 digits. The number of digits may 2,4,6 or 8. Although Maxint is only 32767 and numbers of eight digits are to asked for, a well-versed programmer can keep he numbers in The range of the integers. However efficiency should be given a thought.

Input

The input of your program is a Textflle containing numbers of digits (taken to 2,4,6,8), each number on a line of it ow N.

Output

The output is a textfile consisting of lines containing the quirksome numbers (ordered according to the input numbers and For each input number in increasing order).

Warning:please Note this number of digits in the output was equal to the number in the corresponding input Line:lead ING zeroes May is suppressed.

Sample Input

2
2

Sample Output

of
81 ba
(a)

You can enumerate directly from the 0~9999, because the right side of the equation must be a full square.

Enumeration method:

/*0.015s*/
    
#include <bits/stdc++.h>  
using namespace std;  
const int MAXN = 5;  
const int DIGIT[MAXN] = {0, m, 1000, 10000};  
    
Vector<int> STORE[MAXN];  
int I, II, J, X;  
    
void init ()  
{for  
    (i = 0; i < 10000; ++i)  
    {  
        II = i * i;  
        for (j = 1; j < 5; ++j)  
        {  
            if (I < digit[j])  
            {  
                X = ii/digit[j] + II% digit[j];  
                if (X = = i) store[j].push_back (ii)  
            ;  
        }  
    
}} int main ()  
{  
    init ();  
    int N, M;  
    while (~SCANF ("%d", &n))  
    {  
        M = N >> 1;  
        for (i = 0; i < store[m].size (); ++i)  
            printf ("%0*d\n", N, Store[m][i));  
    return 0;  
}