[解题报告]256 - Quirksome Squares-优快云博客

本文探讨了具有特定性质的数字，即将其等分为两部分并平方和后能得到原数的数字。通过程序实现，本文提供了针对不同位数数字的求解方法。

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Quirksome Squares

The number 3025 has a remarkable quirk: if you split its decimal representation in two strings of equal length (30 and 25) and square the sum of the numbers so obtained, you obtain the original number:

The problem is to determine all numbers with this property having a given even number of digits.

For example, 4-digit numbers run from 0000 to 9999. Note that leading zeroes should be taken into account. This means that 0001 which is equal to is a quirksome number of 4 digits. The number of digits may be 2,4,6 or 8. Although maxint is only 32767 and numbers of eight digits are asked for, a well-versed programmer can keep his 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 from 2,4,6,8), each number on a line of its own.

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 that the number of digits in the output is equal to the number in the corresponding input line : leading zeroes may not be suppressed.

Sample Input

2
2

Sample Output

#include<stdio.h>
int main()
{
    int n;
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==2)
        {
            for(i=0;i<10;i++)
                for(j=0;j<10;j++)
                {
                    if(i+j>=10) break;
                    if ((i+j)*(i+j)==(i*10+j)) printf("%02d\n",i*10+j);
                }
        }
        if(n==4)
        {
            for(i=0;i<100;i++)
                for (j=0;j<100;j++)
                {
                    if(i+j>=100) break;
                    if((i+j)*(i+j)==(i*100+j)) printf("%04d\n",i*100+j);
                }
        }

        if(n==6)
        {
            for(i=0;i<1000;i++)
                for(j=0;j<1000;j++)
                 {
                    if(i+j>=1000) break;
                    if((i+j)*(i+j)==(i*1000+j)) printf("%06d\n",i*1000+j);
                }
        }

        if(n==8)
        {
            for(i=0;i<10000;i++)
                for(j=0;j<10000;j++)
                {
                    if(i+j>=10000) break;
                    if((i+j)*(i+j)==(i*10000+j)) printf("%08d\n",i*10000+j);
                }
        }
    }
    return 0;
}