| 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
00 01 81 00 01 81
略水
#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; }