Cycle-13. Finding positive integer solutions to special equations

Cycle-13. A positive integer solution for a special equation (15) time limit MS Memory limit 65536 KB code length limit 8000 B procedure StandardAuthor Zhang Tong Hongyu (Zhejiang University)

The subject requires that all positive integer solutions of the equation x2+y2=n be obtained for any given positive integer n.

Input format:

The input gives a positive integer n (<=10000) in a row.

Output format:

The output equation x2+y2=n all positive integer solutions, of which x<=y. Each group of solutions accounted for 1 rows, and the two numbers were separated by 1 spaces and output in ascending order of X. If there is no solution, the output is "no solution".

Input Sample 1:

884

Output Example 1:

10 2820 22

Input Sample 2:

11

Output Example 2:

No Solution

1#include <stdio.h>2#include <stdlib.h>3#include <math.h>4 intMain ()5 {6     intN;7scanf"%d", &n);8     intx, y, flag =0;9     Doublem =sqrt (n);Ten      for(x =1; X <= m +1; X + +) One     { A          for(y = x; y <= m +1; y++) -         { -             if(x*x + Y*y = =N) the             { -printf"%d%d\n", x, y); -Flag =1; -             } +         } -     } +     if(!flag) Aprintf"No solution\n"); at     return 0; -}

Cycle-13. Finding positive integer solutions to special equations