UVA 725 Division (Violent enumeration)

UVA 725 Division

Write a program, finds and displays all pairs of 5-digit numbers, between them use the digits0 through 9 once each, such the first number divided by the second are equal to an integer N, where. That's,

Abcde/fghij = N

Where each letter represents a different digit. The first digit of the numerals is allowed to be zero.

InputEach line of the input file consists of a valid integer N. An input of zero was to terminate the program.OutputYour program has to display all qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator ).

Your output should is in the following general form:

xxxxx/xxxxx = N

xxxxx/xxxxx = N

.

.

In case there is no pairs of numerals satisfying the condition, you must write 'there is no solutionsforN .". Separate the output for both different values of N by a blank line.

Sample Input

61620

Sample Output

There is no solutions for 61.79546/01283 = 6294736/01528 = 62

The main idea: 0~9 10 numbers make up two 5 digits (or four digits beginning with 0), requiring the quotient of two numbers equal to the input data N.

problem-solving ideas: Violence enumerates all situations and then determines whether they are established.

#include <stdio.h>int number[15];int Check (int a, int b) {if (a > 98765) return 0;  for (int i = 0; i < i++) {Number[i] = 0;  } if (b < 10000) number[0] = 1;  while (a) {  number[a%] = 1;  a/=;  }  while (b) {  number[b%] = 1;  b/=;  }  int sum = 0;  for (int i = 0; i < i++)  sum + = Number[i];  return sum = = ten;  } int main () {int ans, cnt = 0;while (scanf ("%d", &ans) = = 1, ans) {if (cnt++) printf ("\ n"); int flag = 0;for (int i = 12 34; i < 99999; i++) {if (check (i * ans, i)) {printf ("%05d/%05d =%d\n", I * ans, I, ans); flag = 1;}} if (!flag) {printf ("There is no solutions for%d.\n", ans);}} return 0;}

UVA 725 Division (Violent enumeration)