(Find out the form of ABCDE/fghij = N-Brute Force enumeration)

Division Time limit:3000 Ms Memory limit:0 KB 64bit Io format:% LLD & % LlU

Description

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits0Through9Once each, such that the first number divided by the second is equal to an integerN, Where. That is,

ABCDE/fghij = N

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

Input each line of the input file consists of a valid integer N. An input of zero is to terminate the program.

Output your program have to display all qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator ).

Your output shocould be in the following general form:

XXXXX/XXXXX = N

XXXXX/XXXXX = N

.

.

In case there are no pairs of numerals satisfying the condition, you must write''There are no solutionsN.". Separate the output for two different valuesNBy a blank line.

Sample Input

61620

Sample output

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

Miguel Revilla
2000-08-31

Question:

As shown in the sample.

Solution:

Brute force enumeration. It is faster to start searching for molecules from the denominator, And the sprintf function is also used to make it easier to determine whether or not the numerator is heavy.

Code:

# Include <iostream> # include <cstdio> # include <string> # include <cstring> using namespace STD; int N, num, den; char buffer [20]; bool visited [10]; bool havecommonelement (int A, int B) {int temp; sprintf (buffer, "% 05d % 05d", a, B ); // The buffer here must be char and cannot be string. memset (visited, false, sizeof (visited); For (INT I = 0; I <strlen (buffer); I ++) {temp = buffer [I]-'0'; If (visited [temp]) return true; else visited [temp] = true;} return F Alse;} int main () {int Casen = 0, flag; while (scanf ("% d", & N )! = EOF & N) {If (CASEN> 0) printf ("\ n"); Casen ++; flag = 0; For (INT den = 1234; den <100000; den ++) {// the molecule is controlled within 5 widths, so it must be less than 100000. if (Den % N = 0) {// previously, it takes too much time to obtain the denominator using known molecules. After using the known denominator to obtain the denominator, because only the entire division enters the subfunction, therefore, the efficiency has improved a lot. num = den/N; If (! Havecommonelement (Num, Den) {printf ("% 05d/% 05d = % d \ n", Den, num, n); flag = 1 ;}}} if (flag = 0) printf ("there are no solutions for % d. \ n ", n);} return 0 ;}