uva 725 Division（除法）暴力法！

                       uva 725  Division（除法）

A - 暴力求解

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

 

Description

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, 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 should 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 solutions for N.". Separate the output for two different values of N by a blank line.

 

Sample Input 

61

62

0

 

Sample Output 

There are no solutions for 61.



79546 / 01283 = 62

94736 / 01528 = 62

 

题意：

输入正整数n,按从小到大的顺序输出所有形如abcde/fghij=n的表达式，其中a—j恰好为数字0—9的一个排列（可以有0前导）  2=<n<=79

思路：

可以用五层循环，但是过程有些繁琐

也可只枚举fghij就可以算出abcde,然后判断符不符合条件。

 

#include<stdio.h>

int p[15];

int juge(int a, int b)    //判断出符合条件的a，b，把a，b每一位都标记一次

{

    if (a > 98765)

        return 0;

    for (int i = 0; i < 10; i++)

    {

        p[i] = 0;

    }

    if (b<10000)

        p[0]=1;

    while(a)

    {

        p[a%10] = 1;                      //判断每一位，然后标记

        a/= 10;

    }

    while(b)

    {

        p[b%10] = 1;

        b/=10;

    }

    int total = 0;

    for (int i = 0; i < 10; i++)

        total+=p[i];                          

    return total==10;                      //直到有符合条件的不同的十个数字

}

int main()

{

    int n, m = 0;

    while (scanf("%d", &n) == 1, n)

    {



        if (m>0) printf("\n");m++;                     //输出格式要求

        int f = 1;

        for (int i = 1234; i < 99999; i++)             //范围应该很好想出

        {

            if (juge(i*n,i))

            {

                printf("%d / %05d = %d\n", i*n,i,n);   //格式控制输出，五位数如果缺就用0补上

                f =0;

            }

        }

        if (f)

        {

            printf("There are no solutions for %d.\n",n);

        }

    }

    return 0;

}