UVA 725：Division

题目来源：

UVA 725

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题目如下：
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, where2 ≤ N ≤ 79. 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

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题意：
输入正整数n，按照从小到大的顺序输出所有形如abcde/fghiij = n的表达式，其中a-j是0-9的一个排列（可以有前导0），2<=n<=79.

理解：
这道题没有必要枚举0-9的所有排列，只需要枚举fghij乘上n算出abcde再判断是不是都不相同即可。
那么fghij可以从01234开始到54321结束。

判断重复的方式：
由abcde fghij 这两个数定义一个ar[10]，下标0-9，每次ar[fghij%10]++，最后判断是否都为1就可以。

代码如下：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int main(){    int n, s, j, m = 0;    while(scanf("%d", &n) != EOF && n)    {        if(m > 0)            cout << endl;        m++;        int k = 0;        for(int i = 1234; i < 50000; i++){        s = n * i;        if(s >= 12345 && s <= 98765)        {            int ii = i;            int ss = s;            int ar[10] = {0};            while(ss){                ar[ss % 10]++;                ar[ii % 10]++;                ss /= 10;                ii /= 10;            }            for(j = 0; j < 10; j++){                if(ar[j] != 1)                    break;            }            if(j == 10)            {            if(i < 10000)                cout << s << " / 0" << i << " = " << n << endl;            else                cout << s << " / " << i << " = " << n << endl;                    k = 1;            }        }    }    if(k == 0)    cout << "There are no solutions for " << n << "." << endl;    }    return 0;}