除法（Division ,UVA 725）-ACM集训

参考：http://www.cnblogs.com/xiaobaibuhei/p/3301110.html

算法学到很弱，连这么简单个问题都难到我了。但我偏不信这个邪，终于做出来了。不过，是参照别人的，是 xiaobaibuhei 到博客让我找到到感觉，不过我只看了一遍他到代码，后面都是自己写的，虽然写的很像。。。

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

//============================================================================
// Name        : uva 725
// Author      : lvyahui
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <cstdio>
#include <list>
#include <cstring>
using namespace std;

char str[10];
bool val[10];

bool check(int a,int b){
bool isok = true;
sprintf(str,"%05d%05d",a,b);
memset(val,false,sizeof(val));
for (int i = 0; i < 10; ++i) {
if(val[str[i]-'0']){
isok = false;break;
}
val[str[i]-'0'] = true;
}
return isok;
}
int main() {
int n;

memset(val,0,sizeof(val));
scanf("%d",&n);
int b;
bool hasans = false;
for(int a=1234;a < 49383;a++){// b最大到98765 除2就是49383
// b / a = n
b = a * n;
if(b >= 98765){
break;
}
if(check(a,b)){
printf("%05d / %05d = %d\n",b , a , n);
if(!hasans){
hasans = true;
}
}
}
if(!hasans){
printf("There are no solutions for %d",n);
}
return 0;
}