uva 471 - Magic Numbers(枚举技巧)

Magic Numbers

Write a program that finds and displays all pairs of integers 

 and 

 such
that:
neither 

 nor 

 have
any digits repeated; and


 , where N is a given integer;

Input and Output

The input file consist a integer at the beginning indicating the number of test case followed by  a blank line. Each test case consists of one line of input containing N. Two input are separated
by a blank line.
For each input the output consists of a sequence of zero or more lines each containing 

 / 

 = N,
where 

 and N are the integers described above. When there are two or more solutions, sort them by increasing numerator
values. Two consecutive output set will separated by a blank line. 

Sample Input

1

1234567890

Sample Output

1234567890 / 1 = 1234567890
2469135780 / 2 = 1234567890
4938271560 / 4 = 1234567890
6172839450 / 5 = 1234567890
8641975230 / 7 = 1234567890
9876543120 / 8 = 1234567890

题意： 给一个n，求出有几种s1 / s2方式去得到他，并且s1，s2必须不能有重复的数字。

思路: 感觉这题题目不清不楚啊。没有n的范围啊。如果n很小那么情况会是非常多的肯定就超时了。。说明没有这样的数字，N应该都是很大的。所以去枚举到9876543210就结束了。然后去判断有没有重复的数字。

代码：

#include <stdio.h>
#include <string.h>
const long long MAXN = 9876543210;
int T, vis[10];
long long n, num;

bool check(long long num) {
memset(vis, 0, sizeof(vis));
while (num) {
if (vis[num % 10]) return false;
vis[num % 10]++;
num /= 10;
}
return true;
}

int main() {
scanf("%d", &T);
while (T--) {
scanf("%lld", &n);
for (long long i = 1; i <= MAXN; i ++) {
num = n * i;
if (num > MAXN) break;
if (check(num) && check(i))
printf("%lld / %lld = %lld\n", num, i, n);
}
if (T) printf("\n");
}
return 0;
}