湖南大学第十四届ACM新生赛--E---Easy problem
Easy problem
TimeLimit：C/C++ 1S，JAVA:2S
MemoryLimit：C/C++ 32.000MB，JAVA:64.000MB
题目描述
Zghh likes number, but he doesn’t like writing problem description. So he will just give you a
problem instead of telling a long story for it. Now given a positive integer x and k digits a1,a2,…,ak, can you find a positive integer y such that
y is the multiple of x and in decimal representation y contains all digits of a1,a2,…,ak.
输入描述
The first line contains an integer T (1<=T<=10000) which is the number of test case.The
following T lines each line is a test case, start with two integer x (1<=x<=1e8) and k (1<=k<=10), k
integer a1,a2,…,ak (0<=ai<=9 for i=1…k and ai!=aj for i!=j) is following.
输出描述
For each test case output your answer y. Your answer should be a positive integer without
leading zero and should be no more than 1e18. Every answer that satisfy the conditions
descripted above will be accepted.
示例输入
3
5 3 1 5 7
21 4 2 5 6 9
10 8 0 1 2 3 4 5 6 7 9
输出
175
2592576
976543210
解：
题意为要找出一个数，既是x的倍数，同时又含有a1,a2……ak，这些数字，如果粗暴地来，按顺序依次检验x的倍数是否含有目标数字，肯定超时，需要改进思路。
而由于本题Special Judge可以有多个正确的解，我们只需要找一个很大很特殊的数，同时含有1,2，3,4，5,6，7,8，9，0中的所有数字，再根据x适当调整成他的倍数，而不影响数字的组成，就可以得到想要的“通解”
#include
#include
#include
#include
using namespace std;
int main()
{
long long m,n,i,j,k,p,t,y,chang,ge;
int a[11],x;
scanf("%lld",&t);
for(p=0;p