E.Easy Problem

Description:

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.

Input:

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.

Output:

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.

Sample Input:

3
5 3 1 5 7
21 4 2 5 6 9
10 9 0 1 2 3 4 5 6 7 9

Sample Output:

175
2592576
976543210

題意：

多組資料，每組資料給出一個數x，然後k個0~9的數，現在要你求出一個數y，滿足y%x=0並且y包含這k個數。

題解:

比賽的時候想了半天都沒有想到啊...後來看別人的程式碼恍然大悟。

注意這裡的資料範圍，x只有1e8，然後0~9一共10個數，所以我們可以選取一個大數比如1234567890*1e8，可以將這個作為答案進行待定。

因為要求能夠整除，所以我們用這個大數（假定為n）n%x，令r=n%x，那麼易知r是小於1e8的，我們現在用n加上x-r那麼就可以同時滿足題目中的條件了。

這裡如果用減的話可能會因為借位而對1234567890進行改變，用加就不用擔心這個問題出現了。

感覺思路特別巧妙，主要還是對資料範圍的細心觀察。

程式碼如下：

#include <bits/stdc++.h>
typedef long long ll;
ll n = 123456789000000000;
int main(){
    ll T,k,t,r;
    ll x;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld %lld",&x,&k);
        for(int i=1;i<=k;i++){
            int tmp;
            scanf("%d",&tmp);
        }
        r = n % x;
        t = x - r;
        printf("%lld\n",n+t);
    }
    return 0;
}