湖南大学第十四届ACM程序设计新生杯 E.Easy Problem

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

题意：

多组数据，每组数据给出一个数x，然后k个0~9的数，现在要你求出一个数y，满足y%x=0并且y包含这k个数。

题解:

比赛的时候想了半天都没有想到啊...后来看别人的代码恍然大悟。

注意这里的数据范围，x只有1e8，然后0~9一共10个数，所以我们可以选取一个大数比如1234567890*1e8，可以将这个作为答案进行待定。

因为要求能够整除，所以我们用这个大数（假定为n）n%x，令r=n%x，那么易知r是小于1e8的，我们现在用n加上x-r那么就可以同时满足题目中的条件了。

这里如果用减的话可能会因为借位而对1234567890进行改变，用加就不用担心这个问题出现了。

感觉思路特别巧妙，主要还是对数据范围的细心观察。

代码如下：

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