题意：In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10ai burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.

Let’s denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59)=14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9⋅1+5⋅10=59 burles, and there’s no way to do it with fewer banknotes.

For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s)>k).
Input
The first line contains a single integer t (1≤t≤104) — number of test cases.

The first line of each test case contains two integers n and k (1≤n≤10;1≤k≤109).

The next line contains n integers a1,a2,…,an (0=a1<a2<⋯<an≤9).
Output
For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
主要意思就是给定n种钞票，每种钞票的面值都是10的ai次幂。给出你可以使用的钞票的个数m，问在m个钞票内（包括m），你得不到钞票总面值的最小值。
因为要求总面值的最小值，所以我们就从最小的面值开始加。因为个面值之间相差都是10的ai次幂，所以我们每种面值的钞票最多用c[i]=b[i]/b[i-1]-1张（b[i]表示面值,c[i[表示最多能用的个数），那就能多用就多用，先让sum+=b[i]*c[i]（sum代表总面值）。一旦剩余的钞票数m小于当前面值钞票最多能用的张数，剩下的钞票数全用当前面值的钞票，sum+=(m+1)b[i]（因为之前是从小到大加过来的，之前的小面额全都可以表示了，所以这里要k+1）。这就是最大面值。
根据下面的代码更容易理解：

#include<bits/stdc++.h>
using namespace std;
long long n, k, a[101], b[101], c[101];
int main(){int t;cin>>t;while(t--){cin>>n>>k;for (int i=1;i<=n;i++){cin>>a[i];b[i]=pow(10, a[i]);//面值if (i==1)continue;c[i-1]=b[i]/b[i-1]-1;//最多用的个数}long long sum=0, i=1;while(k>=c[i] && i<n){k-=c[i];sum+=c[i]*b[i];i++;}sum+=(k+1)*b[i];cout<<sum<<endl;}
}