题目：
Suppose you are performing the following algorithm. There is an array v1,v2,…,vn filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:
either choose position pos (1≤pos≤n) and increase vpos by ki;
or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (vj=aj for each j) after some step?

Input
The first line contains one integer T (1≤T≤1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.
The first line of each test case contains two integers n and k (1≤n≤30, 2≤k≤100) — the size of arrays v and a and value k used in the algorithm.
The second line contains n integers a1,a2,…,an (0≤ai≤1016) — the array you’d like to achieve.

Output
For each test case print YES (case insensitive) if you can achieve the array a after some ste***O (case insensitive) otherwise.

样例：

输入：
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810

输出：
YES
YES
NO
NO
YES

题意： 给定一个长度为n的数组，然后判断能否有一个全0的数组加k的次方得到，可以选择不加，但是最多只能加一次，并且k的次方不可重复。
判断是否可以通过变换得到原数组，如果可以输出YES，否则输出NO。

题解： 把原数组中的数写成k进制，判断相同位上的数之和是否大于1

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
bool cmp(int a, int b){
    return a > b;
}
int main()
{
    int T; cin >> T;
    while(T--){
        int n; ll k;
        cin >> n >> k;
        ll a[105];
        for(int i = 0; i < n; i++)
            cin >> a[i];
        int flag = 0, b[105];
        memset(b, 0, sizeof(b));
        for(int i = 0; i < n; i++){
            ll t = a[i];
            int j = 0;
            while(t){
                b[j] += (t % k);
                if(b[j] > 1){
                    flag = 1;
                    break;
                }
                j++;
                t /= k;
            }
            if(flag)
                break;
        }
        if(flag)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
     }
    return 0;
}