C. Adding Powers---------------------------思维

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 step or NO (case insensitive) otherwise.

Example
inputCopy
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
outputCopy
YES
YES
NO
NO
YES
Note
In the first test case, you can stop the algorithm before the 0-th step, or don’t choose any position several times and stop the algorithm.

In the second test case, you can add k0 to v1 and stop the algorithm.

In the third test case, you can’t make two 1 in the array v.

In the fifth test case, you can skip 90 and 91, then add 92 and 93 to v3, skip 94 and finally, add 95 to v2.

题意:
给定一个数组a，有n个数。再给你一个k，让你把b数组的每个元素加上k的幂(加上k的幂不得重复)使得b数组等于a数组

b数组元素都是0

解析:
对于每个数，分解成k的幂，如果出现重复，即输出"NO";

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[10005];
int t,n;
ll k,maxn;
map<int,int> v;
ll find(ll x)
{
	for(int i=0;;i++)
	{
		if(pow(k,i)>x) return i-1;
	}
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %lld",&n,&k);
		v.clear();
		for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
		ll p=find(maxn);
		int f=0,res=0;
		for(int i=1;i<=n;i++)
		{
			if(a[i]==0) continue;
			ll x=a[i];
			while(x)
			{
				int l=find(x);
				if(l<0||v[l]==1)
				{
					f=1;
					break;
				}
				v[l]=1;
				x-=pow(k,l);
			}
		}
		if(f==1)
		{
			cout<<"NO"<<endl;
		}
		else
		{
			cout<<"YES"<<endl;
		}
		
	}
 }