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C. Adding Powers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Suppose you are performing the following algorithm. There is an array v1,v2,…,vnv1,v2,…,vn filled with zeroes at start. The following operation is applied to the array several times — at ii-th step (00-indexed) you can:

• either choose position pospos (1≤pos≤n1≤pos≤n) and increase vposvpos by kiki;
• 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 vv equal to the given array aa (vj=ajvj=aj for each jj) after some step?

Input

The first line contains one integer TT (1≤T≤10001≤T≤1000) — the number of test cases. Next 2T2T lines contain test cases — two lines per test case.

The first line of each test case contains two integers nn and kk (1≤n≤301≤n≤30, 2≤k≤1002≤k≤100) — the size of arrays vv and aa and value kk used in the algorithm.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10160≤ai≤1016) — the array you'd like to achieve.

Output

For each test case print YES (case insensitive) if you can achieve the array aa after some step or NO (case insensitive) otherwise.

Example

input

Copy

54 1000 0 0 01 213 41 4 13 20 1 33 90 59049 810

output

Copy

YESYESNONOYES

Note

In the first test case, you can stop the algorithm before the 00-th step, or don't choose any position several times and stop the algorithm.

In the second test case, you can add k0k0 to v1v1 and stop the algorithm.

In the third test case, you can't make two 11 in the array vv.

In the fifth test case, you can skip 9090 and 9191, then add 9292 and 9393 to v3v3, skip 9494 and finally, add 9595 to v2v2.

【思路】

给出n个数，，减去k的任意次幂，，每一次幂只能使用一次，，问是否最后的序列可以变成0，

#include 
   
    using namespace std;typedef long long ll;ll a[111],s[100011],x,flag,cnt,m;int main(){    ios::sync_with_stdio(false);    int t,n,k;    cin>>t;    while(t--)    {        cin>>n>>k;        flag=0,m=0;        for(int i=1;i<=n;i++)        {            cin>>a[i];            x=a[i],cnt=0;            while(x)            {                ++cnt;                s[cnt]+=x%k;                x/=k;            }            m=max(m,cnt);        }        for(int i=1;i<=m;i++)        {            if(s[i]>1)            {                flag=1;                break;            }        }        for(int i=1;i<=m;i++)        s[i]=0;        if(flag)            puts("NO");        else            puts("YES");    }    return 0;}
   

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