CF1312C Adding Powers

题目链接:https://codeforces.com/contest/1312

 

题目大意:

能否对一个数组执行任意次操作,使得其变为目标数组。

对于第i次操作,我们可以放弃,或给数组中任意一个元素加上k^i

 

想法:

我们不难发现一个数 = k^x + k^y + k^z + ... (x != y != z)

这个形式很像我们的二进制所以我们可以知道就是把这个数转化为 k 进制,然后每一位只可以用一次

 

#pragma GCC optimize(3,"Ofast","inline")//O3优化
#pragma GCC optimize(2)//O2优化
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <cstring>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)

const double eps = 1e-10;
const int maxn = 2e6 + 10;
const LL mod = 1e9 + 7;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

int vis[100];
int main() {
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while (T--) {
        memset(vis,0,sizeof(vis));
        int n;
        LL k;
        cin >> n >> k;
        bool fl = false;
        for (int i = 1;i <= n;i++) {
            int cnt = 0;
            LL a;
            cin >> a;
            if (a == 0 || fl)
                continue;
            while (1) {
                if (a % k == 1) {
                    if (!vis[cnt])
                        vis[cnt] = 1;
                    else {
                        fl = true;
                        break;
                    }
                }
                else if (a % k > 1) {
                    fl = true;
                    break;
                }
                a = a / k;
                cnt++;
                if (a == 0)
                    break;
            }
        }
        if (fl)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}

 

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