CodeForces-1312C. Adding Powers thinking + hex conversion

CodeForces - 1312C. Adding Powers

Original title address:

http://codeforces.com/contest/1312/problem/C

Basic questions:

Give you an array a of n elements and a number k, and ask if you can add an array of all zeros to k ^ i at a time, so that this array becomes an array a, and each i can be used at most once, that is, cannot Use an i multiple times.

The basic idea:

This problem is essentially a hexadecimal replacement. We convert each number in the array a to k hexadecimal, because each i is used at most once, so each bit is enumerated, and the sum of each bit is not greater than 1. Otherwise, it will not work.

Reference Code:

#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define int long long
#define INF 0x3f3f3f3f

const int maxn = 35;
int n,k,a[maxn];
int memo[maxn][64];
void judge(int pos,int m){
    int x = a[pos];
    int top = 0;
    while (x > 0){
        memo[pos][top++] = x % k;
        x /= m;
    }
}
signed main() {
    IO;
    int t;
    cin >> t;
    while (t-- > 0) {
        cin >> n >> k;
        memset(memo,0, sizeof(memo));
        for (int i = 0; i < n; i++) cin >> a[i];
        for (int i = 0; i < n; i++) {
            judge(i, k);
        }
        bool v = true;
        for (int i = 63; i >= 0; i--) {
            int sum = 0;
            for (int j = 0; j < n; j++) {
                sum += memo[j][i];
            }
            if (sum > 1) v = false;
        }
        if (v) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}