标签：index int cf round step test 83 array YES

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.

input Copy
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

output Copy
YES
YES
NO
NO
YES

int t, n,k;
ll a[55];
int index, ji[64];

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n,&k);
		for (int i = 0; i < n; i++)
			scanf("%lld", a + i);
		

		ms(ji, 0);
		bool flag = true;
		for (int i = 0; i < n; i++) {
			index = 0;
			if (!flag) break;

			while (a[i]) {
				if ((a[i] - 1) % k == 0 && ji[index] == 0) {
					a[i]--;
					if(a[i]) a[i] /= k;
					ji[index++] = 1;
				}
				else if (a[i] % k == 0) {
					a[i] /= k;
					index++;
				}
				else {
					flag = false;
					break;
				}
			}
		}

		if (flag)
			printf("YES\n");
		else
			printf("NO\n");
	}

	return 0;
}

标签：index,int,cf,round,step,test,83,array,YES
来源： https://blog.csdn.net/Fawkess/article/details/104774617

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