B. Nezzar and Lucky Number
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
Nezzar’s favorite digit among 1,…,9 is d. He calls a positive integer lucky if d occurs at least once in its decimal representation.
Given q integers a1,a2,…,aq, for each 1≤i≤q Nezzar would like to know if ai can be equal to a sum of several (one or more) lucky numbers.
Input
The first line contains a single integer t (1≤t≤9) — the number of test cases.
The first line of each test case contains two integers q and d (1≤q≤104, 1≤d≤9).
The second line of each test case contains q integers a1,a2,…,aq (1≤ai≤109).
Output
For each integer in each test case, print “YES” in a single line if ai can be equal to a sum of lucky numbers. Otherwise, print “NO”.
You can print letters in any case (upper or lower).
Example
inputCopy
2
3 7
24 25 27
10 7
51 52 53 54 55 56 57 58 59 60
outputCopy
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
NO
Note
In the first test case, 24=17+7, 27 itself is a lucky number, 25 cannot be equal to a sum of lucky numbers.
本题主要是对于不同的情况进行分类讨论，第一种就是对于末尾包含零的数字项，要么就是不包含的情况，或者就是中间位数包含零项，那么我们此时可以用循环去进行相关项的消除，此时这种情况如果能对于幸运数字进行取余并且此时必须有商，才能保证此时的幸运数字的组合是有解的。第二种就是对于数字过小的情况，此时的话能直接被整除或者直接大于等于幸运数字的整数倍，此时本身就是幸运数字。
#include <bits/stdc++.h>
using namespace std;
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
int main() {
FAST;
int T;
cin >> T;
while (T--) {
int n, d;
cin >> n >> d;
for (int i = 1;i <= n; ++i) {
int x;
cin >> x;
bool flag=false;
for (int j = 0;j <= 100; ++j) {
if (j*10 <= x) {
int t = x - j * 10;
if (t % d == 0) {
if (t/d >= 1) flag=true;
}
}
}
if (flag) {
cout << "YES" << endl;
continue;
}
if (x % d == 0 || x >= d*10) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
return 0;
}