题目描述
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
input
2
3 7
24 25 27
10 7
51 52 53 54 55 56 57 58 59 60
output
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.
题目大意
给你一个d,1<=d<=9。任何一个数x中只要有一位是d,则称x就是幸运数字,现在给你一个数判断他是否可以用一个或者多个幸运数字的和表示。
题目分析
先说结论:如果x的个位数等于某个d的倍数的个位,并且x大于等于这个d的倍数,那么x符合要求。
下面证明一下这个结论。
假设x是一个符合要求的数, x > n ∗ d x>n*d x>n∗d并且x%10==n*d%10 。我们来看看x可以分解为哪些幸运数字。
x = [ ( n − 1 ) ∗ d ] + [ ( x / 10 − ( n ∗ d / 10 ) ) ∗ 10 + d ] x=[(n-1)*d]+[(x/10-(n*d/10))*10+d] x=[(n−1)∗d]+[(x/10−(n∗d/10))∗10+d]
x可以分解成n-1个d和 [ ( x / 10 − ( n ∗ d / 10 ) ) ∗ 10 + d ] [(x/10-(n*d/10))*10+d] [(x/10−(n∗d/10))∗10+d] 这n个数。而不符合上述结论的数则无法进行分解。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <queue>
#include <vector>
#include <set>
#include <bitset>
#include <algorithm>
#define LL long long
#define PII pair<int,int>
#define x first
#define y second
using namespace std;
const int N=2e5+5;
bool check(int x,int d) //判断x是否合法的函数
{
if(x>=10*d) return true; //nd(1<=n<=10)的个位数包含了0-9所有的个位数,因此大于等于10*d的x一定是合法的
for(int i=1;i*d<=x;i++) //枚举所有小于等于x的d的倍数
if(i*d%10==x%10) return true; //如果有符合条件的则返回true
return false;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,d;
scanf("%d%d",&n,&d);
while(n--)
{
int x;
scanf("%d",&x);
if(check(x,d)) puts("YES");
else puts("NO");
}
}
return 0;
}