题目:
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.
题解:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int q,d;
scanf("%d %d",&q,&d);
long long a;
for(int i=0;i<q;i++)
{
scanf("%lld",&a);
if(a>=d*10) cout<<"YES"<<endl;
else
{
int s,r=0;
long long b=a;
while(b>0)
{
s=b%10;
if(s==d)
{
r=1;
break;
}
b/=10;
}
if(r) cout<<"YES"<<endl;
else
{
long long sum=a;
b=a/10;
long long z=b*10;
for(long long i=z;i>=10;i-=10)
{
if((sum-i)%d==0&&sum-i!=0)
{
r=1;
break;
}
}
if(r) cout<<"YES"<<endl;
else
{
if(a%d==0) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
}
}
}
return 0;
}