CF1526B I Hate 1111- Codeforces Round #723 (Div. 2)

原文链接Problem - 1526B - Codeforces

B. I Hate 1111

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer xx. Can you make xx by summing up some number of 11,111,1111,11111,…11,111,1111,11111,…? (You can use any number among them any number of times).

For instance,

33=11+11+1133=11+11+11
144=111+11+11+11144=111+11+11+11

Input

The first line of input contains a single integer tt (1≤t≤10000)(1≤t≤10000) — the number of testcases.

The first and only line of each testcase contains a single integer xx (1≤x≤109)(1≤x≤109) — the number you have to make.

Output

For each testcase, you should output a single string. If you can make xx, output "YES" (without quotes). Otherwise, output "NO".

You can print each letter of "YES" and "NO" in any case (upper or lower).

Example

input

Copy

output

Copy

YES
YES
NO

Note

Ways to make 3333 and 144144 were presented in the statement. It can be proved that we can't present 6969 this way.

---------------------------------------------------------------------------------------------------------------------------------

首先11 111可以表示出之后的1111 11111 111111....

也就是说，如果一个数满足条件 n==a*11+b*111

一般做法是，枚举b的个数，然后判断n-111*b是否能够整除11

# include<iostream>
# include<algorithm>
# include<iomanip>
# include<vector>
# include<map>
# include<math.h>
# include<cstring>
# include<queue>
using namespace std;
typedef long long int ll;int main()
{int t;cin>>t;while(t--){ll n;scanf("%lld",&n);int flag=0;for(ll i=0;i*111<=n;i++){if((n-i*111)%11==0){flag=1;cout<<"YES"<<'\n';break;}}if(!flag){cout<<"NO"<<'\n';}}return 0;
}

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