B. New Year's Number
                        time limit per test2 seconds
                      memory limit per test256 megabytes

inputstandard input
outputstandard output
Polycarp remembered the 2020-th year, and he is happy with the arrival of the new 2021-th year. To remember such a wonderful moment, Polycarp wants to represent the number n as the sum of a certain number of 2020 and a certain number of 2021.

For example, if:

n=4041, then the number n can be represented as the sum 2020+2021;
n=4042, then the number n can be represented as the sum 2021+2021;
n=8081, then the number n can be represented as the sum 2020+2020+2020+2021;
n=8079, then the number n cannot be represented as the sum of the numbers 2020 and 2021.
Help Polycarp to find out whether the number n can be represented as the sum of a certain number of numbers 2020 and a certain number of numbers 2021.

Input

The first line contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.

Each test case contains one integer n (1≤n≤106) — the number that Polycarp wants to represent as the sum of the numbers 2020 and 2021.

Output

For each test case, output on a separate line:

“YES” if the number n is representable as the sum of a certain number of 2020 and a certain number of 2021;
“NO” otherwise.
You can output “YES” and “NO” in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).

一开始想的是dfs，结果交上去果断超时，然后换了种思路：直接看n中有多少个2020，再用n对2020取模，若余数小于2020的个数，则输出YES，否则输出NO。

#include<bits/stdc++.h>
using namespace std;
int flag=0;
 
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    flag=0;
    int n;
    scanf("%d",&n);
    
    int cont=0;
    cont=n/2020;
    int sum=n%2020;
    if(sum<=cont)
    {
      flag=1;
    }
    else
    flag=0;
    if(flag==1)
    {
      printf("YES\n");
    }
    else
    {
      printf("NO\n");
    }
  }
}