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");
}
}
}