Codeforces Round #697 (Div. 3), problem: (B) New Year‘s Number,

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).

Example
inputCopy
5
1
4041
4042
8081
8079
outputCopy
NO
YES
YES
YES
NO
暴力循环即可,或者进行第二重的循环优化即可
第一种:
双重循环:

#include<bits/stdc++.h>
using namespace std;
int main(){
    
    
    int n;
    int t;
    cin>>t;
    while(t--)
    {
    
    
        bool flag=false;
        cin>>n;
        for(int i=0;i<=496;i++)
           for(int j=0;j<=495;j++)
             if((i*2020+j*2021)==n)
                {
    
     flag=true; break;}
        if(flag)
          cout<<"YES"<<endl;
        else
        {
    
    
            cout<<"NO"<<endl;
        }
        flag=false;
        
    }
   // system("pause");
}

第二种的优化:

#include<bits/stdc++.h>
using namespace std;
int main(){
    
    
    int n;
    int t;
    cin>>t;
    while(t--)
    {
    
    
        bool flag=false;
        cin>>n;
        for(int i=0;i<=496;i++)
             if((n-i*2020)%2021==0&&(n-i*2020)>=0)
                {
    
     flag=true; break;}
        if(flag)
          cout<<"YES"<<endl;
        else
        {
    
    
            cout<<"NO"<<endl;
        }
        flag=false;
        
    }
    //system("pause");
}

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转载自blog.csdn.net/weixin_46006714/article/details/113178131