题解 | #求 1 - n 之间偶数的和#

#include <iostream>
using namespace std;

int main() {
    
    int n;
    cin >> n;
    int sum = 0;
    // write your code here......
    if(n%2==1)
    {
        sum=(n/2)*(n/2+1);
        printf("%d",sum);
    }
    else 
    {
        sum=(n/2)*(n+2)/2;
        printf("",sum);
    }

    cout << sum << endl;
    
    return 0;
}