题解 | #奇偶统计#

知识点：

循环控制：循环控制

注意：数学公式的积累

#include <iostream>
using namespace std;

int main() {
    int odd = 0;
    int even = 0;
    int n;

    cin >> n;

    if (n % 2 == 0) {
        odd = even = n / 2;
    } else {
        odd = n / 2 + 1;
        even = n / 2;
    }

    cout << odd << " " << even;

    return 0;
}
// 原来解法
/*
int main() {
    int odd = 0;
    int even = 0;
    int n;

    cin >> n;

    for (int i = 1; i <= n; i++) {
        if (i % 2 != 0) {
            odd++;
        } else {
            even++;
        }
    }

    cout << odd << " " << even;

    return 0;
}
*/