Tachibana_Yuki @ 2022-01-06 16:50:51
代码如下 </hr>
import java.util.*;
class Main {
private static Scanner input = new Scanner(System.in);
public static void main(String []args) {
int n = input.nextInt();
int k = input.nextInt();
int sum1 = 0,sum2 = 0,cnt = 0;
for(int i = 1;i <= n;i += 1) {
if(i % k == 0) {
System.out.println(i);
sum1 += i;
cnt += 1;
} else {
sum2 += i;
}
}
double res1 = sum1 / cnt;
double res2 = sum2 / (n - cnt);
// System.out.printf("%d %d\n",cnt,(n - cnt));
System.out.printf("%.1f %.1f\n",res1,res2);
input.close();
}
}
by Tom俩 @ 2022-01-06 16:53:21
@Tachibana_Yuki 好好读题
by _Haoomff_ @ 2022-01-06 17:36:35
@Tachibana_Yuki A 类数可以被 k 整除(也就是说是 k 的倍数),而 B 类数不能,所以还要定一个cnt2,用来存B类数的个数