题目要求
分析
根据排列组合,得到矩形(含正方形)的数量:
而对于一个边长为 i (1 ≤ i ≤ min{n, m}) 的正方形,数量为:(n-i+1)(m-i+1)
所以所有的正方形的数量:
由于本题的长方形不考虑正方形,所以真实数值为上面两个数值做减法:
要记得用long保命诶!
AC代码(Java语言描述)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
long m = scanner.nextLong(), n = scanner.nextLong();
scanner.close();
long rectangle = (m*(m+1)/2)*(n*(n+1)/2), square = 0;
for (long i = 1; i <= Math.min(m, n); i++) {
square += (n-i+1)*(m-i+1);
}
System.out.println(square + " " + (rectangle-square));
}
}