Luogu p2241 statistical square

Background

1997 First question of the year popularization group

Title Description

There is one n×m Checkerboard with squares , Find out how many squares the square contains 、 Rectangle （ Does not contain a square ）.

Input format

a line , Two positive integers n,m（5000n≤5000,m≤5000）.

Output format

a line , Two positive integers , Indicate how many squares the square contains 、 Rectangle （ Does not contain a square ）.

I/o sample

Input

2 3

Output

8 10

The title is easy to understand , But it still needs some tips to do this topic .

Here is an example of the topic , There is one 2 * 3 The chessboard of , Find the number of squares and rectangles

First look at 1 * 1 The number of squares , Obviously there is 6 individual , 2 * 2 There are... In all 2 individual

Look at the rectangle 1 * 2 The rectangle has 3 individual , 2 * 1 The number of rectangles is 4 individual ,1 * 3 The number of rectangles is 2 individual , Finally, only one large rectangle is left 2 * 3.

If you look at it this way , So we can't seem to find any rules . Now we need to find a law that can find a square and a rectangle  . Here's what I think , Let's bring in a coordinate axis , Find the relationship between each point and the number of squares and rectangles

  Now from (1 , 1) Start to look at , The number of squares is 1 , The number of rectangles is 0, Sum to 1

(2 , 1), At this time, compared with (1 , 1) Compare to , There are too many squares 1 , There are too many rectangles 1 , It's too much 2

(3 , 1), At this time, compared with (2 , 1) Compare to , There are too many squares 1 , There are too many rectangles 2 , It's too much 3

Now let's summarize the rules , If you don't understand this Law , Or if you don't understand it, you can test it according to the rules I put forward

The number of squares added should be (x , y) The smaller of the coordinates , The total increase of square and rectangle should be x * y.

It seems that there is such a simple and convenient rule , It seems that we can calculate the number of squares in the chessboard . The number of rectangles, of course, we can use the number of all rectangles minus the number of squares , Nature is the number of rectangles .

The code here is relatively simple

#include<iostream>
#include<algorithm>
using namespace std;
int main() {
	long long n, m;
	long long squ = 0, sum = 0;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			squ += min(i, j);
			sum = sum + i * j;
		}
	}
	cout << squ << " " << sum - squ;
	return 0;
}

, Just a few comments