题目链接

Nauuo is a girl who loves playing chess.
One day she invented a game by herself which needs n chess pieces to play on a m×m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game’s goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (ri,ci), while the following rule is satisfied: for all pairs of pieces i and j, |ri−rj|+|ci−cj|≥|i−j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn’t find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input
The only line contains a single integer n (1≤n≤1000) — the number of chess pieces for the game.

Output
The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers ri and ci (1≤ri,ci≤m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples
inputCopy
2
outputCopy
2
1 1
1 2
inputCopy
4
outputCopy
3
1 1
1 3
3 1
3 3
Note
In the first example, you can’t place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

In the second example, you can’t place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

then |r1−r3|+|c1−c3|=|1−2|+|1−1|=1, |1−3|=2, 1<2; and |r1−r4|+|c1−c4|=|1−2|+|1−2|=2|1−4|=32<3
However, on a 3×3；

（思路 ）（不难）；
题意：
在一个 m×m 的棋盘上放 n 颗棋子，第 i 颗棋子的坐标为 (ri,ci)，需要满足 |ri−rj|+|ci−cj|≥|i−j|，求 m 的最小值以及任意一种摆放方案。
解题思路：

#include<bits/stdc++.h>
using namespace std;
int main() {
	int n;
	scanf("%d", &n);
	printf("%d\n", n/2+1);
	int t = 1;
	for(int i = 1; i <= n/2; i++){
		printf("%d %d\n", i, t);
		t++;
		printf("%d %d\n", i, t);
	}
	if (n%2)
        printf("%d %d\n", n/2+1, t);
	return 0;
}