Codeforces Round #564 (Div. 2)B
B. Nauuo and Chess
题目链接:http://codeforces.com/contest/1173/problem/B
题目
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.
Example
input
2
output
2
1 1
1 2
题意
给你一个数,你在方格中放这1~n这些数,使得任意两个数a->b在方格中走的格数大于等于|a-b|的值
输出1~n每个数的位置
思路
给你一个表格即可发现规律
1 | 2 | ||
3 | 4 | ||
5 | 6 | ||
行走路径为先向下再向右,两数之间步数与差值都相等
#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; cout<<n/2+1<<endl; for(int i=1;i<=n;i++) { cout<<(i+1)/2<<' '<<i/2+1<<endl; } return 0; }
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