題目描述

You are given a chessboard of size?n \times nn×n?. It is filled with numbers from?11?to?n^2n2?in the following way: the first?\lceil \frac{n^2}{2} \rceil?2n2???numbers from?11?to?\lceil \frac{n^2}{2} \rceil?2n2???are written in the cells with even sum of coordinates from left to right from top to bottom. The rest?n^2 - \lceil \frac{n^2}{2} \rceiln2??2n2???numbers from?\lceil \frac{n^2}{2} \rceil + 1?2n2??+1?to?n^2n2?are written in the cells with odd sum of coordinates from left to right from top to bottom. The operation?\lceil\frac{x}{y}\rceil?yx???means division?xx?by?yyrounded up.

For example, the left board on the following picture is the chessboard which is given for?n=4n=4?and the right board is the chessboard which is given for?n=5n=5?.

You are given?qq?queries. The?ii?-th query is described as a pair?x_i, y_ixi?,yi??. The answer to the?ii?-th query is the number written in the cell?x_i, y_ixi?,yi??(?x_ixi??is the row,?y_iyi??is the column). Rows and columns are numbered from?11?to?nn?.

輸入輸出格式

輸入格式：

The first line contains two integers?nn?and?qq?(?1 \le n \le 10^91≤n≤109?,?1 \le q \le 10^51≤q≤105?) — the size of the board and the number of queries.

The next?qq?lines contain two integers each. The?ii?-th line contains two integers?x_i, y_ixi?,yi??(?1 \le x_i, y_i \le n1≤xi?,yi?≤n?) — description of the?ii?-th query.

輸出格式：

For eachuery from?11?to?qq?print the answer to this query. The answer to the?ii?-th query is the number written in the cell?x_i, y_ixi?,yi??(?x_ixi??is the row,?y_iyi??is the column). Rows and columns are numbered from?11?to?nn?. Queries are numbered from?1?to?q?in order of the input.

輸入輸出樣例

輸入樣例#1

4 5 1 1 4 4 4 3 3 2 2 4

輸出樣例#1

1 8 16 13 4

輸入樣例#2

5 4 2 1 4 2 3 3 3 4

輸出樣例#2

16 9 7 20

說明

Answers to the queries from examples are on the board in the picture from the problem statement.

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思路

非常難搞的數學題......CF B題都那么難的么？

大意是按題目中的圖填數，輸入x，y代表圖中的坐標，輸出(x,y)的數值。

第一輪填數的橫縱坐標和為偶數，第二輪填數的橫縱坐標和為奇數。找規律可得數值與坐標的關系是(x-1)*n+(y+1)。

第二輪只要加上(n*n)就好了。

答案最后別忘了除以2，因為第一輪和第二輪我們是分情況討論的。

#include <stdio.h> #include <iostream> using namespace std; long long int n,t,s; signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cin>>n>>t;while(t--){int x,y;s=0;cin>>x>>y;s=(x-1)*n+y+1;if((x+y)%2){s+=n*n;}cout<<s/2<<endl;}return 0; }

?

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