E - 5 CodeForces - 1027B（找规律）

You are given a chessboard of size n×n. It is filled with numbers from 1 to n2 in the following way: the first ⌈n22⌉ numbers from 1 to ⌈n22⌉ are written in the cells with even sum of coordinates from left to right from top to bottom. The rest n2−⌈n22⌉ numbers from ⌈n22⌉+1 to n2 are written in the cells with odd sum of coordinates from left to right from top to bottom. The operation ⌈xy⌉ means division x by y rounded up.

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

You are given q queries. The i-th query is described as a pair xi,yi. The answer to the i-th query is the number written in the cell xi,yi (xi is the row, yi is the column). Rows and columns are numbered from 1 to n.

Input
The first line contains two integers n and q (1≤n≤109, 1≤q≤105) — the size of the board and the number of queries.

The next q lines contain two integers each. The i-th line contains two integers xi,yi (1≤xi,yi≤n) — description of the i-th query.

Output
For each query from 1 to q print the answer to this query. The answer to the i-th query is the number written in the cell xi,yi (xi is the row, yi is the column). Rows and columns are numbered from 1 to n. Queries are numbered from 1 to q in order of the input.

Examples
Input
4 5
1 1
4 4
4 3
3 2
2 4
Output
1
8
16
13
4
Input
5 4
2 1
4 2
3 3
3 4
Output
16
9
7
20
Note
Answers to the queries from examples are on the board in the picture from the problem statement.

题意：图在cf上，就是一个有规律的棋盘，输入坐标，输出坐标对应的数字。
思路：找规律，推公式。打表估计会超时，找规律题炒鸡恶心~~

#include <bits/stdc++.h>//分奇偶找规律

using namespace std;
#define ll long long
int main()
{
    ll n,m;
    cin>>n>>m;
    while(m--)
    {
        ll x,y;
        ll num;
        cin>>x>>y;
        if(n % 2 == 0)
        {
            num = (x - 1) * (n / 2) + ((y + 1) / 2);
            if((x + y) % 2 == 0)
                cout<<num<<endl;
            else
                cout<<num + n * n / 2<<endl;
        }
        else
        {
            num = ((x - 1) * n + y + 1) / 2;
            if((x + y) % 2 == 0)
                cout<<num<<endl;
            else
                cout<<num + n * n / 2 + 1 <<endl;
        }
    }
    return 0;
}