找规律-CodeForces 1027B-Numbers on the Chessboard

• 找规律-CodeForces 1027B-Numbers on the Chessboard

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• 题目链接：

B. Numbers on the Chessboard

• 思路：

题目大意：

填数问题，坐标（1开始）x+y为偶数的填 1~n*n/2 的数，其他数填在x+y为奇数的各种，给定坐标，求数值

         

题解：

考虑矩阵阶的奇偶

n%2==0：

x+y为偶数：num=(x-1)*(n/2)+((y+1)/2 )

x+y为奇数：num+n*n/2

n%2==1：

x+y为偶数：num=((x-1)*n+y+1)/2

x+y为奇数：num+n*n/2+1

• 代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define closeio std::ios::sync_with_stdio(false)
#define mem(a,b) memset(a,b,sizeof(a))
 
int main()
{
	ll n,m,i;
	cin>>n>>m;
	while(m--)
	{
		ll x,y,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;
}