B. Numbers on the Chessboard -codeforces1027 -csdn博客

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

• 題意：給你一個n×n的方框，在里面根據規則填數字，有q個查詢，輸出每個查詢x，y對應位置數字大小。
  規則：行和列之和為偶數是，數字為前n^2/2（上限)，順序從左到右，從上倒下，如果為奇數，則為n^2/2+1到n^2;

• 題解：首先判斷行和列之和奇偶行，如果是奇數，那么加上n^2/2(如果n^2為奇數，那么要加上1)。然后就是你惡不的判斷了。我們首先可以判定如果行滿足是兩行連續時，那么一共有n個數填了進去，因此我用（a-1）/2看看滿足的有多少個滿兩行對。具體操作如下。
  關于這個題的解法大家可以自己畫圖試一試，找一找為什么要這樣寫

#include<bits/stdc++.h> #define endl '\n' #define pb push_back #define mp make_pair #define _ ios::sync_with_stdio(false) bool SUBMIT = 1; typedef long long ll; using namespace std; const double PI = acos(-1); ll n,q; int main() {if(!SUBMIT)freopen("i.txt","r",stdin);else _;cin>>n>>q;for(ll i=0;i<q;i++){ll a,b;cin>>a>>b;ll c;if(n*n%2)c=n*n/2+1;else c=n*n/2;ll ans=0;if(a>2)ans+=n*((a-1)/2);a-=(a-1)/2*2;if(a>1)ans+=n/2;if((a+b)%2){//if(a>1&&n%2)ans++;//此時不用這句話，因為奇數的話加在了前n^2/2內ans+=(b+1)/2;ans+=c;}else{if(a>1&&n%2)ans++;//必須有，n為奇數且前半部分的話，每對第一行還要多一個數ans+=(b+1)/2;}cout<<ans<<endl;}return 0; }

歡迎歡迎，如果大家喜歡的話可以關注一波O(∩_∩)O哈哈~

posted @ 2018-08-19 13:51 i-Curve 閱讀(...) 評論(...) 編輯 收藏

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