hdu 5100

Chessboard

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 176    Accepted Submission(s): 94

Problem Description

Consider the problem of tiling an n×n chessboard by polyomino pieces that are k×1 in size; Every one of the k pieces of each polyomino tile must align exactly with one of the chessboard squares. Your task is to figure out the maximum number of chessboard squares tiled.

 

Input

There are multiple test cases in the input file.
First line contain the number of cases T (T≤10000). 
In the next T lines contain T cases , Each case has two integers n and k. (1≤n,k≤100)

 

Output

Print the maximum number of chessboard squares tiled.

 

Sample Input

26 35 3

 

Sample Output

3624

 

Source

BestCoder Round #17

首先，若n<k，则棋盘连一个1×k的矩形都放不下，输出0。我们只需要考虑n≥k的情况。将棋盘类似于黑白染色，按(i+j)模k划分等价类，给每个格子标一个号。标号之后，会注意到每条从左下到右上的斜线数字都是相同的，那么对于s×s的格子，其内部数字有且恰好有2s−1种，所以当s<=k/2的时候，内部数字有floor(k/2)∗2−1<k种，所以不能有更佳的方案。从而证明最优的方案一定是仅剩下一个s×s的正方形区域没有被覆盖到，其中s≤k/2。而令l=n mod k之后，根据l大小的不同，可以构造出中心为l×l或(k−l)×(k−l)的风车形图案，又通过上面证明这个l（或k−l）就是之前的s，所以是最优的。所以令l=n mod k，如果l≤k/2，最多可覆盖的格子数即为n^2−l^2，否则为n^2−(k−l)^2，显然这样的方案是可以构造出来的（风车形）。

#include<stdio.h>int main(){    int t,n,k,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        if(n<k)        {            printf("0\n");            continue;        }        m=n%k;        if(2*m<=k)        {            printf("%d\n",n*n-m*m);        }        else        {            printf("%d\n",n*n-(k-m)*(k-m));        }    }    return 0;}