bestcoder#17 1001&&HDU 5100 Chessboard(数学)

Chessboard

                                                                 点击打开题目链接

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 85    Accepted Submission(s): 50

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 (). 
In the next T lines contain T cases , Each case has two integers n and k. ()

 

Output

Print the maximum number of chessboard squares tiled.

 

Sample Input

2 6 3 5 3

 

Sample Output

36 24

 

Source

BestCoder Round #17

 

先来吐槽一下题目，我去，就两句话的英语，读了好长时间才读懂，这次1001栽了不少人；

题目很简单，给一个n*n的正方形棋盘，由kX1的小块组成，问用这个小的（k*1）的小块来覆盖棋盘，最多能占多少格（每格是一个1X1的小格）；

有个结论：当n%k!=0时，最多为：n*n-min（a,b）;其中a=(n%k)*(n%k),b=(k-n%k)*(k-n%k);

当n%k==0时，显然是n*n；

至于证明，当时只是感性的认为这个结论是正确的，按照贪心的思想来放的，发现最后又两种情况；

网上有一则证明：附个网址：http://www.matrix67.com/blog/archives/5900

                          http://www.brand.site.co.il/riddles/201403a.html

代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<stdlib.h>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<set>
#define inf 0x3f3f3f3f
#define eps 1e-5
#define max(a,b) a>b?a:b
//#define min(a,b) a<b?a:b
using namespace std;
int main()
{
    int t,n,k,ans;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d",&n,&k);
            if(k>n)
            {
                printf("%d\n",0);
            }
            else
            {
                if(n%k==0)
                    ans=n*n;
                else
                {
                    int a=(n%k)*(n%k);
                    int b=(k-n%k)*(k-n%k);
                    ans=(n*n)-min(a,b);
                }
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}