HDOJ 5100 Chessboard 构造
2014-12-05 00:06
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MATRIX67大神: http://www.matrix67.com/blog/archives/5900
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 538 Accepted Submission(s): 241
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
2
6 3
5 3
Sample Output
36
24
Source
BestCoder Round #17
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
int T_T;
int n,k;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&k);
if(k>n)
{
puts("0");
continue;
}
int r=n%k;
if(r>k/2) r=k-r;
printf("%d\n",n*n-r*r);
}
return 0;
}
Chessboard
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 538 Accepted Submission(s): 241
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
2
6 3
5 3
Sample Output
36
24
Source
BestCoder Round #17
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
int T_T;
int n,k;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&k);
if(k>n)
{
puts("0");
continue;
}
int r=n%k;
if(r>k/2) r=k-r;
printf("%d\n",n*n-r*r);
}
return 0;
}
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