思维 hdu 5100 (chessboard)
2016-08-16 17:06
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chessboard
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
题意:给出n*n的矩阵,问用k*1的矩形覆盖,最多能覆盖多少小矩形。
题解:规律是:分成三种情况,n<k,肯定不能覆盖。n>k时,设m=n%k,如果m<=k/2,不能覆盖的矩形为m*m,相反,不能覆盖为(k-m)*(k-m).详细题解
<span style="font-size:18px;">#include<cstdio>
#include<cstring>
using namespace std;
void solve(int n,int k){
int s=n*n;
if (n<k)
printf ("0\n");
else{
int a=n%k;
if (a<=k/2)
printf ("%d\n",s-a*a);
else{
printf ("%d\n",s-(k-a)*(k-a));
}
}
}
int main(){
int t,n,k;
scanf ("%d",&t);
while (t--){
scanf ("%d %d",&n,&k);
solve(n,k);
}
return 0;
} </span>
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
题意:给出n*n的矩阵,问用k*1的矩形覆盖,最多能覆盖多少小矩形。
题解:规律是:分成三种情况,n<k,肯定不能覆盖。n>k时,设m=n%k,如果m<=k/2,不能覆盖的矩形为m*m,相反,不能覆盖为(k-m)*(k-m).详细题解
<span style="font-size:18px;">#include<cstdio>
#include<cstring>
using namespace std;
void solve(int n,int k){
int s=n*n;
if (n<k)
printf ("0\n");
else{
int a=n%k;
if (a<=k/2)
printf ("%d\n",s-a*a);
else{
printf ("%d\n",s-(k-a)*(k-a));
}
}
}
int main(){
int t,n,k;
scanf ("%d",&t);
while (t--){
scanf ("%d %d",&n,&k);
solve(n,k);
}
return 0;
} </span>
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