Mondriaan's Dream
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 13519 | Accepted: 7876 |
Description
Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways.
Expert as he was in this
material, he saw at a glance that he'll need a computer to calculate the number
of ways to fill the large rectangle whose dimensions were integer values, as
well. Help him, so that his dream won't turn into a nightmare!
Expert as he was in this
material, he saw at a glance that he'll need a computer to calculate the number
of ways to fill the large rectangle whose dimensions were integer values, as
well. Help him, so that his dream won't turn into a nightmare!
Input
The input contains several test cases. Each test case
is made up of two integer numbers: the height h and the width w of the large
rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.
is made up of two integer numbers: the height h and the width w of the large
rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.
Output
case, output the number of different ways the given rectangle can be filled with
small rectangles of size 2 times 1. Assume the given large rectangle is
oriented, i.e. count symmetrical tilings multiple times.
Sample Input
1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0
Sample Output
1
0
1
2
3
5
144
51205
Source
题目大意：给一个n，m，问n乘m的矩形用1*2的矩形填，填满有多少种方式
ac代码
| 196K | 16MS |
#include<stdio.h>
#include<string.h>
#define LL __int64
LL dp[2][1<<12];
int n,m;
LL add;
void dfs(int i,int p,int pos)
{
if(pos==m)
{
dp[i][p]+=add;
return;
}
dfs(i,p,pos+1);
if(pos<m-1&&!(p&(1<<pos))&&!(p&(1<<(pos+1))))
{
dfs(i,p|(1<<pos)|(1<<(pos+1)),pos+2);
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF,n||m)
{
if(n*m%2)
{
printf("0\n");
continue;
}
int rt=(1<<m)-1;
memset(dp,0,sizeof(dp));
add=1;
dfs(0,0,0);
int i,j;
for(i=1;i<n;i++)
{
memset(dp[i%2],0,sizeof(dp[i%2]));
for(j=0;j<=rt;j++)
{
if(dp[(i-1)%2][j])
{
add=dp[(i-1)%2][j];
dfs(i%2,~j&rt,0);
}
}
}
printf("%I64d\n",dp[(n-1)%2][rt]);
}
}