POJ 3411 Mondriaan's Dream 銆愮姸鍘婦p銆?By cellur925

Posted 2020-12-28

鏍囩锛?a href='http://www.mamicode.com/so/1/name' title='name'>name   wing   paper   閫氳繃   mina   sed   material   seve   numbers   

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 鈥榯oilet series鈥?(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鈥榣l 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鈥榯 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.

Output

For each test 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

锛堜竴鍙ヨ瘽棰樻剰锛?/p>

 

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 1 bool check(int x)
 2 {
 3     bool ok=1;
 4     for(int i=0;i<n;i++)
 5     {
 6         if(x&(1<<i)) ok=ok^1;
 7         //褰撳墠浣嶄负1锛岃涓嬩负1鐨勪釜鏁帮紝鏈€缁堝繀椤讳袱涓ょ浉閭讳笖涓暟涓哄伓鏁?
 8         if((!(x&(1<<i)))&&!ok) return 0;
 9     }
10     if(ok) return 1;
11     else return 0;
12 }

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娉ㄦ剰缁嗚妭锛氬紑long long锛圵A浜嗗ソ鍑犳锛?/p>

code

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 typedef long long ll;
 7 
 8 int n,m,cnt;
 9 int bin[3000];
10 ll f[12][2050];
11 
12 bool check(int x)
13 {
14     bool ok=1;
15     for(int i=0;i<n;i++)
16     {
17         if(x&(1<<i)) ok=ok^1;
18         if((!(x&(1<<i)))&&!ok) return 0;
19     }
20     if(ok) return 1;
21     else return 0;
22 }
23 
24 void init()
25 {
26     memset(f,0,sizeof(f));
27     memset(bin,0,sizeof(bin));
28     cnt=0;
29 }
30 
31 int main()
32 {
33 
34     while(scanf("%d%d",&n,&m))
35     {
36         if(n==0) break;
37         int fAKe=(1<<n)-1;
38         for(int i=0;i<=fAKe;i++)
39              if(check(i)) bin[++cnt]=i;
40          for(int i=1;i<=cnt;i++) f[1][bin[i]]=1;
41          for(int i=2;i<=m;i++)
42           for(int j=0;j<=fAKe;j++)
43            for(int k=1;k<=cnt;k++)
44             if((bin[k]|j)==j) f[i][j]+=f[i-1][bin[k]^j^fAKe];
45          printf("%lld
",f[m][fAKe]);
46         init();
47     }
48     return 0;
49 }

View Code

 

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