列表

详情

NC51189. Mondriaan's Dream

描述

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!

输入描述

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, .

输出描述

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.

示例1

输入: 

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

输出: 

1
0
1
2
3
5
144
51205

原站题解

上次编辑到这里，代码来自缓存 点击恢复默认模板

C++(g++ 7.5.0) 解法, 执行用时: 11ms, 内存消耗: 376K, 提交时间: 2023-05-30 22:57:24 

#include<bits/stdc++.h>
using namespace std;
 
int main() {
  int h, w; 
  while (cin >> h >> w && h) {
    //swap(h, w);
    int sz = 1 << w;
    vector<long long> cur(sz);;
    cur[0] = 1;
    while (h--) {
      for (int i = 0; i < w; i++) {
        vector<long long> nxt(sz);;
        for (int s = 0; s < sz; s++){
          nxt[s] += cur[s^(1<<i)];
          if ((s&(1<<i)) == 0 && i + 1 < w && (s & (1<<(i+1)))) nxt[s] += cur[s^(1<<(i+1))];
        }
        cur = nxt;
      }
    }
    cout << cur[0] << endl;
  }
 
}

C++ 解法, 执行用时: 192ms, 内存消耗: 584K, 提交时间: 2022-01-16 15:44:26 

#include<bits/stdc++.h>
using namespace std;
long long f[12][2048],n,m,ins[2048];
void ps(){for(int i=0;i<1<<m;i++){bool cnt=0,ho=0;for(int j=0;j<m;j++){if(i>>j&1)ho|=cnt,cnt=0;else cnt^=1;}ins[i]=(cnt|ho)?0:1;}}
void se(){
	f[0][0]=1;
	for(int i=1;i<=n;i++)for(int j=0;j<1<<m;j++){f[i][j]=0;for(int k=0;k<1<<m;k++)if((j&k)==0&&ins[j|k])f[i][j]+=f[i-1][k];}
	cout<<f[n][0]<<"\n";
}
int main(){
	while(cin>>n>>m&&(n||m))ps(),se();
	return 0;
}

上一题

© 2025 nowcoder.com