POJ 2506 Tiling
Tiling
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7437 | Accepted: 3635 |
Description
In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles?![](http://poj.org/images/2506_1.jpg)
Here is a sample tiling of a 2x17 rectangle.
![](http://poj.org/images/2506_1.jpg)
Input
Input is a sequence of lines, each line containing an integer number 0 <= n <= 250.
Output
For each line of input, output one integer number in a separate line giving the number of possible tilings of a 2xn rectangle.
Sample Input
2 8 12 100 200
Sample Output
3 171 2731 845100400152152934331135470251 1071292029505993517027974728227441735014801995855195223534251
特别注意输入0。输出1!。!!
。!
!
递推公式不难,Ai=Ai-1+2*Ai-2。非常easy就想清楚了
AC代码例如以下:
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; int a[300][3005],b[3005]; int x,y,z; void work(int c) { int i; memset(b,0,sizeof b); for(i=0;i<=x;i++) { b[i]+=(a[c-2][i]*2); if(b[i]>10) { b[i]=b[i]%10; x++; b[i+1]=1; } } z=x+2; for(i=0;i<=z+2;i++) { a[c][i]+=a[c-1][i]+b[i]; if(a[c][i]>=10&&i==z+1) z++; if(a[c][i]>=10) { a[c][i]=a[c][i]%10; a[c][i+1]=1; } } x=y;y=z; } int main() { int n; int i; while(cin>>n) { if(n==0) {cout<<"1"<<endl;continue;} int bj=0; memset(a,0,sizeof a); a[1][0]=1;a[2][0]=3; x=0;y=0; for(i=3;i<=251;i++) work(i); for(i=z-1;i>=0;i--) { if(bj==0&&a[n][i]==0) continue; if(a[n][i]!=0) bj=1; cout<<a[n][i]; } cout<<endl; } return 0; }