poj--2506

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                                                                                                                           Tiling
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8319 Accepted: 4012

Description

In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles?
Here is a sample tiling of a 2x17 rectangle.

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

2812100200

Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251

题目大意:计算要铺一个2xn的地板,只能用2x1,2x2的瓷砖,问有几种方法铺这个地板。

考查点:高精度加法。

核心思想:通过枚举一些情况,我们可以得到一个规律,2xn=2x(n-1)+2倍的 2x(n-2);接下来我们就可以用一个二维数组存储n从0到250每一种情况的答案。

代码如下:

#include<stdio.h>#include<string.h>int a[260][1000];int main(){int n,m,p,i,j,l;memset(a,0,sizeof(a));a[0][0]=1;a[1][0]=1;    a[2][0]=3;a[3][0]=5;p=0;    int k=1;for(i=4;i<=250;i++){   p=0;//p表示进位 l=k;  for(j=0;j<l;j++)  {  a[i][j]=2*(a[i-2][j])+a[i-1][j]+p;  p=a[i][j]/10;  a[i][j]=a[i][j]%10;  }  while(p)  {  a[i][j]=p%10;  p=p/10;  j++;  }  k=j;}while(scanf("%d",&n)!=EOF){for(i=80;i>=0;i--){if(a[n][i]!=0)break;}for(j=i;j>=0;j--){printf("%d",a[n][j]);}printf("\n");}return 0;}

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