poj 2506 Tiling——递推和大数模拟

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7362		Accepted: 3589

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

2
8
12
100
200

Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251

Source
The UofA Local 2000.10.14

这个题是递推，还是比较好分析的，不过看下面的数据，似乎要用到大数模拟...

递推公式:s[0]=1;s[1]=1;while(i>=2) s[i]=s[i-1]+2*s[i-2];

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define mx 2010

int ls[mx];
int s[260][mx];//用于存储250种情况的数

int main()
{
int n,i,j;
memset(ls,0,sizeof(ls));
memset(s,0,sizeof(s));
s[0][0]=1;
s[1][0]=1;
for(i=2;i<=250;i++)
{
for(j=0;j<=mx-10;j++)
{
ls[j]=s[i-1][j]+s[i-2][j]+s[i-2][j];//这里2倍就这么处理：多加一个s[i-2]；
}
for(j=0;j<mx;j++)
{
ls[j+1]+=(ls[j]/10);
ls[j]%=10;
}
for(j=0;j<=mx;j++)
{
s[i][j]=ls[j];
}
memset(ls,0,sizeof(ls));
}
while(~scanf("%d",&n))
{
j=mx-1;
while(s
[j]==0)
j--;
for(;j>=0;j--)
{
printf("%d",s
[j]);
}
printf("\n");
}

return 0;
}


如果你看不懂，就去看看大数吧...链接http://blog.csdn.net/codeblf2/article/details/28908565