poj 2506 Tiling 《大数加法+递推》

Tiling

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 8689 Accepted: 4183

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

神题。。。n=0时，输出1。。。。。。。。。。

代码：

#include<cstdio>
#include<cstring>
int s[50],n;
int ge[310][200],shu[310];
void suan(int xx)
{
int ii=xx;
shu[ii]=shu[ii-1];
for (int i=0;i<shu[ii];i++)
{
ge[ii][i]=ge[ii][i]+ge[ii-1][i]+ge[ii-2][i]*2;
if (ge[ii][shu[ii]-1]>9)
shu[ii]++;
if (ge[ii][i]>9)
{
ge[ii][i+1]=ge[ii][i]/10;
ge[ii][i]=ge[ii][i]%10;
}
}
}
int main()
{
s[0]=1;s[1]=1;s[2]=3;
for (int i=3;i<31;i++)
{
s[i]=s[i-1]+2*s[i-2];
}
memset(ge,0,sizeof(ge));
ge[29][0]=1;ge[29][1]=4;ge[29][2]=9;ge[29][3]=3;ge[29][4]=1;ge[29][5]=9;ge[29][6]=7;ge[29][7]=5;ge[29][8]=3;
ge[30][0]=3;ge[30][1]=8;ge[30][2]=8;ge[30][3]=7;ge[30][4]=2;ge[30][5]=8;ge[30][6]=5;ge[30][7]=1;ge[30][8]=7;
shu[29]=9;shu[30]=9;
for (int i=31;i<305;i++)
{
suan(i);
}
while (~scanf("%d",&n))
{
if (n<31)
{
printf("%d\n",s
);
continue;
}
for (int i=shu
-1;i>=0;i--)
printf("%d",ge
[i]);
printf("\n");
}
return 0;
}