POJ 2506 Tiling（大数）

发布时间：2020-12-14 02:23:30 所属栏目：大数据来源：网络整理

导读：Tiling Time Limit: 1000MS ? Memory Limit: 65536K Total Submissions: 8300 ? Accepted: 4002 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

Tiling

Time Limit: 1000MS	?	Memory Limit: 65536K
Total Submissions: 8300	?	Accepted: 4002

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

Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251


规律题：F(n)=F(n-2)*2+F(n-1)

ac代码：
#include<stdio.h>
#include<string.h>
#define MAXN 10010
int num[255][MAXN+10];
void db()
{
	memset(num,sizeof(num));
	num[0][0]=0;
	num[1][0]=1;
	num[2][0]=3;
	int i,j;
	int c;
	for(i=3;i<=250;i++)
	{
		c=0;
		for(j=0;j<=MAXN;j++)
		{
			num[i][j]=(num[i-2][j]*2+num[i-1][j]+c)%10;
			c=(num[i-2][j]*2+num[i-1][j]+c)/10;
		}
	}
}
int main()
{
	int i,j,n;
	db();
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
		{
			printf("1n");
			continue;
		}
		for(i=MAXN;i>=0;i--)
		if(num[n][i])
		break;
		for(j=i;j>=0;j--)
		printf("%d",num[n][j]);
		printf("n");
	}
	return 0;
}
?

（编辑：李大同）

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