Tiling POJ 2506 【大数】

http://poj.org/problem?id=2506

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

做之前看看这个：http://blog.csdn.net/yuzhiwei1995/article/details/47909743

n==0时输出1 神数据。wa了n次这个地方

//贴两种大数进位方式

#include<stdio.h>
#include<string.h>
int a[300][2010];
void fun()
{
	int i,j;
	memset(a,0,sizeof(a));
	a[0][0]=1;
	a[1][0]=1;
	a[2][0]=3;
	int t;
	for(i=3;i<=260;++i)
	{
		int k=0;
		for(j=0;j<=2000;++j)
		{
			a[i][j] += a[i-1][j] + 2 * a[i-2][j];		
			if(a[i][j]>=10) 
			{
				a[i][j+1]=a[i][j]/10;
				a[i][j]%=10;
			}
		}
	}
}
int main()
{
	int n,i;	
	fun();
	while(~scanf("%d",&n))
	{
		for(i=2000;i>0&&a[n][i]==0;--i);
		for(;i>=0;--i)
			printf("%d",a[n][i]);
		printf("
");
	}
	return 0;
}

#include<stdio.h>
#include<string.h>
int a[300][2010];
void fun()
{
	int i,j;
	memset(a,0,sizeof(a));
	a[0][0]=1;
	a[1][0]=1;
	a[2][0]=3;
	int t;
	for(i=3;i<=260;++i)
	{
		int k=0;
		for(j=0;j<=2000;++j)
		{
			t = a[i-1][j] + 2 * a[i-2][j] + k;
			k=t / 10;
			a[i][j]=t%10;
		}
	}
}
int main()
{
	int n,i;	
	fun();
	while(~scanf("%d",&n))
	{
		for(i=2000;i>0&&a[n][i]==0;--i);
		for(;i>=0;--i)
			printf("%d",a[n][i]);
		printf("
");
	}
	return 0;
}