poj 2506--Tiling

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

假设我们已经铺好了2*（n-1)的，要铺成2*n的只能用2*1的地板
假设我们已经铺好了2*（n-2)的，要铺成2*n的可以用1个2*2的，也可以用两个2*1的
所以得到递推公式f（n）=f（n-1)+2*f（n-2);
根据题目的范围考虑用高精度模拟实现即可

#include <stdio.h>
#include <string.h>
char a[260][1000];
void add(char b[],char c[],char d[])
{
    int i,j,k=0,up=0;
    int x,y,z,t;
    i=strlen(b)-1,j=strlen(c)-1;
    while(i>=0||j>=0)
    {
        if(i<0) x=0;else x=b[i]-'0';
        if(j<0) y=0;else y=c[j]-'0';
        z=x+y+up;
        if(z>9)
        {
            up=1;
            z-=10;
        }
        else
            up=0;
        d[k++]=z+'0';
        i--,j--;
    }
    if(up)
        d[k++]='1';
    d[k]='\0';
    for(i=0;i<k/2;i++)
    {
        t=d[i];
        d[i]=d[k-i-1];
        d[k-i-1]=t;
    }
}
void init()
{
    int i;
    char b[1000];
    strcpy(a[0],"1");
    strcpy(a[1],"1");
    for(i=2;i<=250;i++)
    {
        add(a[i-2],a[i-2],b);
        add(a[i-1],b,a[i]);
    }
}

int main()
{
    init();
    int n;
    while(scanf("%d",&n)!=EOF)
    {
       printf("%s\n",a[n]);
    }
    return 0;
}