poj2506--Tiling

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Tiling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7288 Accepted: 3546

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

2812100200

Sample Output

317127318451004001521529343311354702511071292029505993517027974728227441735014801995855195223534251

Source

The UofA Local 2000.10.14

递推的题目主要还是找到规律，这个题的规律比较明显，p[n]的情况可以由p[n-1]加一个竖的，或是由p[n-2]加两个横的，或是p[n-2]加上一个大块得到由此推出，p[n] = p[n-1] + p[n-2] * 2 ;

还有的就是一个大数问题，很好解决的大数的加法

#include <stdio.h>struct node{    int a[1000] ;    int top ;}p[300];int main(){    int n , i ;    p[0].a[0] = 1 ;    p[0].top = 1 ;    p[1].a[0] = 1 ;    p[1].top = 1 ;    for(i = 2 ; i <= 250 ; i++)    {        int j , s = 0 , k = 0 ;        for(j = 0 ; j < p[i-1].top ; j++)        {            s = p[i-1].a[j] + p[i-2].a[j] * 2 + k ;            p[i].a[j] = s % 10 ;            k = s / 10 ;        }        p[i].top = p[i-1].top ;        if(k != 0)        {            p[i].a[j] = k ;            p[i].top++ ;        }    }    while(scanf("%d", &n) !=EOF)    {        for(i = p[n].top-1 ; i >= 0 ; i--)        {            printf("%d", p[n].a[i]);        }        printf("\n");    }    return 0;}

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