POJ 2506-Tiling(递推+大数)

Tiling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7897 Accepted: 3841

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

50

ps：本以为要用求大数的方法写个函数，然后调用它，那也太麻烦了，然后飞神告诉了我这个方法，太神奇了，总共才250,250*110果断爆不了。用二维数组存储大数的每一位。感觉我还是太年轻 sad~

#include <stdio.h>#include <string.h>#include <stdlib.h>char a[310][120];int main(){        int n,i,j;        memset(a,'0',sizeof(a));        a[0][0]='1';        a[1][0]='1';        a[2][0]='3';        for(i=3; i<=250; i++)        {                for(j=0; j<110; j++)                {                        int sum=2*(a[i-2][j]-'0')+a[i-1][j]-'0'+a[i][j]-'0';                        a[i][j]=sum%10+'0';                        a[i][j+1]=sum/10+'0';                }        }        while(~scanf("%d",&n))        {                int flag=0;                for(i=110; i>=0; i--)                        if(a[n][i]!='0')                        {                                flag=i;                                break;                        }                for(i=flag; i>=0; i--)                        printf("%c",a[n][i]);                printf("\n");        }        return 0;}