poj 2506 Tiling(递推+大数加法）

发布时间：2020-12-14 03:40:02 所属栏目：大数据来源：网络整理

导读：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 in

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

公式为：f[x]=f[x-1]+2*f[x-2];

至于为什么f[0]为1，我也不知道。。。我是根据公式以及给出的案例反推的f[0].

#include <stdio.h>
#include <string.h>
char s[260][200];
int main()
{
    int n,i,a,j,x,b;
    memset(s,'0',sizeof(s));
    s[0][0]='1';
    s[1][0]='1';
    s[2][0]='3';
    for(i=3;i<=250;i++)
    {
        for(j=0;j<100;j++)
        {
            a=(2*(s[i-2][j]-'0')+s[i-1][j]-'0'+s[i][j]-'0');
            s[i][j]=a%10+'0';
            s[i][j+1]=a/10+'0';
        }
    }
    while(scanf("%d",&n)!=EOF)
    {
        x=0;
        for(i=100;i>=0;i--)
        {
            if(s[n][i]!='0')
            {
                x=i;
                break;
            }
        }
        for(i=x;i>=0;i--)
        {
            printf("%c",s[n][i]);
        }
        printf("n");
    }
    return 0;
}

（编辑：李大同）

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