POJ-2506-Tiling

最新推荐文章于 2021-10-01 21:52:49 发布

Stone_____

最新推荐文章于 2021-10-01 21:52:49 发布

阅读量187

点赞数

CC 4.0 BY-SA版权

分类专栏： POJ

本文链接：https://blog.youkuaiyun.com/zcy_2016/article/details/52033364

POJ 专栏收录该内容

82 篇文章

订阅专栏

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9194		Accepted: 4372

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

Source

The UofA Local 2000.10.14

n=0的时候是1

#include <stdio.h>
#include<string.h>
using namespace std;
char sum[1010];
void f(char s1[], char s2[])
{
    int l1 = strlen(s1),l2=strlen(s2);
    memset(sum, 0,sizeof(sum));
    int len;
    if(l1>l2)len=l1;
    else len = l2;
    int ksum = 0,k = 0;
    for(int i = 0; i<=len;++i)
    {
        ksum = k;
        if(i<l1)ksum+=(s1[i]-'0');
        if(i<l2)ksum+=2*(s2[i]-'0');
        sum[i] = ksum%10+'0';
        k = ksum/10;
    }
    if(sum[len]=='0')sum[len] = '\0';
    else sum[len+1] = '\0' ;
    strcpy(sum,sum);
}
int main()
{
    int n;
    char s1[1000], s2[1000];
    char k1[] ={"3"};
    char k2[] ={"1"};
    while(~scanf("%d", &n))
    {
        if(n<=1)printf("1\n");
        else if(n==2)printf("3\n");
        else
        {
            strcpy(s1,k1);
            strcpy(s2,k2);
            for(int i = 3; i<=n; ++i)
            {
                f(s1, s2);
                strcpy(s2,s1);
                strcpy(s1,sum);
            }
            int len = strlen(s1);
            for(int i = len-1;i>=0;--i)printf("%c",s1[i]);
            printf("\n");
        }

    }
    return 0;
}