POJ 2506 Tiling(递归+高精度)

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5935		Accepted: 2893

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

Source

The UofA Local 2000.10.14

解题报告：此题就是求递推式，其实通过前几个的数，很容易推出递推式为F(n) = 2 * F(n-2) + F(n - 1)；但是最重要的是对大数的处理，我是参考了一下网上的方法，开一个大的数组不断的更新数据，例如：F(6) = 43； F(7) = 85，求F(8)时；让ans[8][0] = 2 * 3 + 5；即 ans[8][0] = 2 * ans[6][0] + ans[7][0] = 11; 若超过十则用一个数记录超过的部分beyond =  = ans[8][0]/10;而ans[8][0] = ans[8][0] %10;依此类推就可以得出F(8) = 171；注意的是n = 0时应该输出1；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 260;
int ans[N][N];//存储每位结果
int main()
{
    int n, i, j, count, beyond, p;
    while (scanf("%d", &n) != EOF)
    {
        memset(ans, 0, sizeof(ans));
        ans[0][0] = 1;//初始化
        ans[1][0] = 1;
        ans[2][0] = 3;
        if (n <= 2)
        {
            printf("%d\n", ans[n][0]);
        }
        else
        {
            count = 1;
            for (i = 3; i <= n; ++i)
            {
                beyond = 0; //记录是否超过十
                p = 0;
                for (j = 0; j < count; ++j)
                {
                    p = ans[i - 2][j] * 2 + ans[i - 1][j] + beyond;
                    ans[i][j] = p % 10;
                    beyond = p / 10;
                }
                if (beyond)//位数增加一位时的处理
                {
                    ans[i][count] = beyond;
                    count ++;
                }
            }
            for (i = count - 1; i >= 0; --i)
            {
                printf("%d", ans[n][i]);
            }
            printf("\n");
        }
    }
    return 0;
}