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.
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
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;
}