Tiling
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7965 | Accepted: 3866 |
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 算法分析:递推公式:f[i]=f[i-1]+f[i-2]*2; 此公式也不是我自己推导出来的,我也没推导出来,我从ACM之家上的java代码看出的公式。代码:
#include <stdio.h> #include <string.h> int a[1001][501]={0}; int main() { int n; int i, j, h, e; a[0][500] = 1; a[1][500] = 1; a[2][500] = 3; for(i=3; i<=250; i++) { h = 0; for(j=500; j>=0; j--) { e=a[i-2][j]*2+a[i-1][j]+h; a[i][j]=e%10; h=e/10; } } while(scanf("%d", &n)!=EOF) { if(n==0) { printf("1\n"); continue; } i = 0; while(a[n][i]==0) { i++; } for(i; i<=500; i++) { printf("%d", a[n][i] ); } printf("\n"); } return 0; }
这还有一份java代码,但是提交WA,不知道哪里出了问题。
import java.util.*; import java.math.*; public class Main{ static BigInteger[] ans; // public static void main(String[] args){ Scanner reader=new Scanner(System.in); ans = new BigInteger[251]; ans[0]=BigInteger.valueOf(1); ans[1]=BigInteger.valueOf(2); ans[2]=BigInteger.valueOf(3); for(int i=3; i<=250; i++) { ans[i] = ans[i-1].add(ans[i-2].multiply(BigInteger.valueOf(2))); } int n; while(reader.hasNextInt()){ n=reader.nextInt(); System.out.println(ans[n]); } } }
时间: 2024-10-17 10:00:48