POJ 2663 Tri Tiling
2016-07-05 23:24
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Description
In how many ways can you tile a 3xn rectangle with 2x1 dominoes?
Here is a sample tiling of a 3x12 rectangle.
Input
Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.
Output
For each test case, output one integer number giving the number of possible tilings.
Sample Input
Sample Output
分析:这是一道思维训练的好题。分两种情况考虑
1.在此处放一单位的积木。等于 3 * DP[i-1];
2.与前面相连接。连接的方案有 2 * (DP[i-2] + DP[i-3] +......+ DP[0]);
因此有:DP[i] = 3 * DP[i-1] + 2 * (DP[i-2] + DP[i-3] +......+ DP[0]);
理解到这里应该就差不多了,当然也可以再化简为 DP[i] = 4 * DP[i-1] - DP[i-2];
In how many ways can you tile a 3xn rectangle with 2x1 dominoes?
Here is a sample tiling of a 3x12 rectangle.
Input
Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.
Output
For each test case, output one integer number giving the number of possible tilings.
Sample Input
2 8 12 -1
Sample Output
3 153
分析:这是一道思维训练的好题。分两种情况考虑
1.在此处放一单位的积木。等于 3 * DP[i-1];
2.与前面相连接。连接的方案有 2 * (DP[i-2] + DP[i-3] +......+ DP[0]);
因此有:DP[i] = 3 * DP[i-1] + 2 * (DP[i-2] + DP[i-3] +......+ DP[0]);
理解到这里应该就差不多了,当然也可以再化简为 DP[i] = 4 * DP[i-1] - DP[i-2];
#include <iostream> using namespace std; int main() { long long a[32]; int n; a[0] = 1; a[2]= 3;a[1] = 0; for(long long i =3 ;i<31;i++) { if(i&1) a[i] = 0; else { a[i] = 4*a[i-2] - a[i-4]; } //cout << i<< " "<< a[i] <<endl; } while(cin>>n) { if(n == -1) break; cout << a <<endl; } return 0; }
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