We wish to tile a grid 4 units high and N units long with rectangles (dominoes) 2 units by one unit (in either orientation). For example, the figure shows the five different ways that a grid 4 units high and 2 units wide may be tiled.
Write a program that takes as input the width, W, of the grid and outputs the number of different ways to tile a 4-by-W grid.
Input
The first line of input contains a single integer N, (1 <= N <= 1000) which is the number of datasets that follow.
Each dataset contains a single decimal integer, the width, W, of the grid for this problem instance.
Output
For each problem instance, there is one line of output: The problem instance number as a decimal integer (start counting at one), a single space and the number of tilings of a 4-by-W grid. The values of W will be chosen so the count will fit in a 32-bit integer.
Sample Input
3
2
3
7
Sample Output
1 5
2 11
3 781
#include<stdio.h> #include<string> #include<iostream> #include<map> #include<string.h> #include<algorithm> using namespace std; int dp[1200][20];// 1表示 需要下一行一个位子 int okling(int z)//判断z状态 是否 0 都是连续成对出现 { int flag=0; int yiyi=3; for(int i=0;i<3;i++) { if(((z>>i)&yiyi)==0) { z|=(yiyi<<i); } else if(!(((z>>i)&yiyi)==3||(((z>>i)&yiyi)==1&&i!=2))) return 0; } return 1; } int ok(int z1,int z2)//左边列状态z1 右边列状态z2 可以匹配返回1 { for(int i=0;i<4;i++) { if(((z1>>i)&1)==1&&((z2>>i)&1)==1) return 0; else if(((z1>>i)&1)==1) z2=(z2|(1<<i)); } if(okling(z2)) return 1; else return 0; } int find(int wei,int m)//wei 这行 满足 wei+1行状态m的方法数。 { int ans=0; for(int i=0;i<16;i++) { if(ok(i,m)) ans+=dp[wei][i]; } return ans; } void init() { memset(dp,0,sizeof dp); dp[0][0]=1; for(int i=1;i<=1000;i++) { for(int j=0;j<16;j++)//本行 { dp[i][j]+=find(i-1,j); } } } int main() { int t; int cas=1; int n; int op; init(); scanf("%d",&t); while(t--) { scanf("%d",&n); printf("%d %d\n",cas++,dp[n][0]); } return 0; }