2011年全国大学生程序设计邀请赛（福州）——Tiling

Tiling

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on FZU. Original ID:  2040
64-bit integer IO format:  %I64d      Java class name:  Main

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We can perfectly tile one m*n rectangle by any size of tiles. The following picture shows all the 34 possible ways to perfectly title a 3*2 rectangle. Now you are given one m*n rectangle, your task is to calculate the total number of the ways to perfectly title the given rectangle. Please mod the answer with 1,000,000,007 when you output it.

Input

The first line of the input contains an integer T(T≤100), indicating the number of test cases. Following is T lines, each containing two integer numbers m(1≤m≤6) and n(1≤n≤1,000,000,000).

Output

For each test case, print a line containing the test case number (beginning with 1) and the answer.

Sample Input

2
2 2
3 2

Sample Output

Case 1: 8
Case 2: 34

矩阵乘法，借用bianchengla的模板

矩阵表示最后两行的状态转移，即上一行的某个状态可以通过几种方式转移到最后一行的状态，构成一个矩阵。由于每一行最多六个格子，状态只需要考虑5个分割线是否存在，所以最大是32*32。如果上一行某两个位置有分割线，下一行相同的两个位置也有，（且中间没有其他的隔断），那么此处有两种可能，即上下联通或者不联通，结果mat[i][j]*2，上代码：

#include <iostream>
#include <cstdio>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn = 1 << 5;
const LL MOD = 1000000007;

struct mat {
    int n;
    LL data[maxn][maxn];
    friend mat operator * (const mat& a, const mat& b) {
        mat r;
        r.n = a.n;
        for (int i = 0; i < r.n; ++i) {
            for (int j = 0; j < r.n; ++j) {
                r.data[i][j] = 0;
                for (int k = 0; k < r.n; ++k) {
                    r.data[i][j] += a.data[i][k] * b.data[k][j] % MOD;
                }
                r.data[i][j] %= MOD;
            }
        }
        return r;
    }
    friend mat pow(mat a, int b) {
        mat r; r.n = a.n;
        for (int i = 0; i < r.n; ++i) {
            for (int j = 0; j < r.n; ++j) {
                r.data[i][j] = 0;
            }
            r.data[i][i] = 1;
        }
        for (;b;) {
            if ( b & 1 ) r = r * a;
            if ( b >>= 1 ) a = a * a;
        }
        return r;
    }
} ;
//上面是bianchengla的矩阵乘法和快速幂模板
///////////////////////////////////////////////////////////

mat m[6];
void buildmat(int n){
	m[n].n = 1<<n;
	for(int x=0; x<(1<<n); x++){
		for(int y=0; y<(1<<n); y++){	//填充x到y的状态转移方式
			//
			m[n].data[x][y] = 1;

			int t[8] = {0};
			for(int i=0; i<n; i++){
				if( x&(1<<i) )  t[i+1] ++;
				if( y&(1<<i) )  t[i+1] ++;
			}
			t[0] = 2; t[n+1] = 2;		//t[]存储两行中这个位置有几条分割线，头尾要各补充一条
			int re = 0;
			for(int i=0; i<=n+1; i++){
				if(t[i]==2)  re++;		//==2就+1
				if(t[i]==1)  re=0;		//==1就中断了
				if(re==2)  m[n].data[x][y] *= 2, re=1;	//凑齐两个就×2，re变成1
			}
		}
	}
}

int main()
{
	for(int ii=0; ii<6; ii++)
		buildmat(ii);		// built mat in m[i]

	int T; scanf("%d", &T);
	int r, n;

	for(int ti=1; ti<=T; ti++){
		scanf("%d%d", &r, &n);
		mat res = pow(m[r-1], n-1);		//我的习惯所有的数组都从0处开始储存，所以这儿会出现-1
		LL sum = 0;
		for(int i=0; i<res.n; i++)
			for(int j=0; j<res.n; j++)
				sum = (sum+res.data[i][j]) % MOD;
		cout << "Case " << ti << ": " << (sum%MOD) << endl;
	}
}

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