2017 ICPC 广西邀请赛1004 Covering

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签： java

原文链接：http://www.cnblogs.com/wuwangchuxin0924/p/7462479.html

探讨使用1×2和2×1木块完全覆盖4×n矩阵的方案数量，通过状态压缩动态规划与矩阵快速幂求解。

Covering

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 187 Accepted Submission(s): 107

Problem Description

Bob's school has a big playground, boys and girls always play games here after school.

To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carpets.

Meanwhile, Bob is a mean boy, so he acquired that his carpets can not overlap one cell twice or more.

He has infinite carpets with sizes of

Input

There are no more than 5000 test cases.

Each test case only contains one positive integer n in a line.

Output

For each test cases, output the answer mod 1000000007 in a line.

Sample Input

1 2

Sample Output

1 5

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

打表

/*
* @Author: Administrator
* @Date:   2017-08-31 17:40:04
* @Last Modified by:   Administrator
* @Last Modified time: 2017-09-01 11:03:00
*/
/*
 题意：给你一个4*n的矩阵，然后让你用1*2和2*1的木块放，问你完美覆盖的
    方案数

 思路：状压DP找规律
*/

#include <bits/stdc++.h>

#define MAXN 100
#define MAXM 20
#define MAXK 15
using namespace std;

int dp[MAXN][MAXM];//dp[i][j]表示前ihang 
int n;

inline bool ok(int x){
    //判断是不是有连续个1的个数是奇数
    int res=0;
    while(x){
        if(x%2==1){
            res++;
        }else{
            if(res%2==1) return false;
            else res=0;
        }
        x/=2;
    }
    if(res%2==1) return false;
    else return true;
}

inline void init(){
    memset(dp,0,sizeof dp);
}

int main(){
    freopen("in.txt","r",stdin);
    for(int n=1;n<=50;n++){
        init();
        for(int i=0;i<=MAXK;i++){//初始化第一行的没种状态
            if(ok(i)==true)
                dp[1][i]=1;
        }
        for(int i=1;i<n;i++){
            for(int j=0;j<=MAXK;j++){
                if(dp[i][j]!=0){
                    for(int k=0;k<=MAXK;k++){
                        if( (j|k)==MAXK && ok(j&k) )
                            ///j|k==tot-1的话就是能拼起来组成
                            dp[i+1][k]+=dp[i][j];
                    }
                }
            }
        }
        printf("%d\n",dp[n][MAXK]);
    }
    return 0;
}

/*
* @Author: Administrator
* @Date:   2017-09-01 11:17:37
* @Last Modified by:   Administrator
* @Last Modified time: 2017-09-01 11:28:09
*/
#include <bits/stdc++.h>

#define MAXN 5
#define mod 1000000007
#define LL long long

using namespace std;

/********************************矩阵快速幂**********************************/
class Matrix {
    public:
        LL a[MAXN][MAXN];
        LL n;  

        void init(LL x) {
            memset(a,0,sizeof(a));
            if (x)  
                for (int i = 0; i < MAXN ; i++)
                    a[i][i] = 1LL;
        }
        
        Matrix operator +(Matrix b) {
            Matrix c;
            c.n = n;
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;
            return c;
        }

        Matrix operator +(LL x) {
            Matrix c = *this;
            for (int i = 0; i < n; i++)
                c.a[i][i] += x;
            return c;
        }

        Matrix operator *(Matrix b)
        {
            Matrix p;
            p.n = b.n;   
            p.init(0);
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                for (int k = 0; k < n; k++)
                    p.a[i][j] = (p.a[i][j] + (a[i][k]*b.a[k][j])%mod) % mod;
            return p;
        }

        Matrix power(LL t) {
            Matrix ans,p = *this;
            ans.n = p.n;  
            ans.init(1);
            while (t) {
                if (t & 1)
                    ans=ans*p;
                p = p*p;
                t >>= 1;
            }
            return ans;
        }
}init,unit;
/********************************矩阵快速幂**********************************/

LL n;

int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%lld",&n)!=EOF){
        if(n<=4){
            switch(n){
                case 1:
                    puts("1");
                    break;
                case 2:
                    puts("5");
                    break;
                case 3:
                    puts("11");
                    break;
                case 4:
                    puts("36");
                    break;
            }
            continue;
        }
        init.init(0);
        init.n=4;
        init.a[0][0]=36;
        init.a[0][1]=11;
        init.a[0][2]=5;
        init.a[0][3]=1;
        unit.init(0);
        unit.n=4;
        unit.a[0][0]=1;
        unit.a[1][0]=5;
        unit.a[2][0]=1;
        unit.a[3][0]=-1;
        unit.a[0][1]=1;
        unit.a[1][2]=1;
        unit.a[2][3]=1;
        unit=unit.power(n-4);
        init=init*unit;
        printf("%lld\n",(init.a[0][0]+mod)%mod);
    }
    return 0;
}