HDU 5318(The Goddess Of The Moon-矩阵乘法)[Template:矩阵]

The Goddess Of The Moon

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1086    Accepted Submission(s): 489

Problem Description

Chang’e （嫦娥） is a well-known character in Chinese ancient mythology. She’s the goddess of the Moon. There are many tales about Chang'e, but there's a well-known story regarding the origin of the Mid-Autumn Moon Festival. In a very distant past, ten suns had risen together to the heavens, thus causing hardship for the people. The archer Yi shot down nine of them and was given the elixir of immortality as a reward, but he did not consume it as he did not want to gain immortality without his beloved wife Chang'e.



However, while Yi went out hunting, Fengmeng broke into his house and forced Chang'e to give up the elixir of immortality to him, but she refused to do so. Instead, Chang'e drank it and flew upwards towards the heavens, choosing the moon as residence to be nearby her beloved husband.



Yi discovered what had transpired and felt sad, so he displayed the fruits and cakes that his wife Chang'e had liked, and gave sacrifices to her. Now, let’s help Yi to the moon so that he can see his beloved wife. Imagine the earth is a point and the moon is also a point, there are n kinds of short chains in the earth, each chain is described as a number, we can also take it as a string, the quantity of each kind of chain is infinite. The only condition that a string A connect another string B is there is a suffix of A , equals a prefix of B, and the length of the suffix(prefix) must bigger than one(just make the joint more stable for security concern), Yi can connect some of the chains to make a long chain so that he can reach the moon, but before he connect the chains, he wonders that how many different long chains he can make if he choose m chains from the original chains.

 

Input

The first line is an integer T represent the number of test cases.
Each of the test case begins with two integers n, m.
(n <= 50, m <= 1e9)
The following line contains n integer numbers describe the n kinds of chains.
All the Integers are less or equal than 1e9.

 

Output

Output the answer mod 1000000007.

 

Sample Input

   
   
   
   

    
    
    
    2
10 50
12 1213 1212 1313231 12312413 12312 4123 1231 3 131
5 50
121 123 213 132 321
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    86814837
797922656

    
    
    
    
 
     
 
      Hint 
     
 
     11 111 is different with 111 11 
    
    
    
    
     
   
   
   
   

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

 

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求出邻接矩阵

然后矩阵快速幂

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXN (60)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int a[MAXN];

bool conn(ll a,ll b)
{ 
	while (a<b) b/=10;
	
	ll p=1;
	while (p*10<=a) p*=10;
	
	while (b)
	{
		if (a<b) b/=10;
		else if (a>b) {
			a-=a/p*p;
			p/=10;
		}
		else return abs(a)>9;
	}
	return 0;
	
}	
struct M  
{  
    int n,m;  
    ll a[MAXN][MAXN];  
    M(int _n=0){n=m=_n;MEM(a);}
    M(int _n,int _m){n=_n,m=_m;MEM(a);}
    void mem (int _n=0){n=m=_n;MEM(a);}
    void mem (int _n,int _m){n=_n,m=_m;MEM(a);}
    
	friend M operator*(M a,M b)  
    {  
        M c(a.n);  
	    For(k,a.m)
		    For(i,a.n)  
	            For(j,b.m)  
	                c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%F;  
		return c;     
    }  
    void pri()
	{
	    For(i,n) 
	    { 
	        For(j,m) cout<<a[i][j]<<' ';cout<<endl; 
	    }  		
	} 
	void make_I(int _n)  
    {  
    	n=m=_n; MEM(a)
        For(i,n) a[i][i]=1;  
    }  
	
	
}A,I;

M pow2(M a,ll b)
{
	M c=I;  
    static bool a2[1000000];  
    int n=0;while (b) a2[++n]=b&1,b>>=1;  
    For(i,n)  
    {  
        if (a2[i]) c=c*a;  
        a=a*a;  
    }  
    return c;  
}


int n,m;
int main()
{
 //	freopen("C.in","r",stdin);
	
	int T;cin>>T;
	
	
	while(T--)
	{
		scanf("%d%d",&n,&m);
		For(i,n) scanf("%d",&a[i]);
		sort(a+1,a+1+n);
		n=unique(a+1,a+1+n)-(a+1);
		if (m==0)
		{
			cout<<"1"<<endl;
			continue;
		}
		
		
		I.make_I(n); A.mem(n);
		For(i,n) 
			For(j,n) A.a[i][j]=conn(a[i],a[j]);
		
	//	A.pri();
		A=pow2(A,m-1);
		ll ans=0;
		For(i,n) For(j,n) upd(ans,A.a[i][j]);
		printf("%lld\n",ans);
		
	}
	
	return 0;
}