矩阵快速幂 HDU5015

  1 #include <iostream>

  2 #include <cstring>

  3 

  4 using namespace std;

  5 

  6 long long tmp[11];

  7 

  8 //矩阵大小上限

  9 const int SIZ=100;

 10 int MOD=10000007;

 11 

 12 //矩阵大小为n*m，初始化全部为0

 13 struct mat

 14 {

 15     int n,m;

 16     long long ar[SIZ][SIZ];

 17     mat()

 18     {

 19         memset(ar,0,sizeof(ar));

 20         n=m=SIZ;

 21     };

 22 };

 23 

 24 //矩阵乘法

 25 mat operator *(mat a,mat b)

 26 {

 27     mat c;

 28     c=mat();

 29     c.n=a.n;

 30     c.m=b.m;

 31     for(int i=1;i<=a.n;i++)

 32         for(int j=1;j<=b.m;j++)

 33             for(int k=1;k<=a.m;k++)

 34             {

 35                 c.ar[i][j]+=(a.ar[i][k]*b.ar[k][j])%MOD;

 36                 c.ar[i][j]%=MOD;

 37             }

 38     return c;

 39 }

 40 

 41 //矩阵加法

 42 mat operator +(mat a,mat b)

 43 {

 44     mat c;

 45     c=mat();

 46     c.n=a.n;

 47     c.m=a.m;

 48     for(int i=1;i<=a.n;i++)

 49         for(int j=1;j<a.m;j++)

 50             c.ar[i][j]=a.ar[i][j]+b.ar[i][j];

 51     return c;

 52 }

 53 

 54 //矩阵快速幂

 55 mat operator ^(mat a,int k)

 56 {

 57     mat c;

 58     c=mat();

 59     c.n=a.n;

 60     c.m=a.m;

 61     for(int i=1;i<=a.n;i++)

 62         c.ar[i][i]=1;

 63     while(k)

 64     {

 65         if(k&1)

 66             c=c*a;

 67         a=a*a;

 68         k/=2;

 69     }

 70     return c;

 71 }

 72 

 73 int main()

 74 {

 75     int n,m;

 76     while(cin>>n>>m)

 77     {

 78         for(int i=1;i<=n;i++)

 79         {

 80             cin>>tmp[i];

 81         }

 82         mat mat1;

 83         mat1=mat();

 84         mat1.m=mat1.n=n+2;

 85         for(int i=1;i<mat1.m;i++)

 86         {

 87             for(int t=2;t<=i;t++)

 88             {

 89                 mat1.ar[i][t]=1;

 90             }

 91             mat1.ar[i][1]=10;

 92             mat1.ar[i][mat1.m]=1;

 93         }

 94         mat1.ar[mat1.m][mat1.m]=1;

 95         mat mat2 ;

 96         mat2 =mat();

 97         mat2.n=n+2;

 98         mat2.m=1;

 99         for(int i=2;i<mat2.n;i++)

100         {

101             mat2.ar[i][1]=tmp[i-1];

102         }

103         mat2.ar[1][1]=23;

104         mat2.ar[mat2.n][1]=3;

105         mat mat3;

106         mat3=(mat1^(m))*mat2;

107         cout<<mat3.ar[n+1][1]<<endl;

108     }

109     return 0;

110 }

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