HDU 5015 233 Matrix （数论——矩阵快速幂）

233 Matrix

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0 Accepted Submission(s): 0



Problem Description

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means
a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell
me an,m in the 233 matrix?

Input

There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).

Output

For each case, output an,m mod 10000007.

Sample Input

1 1
1
2 2
0 0
3 7
23 47 16

Sample Output

234
2799
72937

Hint

Source

2014 ACM/ICPC Asia Regional Xi'an Online

题目大意：

对于一个n+1行，m+1列的矩阵A[n+1][m+1]，已知a[0][1]=233，a[0][2]=2333，a[0][3]=23333，…………并且对于矩阵当中元素的值，有a[i][j]=a[i-1][j]+a[i][j-1](i,j!=0)。给出a[1][0]，a[2][0]，…………，a[n-1][0]，a
[0]以及n，m，求解a
[m]（矩阵第n行第m列元素的值）。

解题思路：

因为a[0][1]=233，a[0][2]=2333，a[0][3]=23333，…………，所以有a[0][i]=10*a[0][i-1]+3（i>1，a[0][1]=233）

如果令a[0][0]=23，那么a[0][i]=10*a[0][i-1]+3（i>0，a[0][0]=23）

则对于n+1行，m+1列的矩阵A，有

第一列（下标j=0）



第二列（下标j=1）



第三列（下标j=2）



对上述矩阵的各列适当变形得：

第一列（下标j=0）



第二列（下标j=1）



第三列（下标j=2）

所以有



根据上面的递推公式，我们可以构造矩阵得，





（其中

）

矩阵快速幂

代码如下：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
#include <limits.h>
#define debug "output for debug\n"
#define pi (acos(-1.0))
#define eps (1e-6)
#define inf (1<<28)
#define sqr(x) (x) * (x)
#define mod 10000007
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
#define MAX 25
int n;
struct Matrix
{
ll a[MAX][MAX];
Matrix()
{
memset(a,0,sizeof(a));
}
};
Matrix operator *(Matrix a,Matrix b)
{
Matrix c;
ll i,j,k;
memset(c.a,0,sizeof(c.a));
for(k=0;k<n+2;k++)
{
for(i=0;i<n+2;i++)
{
if(a.a[i][k]==0)
continue;
for(j=0;j<n+2;j++)
{
if(b.a[k][j]==0)
continue;
c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%mod;
c.a[i][j]%=mod;
}
}
}
return c;
}
Matrix operator ^(Matrix a,int k)
{
Matrix c;
ll i;
for(i=0;i<n+2;i++)
c.a[i][i]=1;
while(k)
{
if(k&1)
c=c*a;
a=a*a;
k>>=1;
}
return c;
}
int main()
{
Matrix a,b,c;
ll i,j,k,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
a.a[0][0]=23;
for(i=1;i<=n;i++)
{
scanf("%d",&a.a[i][0]);
}
a.a[n+1][0]=3;
for(i=0;i<n+2;i++)
{
for(j=0;j<n+2;j++)
{
b.a[i][j]=0;
if(j==0&&i!=(n+1))
b.a[i][j]=10;
if(j==(n+1))
b.a[i][j]=1;
if(j==0&&i==(n+1))
b.a[i][j]=0;
if(i>=1&&i<=n&&j>=1&&j<=n)
{
if(i>=j)
b.a[i][j]=1;
}
}
}
b=b^m;
long long ans=0;
for(i=0;i<n+2;i++)
ans=(ans+b.a
[i]*a.a[i][0])%mod;
printf("%I64d\n",ans%mod);
}
return 0;
}