hdu 5015（矩阵快速幂）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5015

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 823    Accepted Submission(s): 493

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 a _0,1 = 233,a _0,2 = 2333,a _0,3 = 23333...) Besides, in 233 matrix, we got a _i,j = a _i-1,j +a _i,j-1( i,j ≠ 0). Now you have known a _1,0,a _2,0,...,a _n,0, could you tell me a _n,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 ≤ 10 ⁹). The second line contains n integers, a _1,0,a _2,0,...,a _n,0(0 ≤ a _i,0 < 2 ³¹).

 

Output

For each case, output a _n,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

 

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思路：矩阵快速幂，由第一列推出剩下的列~有点得注意，要用I64d,如果用int 会wrong~~~可以从题目给的时间5000ms和题目给的数据看很出来~

#include 
#include 
#include 
#include 
#include 
#include 
const int mod=10000007;
typedef long long ll;
using namespace std;
int a[12];

struct matrix
{
 ll f[12][12];
};
//两矩阵相乘;
matrix mul(matrix a,matrix b)
{
  matrix c;
  memset(c.f,0,sizeof(c.f));
  for(int i=0;i<12;i++)
   for(int j=0;j<12;j++)
    for(int k=0;k<12;k++)
    {
     c.f[i][j]+=a.f[i][k]*b.f[k][j]%mod;  //再次提醒自己  是乘不是加，之所以要mod 是因为两者相乘会溢出
     c.f[i][j]%=mod;
    }
  return c;
}
matrix pow_mod(matrix a,int b)
{
  matrix s;
  memset(s.f,0,sizeof(s.f));
  for(int i=0;i<12;i++)
    s.f[i][i]=1;
  while(b)
  {
    if(b&1)
    {
     s=mul(s,a);
    }
    b=b/2;
    a=mul(a,a);
  }
  return s;
}
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
      matrix e;
      memset(e.f,0,sizeof(e.f));
      for(int i=1;i<11;i++)
        for(int j=0;j<=i;j++)
          e.f[i][j]=1;
      e.f[0][0]=10; e.f[0][11]=1,e.f[11][11]=1;

      memset(a,0,sizeof(a));
      for(int i=1;i<=n;i++)cin>>a[i];
      a[0]=233,a[11]=3;
      e=pow_mod(e,m);
      ll sum=0;
      for(int i=0;i<12;i++)
      {
       sum=(sum+(e.f[n][i]*a[i]))%mod;
      }
      printf("%I64d\n",sum%mod);
   }
    return 0;
}