hdu A Simple Math Problem

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

Output

For each case, output f(k) % m in one line.

Sample Input

10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0

Sample Output

45 104

Author

linle

Source

2007省赛集训队练习赛（6）_linle专场

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下面我们研究一下这道题如何运用矩阵。

假设我们已有一个1*10的矩阵f0 f1 f2...f9 只要将其乘以一下矩阵，就可以推导至矩阵f1 f2 f3 ...f10：
                    0 0 0 0 0 0 0 0 0 a(9)
                           1 0 0 0 0 0 0 0 0 a(8)
                           0 1 0 0 0 0 0 0 0 a(7)
                           0 0 1 0 0 0 0 0 0 a(6)
                           0 0 0 1 0 0 0 0 0 a(5)
                           0 0 0 0 1 0 0 0 0 a(4)
                           0 0 0 0 0 1 0 0 0 a(3)
                           0 0 0 0 0 0 1 0 0 a(2)
                           0 0 0 0 0 0 0 1 0 a(1)
                           0 0 0 0 0 0 0 0 1 a(0)
     设这个矩阵为D。

f1,f2,f3...f10再乘以这个矩阵就可以推导至f2,f3,f4...f11。以此类推，递推的过程就变成了不断乘以这个矩阵的过程。这样，如果我们要得到fk，我们就得将初始矩阵（f0,f1,f2...f10）乘以D^(k-9)，从而得出目标矩阵（目标矩阵的最后一项即为所求）。而D^(k-9)我们可以用10^3*log(k-9)的时间计算出来，此题圆满解决。

//看懂这个题解之后就可以知道以后再给出这样的就该怎么地推求出矩阵表达式

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
struct node
{long long martix[15][15];
}a,e;
int k,m;
void init()
{memset(a.martix,0,sizeof(a.martix));memset(e.martix,0,sizeof(e.martix));for(int i=0;i<10;++i)e.martix[i][i]=1;for(int i=9;i>=0;--i){cin>>a.martix[i][9];if(i>=1)a.martix[i][i-1]=1;}
}
node POW(node a,node b)
{node ans;for(int i=0;i<10;++i)for(int j=0;j<10;++j){ans.martix[i][j]=0;for(int k=0;k<10;++k){if(a.martix[i][k]&&b.martix[k][j]){ans.martix[i][j]+=(a.martix[i][k]%m*b.martix[k][j]%m)%m;}}}return ans;
}
void martix_pow(int x)
{while(x){if(x&1)e=POW(e,a);a=POW(a,a);x>>=1;}
}
int main()
{while(cin>>k>>m){init();martix_pow(k-9);//long long sum=0;for(int i=0;i<10;++i){sum+=e.martix[i][9]*i%m;}cout<<sum%m<<endl;//注意最后还要%一次，}return 0;
}

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