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
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
typedef long long int LL;
LL mod;
struct Matrix
{
LL mat[15][15];
Matrix()
{
memset(mat,0,sizeof(mat));
}
};
Matrix mul (Matrix a,Matrix b)
{
Matrix res;
for(int i=0;i<10;i++)
{
for(int j=0;j<10;j++)
{
for(int k=0;k<10;k++)
{
res.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
res.mat[i][j]%=mod;
}
}
}
return res;
}
Matrix pow_matrix(Matrix a,LL n)
{
Matrix res;
for(int i=0;i<10;i++)
{
res.mat[i][i]=1;
}
while(n)
{
if(n&1)
res=mul(res,a);
a=mul(a,a);
n>>=1;
}
return res;
}
int main()
{
LL k;
while(~scanf("%lld%lld",&k,&mod))
{
Matrix A,B;
LL f[11];
for(int i=0;i<10;i++)
{
scanf("%lld",&f[i]);
}
if(k<10)
{
printf("%lld\n",k%mod);
}
else
{
for(int i=0;i<10;i++)
{
A.mat[i][0]=9-i;
}
for(int i=0;i<10;i++)
{
B.mat[0][i]=f[i];
}
for(int i=1;i<10;i++)
{
B.mat[i][i-1]=1;
}
B=pow_matrix(B,k-9);
A=mul(B,A);
printf("%lld\n",(A.mat[0][0])%mod);
}
}
}