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
代码:递推式套矩阵快速幂即可,注意多组输入
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N=10;
int n,mod;
int temp[N][N];
int res[N][N],a[N][N];
void mul(int a[][N],int b[][N]) {
memset(temp,0,sizeof(temp));
for(int i=0; i<N; i++)
for(int j=0; j<N; j++)
for(int k=0; k<N; k++)
temp[i][j]=(temp[i][j]+a[i][k]*b[k][j]%mod)%mod;
for(int i=0; i<N; i++)
for(int j=0; j<N; j++)
a[i][j]=temp[i][j];
return ;
}
void QuickPow(int nn) {
memset(res,0,sizeof(res));
for(int i=0; i<N; i++)
res[i][i]=1;
while(nn) {
if(nn&1)
mul(res,a);
mul(a,a);
nn>>=1;
}
return ;
}
int main() {
while(~scanf("%d %d",&n,&mod)) {
memset(a,0,sizeof(a));
for(int i=0; i<N; i++)
scanf("%d",&a[0][i]);
for(int i=1; i<N; i++)
a[i][i-1]=1;
if(n<10) printf("%d
",n%mod);
else {
QuickPow(n-9);
int ans=0;
for(int i=0; i<N; i++)
ans+=res[0][i]*(9-i)%mod;
printf("%d
",ans%mod);
}
}
return 0;
}