CodeForcesGym 100735B Retrospective Sequence

Retrospective Sequence

Time Limit: Unknown ms
Memory Limit: 65536KB
This problem will be judged on CodeForcesGym. Original ID: 100735B
64-bit integer IO format: %I64d Java class name: (Any)

Retrospective sequence is a recursive sequence that is defined through itself. For example Fibonacci specifies the rate at which a population of rabbits reproduces and it can be generalized to a retrospective sequence. In this problem you will have to find the n-th Retrospective Sequence modulo MOD = 1000000009. The first (1 ≤ N ≤ 20) elements of the sequence are specified. The remaining elements of the sequence depend on some of the previous N elements. Formally, the sequence can be written as Fm = Fm - k1 + Fm - k2 + ... + Fm - ki + ... + Fm - kC - 1 + Fm - kC. Here, C is the number of previous elements the m-th element depends on, 1 ≤ ki ≤ N.

Input

The first line of each test case contains 3 numbers, the number (1 ≤ N ≤ 20) of elements of the retrospective sequence that are specified, the index (1 ≤ M ≤ 1018) of the sequence element that has to be found modulo MOD, the number (1 ≤ C ≤ N) of previous elements the i-th element of the sequence depends on.

The second line of each test case contains N integers specifying 0 ≤ Fi ≤ 10, (1 ≤ i ≤ N).

The third line of each test case contains C ≥ 1 integers specifying k1, k2, ..., kC - 1, kC (1 ≤ ki ≤ N).

Output

Output single integer R, where R is FM modulo MOD.

Sample Input

Input

2 2 2
1 1
1 2

Output

1

Input

2 7 2
1 1
1 2

Output

13

Input

3 100000000000 3
0 1 2
1 2 3

Output

48407255

Source

KTU Programming Camp (Day 1)

解题：矩阵快速幂

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1000000009;
LL n,N,M,C;
struct Matrix{
LL m[60][60];
void init(){
memset(m,0,sizeof m);
}
void setOne(){
init();
for(int i = 0; i < 60; ++i) m[i][i] = 1;
}
Matrix(){
init();
}
Matrix operator*(const Matrix &rhs) const{
Matrix ret;
for(int k = 0; k <= n; ++k)
for(int i = 0; i <= n; ++i)
for(int j = 0; j <= n; ++j)
ret.m[i][j] = (ret.m[i][j] + m[i][k]*rhs.m[k][j]%mod)%mod;
return ret;
}
void print(){
for(int i = 0; i <= n; ++i){
for(int j = 0; j <= n; ++j)
cout<<m[i][j]<<" ";
cout<<endl;
}
cout<<endl;
}
};
Matrix a,b;
void quickPow(LL index){
//Matrix ret;
//ret.setOne();
while(index){
if(index&1) a = a*b;
index >>= 1;
b = b*b;
}
//a = a*ret;
}
int main(){
while(~scanf("%I64d%I64d%I64d",&N,&M,&C)){
a.init();
b.init();
n = N;
for(int i = 1; i <= N; ++i){
scanf("%I64d",&a.m[0][i]);
b.m[i+1][i]++;
}
for(int i = 1,tmp; i <= C; ++i){
scanf("%d",&tmp);
b.m[N + 1 - tmp]
++;
}
if(M <= N){
printf("%I64d\n",a.m[0][M]%mod);
continue;
}
quickPow(M - N);
printf("%I64d\n",a.m[0]
%mod);
}
return 0;
}
/*
2 3 2
1 1
1 2

3 5 3
0 1 2
1 2 3
*/

View Code