HDU 1757 A Simple Math Problem（矩阵快速幂）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4544 Accepted Submission(s): 2741

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

9999
1 1 1 1 1 1 1 1 1
500
0 1 0 1 0 1 0 1 0

Sample Output

45
104

Author

linle

/*
{ f[n] f[n-1]...f[n-9] } = { [ f[n-1] f[n-2] ...f[n-10] } *  
a[0] 1 0 ...0
a[1] 0 1 ...0
............1
a[9] 0 0 ...0
*/  
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
using namespace std;
typedef long long ll;

const long long N = 10;
ll f1,f2,k;
ll mod;
int n;
struct Fast_Matrax {
    ll a[N][N];
    Fast_Matrax() {
        memset(a,0,sizeof(a));
    }
    void init() {
        for(int i=0; i<N; i++)
            for(int j=0; j<N; j++)
                a[i][j]=(i==j);
    }
    Fast_Matrax operator * (const Fast_Matrax &B)const {
        Fast_Matrax C;
        for(int i=0; i<N; i++)
            for(int k=0; k<N; k++)
                for(int j=0; j<N; j++)
                    C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j]%mod+mod)%mod;
        return C;
    }
    Fast_Matrax operator ^ (const ll &t)const {
        Fast_Matrax A=(*this),res;
        res.init();
        ll p=t;
        while(p) {
            if(p&1)res=res*A;
            A=A*A;
            p>>=1;
        }
        return res;
    }
} ans,tmp,x;
int main() {
    for(int i=0; i<10; i++) {
        x.a[0][i]=9-i;
    }
    while(~scanf("%lld%lld",&k,&mod)) {
        //tmp.init();
        for(int i=0; i<10; i++) {
            scanf("%lld",&tmp.a[i][0]);
            if(i<9)
                tmp.a[i][i+1]=1;
        }
        if(k<10) {
            printf("%lld
",k%mod);
            continue;
        }
        ans=x*(tmp^(k-9));
        printf("%lld
",ans.a[0][0]);
    }
    return 0;
}