HDU1757 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): 4331    Accepted Submission(s): 2603

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

 

题意：给出递推公式求第n项

题解：矩阵快速幂，注意初始矩阵为[9,8,7,6,5,4,3,2,1,0]

#include<bits/stdc++.h>
#define N 10
#define mes(x) memset(x, 0, sizeof(x));
#define ll __int64
long long mod = 1e9+7;
const int MAX = 0x7ffffff;
using namespace std;
struct mat{
    ll a[N][N];
    mat(){
        memset(a, 0, sizeof(a));
    }
    mat operator *(mat b){
        mat c;
        for(int i=0;i<N;i++)
            for(int j=0;j<N;j++)
            for(int k=0;k<N;k++)
            c.a[i][j] = (c.a[i][j] + a[i][k]*b.a[k][j])%mod;
        return c;
    }
};
mat f(mat x,ll m){
    mat t;
    for(int i=0;i<N;i++)
        t.a[i][i] = 1;
    while(m){
        if(m&1) t = t*x;
        x = x*x;
        m >>= 1;
    }
    return t;
}
int main()
{
    ll k, i;
    while(~scanf("%I64d%I64d", &k, &mod)){
        if(k<10){
            printf("%I64d
", k);
            continue;
        }
        mat A, s;
        for(i=0;i<10;i++)
            s.a[0][i] = 9-i;
        for(i=0;i<10;i++)
            scanf("%I64d", &A.a[i][0]);
        for(i=0;i<9;i++)
            A.a[i][i+1] = 1;    
        s = s*f(A, k-9);
        printf("%I64d
", s.a[0][0]);
    }
    
    return 0;
}