HDU

A Simple Math Problem

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

SubmitStatus

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： f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)，ai(0<=i<=9)为0或1

思路：能够用递推做。只是太耗时了，准TLE。用转化为矩阵，再用高速幂，复杂度大大的降低。

盗用一张图：

把问题转化为求矩阵的n-9次幂即可了；

<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
int m, k;
struct Matrix
{
    int m[10][10];
    void clear()
    {
        memset(m, 0, sizeof(m));
    }
};

Matrix multi(Matrix a, Matrix b)
{   //矩阵乘法
    Matrix t;
    t.clear();
    for(int i = 0; i < 10; i++)
    {
        for(int j = 0; j < 10; j++)
        {
            for(int k = 0; k < 10; k++)
            {
                t.m[i][j] += a.m[i][k]*b.m[k][j];
            }
            t.m[i][j] %= m;
        }
    }
    return t;
}

Matrix pow_mod(Matrix a, Matrix b)
{   //高速幂（重复平发法）
    while(k)
    {
        if(k&1)
            b = multi(a, b);
        a = multi(a, a);
        k >>= 1;
    }
    return b;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while(cin>>k>>m)
    {
        Matrix a, b;
        a.clear();
        b.clear(); //构造矩阵a
        for(int i = 1; i < 10; i++)
        {
            a.m[i][i-1] = 1;
        }
        //b为单元矩阵，相当于整数的1
        for(int i = 0; i < 10; i++)
        {
            b.m[i][i] = 1;
        }
        for(int i = 0; i < 10; i++)
        {
            scanf("%d", &a.m[0][i]);
        }
        if(k < 10)
        {
            printf("%d
", k%m);
            continue;
        }
        k -= 9;
        b = pow_mod(a, b); //求a的 n-9 次幂并保存在 b 中
        int ans = 0;
        for(int i = 0; i < 10; i++)
            ans = (ans+b.m[0][i]*(9-i))%m; //最后乘上f[9],f[8],f[7]...f[0]
        cout<<ans<<endl;
    }
    return 0;
}</span>