HDU 1757 A Simple Math Problem （矩阵快速幂）

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本文详细介绍了矩阵快速幂的模板实现，并通过实例演示了如何解决特定问题，包括输入处理、矩阵运算以及最终结果输出。

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3093 Accepted Submission(s): 1874

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

【思路分析】

矩阵快速幂纯模板- -

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int SIZE = 10;
int k,mod;
struct Matrix
{
    long long value[SIZE][SIZE];
    Matrix()
    {
        memset(value,0,sizeof(value));
    }
    void init(int n)
    {
        for(int i = 0;i < SIZE;i++)
        {
            value[i][i] = n;
        }
    }
};

Matrix operator * (Matrix a,Matrix b)
{
    Matrix c;
    memset(c.value,0,sizeof(c.value));
    for(int i = 0;i < SIZE;i++)
    {
        for(int j = 0;j < SIZE;j++)
        {
            for(int k = 0;k < SIZE;k++)
            {
                c.value[i][j] += (a.value[i][k] * b.value[k][j]) % mod;
                c.value[i][j] %= mod;
            }
        }
    }
    return c;
}

Matrix operator ^ (Matrix a,long long t)
{
    Matrix c;
    c.init(1);
    while(t)
    {
        if(t & 1)
            c = a * c;
        a = a * a;
        t = t >> 1;
    }
    return c;
}
int main()
{
    Matrix f,temp,res;
    for(int i = 0;i < SIZE;i++)
    {
        f.value[i][0] = 9 - i;//注意顺序
    }
    for(int i = 1;i < SIZE;i++)
    {
        temp.value[i][i - 1] = 1;
    }
    //freopen("1.txt","r",stdin);
    while(scanf("%d %d",&k,&mod) != EOF)
    {
        if(k < 10)
        {
            printf("%d\n",k % mod);
        }
        else
        {
            for(int i = 0;i < 10;i++)
            {
                scanf("%lld",&temp.value[0][i]);
            }
            res = temp ^ (k - 9);
            res = res * f;
            printf("%lld\n",res.value[0][0] % mod);
        }
    }
    return 0;
}