hdu 1757 A Simple Math Problem 矩阵快速幂

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1757

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.

题意描述：这题题意没什么说的吧，如那个公式所示，给出k和m，然后求f(k)%m。

算法分析：看到k这么大，肯定暴力是不行的，这里得运用到矩阵快速幂。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;

int K,m;
struct matrix
{
    int an[12][12];
}temp;

matrix multiply(matrix a,matrix b)
{
    matrix c;
    memset(c.an,0,sizeof(c.an));
    for (int i=1 ;i<=10 ;i++)
    {
        for (int j=1 ;j<=10 ;j++)
        {
            for (int k=1 ;k<=10 ;k++)
            {
                c.an[i][j]=(c.an[i][j]+a.an[i][k] * b.an[k][j])%m;
                c.an[i][j] %= m;
            }
        }
    }
    return c;
}

int calc(int u)
{
    matrix x;
    memset(x.an,0,sizeof(x.an));
    for (int j=1 ;j<=10 ;j++) x.an[j][1]=10-j;
    while (u)
    {
        if (u&1) x=multiply(temp,x);
        u >>= 1;
        temp=multiply(temp,temp);
    }
    int sum=0;
    return x.an[1][1]%m;
}

int main()
{
    while (scanf("%d%d",&K,&m)!=EOF)
    {
        for (int i=1 ;i<=10 ;i++) scanf("%d",&temp.an[1][i]);
        if (K<=9) {printf("%d
",K);continue; }
        for (int i=2 ;i<=10 ;i++)
        {
            for (int j=1 ;j<=10 ;j++)
                temp.an[i][j]= i==j+1 ? 1 : 0 ;
        }
        printf("%d
",calc(K-9));
    }
    return 0;
}