Hdu 1757 A Simple Math Problem【矩阵快速幂】

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A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3778 Accepted Submission(s): 2280

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 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

Sample Output

矩阵快速幂，原本是为求解矩阵的高次幂运算的，但是在使用中，可以构造出自己需要的矩阵，来高速解决很多种计算问题，比如递推式求值，研究了这一道题，大概明白了利用矩阵求解的原理，不过理解比较浅显，需要继续努力！

矩阵的难点就在于构造矩阵了

#include<stdio.h>#include<string.h>int k,mod,f[10];struct matrix{    int m[15][15];    matrix operator * (const matrix &a) const//定义矩阵的乘法     {        matrix tp;        memset(tp.m,0,sizeof(tp));        for(int i=0;i<10;++i)        {            for(int j=0;j<10;++j)            {                if(!m[i][j])//加速矩阵的运算                 {                    continue;                }                for(int k=0;k<10;++k)                {                    tp.m[i][k]+=m[i][j]*a.m[j][k]%mod;                    tp.m[i][k]%=mod;                }            }        }        return tp;    }}base,ans;void init()//base为构造矩阵，ans初始化为单位矩阵， {    memset(base.m,0,sizeof(base.m));    memset(ans.m,0,sizeof(ans.m));    for(int i=0;i<10;++i)    {        ans.m[i][i]=1;        base.m[i+1][i]=1;        base.m[0][i]=f[i];    }}int slove(){    if(k<10)    {        return k%mod;    }     init();k-=9;    while(k)    {        if(k&1)        {            ans=ans*base;        }        base=base*base;        k>>=1;    }    int sum=0;    for(int i=0;i<10;++i)    {        sum+=(9-i)*ans.m[0][i]%mod;        sum%=mod;    }    return sum;}int main(){    //freopen("shuju.txt","r",stdin);    while(~scanf("%d%d",&k,&mod))    {        for(int i=0;i<10;++i)        {            scanf("%d",&f[i]);        }        printf("%d\n",slove());    }    return 0;}

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