HDU 1757 A Simple Math Problem 矩阵快速幂

点击打开链接

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2936    Accepted Submission(s): 1767

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

45104

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场

 

f(x) = x.  x < 10 
 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)   x >= 10

求f(k)%m

递推一般都可以用矩阵快速幂来解决，关键在于构造矩阵。

/*                                                                            k-9f(n)           a0  a1  a2  a3  a4  a5  a6  a7  a8  a9                     f9f(n-1)        1    0    0   0    0    0    0   0     0     0                     f8f(n-2)        0    1    0   0    0    0    0   0     0     0                     f7f(n-3)        0    0    1   0    0    0    0   0     0     0                     f6f(n-4)  =   0    0    0    1    0   0     0   0     0    0           *         f5f(n-5)        0    0    0    0   1    0    0    0     0    0                     f4f(n-6)        0    0    0    0   0    1    0    0     0    0                     f3f(n-7)        0    0    0    0   0    0    1    0     0    0                     f2f(n-8)        0    0    0    0   0    0    0    1     0    0                     f1f(n-9)        0    0    0    0   0    0    0    0     1    0                     f0*///15MS1112K#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#define mod 10000007#define ll __int64using namespace std;ll one[10];struct Matrax{    ll m[11][11];}a,per,tmp;void init()//建立矩阵{    memset(a.m,0,sizeof(a.m));    memset(per.m,0,sizeof(per.m));    for(int i=0;i<10;i++)a.m[0][i]=one[i];    for(int i=0;i<10;i++)a.m[i+1][i]=1;    for(int i=0;i<10;i++)        for(int j=0;j<10;j++)            if(i==j)per.m[i][i]=1;}Matrax multi(Matrax a,Matrax b,ll m)//矩阵相乘{    Matrax c;    for(int i=0;i<10;i++)        for(int j=0;j<10;j++)        {            c.m[i][j]=0;            for(int k=0;k<10;k++)                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%m;        }        return c;}Matrax power(ll k,ll m)//矩阵快速幂{    Matrax pp=a,ans=per;    while(k)    {        if(k&1){ans=multi(ans,pp,m);k--;}        else {k>>=1;pp=multi(pp,pp,m);}    }    return ans;}int main(){    ll k,m;    while(scanf("%I64d%I64d",&k,&m)!=EOF)    {        ll sum=0;        for(int i=0;i<10;i++)            scanf("%I64d",&one[i]);        init();        Matrax ans=power(k-9,m);        for(int i=0;i<10;i++)            sum=(sum+ans.m[0][i]*(10-i-1))%m;        printf("%I64d\n",sum);    }    return 0;}