HDU 1757 A Simple Math Problem（矩阵快速幂）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2930    Accepted Submission(s): 1762

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

45104ＣＯＤＥ：
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<string>#include<algorithm>#include<cstdlib>#include<set>#include<queue>#include<stack>#include<vector>#include<map>#define N 100010#define lson l,mid,idx<<1#define rson mid+1,r,idx<<1|1#define lc idx<<1#define rc idx<<1|1const double EPS = 1e-11;const double PI = acos ( -1.0 );const double E = 2.718281828;typedef long long ll;const int INF = 1000010;using namespace std;typedef vector<int>vec;typedef vector<vec>mat;int m,fac[11];ll k;void init(){    for(int i=0;i<10;i++)        fac[i]=i;}mat mul(mat &A,mat &B,int Mod){    mat C(A.size(),vec(B[0].size()));    for(int i=0; i<A.size(); i++)    {        for(int k=0; k<B.size(); k++)        {            for(int j=0; j<B[0].size(); j++)            {                C[i][j]=(C[i][j]+A[i][k]*B[k][j])%Mod;            }        }    }    return C;}mat pow(mat A,ll n,int Mod){    mat B(A.size(),vec(A.size()));    for(int  i=0; i<A.size(); i++)    {        B[i][i]=1;    }    while(n>0)    {        if(n&1)            B=mul(B,A,Mod);        A=mul(A,A,Mod);        n>>=1;    }    return B;}int main(){    init();    while(cin>>k>>m)    {        if(k<10)        {            printf("%d\n",fac[k]%m);            continue;        }        mat A(10,vec(10));        for(int i=0; i<10; i++)        {            scanf("%d",&A[0][i]);        }        for(int i=1;i<10;i++)        {            for(int j=0;j<10;j++)            {                if(i==j+1)                    A[i][j]=1;                else                    A[i][j]=0;            }        }        A=pow(A,k-9,m);        int ans=0;        for(int i=0;i<10;i++)        {            ans+=A[0][i]*fac[9-i];            if(ans>=m)                ans%=m;        }        cout<<ans<<endl;    }    return 0;}