HDU 1757（数论，矩阵）

 

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 440    Accepted Submission(s): 253

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

此题矩阵的幂求解要用到快速求幂~ 有点类似求斐波那契数列！
#include <iostream>using namespace std;#define ll __int64struct Mat {int martix[10][10];};ll n,m;Mat init,q,tp,res;Mat Mul(Mat a,Mat b){int i,j,k;Mat c;for(i=0;i<10;i++){for(j=0;j<10;j++){c.martix[i][j]=0;for(k=0;k<10;k++){c.martix[i][j]+=(a.martix[i][k]*b.martix[k][j])%m;c.martix[i][j]%=m;}}}return c;}void mat_pow(ll x)  {tp=q;q=res;while(x){if(x&1)q=Mul(q,tp);tp=Mul(tp,tp);x=x>>1;}}int main(){int i,j,sum;memset(init.martix,0,sizeof(init.martix));init.martix[1][0]=1;init.martix[2][1]=1;init.martix[3][2]=1;init.martix[4][3]=1;init.martix[5][4]=1;init.martix[6][5]=1;init.martix[7][6]=1;init.martix[8][7]=1;init.martix[9][8]=1;for(i=0;i<10;i++){for(j=0;j<10;j++){if(i==j)res.martix[i][j]=1;elseres.martix[i][j]=0;}}while (scanf("%I64d%I64d",&n,&m)!=EOF){q=init;for(i=0;i<10;i++)scanf("%d",&q.martix[0][i]);if(n>=10){sum=0;mat_pow(n-9);for(i=0;i<10;i++)sum+=q.martix[0][i]*(9-i);printf("%d/n",sum%m);}elseprintf("%d/n",n%m);}return 0;}