HDU - 1757 - A Simple Math Problem ( 矩阵快速幂 )

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

思路：和UVa的1087是一样的  题目链接

#include<cstdio>#include<cstring>#define LL long longusing namespace std;int n,k,m;struct Matrix{LL m[10][10];Matrix(){memset(m,0,sizeof(m));}};Matrix Mul(Matrix x ,Matrix y){Matrix ans;for(int i=0 ;i<10 ;i++){for(int j=0 ;j<10 ;j++){for(int k=0 ;k<10 ;k++){ans.m[i][j] = ( ans.m[i][j] +  x.m[i][k] * y.m[k][j] )%m;}}}return ans;}Matrix q_pow(Matrix x,int k){Matrix ans;for(int i=0 ;i<10 ;i++) ans.m[i][i]=1;while(k){if(k&1) ans = Mul(ans,x);x = Mul(x,x);k >>= 1;}return ans;}int main(){LL a0,a1,a2,a3,a4,a5,a6,a7,a8,a9;while(scanf("%d%d",&k,&m)!=EOF){scanf("%lld%lld%lld%lld%lld%lld%lld%lld%lld%lld",&a0,&a1,&a2,&a3,&a4,&a5,&a6,&a7,&a8,&a9);if(k<10){printf("%d\n",k);}else{Matrix T ;T.m[0][0] = a0;T.m[0][1] = a1;T.m[0][2] = a2;T.m[0][3] = a3;T.m[0][4] = a4;T.m[0][5] = a5;T.m[0][6] = a6;T.m[0][7] = a7;T.m[0][8] = a8;T.m[0][9] = a9;for(int i=1 ;i<10 ;i++) T.m[i][i-1]=1;Matrix ans = q_pow(T,k-9);LL res = 0;for(int i=0 ;i<10 ;i++){res = (res+ans.m[0][i]*(9-i))%m;}printf("%lld\n",res);}}return 0;}