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 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0
Sample Output
45 104
把问题转化为求矩阵的n-9次幂就行了；
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int k,MOD; 6 int a[10]; 7 int f[10]; 8 struct Matrix 9 { 10 int m[10][10]; 11 }matrix; 12 13 Matrix Mul(Matrix a,Matrix b) 14 { 15 Matrix res; 16 int i,j,k; 17 for(i=0;i<10;i++) 18 { 19 for(j=0;j<10;j++) 20 { 21 res.m[i][j] = 0; 22 for(k=0;k<10;k++) 23 res.m[i][j] = (res.m[i][j]+(a.m[i][k]*b.m[k][j]))%MOD; 24 } 25 } 26 return res; 27 } 28 29 Matrix fastm(Matrix a,int b) 30 { 31 Matrix res; 32 memset(res.m,0,sizeof(res.m)); 33 for(int i=0;i<10;i++) 34 res.m[i][i] = 1; 35 while(b) 36 { 37 if(b&1) 38 res = Mul(res,a); 39 a = Mul(a,a); 40 b >>= 1; 41 } 42 return res; 43 } 44 void init() 45 { 46 for(int i=0;i<=9;i++) 47 { 48 f[i]=i; 49 } 50 } 51 int main() 52 { 53 init(); 54 while(scanf("%d%d",&k,&MOD)==2) 55 { 56 for(int i=0;i<=9;i++) 57 { 58 scanf("%d",&a[i]); 59 } 60 if(k<10) 61 { 62 printf("%d ",k%MOD); 63 continue; 64 } 65 66 memset(matrix.m,0,sizeof(matrix.m)); 67 for(int i=0;i<=9;i++) 68 matrix.m[0][i]=a[i]; 69 for(int i=1;i<=9;i++) 70 matrix.m[i][i-1] = 1; 71 Matrix ans=fastm(matrix,k-9); 72 Matrix cnt; 73 for(int i=0;i<10;i++) 74 { 75 cnt.m[i][0]=f[9-i]; 76 } 77 Matrix p=Mul(ans,cnt); 78 printf("%d ",p.m[0][0]%MOD); 79 } 80 return 0; 81 }