f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Sample Input
1 1 3 //a b n
1 2 10
0 0 0
Sample Output
2
5
矩阵A * 矩阵B = 矩阵C
a b f(n-1) f(n)
1 0 f(n-2) f(n-1)
1 # include <iostream> 2 # include <cstdio> 3 # include <algorithm> 4 # include <map> 5 # include <cmath> 6 # define LL long long 7 using namespace std ; 8 9 const int MOD = 7 ; 10 11 struct Matrix 12 { 13 int mat[2][2]; 14 }; 15 16 Matrix mul(Matrix a,Matrix b) //矩阵乘法 17 { 18 Matrix c; 19 for(int i=0;i<2;i++) 20 for(int j=0;j<2;j++) 21 { 22 c.mat[i][j]=0; 23 for(int k=0;k<2;k++) 24 { 25 c.mat[i][j]=(c.mat[i][j] + a.mat[i][k]*b.mat[k][j])%(MOD); 26 } 27 } 28 return c; 29 } 30 Matrix pow_M(Matrix a,int k) //矩阵快速幂 31 { 32 Matrix ans; 33 memset(ans.mat,0,sizeof(ans.mat)); 34 for (int i=0;i<2;i++) 35 ans.mat[i][i]=1; 36 Matrix temp=a; 37 while(k) 38 { 39 if(k&1)ans=mul(ans,temp); 40 temp=mul(temp,temp); 41 k>>=1; 42 } 43 return ans; 44 } 45 46 47 48 int main () 49 { 50 // freopen("in.txt","r",stdin) ; 51 int a,b,n; 52 while(scanf("%d%d%d" , &a,&b,&n) != EOF) 53 { 54 if (a==0 && b==0 && n==0) 55 break ; 56 if (n <= 2) 57 { 58 printf("1\n") ; 59 continue ; 60 } 61 Matrix t ; 62 t.mat[0][0] = a ; 63 t.mat[0][1] = b ; 64 t.mat[1][0] = 1 ; 65 t.mat[1][1] = 0 ; 66 Matrix ans = pow_M(t,n-2) ; 67 printf("%d\n" , (ans.mat[0][0] + ans.mat[0][1])%MOD) ; 68 69 } 70 71 72 return 0 ; 73 }
时间: 2024-09-29 09:48:32