HDU 4549 M Fibonacci series (Rapid matrix power, rapid power reduction)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4549

F [0] = a ^ 1 * B ^ 0% P, F [1] = a ^ 0 * B ^ 1% P, f [2] = a ^ 1 * B ^ 1% p ..... f [N] = a ^ fib [n-1] * B ^ fib [N-2] % P.

Here, P is a prime number and a and P are mutually Prime, So we calculate a ^ B % P. When B is very large, we need to reduce the power of B.

Because a and P are mutually unique, a ^ (p-1) % P = 1 is known by the Fermat theorem. Make B = K * (PM) + B ', a ^ B % P = a ^ (K * (PM) + B ') % P = a ^ (K P-1) % P * a ^ B '% P

= A ^ B '% P. Here, p' = B % (p-1 ).

Then, the FIB [n-1] and FIB [N-2] of the above question can be obtained through the matrix Rapid power pair P-1) modulo, realizing power reduction and then using the rapid power.

In fact, a ^ B % C is a ^ (B % Phi [c] + Phi [c]) % C (B> = Phi [c]), when C is a prime number or A and C are mutually Prime, it can be simplified to a ^ (B % (c-1) % C.

At noon, the result was that the FIB series was wrong. F [0] = 0, F [1] = 1, F [N] = f [n-1] + F [N-2] (n> = 2 ). The 0th items are 0, sad ....

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL long long#define _LL __int64#define eps 1e-12#define PI acos(-1.0)#define C 240#define S 20using namespace std;const int maxn = 110;const int mod = 1000000007;struct matrix{LL mat[3][3];void init(){memset(mat,0,sizeof(mat));for(int i = 0; i < 2; i++)mat[i][i] = 1;}}m;LL a,b,n;matrix mul_matrix(matrix x, matrix y){matrix ans;memset(ans.mat,0,sizeof(ans.mat));for(int i = 0; i < 2; i++){for(int k = 0; k < 2; k++){if(x.mat[i][k] == 0) continue;for(int j = 0; j < 2; j++)ans.mat[i][j] = (ans.mat[i][j] + x.mat[i][k]*y.mat[k][j])%(mod-1);}}return ans;}matrix pow_matrix(matrix m, LL n){matrix ans;ans.init();while(n){if(n&1)ans = mul_matrix(ans,m);m = mul_matrix(m,m);n >>= 1;}return ans;}LL pow_mod(LL a, LL b){LL res = 1;a = a%mod;while(b){if(b&1)res = res*a%mod;a = a*a%mod;b >>= 1;}return res;}int main(){while(~scanf("%lld %lld %lld",&a,&b,&n)){m.mat[0][0] = m.mat[0][1] = m.mat[1][0] = 1;m.mat[1][1] = 0;if(n == 0){printf("%lld\n",a%mod);continue;}if(n == 1){printf("%lld\n",b%mod);continue;}else if(n == 2){printf("%lld\n",a*b%mod);continue;}matrix ans = pow_matrix(m,n);LL res = pow_mod(a,ans.mat[1][1]) * pow_mod(b,ans.mat[0][1]) %mod;printf("%lld\n",res);}return 0;}