题意：M斐波那契数列F[n]是一种整数数列，它的定义如下：
F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )
现在给出a, b, n，你能求出F[n]的值吗？
思路：观察发现。。。F[n] = a^(fib(n-1)) * b ^ (fib(n))
此处要用到指数循环节的知识：
a^n ≡ a^(n % Phi(M) + Phi(M)) (mod M) (n >= Phi(M))
然后 因为1000000007是质数，对于任意的x,有gcd(x,1000000007) = 1，所以可以结合费马小定理化简上式：
a^n ≡ a^(n%(m-1)) * a^(m-1)≡ a^(n%(m-1)) (mod m)
记得特判一下n为0和1的情况。
xiaodingli
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/* *********************************************** Author :111qqz Created Time :Wed 26 Oct 2016 09:16:22 AM CST File Name :code/hdu/4549.cpp ************************************************ */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <cstdlib> #include <ctime> #define fst first #define sec second #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define ms(a,x) memset(a,x,sizeof(a)) typedef long long LL; #define pi pair < int ,int > #define MP make_pair using namespace std; const double eps = 1E-8; const int dx4[4]={1,0,0,-1}; const int dy4[4]={0,-1,1,0}; const int inf = 0x3f3f3f3f; const LL mod = 1E9+7; LL a,b,n; struct Mat { LL mat[2][2]; void clear() { ms(mat,0); } }M,M1; Mat operator * ( Mat a,Mat b) { Mat res; res.clear(); for ( int i = 0 ; i < 2 ; i ++) for ( int j = 0 ; j < 2 ; j++) for ( int k = 0 ; k < 2 ; k++) res.mat[i][j] = (res.mat[i][j] + a.mat[i][k] * b.mat[k][j] ) % (mod - 1); return res; } Mat operator ^ (Mat a,LL b) { Mat res; res.clear(); for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1; while (b>0) { if (b&1) res = res * a; b = b >> 1LL; a = a * a ; } return res; } LL ksm( LL a,LL b) { LL res = 1LL; while (b>0) { if (b&1){ res = (res * a) %mod; } b = b >> 1LL; a = ( a * a ) % mod; } return res; } void init() { M.clear(); M.mat[0][1] = 1; M.mat[1][0] = 1; M.mat[1][1] = 1; M1.clear(); M1.mat[0][0] = 0; M1.mat[1][0] = 1; } int main() { #ifndef ONLINE_JUDGE freopen("code/in.txt","r",stdin); #endif while (~scanf("%lld %lld %lld",&a,&b,&n)) { if (n==0) { printf("%lld\n",a); continue; } if (n==1) { printf("%lld\n",b); continue; } init(); Mat ans; ans.clear(); ans = (M ^(n-1))*M1; LL x = ans.mat[0][0]; LL y = ans.mat[1][0]; LL ret =(ksm(a,x)*ksm(b,y))%mod; printf("%lld\n",ret); } #ifndef ONLINE_JUDGE fclose(stdin); #endif return 0; } |
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