其中 f(1)=x0,f(2)=y0, f( i )=f( i-1 )+f( i+1 ),求 f(n)
#include <iostream> #include <cmath> #include <algorithm> using namespace std; #define N 2 int mod=1e9+7; #define int long long struct matrix { int a[N+2][N+2]; }; int n; matrix m1; void init_e(matrix &x){ int i,j; for(i=1;i<=N;i++) for(j=1;j<=N;j++) { x.a[i][j]=0; if(i==j) x.a[i][j]=1; } } matrix mul(matrix &x,matrix &y){ int i,j,k; matrix z; for(i=1;i<=N;i++) for(j=1;j<=N;j++){ z.a[i][j]=0; for(k=1;k<=N;k++) z.a[i][j]+=x.a[i][k]*y.a[k][j], z.a[i][j]%=mod; } return z; } matrix ksm(matrix &x,int k){ matrix tmp=x, ans; init_e(ans); while(k){ if(k&1) ans=mul(ans,tmp); tmp=mul(tmp,tmp); k/=2; } return ans; } signed main(){ std::ios::sync_with_stdio(0); int x0,y0; cin>>x0>>y0>>n; if(n==1){ cout<<(x0%mod+mod)%mod<<endl; return 0; } m1.a[1][1]=1,m1.a[1][2]= -1; m1.a[2][1]=1,m1.a[2][2]= 0; matrix ans =ksm(m1,n-2); cout<< ((ans.a[1][1]*y0+ans.a[1][2]*x0)%mod+mod)%mod<<endl; cout<<endl; }
标签:matrix,int,Jzzhu,long,Sequences,CF450B,include From: https://www.cnblogs.com/towboa/p/17156802.html