51Nod——T 1242 斐波那契数列的第N项

 1 #include <cstdio>
 2 
 3 const int mod(1000000009);
 4 #define LL long long
 5 inline void read(LL &x)
 6 {
 7     x=0; register char ch=getchar();
 8     for(; ch>‘9‘||ch<‘0‘; ) ch=getchar();
 9     for(; ch>=‘0‘&&ch<=‘9‘; ch=getchar()) x=x*10+ch-‘0‘;
10 }
11 LL n;
12 
13 struct Matrix {
14     LL e[2][2];
15     Matrix operator * (const Matrix x) const
16     {
17         Matrix tmp;
18         for(int i=0; i<2; ++i)
19             for(int j=0; j<2; ++j)
20             {
21                 tmp.e[i][j]=0;
22                 for(int k=0; k<2; ++k)
23                     tmp.e[i][j]+=e[i][k]*x.e[k][j],tmp.e[i][j]%=mod;
24             }
25         return tmp;
26     }
27 }ans,base;
28 
29 int Presist()
30 {
31     read(n);
32     ans.e[0][0]=ans.e[1][1]=1;
33     base.e[0][0]=base.e[0][1]=base.e[1][0]=1;
34     for(; n; n>>=1, base=base*base)
35         if(n&1) ans=ans*base;
36     printf("%lld",ans.e[0][1]);
37     return 0;
38 }
39 
40 int Aptal=Presist();
41 int main(){;}