HDU 3117（数论，斐波那契+矩阵连乘）

Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 378    Accepted Submission(s): 184

Problem Description

The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one.

What is the numerical value of the nth Fibonacci number?

 

 

Input

For each test case, a line will contain an integer i between 0 and 10 ⁸ inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”).

There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF).

 

 

Sample Input

   
   
   
   

    
    
    
    0
1
2
3
4
5
35
36
37
38
39
40
64
65
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    0
1
1
2
3
5
9227465
14930352
24157817
39088169
63245986
1023...4155
1061...7723
1716...7565
   
   
   
   

 

 

/************************************************************************ 此题前四位要用到斐波那契的通项公式 Fibonacci Numbers 的通项公式： f(n)=1/sqrt(5.0)(((1+sqrt(5.0))/2)^n+((1-sqrt(5.0))/2)^n); 假设F[n]可以表示成 t * 10^k（t是一个小数），那么对于F[n]取对数log10， 答案就为log10 t + K，此时很明显log10 t<1，于是我们去除整数部分， 就得到了log10 t ，再用pow（10，log10 t)我们就还原回了t。 将t×1000就得到了F[n]的前四位。 log10(fn)=n* log10((1.0+sqrt(5.0))/2)+log10(1/sqrt(5.0)); 后四位用矩阵连乘，当然要用到快速幂了 别忘了后四位求出来可能没有四位数哦，有可能是0123这种情况 ************************************************************************/ #include <iostream> #include <math.h> using namespace std; #define N 2 #define ll __int64 #define MOD 10000 struct Mat { int martix[N][N]; }; Mat q,res,tp,tp1,tp2,tp3; int Fibonacci[40]; void er_fun(int x) { int i,j,k,flag=0; tp1=q; tp=res; while (x) { if(x&1) { flag=1; memset(tp2.martix,0,sizeof(tp2.martix)); for (i=0;i<N;i++) { for (j=0;j<N;j++) { for(k=0;k<N;k++) { tp2.martix[i][j]+=(tp.martix[i][k]*tp1.martix[k][j])%MOD; tp2.martix[i][j]%=MOD; } } } } memset(tp3.martix,0,sizeof(tp3.martix)); for (i=0;i<N;i++) { for (j=0;j<N;j++) { for(k=0;k<N;k++) { tp3.martix[i][j]+=(tp1.martix[i][k]*tp1.martix[k][j])%MOD; } } } if(flag) tp=tp2; tp1=tp3; x>>=1; } } void init() { int i; Fibonacci[0]=0; Fibonacci[1]=1; for(i=2;i<=39;i++) { Fibonacci[i]=Fibonacci[i-1]+Fibonacci[i-2]; } } int main() { int i,j; int res_back; double res_front; int n; for(i=0;i<N;i++) { for (j=0;j<N;j++) { res.martix[i][j]=(i==j); } } q.martix[0][0]=q.martix[0][1]=q.martix[1][0]=1; q.martix[1][1]=0; init(); while (scanf("%d",&n)!=EOF) { if(n<40) { printf("%d/n",Fibonacci[n]); continue; } er_fun(n-1); res_back=tp.martix[0][0]; res_front=n* log10((1.0+sqrt(5.0))/2)+log10(1/sqrt(5.0)); res_front-=(int)res_front; res_front=pow(10.0,res_front); while(res_front<1000) res_front*=10.0; printf("%d...%4.4d/n",(int)res_front,res_back); } return 0; }