hdu 3117 Fibonacci Numbers
Fibonacci Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1766 Accepted Submission(s): 695
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.
![](http://acm.hdu.edu.cn/data/images/C236-1008.JPG)
What is the numerical value of the nth Fibonacci number?
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 108 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).
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
题解及代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #define s_5 sqrt(5.0) #define sl_5 (0.5+sqrt(5.0)/2.0) #define sr_5 (0.5-sqrt(5.0)/2.0) using namespace std; int fibo[50]; void init() { fibo[0]=0; fibo[1]=1; int i; for(i=2;i<=50;i++) { fibo[i]=fibo[i-1]+fibo[i-2]; if(fibo[i]>=100000000) { break; } } } struct mat { long long t[2][2]; void set() { memset(t,0,sizeof(t)); } }a,b; mat multiple(mat a,mat b,int n,int p) { int i,j,k; mat temp; temp.set(); for(i=0;i<n;i++) for(j=0;j<n;j++) { if(a.t[i][j]) for(k=0;k<n;k++) temp.t[i][k]=(temp.t[i][k]+a.t[i][j]*b.t[j][k])%p; } return temp; } mat quick_mod(mat b,int n,int p) { mat t; t.t[0][0]=1; t.t[0][1]=0; t.t[1][0]=0; t.t[1][1]=1; while(n) { if(n&1) { t=multiple(t,b,2,p); } n>>=1; b=multiple(b,b,2,p); } return t; } void init1() { b.t[0][0]=1; b.t[0][1]=1; b.t[1][0]=1; b.t[1][1]=0; } int main() { int n; init(); while(cin>>n) { if(n<40) { cout<<fibo[n]<<endl; continue; } double s=log10(1.0/s_5)+n*log10(sl_5); int len=(int)s+1; double t=s-len+4; cout<<(int)pow(10.0,t); init1(); a=quick_mod(b,n,10000); printf("...%04d\n",a.t[1][0]); } return 0; } /* 一道比較简单的数学题,输出斐波那契数的前四位和后四位。 对于前40项,直接打表输出就能够了。 接下来我们讲一下大于40位的时候怎样计算: 对于后四位,我们能够想到,我们进行滚动数组取余就能求得答案。 可是数据量有点大,这样可定会超时,所以仅仅能使用矩阵高速幂来加速。 对于前四位。我们要使用到斐波那契数的封闭公式: f[n]=1/sqrt(5)*{[0.5+sqrt(5)/2]^n-[0.5-sqrt(5)/2]^n}; 由于n非常大时,[0.5-sqrt(5)/2]^n非常小。差点儿能够忽略。又由于我们计算 的是前四位,所以不必管后面的精度,所以这一项能够省略。 f[n]=1/sqrt(5)*[0.5+sqrt(5)/2]^n; 我们假定t=f[n],k为其前四位,想要求t的前四位数k。我们能够使用科学 计数法来写一下t,t=k.xxxx……*10^(len-4);这里len指的是t的位数。 而计算一个数的位数能够使用log10(t)+1来计算得到。那么知道了这些就非常好计算了:首先我们先求出t的位数 len=log10(1/sqrt(5)*[0.5+sqrt(5)/2]^n); 化简一下:len=log10(1/sqrt(5))+n*log10(0.5+sqrt(5)/2); 对于t=k.xxxx……*10^(len-4);我们两边进行log10取对数, 得到log10(t)=log10(k.xxxx……)+len-4; 化简一下得到log10(k.xxxx……)=log10(t)-(len-4); 那么我们为了得到k.xxx……,我们能够求10^[log10(t)-(len-4)]即可了。 最后一项,对k取整。
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