HDU 3117 矩陣

Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 471    Accepted Submission(s): 233

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 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).

 

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

#include<stdio.h> #include<math.h> #define aa (sqrt(5.0)+1.0)/2 struct node { __int64 num[2][2]; }; struct node tp,res,p; struct node mut(struct node a,struct node b) { struct node c; int i,j,k; for(i=0;i<2;i++) for(j=0;j<2;j++) { c.num[i][j]=0; for(k=0;k<2;k++) c.num[i][j]+=(a.num[i][k]*b.num[k][j])%10000; c.num[i][j]%=10000; //爲了得到後4位，每次乘都mod去10000 } return c; } void abc(int temp) { tp=p; p=res; while(temp) { if(temp&1) p=mut(p,tp); tp=mut(tp,tp); temp>>=1; } } int main() { int n,i; int f[40]={0,1}; for(i=2;i<40;i++) f[i]=f[i-1]+f[i-2]; res.num[0][0]=res.num[1][1]=1; res.num[0][1]=res.num[1][0]=0; while(scanf("%d",&n)!=EOF) { double t=0; p.num[0][0]=p.num[0][1]=p.num[1][0]=1; p.num[1][1]=0; if(n<40) { printf("%d/n",f[n]); continue; } abc(n); t=-0.5*(log10(5.0))+n*log10(aa); // 有一個公式可以得到前4位，百度 t=t-(int)t; //減去整數部分，得到小數 t=pow(10.0,t); while(t<1000) t=t*10.0; printf("%d...",(int)t); printf("%04d/n",p.num[0][1]); //這裏記住，如果後面不夠4位，用0補 } return 0; }