poj 3070

 

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4686		Accepted: 3241

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

Source

Stanford Local 2006

分析：简单题~~~

代码：

#include<cstdio> int main() { int n,i,j,k,p,q,g[31][2][2],cur[2][2]; g[1][0][0]=0,g[1][0][1]=g[1][1][0]=g[1][1][1]=p=q=1; while((q<<1)<=1000000000) { k=p+1; for(i=0;i<2;++i) for(j=0;j<2;++j) { g[k][i][j]=g[p][i][0]*g[p][0][j]+g[p][i][1]*g[p][1][j]; if(g[k][i][j]>=10000)g[k][i][j]%=10000; } ++p,q<<=1; } while(scanf("%d",&n)!=EOF) { if(n<0)break; if(n>1) { --n; g[0][0][1]=g[0][1][0]=0,g[0][0][0]=g[0][1][1]=1; while(n>0) { q=p=1; while((q<<1)<=n)++p,q<<=1; n-=q; for(i=0;i<2;++i) for(j=0;j<2;++j) { for(cur[i][j]=k=0;k<2;++k) cur[i][j]+=g[0][i][k]*g[p][k][j]; if(cur[i][j]>=10000)cur[i][j]%=10000; } for(i=0;i<2;++i) for(j=0;j<2;++j) g[0][i][j]=cur[i][j]; } n=g[0][1][1]; } printf("%d/n",n); } }