POJ3070 Fibonacci

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

 

 

正解：矩阵乘法+矩阵快速幂

解题报告：

　　就是一个斐波那契数列求第几项是多少的题，考虑用矩阵乘法优化递推公式，显然要用矩阵快速幂（尽管矩阵不满足交换律但是满足结合律）。

　　因为只有2X2，直接打不用for也可以。

 

//It is made by jump~
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#ifdef WIN32   
#define OT "%I64d"
#else
#define OT "%lld"
#endif
using namespace std;
typedef long long LL;
const int MOD = 10000;
int n;

struct Fibn{
    LL s[5];
}a,b;

inline int getint()
{
       int w=0,q=0;
       char c=getchar();
       while((c<'0' || c>'9') && c!='-') c=getchar();
       if (c=='-')  q=1, c=getchar();
       while (c>='0' && c<='9') w=w*10+c-'0', c=getchar();
       return q ? -w : w;
}

inline Fibn cheng(Fibn p,Fibn q){
    Fibn tmp;
    tmp.s[1]=p.s[1]*q.s[1]+p.s[2]*q.s[3]; tmp.s[1]%=MOD;
    tmp.s[2]=p.s[1]*q.s[2]+p.s[2]*q.s[4]; tmp.s[2]%=MOD;    
    tmp.s[3]=p.s[3]*q.s[1]+p.s[4]*q.s[3]; tmp.s[3]%=MOD;
    tmp.s[4]=p.s[3]*q.s[2]+p.s[4]*q.s[4]; tmp.s[4]%=MOD;
    return tmp;
}

inline void fib(){
    a.s[1]=1; a.s[2]=0; a.s[3]=0; a.s[4]=1;
    b.s[1]=0; b.s[2]=1; b.s[3]=1; b.s[4]=1;
    while(n) {
    if(n&1) a=cheng(a,b);
    b=cheng(b,b);
    n=n/2;
    }
    //printf("%d %d %d %d
",a.s[1],a.s[2],a.s[3],a.s[4]);
    printf("%d
",(int)a.s[3]);
}

inline void work(){
    while(1) {
    n=getint();
    if(n==-1) break;
    if(n==0) { printf("0
"); continue; }
    fib();
    }
}

int main()
{
  work();
  return 0;
}