poj 3070 矩阵快速幂

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536K

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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:

思路：裸地矩阵快速幂

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define maxn 0x3f3f3f3f#define MAX 1000100///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef unsigned long long ull;#define INF (1ll<<60)-1using namespace std;const ll mod=10000;ll n;struct Matrix{    ll ma[2][2];    Matrix(){        for(int i=0;i<2;i++)            for(int j=0;j<2;j++)                ma[i][j]=0;    }};Matrix Mult(Matrix a,Matrix b){    Matrix c;    for(int i=0;i<2;i++){        for(int j=0;j<2;j++){            for(int k=0;k<2;k++){                c.ma[i][j]=(c.ma[i][j]+a.ma[i][k]*b.ma[k][j]%mod)%mod;            }        }    }    return c;}Matrix MPow(Matrix a,ll b){    Matrix tmp;    for(int i=0;i<2;i++) tmp.ma[i][i]=1;    while(b){        if(b%2) tmp=Mult(tmp,a);        a=Mult(a,a);        b/=2;    }    return tmp;}int main(){    while(scanf("%I64d",&n)!=EOF){        if(n==-1) break;        if(n==0){            cout<<0<<endl;            continue;        }        Matrix a,b;        a.ma[0][0]=1;a.ma[0][1]=1;a.ma[1][0]=1;        b=MPow(a,n);        printf("%I64d\n",b.ma[1][0]%mod);    }    return 0;}