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:

.

/*
矩阵快速幂裸题
*/
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 30;
ll sz,mod,n,f[maxn],a[maxn],ans[maxn];
struct mtx{
    ll v[maxn][maxn];
    void cler(){
        memset(v,0,sizeof(v));
    }
    mtx mul(mtx A,mtx B){
        mtx C;
        C.cler();
        for(int i = 1;i <= sz;i++){
            for(int j = 1;j <= sz;j++){
                for(int k = 1;k <= sz;k++){
                    C.v[i][j] = (C.v[i][j] + A.v[i][k] * B.v[k][j]) % mod;
                }
            }
        }
        return C;
    }
    mtx mi(mtx A,int n){
        mtx R;
        R.cler();
        for(int i = 1;i <= sz;i++) R.v[i][i] = 1;
        while(n){
            if(n&1) R = R.mul(R,A);
            n >>= 1;
            A = A.mul(A,A);
        }
        return R;
    }
    void get_tr(mtx A){
        memset(ans,0,sizeof(ans));
        for(int i = 1;i <= sz;i++){
            for(int j = 1;j <= sz;j++){
                ans[j] = (ans[j] + f[i] * A.v[i][j]) % mod;
            }
        }
    }
};
int main(){
    sz = 2;
    mod = 10000;
    f[1] = f[2] = 1;
    a[1] = a[2] = 1;
    mtx A;
    while(1){    
        cin>>n;
        if(n == -1){
            return 0;
        }
        if(n <= 2){
            if(n==0) cout<<0<<endl;
            if(n==1||n==2) cout<<1<<endl;
            continue;
        }
        A.cler();
        for(int i = 1;i <= sz;i++) A.v[i][1] = a[i];
        for(int i = 2;i <= sz;i++) A.v[i-1][i] = 1;
        A = A.mi(A,n-sz);
        A.get_tr(A);
        cout<<ans[1]<<endl;
    }
    return 0;
}