POJ 3070 Fibonacci 【矩阵快速幂】

<题目链接>

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

 解题分析：

由于n很大，所以直接计算时不可行的，可以用矩阵快速幂来加速，并且，此题直接给出了矩阵的递推式，于是，我们只要按照题意构造矩阵即可。

#include <cstdio>
#include <cstring>

const int mod=;

struct Matrix{
    int m[][];
    Matrix(){}
    Matrix(int x,int y,int z,int k){
        m[][]=x;
        m[][]=y;
        m[][]=z;
        m[][]=k;
    }
};

Matrix Mul(Matrix a,Matrix b){
    Matrix temp;
    for(int i=;i<;i++){
        for(int j=;j<;j++){
            temp.m[i][j]=;
            for(int k=;k<;k++){
                temp.m[i][j]=(temp.m[i][j]+a.m[i][k]%mod*b.m[k][j]%mod)%mod;
            }
        }
    }
    return temp;
}

Matrix pow_Mul(Matrix x,int c){
    Matrix tmp(,,,);
    while(c>){
        if(c&){
            tmp=Mul(tmp,x);
        }
        c>>=;
        x=Mul(x,x);
    }
    return tmp;
}

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==-)break;
        int f[]={,,,};
        if(n<=){
            printf("%d\n",f[n]);
            continue;
        }
        Matrix init(f[],f[],f[],f[]);
        Matrix tmp(,,,);
        Matrix ans=Mul(pow_Mul(tmp,n-),init);
        printf("%d\n",ans.m[][]%mod);
    }
    return ;
}

2018-08-21