HDU 3117 Fibonacci Numbers（围绕四个租赁斐波那契，通过计++乘坐高速动力矩阵） HDU 3117 Fibonacci Numbers（斐波那契前后四位，打表+取对+矩阵高速幂）

ACM

题目地址：HDU 3117 Fibonacci Numbers

题意： 
求第n个斐波那契数的前四位和后四位。 
不足8位直接输出。

分析： 
前四位有另外一题HDU 1568，用取对的方法来做的。 
后四位能够用矩阵高速幂，MOD设成10000即可了。

代码：

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  Blog:        http://blog.csdn.net/hcbbt
*  File:        3117.cpp
*  Create Date: 2014-08-04 10:25:26
*  Descripton:   
*/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define repf(i,a,b) for(int i=(a);i<=(b);i++)

typedef long long ll;

const int N = 41;
const int SIZE = 2;        // max size of the matrix
const int MOD = 10000;

ll n;
ll tab[N];
double ans;

struct Mat{
    int n;
    ll v[SIZE][SIZE];    // value of matrix

    Mat(int _n = SIZE) {
        n = _n;
        memset(v, 0, sizeof(v));
    }

    void init(ll _v) {
        repf (i, 0, n - 1)
            v[i][i] = _v;
    }

    void output() {
        repf (i, 0, n - 1) {
            repf (j, 0, n - 1)
                printf("%lld ", v[i][j]);
            puts("");
        }
        puts("");
    }
} a, b;

Mat operator * (Mat a, Mat b) {
    Mat c(a.n);
    repf (i, 0, a.n - 1) {
        repf (j, 0, a.n - 1) {
            c.v[i][j] = 0;
            repf (k, 0, a.n - 1) {
                c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;
                c.v[i][j] %= MOD;
            }
        }
    }
    return c;
}

Mat operator ^ (Mat a, ll k) {
    Mat c(a.n);
    c.init(1);
    while (k) {
        if (k&1) c = a * c;
        a = a * a;
        k >>= 1;
    }
    return c;
}

double fib(int x) {
    return -0.5 * log(5.0) / log(10.0) + ( (double)n) * log((sqrt(5.0) + 1) / 2) / log(10.0);
}

void table() {
    // table
    tab[0] = 0;
    tab[1] = 1;
    repf (i, 2, 40)
        tab[i] = tab[i - 1] + tab[i - 2];
}

void pre4(int n) {
    ans = fib(n);
    ans -= floor(ans);
    ans = pow(10.0, ans);
    while (ans < 1000)
        ans *= 10;
    printf("%d", (int)ans);
}

void last4(int n) { 
    a.init(0);
    a.v[0][0] = a.v[0][1] = a.v[1][0] = 1;

    b = a ^ (n - 1);
    printf("%04lld
", b.v[0][0]);
}

int main() {
    table();
    while (~scanf("%lld", &n)) {
        if (n < 40) {
            printf("%lld
", tab[n]);
            continue;
        }
        pre4(n);
        printf("...");
        last4(n);
    }
    return 0;
}