题目链接

题意：斐波那契数列，当长度大于8时，要输出前四位和后四位

思路：后四位很简单，矩阵快速幂取模，难度在于前四位的求解。 
已知斐波那契数列的通项公式：f(n) = (1 / sqrt(5)) * (((1 + sqrt(5)) / 2) ^ n - ((1 + sqrt(5)) / 2) ^ n)，当n >= 40时((1 + sqrt(5)) / 2) ^ n近似为0。所以我们假设f(n) = t * 10 ^ k(t为小数)，所以当两边同时取对数时，log10(t * 10 ^ k) = log10(t) + k = log10((1 / sqrt(5)) * (((1 + sqrt(5)) / 2))) = log10(1 / sqrt(5)) + n * log10(((1 + sqrt(5)) / 2)))，然后减掉整数k，就可以得到log10(t)，进而得到t值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

//typedef long long ll;
typedef __int64 ll;

const int MOD = 10000;

struct mat{
    ll s[2][2];
    mat(ll a = 0, ll b = 0, ll c = 0, ll d = 0) {
        s[0][0] = a;
        s[0][1] = b;
        s[1][0] = c;
        s[1][1] = d;
    }
    mat operator * (const mat& c) {
        mat ans; 
        memset(ans.s, 0, sizeof(ans.s));
        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 2; j++)
                for (int k = 0; k < 2; k++) {
                    ans.s[i][j] = (ans.s[i][j] + s[i][k] * c.s[k][j]);
                    if (ans.s[i][j] >= 100000000)
                        ans.s[i][j] %= MOD;
                }
        return ans;
    }
}c(1, 1, 1, 0), tmp(1, 0, 0, 1);

ll n;

mat pow_mod(ll k) {
    if (k == 0) 
        return tmp;
    else if (k == 1)
        return c;
    mat a = pow_mod(k / 2);
    mat ans = a * a;
    if (k % 2)
        ans = ans * c;
    return ans;
}

int main() {
    while (scanf("%I64d", &n) != EOF) {
        if (n == 0) 
            printf("0\n");
        else {
            mat ans = pow_mod(n - 1); 
            if (n >= 40) {
                double k = log10(1.0 / sqrt(5.0)) + (double)n * log10((1.0 + sqrt(5.0)) / 2.0);
                double temp = k;
                temp = k - (int)temp;
                printf("%d...%.4I64d\n", (int)(1000.0 * pow(10.0, temp)), ans.s[0][0] % MOD);
            }
            else 
                printf("%I64d\n", ans.s[0][0]);
        }
    } 
    return 0;
}

HDU3117-Fibonacci Numbers(矩阵快速幂+log)