鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3117

題意：

1. n < 40 時不會超過8位，直接打表輸出
2. n ≥ 40 時：

• • 求一個數的前四位：
    • 推導公式：f(n) =\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n}(1-(\frac{1-\sqrt{5}}{1+\sqrt{5}})^{n}]
    • s = d.xxx * 10^{len - 4}
    • 兩邊取對數得 \log_{10}s = \log_{10}d.xxx + len -4
    • 再得d.xxxx=10^{\log_{10}s-len+4}
    • s =\frac{1}{ \sqrt{5}}*(\frac{1+\sqrt{5}}{2})^{n} （s後面的數值太小可以忽略）
    • len = {\log_{10}s + 1
  • 後四位：
  • 矩陣相乘：\begin{bmatrix}f(n)\\f(n-1) \end{bmatrix}=\begin{bmatrix}f(1)\\f(0) \end{bmatrix}*(\begin{bmatrix}1 & 1\\ 1 & 0\end{bmatrix})^{n-1}

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long LL;

int fib[40
 
];
struct mat {
    int m[2][2];
};

int flag;
int mod = 10000;


void f() {
    fib[0] = 0;
    fib[1] = 1;
    for(int i = 2; ; i++) {
        fib[i] = fib[i-1] + fib[i-2];
        if(fib[i] >= pow(10, 8)) {
            flag = i - 1;
            break;
        }
    }
}

int prev(LL n){
  return pow(
      
 
10, (log10(1 / sqrt(5.0)) + n * log10((1 + sqrt(5.0)) / 2.0) -
           (int)(log10(1 / sqrt(5.0)) + n * log10((1 + sqrt(5.0)) / 2.0)) + 3));
}

mat multiply(mat a, mat b) {
    mat c;
    for(int i = 0; i < 2; i++) {
        for(int j = 0; j < 2; j++) {
            c.m[i][j] = 0;
            for(int k = 0; k < 2; k++) {
                c.m[i][j] += (a.m[i][k] * b.m[k][j]) % mod;
                c.m[i][j] %= 10000;
            }
        }
    }
    return c;
}

mat p = {1, 1, 1, 0};
mat q = {1, 0, 0, 1};

mat quickPower(LL n) {
    mat m = p, b = q;
    while(n) {
        if(n & 1) {
            b = multiply(b, m);
        }
        n >>= 1;
        m = multiply(m, m);
    }
    return b;
}

int main() {
    LL n;
    f();

    while (~scanf("%lld", &n)) {
        if(n <= flag) printf("%d\n", fib[n]);
        else {
            printf("%d...%04d\n", prev(n), quickPower(n).m[0][1]);
        }
    }
    return 0;
}

HDU 3117 Fibonacci Numbers