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
");
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
", (int)(1000.0 * pow(10.0, temp)), ans.s[0][0] % MOD);
}
else
printf("%I64d
", ans.s[0][0]);
}
}
return 0;
}