poj - problem 3070 Fibonacci 【矩阵 +快速幂】

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13732		Accepted: 9728

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

Sample Output

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

这道题就是根据题意算快速幂。裸，但是这个规律比较重要了。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 100 + 10;
int n, mod;
LL m;
struct Matrix {LL m[3][3];int row, col;
};
Matrix ori, res;
void init() {memset(res.m, 0, sizeof(res.m));ori.m[1][1] = ori.m[1][2] = ori.m[2][1] = 1;
}
Matrix multi(Matrix x, Matrix y) {Matrix z;memset(z.m, 0, sizeof(z.m));for (int i = 1; i <= 2; i++) {for (int k = 1; k <= 2; k++) {for (int j = 1; j <= 2; j++) {z.m[i][j] += x.m[i][k]*y.m[k][j]%mod;}z.m[i][k] %= mod;}}return z;
}
Matrix pow_mod(Matrix a, LL x){Matrix b;memset(b.m, 0, sizeof(b.m));for (int i = 1; i <= 2; i++) {b.m[i][i] = 1;}while(x){if(x&1) b = multi(a,b);a = multi(a, a);x >>= 1;}return b;
}
int main() {while (scanf("%lld", &m), m != -1) {init(); mod = 10000;res = pow_mod(ori, m);printf("%lld\n", res.m[1][2]%mod);}return 0;
}

转载于:https://www.cnblogs.com/cniwoq/p/6770759.html

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