【POJ 3070 Fibonacci】+ 矩阵快速幂

最新推荐文章于 2021-10-08 19:54:56 发布

楚江枫

最新推荐文章于 2021-10-08 19:54:56 发布

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分类专栏：矩阵快速幂

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本文介绍了一种计算斐波那契数列第n项最后四位数字的有效方法，使用了矩阵快速幂算法来提高计算效率。

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Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13752 Accepted: 9742

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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

. 这里写图片描述

Given an integer n, your goal is to compute the last 4 digits of Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

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:

. 这里写图片描述

矩阵都给来了～～～直接算～～矩阵还是很好用的～～一个快速计算卡特兰序列的方法～～

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int mod = 10000;
struct node{
   int m[2][2];
}dw,ans;
node jz(node a,node b){ // 矩阵构造
   node sum;
   for(int i = 0 ; i < 2 ; i++)
     for(int j = 0 ; j < 2 ; j++){
       sum.m[i][j] = 0;
       for(int k = 0 ; k < 2 ; k++)
        sum.m[i][j] = (sum.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;
   }
   return sum;
}
int solve(int n){ // 矩阵快速幂
   dw.m[0][0] = dw.m[1][1] = 1;
   dw.m[0][1] = dw.m[1][0] = ans.m[1][1] = 0;
   ans.m[0][0] = ans.m[0][1] = ans.m[1][0] = 1;
   while(n){
    if(n & 1)
        dw = jz(dw,ans);
    ans = jz(ans,ans);
    n >>= 1;
   }
   return dw.m[0][1];
}
int main()
{
    int N;
    while(scanf("%d",&N) != EOF && N != - 1){
        printf("%d\n",solve(N));
    }
    return 0;
}