POJ 3070 Fibonacci-优快云博客

本文链接：https://blog.youkuaiyun.com/lizhaowei213/article/details/47860997

本文介绍了一种解决C-Fibonacci问题的有效方法——利用矩阵快速幂算法。该算法可以高效地计算出Fibonacci数列中指定项的最后四位数字，适用于处理大规模的数据输入。文章提供了一个C语言实现的示例代码。

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C - Fibonacci

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n = F_n_{− 1} + F_n_{− 2} forn ≥ 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 ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_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:

矩阵快速幂模板题。一次过。

#include <stdio.h>
#include <string.h>
#define N 2
#define MOD 10000
int n=2,s;
typedef struct Matrix{
    long long mat[N][N];
    Matrix operator*(const Matrix m)const{
        Matrix tmp;
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                tmp.mat[i][j] = 0;
                for(int k = 0;k < n;k++){
                    tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%MOD;
                    tmp.mat[i][j] %= MOD;
                }
            }
        }
        return tmp;
    }
}Matrix;
long long Pow(Matrix &m,int k){
    Matrix ans;
    ans.mat[0][0]=ans.mat[0][1]=ans.mat[1][0]=1;
                ans.mat[1][1]=0;
    while(k){
        if(k&1) ans = ans*m;
        k >>= 1;
        m = m*m;
    }
    return ans.mat[0][1];
}
Matrix a;
int main()
{
        while(scanf("%d",&s)>0&&s!=-1)
        {
                if (!s)
                {
                       printf("0\n");
                        continue;
                }
                a.mat[0][0]=a.mat[0][1]=a.mat[1][0]=1;
                a.mat[1][1]=0;
                printf("%lld\n",Pow(a,s-1));
        }
        return 0;
}