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  • 【poj3070】Fibonacci

    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

    .

    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:

    .

    Source

     
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    void mul(int a[2][2],int b[2][2]){
        int c[2][2];
        memset(c,0,sizeof(c));
        for (int i=0;i<2;i++)
            for (int j=0;j<2;j++)
                for (int k=0;k<2;k++){
                    c[i][j]=(c[i][j]+a[i][k]*b[k][j])%10000;
                }
        memcpy(a,c,sizeof(c));
    }
    int main(){
        int n;
        while (cin>>n && n!=-1){
            int a[2][2]={{0,1},{1,1}};
            int f[2][2]={{0,1},{0,0}}; //为了好算,写成2*2的 
            while (n>0){
                if (n&1) mul(f,a);
                mul(a,a);
                n>>=1;
            }
            cout<<f[0][0]<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/liumengyue/p/5172034.html
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