poj 3070 Fibonacci(简单矩阵连乘)

poj 3070 Fibonacci(简单矩阵连乘)

题目：​​class="data-table" data-id="t7a7e9d1-IYGbQabU" data-transient-attributes="class" data-width="1162px" style="width: 100%; outline: none; border-collapse: collapse;">

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 10994

Accepted: 7823

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

09 999999999 1000000000 -1

Sample Output

034 626 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:

题意如上，没有什么陷阱，简单的矩阵连乘取模。

#include #include using namespace std;const int mod=1e4;struct matrie{ int m[2][2];};matrie A={ 1,1, 1,0};matrie I={ 1,0, 0,1};matrie multi(matrie a,matrie b){ matrie c; for(int i=0;i<2;i++){ for(int j=0;j<2;j++){ c.m[i][j]=0; for(int k=0;k<2;k++){ c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod; } c.m[i][j]%=mod; } } return c;}matrie power(int n){ matrie ans=I,tmp=A; while(n){ if(n&1)ans=multi(ans,tmp); tmp=multi(tmp,tmp); n>>=1; } return ans;}int main(){ int n; while(cin>>n&&n!=-1){ matrie q=power(n); printf("%d\n",q.m[1][0]); } return 0;}

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