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.
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:
.
输入 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. 输出 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). 样例输入 0 9 1000000000 -1 样例输出 0 34 6875矩阵a=.所以只要求得a[0][1]或a[1][0]即可。
对于矩阵乘法与递推式之间的关系：
如：在斐波那契数列之中
f[i] = 1*f[i-1]+1*f[i-2]
f[i-1] = 1*f[i-1] + 0*f[i-2];
即
所以
一个理解博客
#include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #include<iostream> #include<vector> #include<map> using namespace std; typedef long long ll; const int mod=10000; struct mat { ll a[2][2]; }; mat matrix(mat x,mat y) { mat res; memset(res.a,0,sizeof(res.a)); for(int i=0;i<2;i++) { for(int j=0;j<2;j++) { for(int k=0;k<2;k++) { res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod; } } } return res; } void pow(int n) { mat res,c; memset(res.a,0,sizeof(res.a)); for(int i=0;i<2;i++) res.a[i][i]=1; c.a[0][0]=c.a[0][1]=c.a[1][0]=1; c.a[1][1]=0; while(n) { if(n&1) res=matrix(res,c); c=matrix(c,c); n=n>>1; } printf("%I64d\n",res.a[1][0]); } int main() { int n; while(~scanf("%d",&n)&&n!=-1) { pow(n); } }