Fibonacci(矩阵快速幂模板)

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:

.

#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <algorithm>#define maxn 1000#define kuma 10#define mod 10000using namespace std;struct matrix{    int a[kuma][kuma];    void init()    {        memset(a,0,sizeof(a));        for(int i=0;i<kuma;i++)        {            a[i][i]=1;        }    }};matrix mul(matrix x,matrix y){    matrix ans;    for(int i=0;i<kuma;i++)    {        for(int j=0;j<kuma;j++)        {            ans.a[i][j]=0;            for(int k=0;k<kuma;k++)            {                ans.a[i][j]=(ans.a[i][j]+((x.a[i][k]%mod)*(y.a[k][j]%mod))%mod)%mod;            }        }    }    return ans;}matrix add(matrix x,matrix y){    matrix ans;    for(int i=0;i<kuma;i++)    {        for(int j=0;j<kuma;j++)        {            ans.a[i][j]=(x.a[i][j]%mod+y.a[i][j]%mod)%mod;        }    }    return ans;}matrix quickpow(matrix a,int p){    matrix ans;    ans.init();    matrix t;    t=a;    while(p)    {        if(p&1)ans=mul(ans,t);        p>>=1;        t=mul(t,t);    }    return ans;}int main(){    matrix nico;    nico.init();    nico.a[0][0]=1;    nico.a[1][0]=1;    nico.a[0][1]=1;    nico.a[0][0]=0;    int n;    while(scanf("%d",&n)==1)    {        if(n==-1)break;        if(n==0){printf("0\n");continue;}        if(n==1||n==2){printf("1\n");continue;}        matrix ans=quickpow(nico,n-2);        printf("%d\n",(ans.a[1][0]%mod+ans.a[1][1]%mod)%mod);    }    return 0;}