Fibonacci_牛客博客

题解：矩阵快速幂
/*
*@Author:   STZG
*@Language: C++
*/
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int b[2][2];
int a[2][2];
char str;
struct node{};
void Matrix(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;
            }
        }
    }
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            a[i][j]=c[i][j];
        }
    }
}
int power(int k){
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){
            a[i][j]=1;
            b[i][j]=1;
        }
    }
    a[0][0]=b[0][0]=0;
    while(k){
        if(k&1)Matrix(a,b);
        Matrix(b,b);
        k>>=1;
    }
    return a[0][0];
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //
    while(~scanf("%d",&n)&&n!=-1){
        cout<<power(n)<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}