hdu 1021 Fibonacci Again(矩阵连乘 || 循环节)

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题目：http://acm.hdu.edu.cn/showproblem.php?pid=1021

Description

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

Sample Input

Sample Output

nonoyesnonono

矩阵连乘：

#include <iostream>#include <cstdio>using namespace std;struct matrie{    int m[2][2];};matrie A={  // f[n],f[n-1] ~ f[1],f[0] ~ pow=n-1    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])%3;            }            c.m[i][j]%=3;        }    }    return c;}matrie power(int p){    matrie ans=I,tmp=A;  // 矩阵结构体可以直接相等    while(p){       if(p&1) ans=multi(ans,tmp);       tmp=multi(tmp,tmp);       p>>=1;    }    return ans;}int main(){    //freopen("cin.txt","r",stdin);    int n;    while(cin>>n){        if(n==0){             puts("no");             continue;        }        matrie w=power(n-1);        int ans=(w.m[0][0]*7+w.m[0][1]*11)%3;        if(ans==0)puts("yes");        else puts("no");    }    return 0;}

循环节：

#include <iostream>#include <cstdio>using namespace std;typedef long long LL;LL f[100];int main(){    /*f[0]=7;    f[1]=11;    for(int i=2;i<40;i++){        f[i]=(f[i-1]+f[i-2])%3;        cout<<f[i]<<" ";    }*/    int n;    while(cin>>n){         if((n-2)%4==0)printf("yes\n");         else puts("no");    }    return 0;}

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