Happy Necklace

Problem Description

Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads.
Little Q desperately wants to impress his girlfriend, he knows that she will like the necklace only if for every prime length continuous subsequence in the necklace, the number of red beads is not less than the number of blue beads.
Now Little Q wants to buy a necklace with exactly

Input

The first line of the input contains an integer

Output

For each test case, print a single line containing a single integer, denoting the answer modulo

Sample Input

Sample Output

题意：每个素数区间的红色个数大于蓝色（0,1 都不是素数）

　　　　F（n）=F（n-1）+F（n-3）；

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=3;  //矩阵大小 
const int mod=1e9+7;
struct Matrix{
    ll m[maxn][maxn];  
    Matrix(){
        memset(m,0,sizeof(m));
    }
};
Matrix Multi(Matrix a,Matrix b){
    Matrix res;
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            for(int k=0;k<maxn;k++){
                res.m[i][j]=(res.m[i][j]+ (a.m[i][k]%mod) *( b.m[k][j]%mod)%mod ) %mod;
            } 
        }
    }
    return res;
} 
Matrix fastm(Matrix a,ll num){
    Matrix res;
    res.m[0][0]=6;
    res.m[1][0]=4;
    res.m[2][0]=3;
    while(num){
        if(num&1){
            res=Multi(a,res);
        }
        a=Multi(a,a);
        num>>=1;
    }
    return res;
}
int ans[3]={3,4,6};
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        ll n;
        scanf("%lld",&n);
        if(n<=4){
            printf("%d\n",ans[n-2]);
        }else{
            Matrix a;
            a.m[0][0]=1;
            a.m[0][1]=0;
            a.m[0][2]=1;
            a.m[1][0]=1;
            a.m[1][1]=0;
            a.m[1][2]=0;
            a.m[2][0]=0;
            a.m[2][1]=1;
            a.m[2][2]=0;
            a=fastm(a,n-4);
            printf("%lld\n",a.m[0][0]);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/lyj1/p/11945505.html

hdu6030 矩阵快速幂

Happy Necklace