HDU6030 Happy Necklace（推导+矩阵快速幂）

HDU6030 Happy Necklace

推导或者可以找规律有公式：\(f[n] = f[n-1] + f[n-3]\) 。

构造矩阵乘法：
\[
\begin{pmatrix} f_i \\ f_{i-1} \\ f_{i-2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} f_{i-1} \\ f_{i-2} \\ f_{i-3} \end{pmatrix}
\]

#include<bits/stdc++.h>

using namespace std;

const int mod = 1e9 + 7;
int t;
long long n;
struct Matrix{
    long long a[5][5];
};

Matrix mul(Matrix M1, Matrix M2)
{
    Matrix ret;
    memset(ret.a, 0, sizeof(ret.a));
    for(int i = 0; i < 3; i++){
        for(int j = 0; j < 3; j++){
            for(int k = 0; k < 3; k++){
                ret.a[i][j] = (M1.a[i][k] * M2.a[k][j] + ret.a[i][j]) % mod;
            }
        }
    }
    return ret;
}
void matrix_pow(long long x)
{
    Matrix ret;
    memset(ret.a, 0, sizeof(ret.a));
    for(int i = 0; i < 3; i++) ret.a[i][i] = 1;
    Matrix tmp;
    memset(tmp.a, 0, sizeof(tmp.a));
    tmp.a[0][0] = tmp.a[0][2] = tmp.a[1][0] = tmp.a[2][1] = 1;
    while(x){
        if(x & 1LL) ret = mul(ret, tmp);
        tmp = mul(tmp, tmp);
        x >>= 1LL;
    }
    long long ans = (ret.a[0][0] * 4 + ret.a[0][2] * 2 + ret.a[0][1] * 3) % mod;
    cout << ans << endl;
}
int main()
{
    for(scanf("%d", &t); t--; ){
        scanf("%lld", &n);
        if(n == 1) {puts("2"); continue;}
        else if(n == 2) {puts("3"); continue;}
        else if(n == 3) {puts("4"); continue;}
        else{
            matrix_pow(n - 3);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/solvit/p/11445300.html