2017多校第10场 HDU 6172 Array Challenge 猜公式，矩阵幂

Posted 2020-10-04 Lsxxxxxxxxxxxxx

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6172

题意：如题。

解法：

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
struct Matrix{
    LL a[3][3];
    void set1(){
        memset(a, 0, sizeof(a));
    }
    void set2(){
        memset(a, 0, sizeof(a));
        for(int i=0; i<3; i++) a[i][i]=1;
    }
    void set3()
    {
        a[0][0]=4,a[0][1]=17,a[0][2]=-12;
        a[1][0]=1,a[1][1]=0,a[1][2]=0;
        a[2][0]=0,a[2][1]=1,a[2][2]=0;
    }
};
Matrix operator * (const Matrix &a, const Matrix &b){
    Matrix res;
    res.set1();
    for(int i=0; i<3; i++){
        for(int j=0; j<3; j++){
            for(int k=0; k<3; k++){
                res.a[i][j] = (res.a[i][j]%mod+a.a[i][k]*b.a[k][j]%mod + mod)%mod;
            }
        }
    }
    return res;
}
Matrix qsm(Matrix a, LL n){
    Matrix res;
    res.set2();
    while(n){
        if(n&1) res = res*a;
        a = a*a;
        n/=2LL;
    }
    return res;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        LL n;
        scanf("%lld", &n);
        if(n==2){
            puts("31");
        }
        else if(n==3){
            puts("197");
        }
        else if(n==4){
            puts("1255");
        }
        else{
            Matrix A,B;
            A.set3();
            B = qsm(A, n-4);
            LL ans = (B.a[0][0]*1255%mod+B.a[0][1]*197%mod+B.a[0][2]*31%mod+mod)%mod;
            printf("%lld\\n", ans%mod);
        }
    }
    return 0;
}