Luogu U15118 萨塔尼亚的期末考试（fail）

感觉...昨天是真的傻...

题意

T个询问，每个询问给一个n，求

$ \frac{\sum_{n}^{i = 1}Fib_{i} * i}{n * (n + 1) / 2} $

Fib是斐波那契数列，对998244353取模

然后...我昨天把n乘了之后就忘记取模了...

30分做法

设$ a_{i} = i * Fib_{i} $, 有$ a_{i} = a_{i - 1} + a_{i - 2} + Fib_{i - 1} + Fib_{i - 2} $，然后就写了一个5*5的转移矩阵，感觉很稳……

长这样：

\begin{bmatrix}
Fib_{i} & Fib{i - 1} & a_{i} & a_{i - 1} & sum_{i - 1}
\end{bmatrix}

转移矩阵长这样：

\begin{bmatrix}
1 & 1 & 1 & 0 & 0 \\
1 & 0 & 2 & 0 & 0\\
0 & 0 & 1 & 1 & 1\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}

虽然这个东西也过了几万组对拍，但是我喜闻乐见地把std也顺便写挂了（捂脸），所以这两个东西都只有10分...

Code：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const ll mod = ;

int testCase;
ll n;

struct Matrix {
    ll l1, l2, s[][];

    inline void init() {
        l1 = l2 = ;
        memset(s, , sizeof(s));
    }

    friend Matrix operator * (const Matrix a, const Matrix b) {
        Matrix res;
        res.init();
        res.l1 = a.l1, res.l2 = b.l2;
        for(int i = ; i < a.l1; i++)
            for(int j = ; j < b.l2; j++)
                for(int k = ; k < a.l2; k++)
                    res.s[i][j] = (res.s[i][j] + a.s[i][k] * b.s[k][j] % mod) % mod;
        return res;
    }

    inline Matrix pow(Matrix a, ll b) {
        Matrix res = *this;
        for(; b > ; b >>= ) {
            if(b & ) res = res * a;
            a = a * a;
        }
        return res;
    }

    inline void print() {
        for(int i = ; i < l1; i++, printf("\n"))
            for(int j = ; j < l2; j++)
                printf("%lld ", s[i][j]);
    }

} f;

inline ll pow(ll x, ll y) {
    ll res = ;
    for(x %= mod; y > ; y >>= ) {
        if(y & ) res = res * x % mod;
        x = x * x % mod;
    }
    return res;
}

inline ll solve() {
    if(n == ) return 1LL;
    if(n == ) return 3LL;

    Matrix g;
    g.init();
    g.l1 = , g.l2 = ;
    g.s[][] = g.s[][] = g.s[][] = ;
    g.s[][] = ;
    g.s[][] = ;
//    g.print();

    g = g.pow(f, n - );

//    g.print();

    return g.s[][];
}

int main() {
    f.init();
    f.l1 = f.l2 = ;
    f.s[][] = f.s[][] = f.s[][] = ;
    f.s[][] = f.s[][] = f.s[][] = f.s[][] = ;
    f.s[][] = f.s[][] = ;//f.s[4][2] = 1;
    f.s[][] = ;

//    f.print();

    for(scanf("%d", &testCase); testCase--; ) {
        scanf("%lld", &n);
        ll tmp = n * (n + ) / ;
        ll inv = pow(tmp, mod - );
        ll sum = solve();
        printf("%lld\n", inv * sum % mod);
    }
    return ;
}

100分做法  

找规律题惹不起a...

$sum_{n} = n * Fib_{n + 2} - Fib_{n + 3} + 2$

所以就变成一个斐波那契了
并不能推导出来...

Code:

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const ll mod = ;

int testCase;
ll n;

struct Matrix {
    ll s[][];

    inline void init() {
        memset(s, , sizeof(s));
    }

    friend Matrix operator * (const Matrix a, const Matrix b) {
        Matrix res;
        res.init();
        for(int i = ; i <= ; i++)
            for(int j = ; j <= ; j++)
                for(int k = ; k <= ; k++)
                    res.s[i][j] = (res.s[i][j] + a.s[i][k] * b.s[k][j] % mod) % mod;
        return res;
    }

    inline Matrix pow(ll b) {
        Matrix res, a = *this;
        res.init();
        for(int i = ; i <= ; i++) res.s[i][i] = ;
        for(; b > ; b >>= ) {
            if(b & ) res = res * a;
            a = a * a;
        }
        return res;
    }

} f;

inline ll pow(ll a, ll b) {
    ll res = ;
    for(; b > ; b >>= ) {
        if(b & ) res = res * a % mod;
        a = a * a % mod;
    }
    return res;
}

inline ll solve() {
    Matrix res = f.pow(n + );
    return (res.s[][] * n % mod - res.s[][] +  + mod) % mod;
}

int main() {
    f.init();
    f.s[][] = f.s[][] = f.s[][] = ;
    for(scanf("%d", &testCase); testCase--; ) {
        scanf("%lld", &n);
        ll inv = pow((n * (n + ) / )% mod, mod - );
        ll sum = solve();
//        printf("%lld %lld\n", sum, inv);
        printf("%lld\n", sum * inv % mod);
    }
    return ;
}

感觉还挺有趣的……