Codeforces 514E Darth Vader and Tree 矩阵快速幂

Darth Vader and Tree

感觉是个很裸的矩阵快速幂， 搞个100 × 100 的矩阵， 直接转移就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

int n, x, c[];

struct Matrix {
    int a[][];
    Matrix() {
        memset(a, , sizeof(a));
    }
    void init() {
        for(int i = ; i < ; i++)
            a[i][i] = ;
    }
    Matrix operator * (const Matrix &B) const {
        Matrix C;
        for(int i = ; i < ; i++)
            for(int j = ; j < ; j++)
                for(int k = ; k < ; k++)
                    C.a[i][j] = (C.a[i][j] + 1ll * a[i][k] * B.a[k][j]) % mod;
        return C;
    }
    Matrix operator ^ (int b) {
        Matrix C; C.init();
        Matrix A = (*this);
        while(b) {
            if(b & ) C = C * A;
            A = A * A; b >>= ;
        }
        return C;
    }
} M;

int main() {
    scanf("%d%d", &n, &x);
    for(int i = ; i <= n; i++) {
        int v; scanf("%d", &v);
        c[v]++;
    }
    for(int j = ; j < ; j++) M.a[][j] = c[j + ]; M.a[][] = ;
    for(int i = ; i < ; i++) M.a[i][i - ] = ; M.a[][] = ;
    Matrix mat = M ^ (x);
    printf("%d\n", (mat.a[][] + mat.a[][]) % mod);
    return ;
 }

/*
*/