E: number number number

E: number number number


比赛时我是直接找规律的
ans[i] = 3*ans[i-1]-ans[i-2]+1
然后矩阵乘法
后来听说是直接找第2n+3个斐波那契数

#include <bits/stdc++.h>using namespace std;#define ll long long/*const int maxn = 100000;int f[21], a[100001], n;*/const int mo = 998244353;struct Matrix {    ll a[3][3];    Matrix() {memset(a, 0, sizeof(a));}    Matrix operator*(const Matrix &other) const {        Matrix t;        for (int i = 0; i < 3; i++)            for (int j = 0; j < 3; j++)                for (int k = 0; k < 3; k++)                    t.a[i][k] = (t.a[i][k]+a[i][j]*other.a[j][k])%mo;        return t;    }};int n;int main() {    Matrix t;    t.a[1][0] = t.a[2][1] = t.a[2][2] = 1;    t.a[1][1] = 3;    t.a[0][1] = -1;    while (scanf("%d", &n) != EOF) {        n--;        Matrix x = t, y;        y.a[0][0] = y.a[1][1] = y.a[2][2] = 1;        while (n) {            if (n%2) {                y = y*x;            }            n = n/2;            x = x*x;        }        ll ans = (y.a[0][0]*4+y.a[1][0]*12+y.a[2][0])%mo;        ans = (ans+mo)%mo;        printf("%lld\n", ans);    }    /*f[0] = 0; f[1] = 1;    for (int i = 2; i <= 20; i++) f[i] = f[i-1]+f[i-2];    scanf("%d", &n);    a[0] = 1;    while (n--) {        for (int i = maxn; i >= 0; i--) if (a[i]) {            for (int j = 1; j <= 20; j++) {                a[i+f[j]] = 1;                //if (i+f[j] == 24) printf("%d %d\n", i, f[j]);            }        }    }    for (int i = 0; i <= maxn; i++) if (!a[i]) {        printf("%d\n", i); break;    }*/}