题解 | 串

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 10;
int f[MAXN] = {0,26};

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int n; cin >> n;
    for(int i = 2; i <= n; i++) f[i] = (f[i-1] * 26) % MOD;
    //小学生应用题,当长度为1时:
    int x = 25,y = 1;//先设x为不含u的串，然后y为含u的串
    //设u为未知是因为要符合子序列us

    int ans = 0;
    for(int i = 2; i <= n; i++)
    {
        int new_x= (x * 25) % MOD; //不含u的串为不合法的
        int new_y = (y * 25 + x) % MOD; //含u但不含s
        int legal = (f[i] - (new_x + new_y + MOD)%MOD  + MOD) % MOD;//出去不合法的就是合法的了
        ans = (ans + legal) % MOD;
        //继承状态为下一层做准备
        x = new_x;
        y = new_y;
    }
    cout << ans << endl;

    return 0;
}