1004.Distinct Sub-palindromes

签到

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi acos(-1)
const int maxn=1e6+10;
const int mod=998244353;
int f[maxn],n,t;
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        if(n==1)
            cout<<26<<endl;
        else if(n==2)
            cout<<676<<endl;
        else if(n==3)
        {
            cout<<26*26*26<<endl;
        }
        else
        {
            cout<<26*25*24<<endl;
        }
    }
    return 0;
}

1005. Fibonacci Sum

由于赛后评测机，额。。。反正找了几个赛时能过的代码，还是过不了，最后用标程过的。。。

#include <bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 9;
const int C = 276601605; // 1 / sqrt(5)
const int A = 691504013; // (1 + sqrt(5)) / 2
const int AA = 691504012; // A ^ -1
const int B = 308495997; // (1 - sqrt(5)) / 2

inline int modExp(int a, long long n) {
    int ret = 1;
    for (; n; n >>= 1, a = (long long)a * a % mod) if (n & 1) ret = (long long)ret * a % mod;
    return ret;
}

inline void add(int &u, int v) {
    u += v;
    if (u >= mod) u -= mod;
}

const int N = 110000;

int f[N], g[N], r[N];

void init() {
    f[0] = g[0] = f[1] = g[1] = r[1] = 1;
    for (int i = 2; i < N; i++) {
        f[i] = (long long)f[i - 1] * i % mod;
        r[i] = (long long)(mod -  mod / i) * r[mod % i] % mod;
        g[i] = (long long)g[i - 1] * r[i] % mod;
    }
}

inline int nCm(int n, int m) {
    if (m < 0 || m > n) return 0;
    return (long long)f[n] * g[m] % mod * g[n - m] % mod;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    init();
    int tcase; cin >> tcase;
    while (tcase--) {
        long long n, c; int k; cin >> n >> c >> k;
        int DD = (long long)modExp((long long)AA * B % mod, c) % mod;
        int q = modExp(modExp(A, c), k);
        int n1 = (n + 1) % mod;
        int n2 = (n + 1) % (mod - 1);
        int Q = modExp(q, n2);
        int D = modExp(DD, n2);
        int ans = 0;
        for (int i = 0; i <= k; i++) {
            int cur = nCm(k, i);
            if (i & 1) cur = mod - cur;
            if (q == 1) add(ans, (long long)cur * n1 % mod);
            else add(ans, (long long)cur * (Q + mod - 1) % mod * modExp(q-1, mod-2) % mod);
            q = (long long)q * DD % mod;
            Q = (long long)Q * D % mod;
        }
        cout << (long long)ans * modExp(C, k) % mod << "\n";
    }
    return 0;
}