文章目录

• • C题
  • D题

C题

Hakase and Nano
博弈论

// #pragma GCC optimize(3, "Ofast", "inline")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
const int N = 1e6 + 10;
const int M = 2e6 + 1000;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const int pi = acos(-1);
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
ll T, m, f, tot;
ll a[N];
void init()
{
    memset(a, 0, sizeof(a));
    tot = 0;
}
int main()
{
    // IOS;
    scanf("%lld", &T);
    while (T--)
    {
        init();
        scanf("%lld %lld", &m, &f);
        for (int i = 1; i <= m; i++)
        {
            scanf("%lld", &a[i]);
            if (a[i] == 1)
                tot++;
        }
        if (f == 1)
        {
            if (tot % 3 == 0 && tot == m)
                puts("No");
            else
                puts("Yes");
        }
        else
        {
            if ((tot == m - 1 && m % 3 == 0) || (m % 3 == 1 && tot >= m - 1))
                puts("No");
            else
                puts("Yes");
        }
    }
    return 0;
}

D题

Master of Random
数论
越界问题蚌埠住了，一直RE

// #pragma GCC optimize(3, "Ofast", "inline")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
const int N = 1e6 + 10;
const int M = 2e6 + 1000;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const int pi = acos(-1);
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
ll T, n, res, cnt;
ll a[N], fac[N], inv[N]; //1e8运行错误，内存会爆
ll ksm(ll a, ll b)
{
    ll res = 1;
    for (; b; b >>= 1)
    {
        if (b & 1)
            res = res * a % mod; //快速幂能写错giao
        a = a * a % mod;
    }
    return res % mod;
}
void init()
{
    fac[0] = 1; //
    for (ll i = 1; i < N; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = ksm(i, mod - 2);
    }
}
int main()
{
    // IOS;
    init();
    scanf("%lld", &T);
    while (T--)
    {
        scanf("%lld", &n);
        for (ll i = 0; i < n; i++)
            scanf("%lld", &a[i]);
        cnt = fac[n - 1];
        res = a[0] * fac[n - 1] % mod;
        for (ll i = 1; i < n; i++)
        {
            cnt = (cnt + fac[n - 1] * inv[i] % mod) % mod;
            res = (res + cnt * a[i] % mod) % mod;
        }
        res = res * ksm(fac[n], mod - 2) % mod;
        // res = res *inv[fac[n]] % mod;//inv[fac[n]]会越界
        printf("%lld\n", res % mod);
    }
    return 0;
}