2020 Multi-University Training Contest 5（待补

2020 Multi-University Training Contest 5

1001 Tetrahedron

• 思路：可以根据类比平面几何的直角三角形 得到三个直角面面积和等于底面面积 然后得到式子 由于是离散型随机变量的期望 求和即可

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int mod = 998244353;
const int N = 6e6 + 10;

int t, n;
int inv[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    inv[1] = 1;
    for (int i = 2; i < N; i ++ )
        inv[i] = (1ll * (mod -(mod / i)) * inv[mod % i] % mod + mod) % mod;
    for (int i = 2; i < N; i ++ )
        inv[i] = 1ll * inv[i] * inv[i] % mod;
    for (int i = 2; i < N; i ++ )
        inv[i] = (1ll * inv[i] + 1ll * inv[i - 1]) % mod;
    cin >> t;
    while (t -- ){
        cin >> n;
        cout << mult_mod(3, mult_mod(pow_mod(n, mod - 2, mod), inv[n], mod), mod) << "
";
    }
    return 0;
}

1003 Boring Game

• 思路：模拟

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll Pow(ll a, ll b){
    ll res = 1;
    while (b){
        if (b & 1)
            res *= a;
        a *= a;
        b >>= 1;
    }
    return res;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e6 + 10;

int t, n, k, m, pos, tmp;
vector<int> vec[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        pos = 1;
        cin >> n >> k;
        m = 2 * n * Pow(2, k);
        for (int i = 1; i <= m; i ++ ){
            int x;
            cin >> x;
            vec[i].clear();
            vec[i].push_back(x);
        }
        while (k -- ){
            tmp = (pos + m) >> 1;
            for (int i = pos; i <= tmp; i ++ ){
                for (int j = vec[i].size() - 1; j >= 0; j -- )
                    vec[pos + m - i].push_back(vec[i][j]);
            }
            pos = tmp + 1;
        }
        for (int i = m - 2 * n + 1; i <= m; i ++ ){
            for (int j = vec[i].size() - 1; j >= 0; j -- ){
                if (i == m && j == 0)
                    cout << vec[i][j] << "
";
                else
                    cout << vec[i][j] << " ";
            }
        }
    }
    return 0;
}

1009 Paperfolding

• 思路：(frac{3^n}{2^{n-1}}+2^n+1)

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int mod = 998244353;

int t;
ll n;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        cin >> n;
        cout << ((1ll * pow_mod(2, n, mod) + mult_mod(1ll * 2, mult_mod(pow_mod(3, n, mod), pow_mod(pow_mod(2, n, mod), mod - 2, mod), mod), mod)) % mod + 1) % mod << "
";
    }
    return 0;
}

1012 Set1

• 思路：前(lfloor frac{n}{2} floor)都为0 (ans_frac{n+1}{2}=inv(2^{frac{n}{2}})) 之后到第(n-1)个数都有(ans_i = ans_{i-1}*(frac{n}{2}+i)*inv(2*i)) (ans_n=ans_{n-1})

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int mod = 998244353;

int t, n;
ll ans;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        cin >> n;
        if (n == 1){
            cout << "1
";
            continue;
        }
        for (int i = 1; i <= n / 2; i ++ )
            cout << "0 ";
        ans = pow_mod(pow_mod(2, n / 2, mod), mod - 2, mod);
        cout << ans << " ";
        for (int i = 1; i < n / 2; i ++ ){
            ans = mult_mod(ans, mult_mod(n / 2 + i, pow_mod(2 * i, mod - 2, mod), mod), mod);
            cout << ans << " ";
        }
        cout << ans << "
";
    }
    return 0;
}