2020 Multi-University Training Contest 9（待补

2020 Multi-University Training Contest 9

1001 Tree

• 思路：把叶子节点和根节点连接起来形成一个环即可满足题意 先用dfs求出每个节点子树大小 再用dfs求出到叶子节点路径取个最大值$

• 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 N = 5e5 + 10;

int t, n, tot;
int edge[N], nxt[N], head[N], sz[N];
ll ans;

inline void add(int u, int v){
    edge[ ++ tot ] = v;
    nxt[tot] = head[u];
    head[u] = tot;
}

inline int dfs1(int u){
    sz[u] = 1;
    for (int i = head[u]; i; i = nxt[i])
        sz[u] += dfs1(edge[i]);
    return sz[u];
}

inline void dfs2(int u, ll res){
    ans = max(ans, res);
    for (int i = head[u]; i; i = nxt[i])
        dfs2(edge[i], res + n - sz[edge[i]]);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
    freopen("my_out.txt", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        ans = 0, tot = 0;
        cin >> n;
        for (int i = 0; i <= n; i ++ )
            head[i] = 0;
        for (int i = 2; i <= n; i ++ ){
            int u;
            cin >> u;
            add(u, i);
        }
        dfs1(1);
        dfs2(1, 0);
        for (int i = 1; i <= n; i ++ )
            ans += sz[i];
        cout << ans << "
";
    }
    return 0;
}

1003 Slime and Stones

• 思路：

[frac{1}{a} + frac{1}{a+k+1}=1 \ a^2+(k-1)a-(k+1)=0 ]

得到必败态

[a_n=frac{sqrt{k^2+2k+5}+1-k}{2}*n quad b_n=a_n+(k+1)*n \ Longrightarrow quad frac{sqrt{k^2+2k+5}+1-k}{2} * frac{b-a}{k+1}=a ]

• 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 double tmp = (sqrt(5.0) + 1) / 2.0;

int t, a, b, k;

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 >> a >> b >> k;
        if (a > b)
            swap(a, b);
        if (b - a - k <= 0){
            cout << "1
";
            continue;
        }
        // cout << (b - a - k) * tmp << " " << a << "
";
        // if (int(double(b - a - k) * tmp) == a)
        //     cout << "0
";
        // else
        //     cout << "1
";
        if ((b - a) % (k + 1) == 0){
            double tmp = (sqrt(1.0 * k * k + 2.0 * k + 5.0) + 1.0 - 1.0 * k) / 2.0;
            // cout << (b - a) / (k + 1) << " " << tmp << " " << a << "
";
            if (int(tmp * ((b - a) / (k + 1))) == a)
                cout << "0
";
            else
                cout << "1
";
        }
        else
            cout << "1
";
    }
    return 0;
}