ICPC武汉邀请赛(补题)

ICPC武汉邀请赛

比赛链接

B

思路

从高到低位枚举，如果高位的所有1都能放到低位，那么就放到低位，这样一定是优的；反之，如果不行，就把这一位的1填满

代码

#include <algorithm>
#include <iostream>
#include <set>
#include<map>
using namespace std;
#define endl '\n'
using ll = long long;

void solve(){
    int n;
    cin>>n;
    ll sum=0;
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        sum+=x;
    }
    ll ans=0;
    for(int i=31;i>=0;i--){
        ll t=1ll<<i;
        ll w=t-1;
        if(w*n>=sum) continue;
        ans+=t;
        ll p=min((ll)n,sum/t);
        sum-=p*t;
    }
    cout<<ans<<endl;
}

int main() {
    cin.tie(nullptr)->ios::sync_with_stdio(false);
    int _ = 1;
    //cin>>_;
    while (_--) {
        solve();
    }
    return 0;
}

E

思路

不会，赛后看是动态求树的直径（lca），具体可以看佬的博客
传送门

代码

//佬的代码，太优美了直接放这了
#include<bits/stdc++.h>
#define fore(i,l,r)	for(int i=(int)(l);i<(int)(r);++i)
#define fi first
#define se second
#define endl '\n'
#define ull unsigned long long
#define ALL(v) v.begin(), v.end()
#define Debug(x, ed) std::cerr << #x << " = " << x << ed;

const int INF=0x3f3f3f3f;
const long long INFLL=1e18;

typedef long long ll;

const int N = 200050;

std::vector<int> g[N];
int fa[N][20];
int dep[N];

void dfs(int u, int father){
    dep[u] = dep[father] + 1;
    fa[u][0] = father;
    for(int i = 1; (1 << i) <= dep[u]; ++i)
        fa[u][i] = fa[fa[u][i - 1]][i - 1];
    for(auto v : g[u])
        if(v ^ father)
            dfs(v, u);
}

int lca(int u, int v){
    if(dep[u] < dep[v]) std::swap(u, v);
    for(int i = 19; i >= 0; --i)
        if(dep[u] - (1 << i) >= dep[v])
            u = fa[u][i];
    if(u == v) return v;
    for(int i = 19; i >= 0; --i)
        if(fa[u][i] != fa[v][i]){
            u = fa[u][i];
            v = fa[v][i];
        }
    return fa[u][0];
}

int dis(int u, int v){
    int L = lca(u, v);
    return dep[u] + dep[v] - 2 * dep[L];
}

int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    int n;
    std::cin >> n;
    fore(i, 1, n){
        int u, v;
        std::cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    int r, t0;
    std::cin >> r >> t0;

    dfs(1, 0);

    std::vector<int> ans(n + 1, t0 + n);
    int A = r, B = r; //直径的两个端点
    int D = 0; //直径
    std::priority_queue<std::pair<int, int>, std::vector<std::pair<int, int>>, 
                std::greater<std::pair<int, int>>> que;
    fore(i, 1, n + 1){
        if(i == r) continue;
        int d = dis(i, r);
        if(d <= t0){
            int d1 = dis(A, i), d2 = dis(B, i);
            if(D >= std::max(d1, d2)) continue;
            if(d1 > d2){
                D = d1;
                B = i;
            }
            else{
                D = d2;
                A = i;
            }
        }
        else que.push({d, i});
    }

    fore(t, t0 + 1, t0 + n + 1){
        while(!que.empty() && que.top().fi == t){
            auto [d, i] = que.top();
            que.pop();
            int d1 = dis(A, i), d2 = dis(B, i);
            // Debug(i, endl)
            if(D >= std::max(d1, d2)) continue;
            /* 更新直径 */
            if(d1 > d2){
                D = d1;
                B = i;
            }
            else{
                D = d2;
                A = i;
            }
        }
        int k = ((D + 1) / 2 + t - t0 - 1) / (t - t0);
        ans[k] = std::min(ans[k], t);
    }

    fore(i, 1, n + 1){
        ans[i] = std::min(ans[i], ans[i - 1]);
        std::cout << ans[i] << ' ';
    } 

    return 0;
}

F

思路

交互题，二分优化一下就过了

代码

#include <bits/stdc++.h>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <map>
#include <set>
#include <vector>
using namespace std;
#define endl '\n'
using ll = long long;
using pii = pair<int, int>;
using piii = pair<int, pii>;
#define fi first
#define se second

int n, k;

bool query(int a, int b, int x) {
    cout << '?' << " " << a << " " << b << " " << x <<flush;
    int t;
    cin >> t;
    return t;
}

void solve() {
    cin >> n >> k;
    int l = 1, r = n * n;
    k = n * n - k + 1;
    while (l < r) {
        int mid = l + r >> 1;
        int now = n;
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            while (now >= 1 && query(i, now, mid) == 0)
                now--;
            cnt += now;
            if (cnt > k - 1)
                break;
        }
        if (cnt > k - 1)
            r = mid;
        else
            l = mid + 1;
    }
    cout << "!" << " " << l << endl;
}

int main() {
     cin.tie(nullptr);
     ios::sync_with_stdio(false);
    int _ = 1;
    //cin >> _;
    while (_--) {
        solve();
    }
    return 0;
}

K

思路

找规律题，发现每4个为一个周期，前两个都是会胜利的

代码

#include <algorithm>
#include <iostream>
#include <set>
#include<map>
using namespace std;
#define endl '\n'
using ll = long long;

void solve(){
    int n;
    cin>>n;
    if((n%4)<=1) cout<<"Fluttershy"<<endl;
    else cout<<"Pinkie Pie"<<endl;
}

int main() {
    cin.tie(nullptr)->ios::sync_with_stdio(false);
    int _ = 1;
    //cin>>_;
    while (_--) {
        solve();
    }
    return 0;
}