Educational Codeforces Round 175（部分题解）

A. FizzBuzz Remixed

思路：加上0后每15个数的前三个作出贡献，再仔细讨论一下就行了。

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int n, m, k;
    cin >> n;
    int res = (n+1)/15*3;
    res += min((n+1)%15, 3ll);
    cout << res << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

B. Robot Program

思路：最开始我是按照题意模拟的，但肯定会超时的，所以得优化一下，首先我们必须确定能在min(n,k)秒内能否到达过pos = 0,如果能我们就从pos=0遍历一下看几秒回到pos=0，用剩下的k除以它(向下取整)就可以了。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int n, x, k;
    cin >> n >> x >> k;
    string s;
    cin >> s;
    int num = 0;
    int cost = 0;
    for (int i=0, pos = x; i<n; i++){
        k--;
        pos += (s[i]=='L' ? -1:+1);
        if (pos == 0){
            num++;
            break;
        }
        if (!k) break;
    }
    for (int i = 0, pos = 0; i<n; i++){
        pos += (s[i]=='L' ? -1:1);
        if (pos == 0) {
            cost = i+1;
            break;
        }
    }
    if (cost && num){
        num += (k/cost);
    }
    cout << num << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

C. Limited Repainting

思路：这题最开始看错题了，以为把不满足的惩罚都要加上，而只要的是最大的惩罚项。所以这题就很明显是一个二分，其次只要就是写好check函数里的约束，对于其中如果是‘R’的值大于x肯定不动它，而是‘B’的值大于x就必须动他它，最优的话的可以动连续的‘B’。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define endl '\n'

void solve()
{
    int n, ma=0, k;
    cin >> n >> k;
    string s;
    cin >> s;
    vi a(n);
    for (int i=0; i<n; i++){
        cin >> a[i];
        ma = max(ma, a[i]);
    }
    auto check = [&](int x)->bool
    {
        int cost = 0, flag = 0;
        for (int i = 0; i<n; i++){
            if (a[i] <= x) continue;
            else if (s[i] == 'B') cost += 1-flag, flag = 1;
            else flag = 0;
        }
        return cost <= k;
    };
    int l = -1, r = ma+1;
    while (l+1 < r){
        int mid = (l+r)/2;
        if (check(mid)){
            r = mid;
        }
        else l = mid;
    }
    cout << r << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

 D. Tree Jumps

思路：这道题感觉用BFS跑比较简单，这里深度比较重要，对于深度不为1，2的点，它的贡献等于它父节点的深度的一系列点减去它父节点的贡献，所以我们只要以动态的思想去更新结果就可以了。

#include<bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;

constexpr int mod = 998244353, N = 2e5 + 5, M = 2e5 + 5;

void solve()
{
	int n;
    cin >> n;
    vector<vector<int>> adj(n+1);
    vector<int> fa(n+1,0);
    for (int i = 2; i<=n; i++){
        cin >> fa[i];
        adj[fa[i]].push_back(i);
    }
    vector<int> dep(n+1, 0);
    int ans = 0;
    vector<int> sum(n+1, 0), now(n+1, 0);
    queue<int> q;
    q.push(1);
    dep[1] = 1;
    while (!q.empty()){
        int u = q.front();
        q.pop();
        if (u == 1 || fa[u] == 1) now[u] = 1;
        else {
            now[u] = (sum[dep[fa[u]]] - now[fa[u]] + mod) % mod;
        }
        sum[dep[u]] = (sum[dep[u]] + now[u]) % mod;
        ans = (ans + now[u]) % mod;
        for (int v : adj[u]){
            dep[v] = dep[u] + 1;
            q.push(v);
        }
    }
    cout << ans << endl;
}

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

下面是DFS的跑法：

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3f
#define vi vector <int>
#define vvi vector <vi>
#define endl '\n'
#define int long long
using namespace std;

constexpr int mod = 998244353, N = 2e5 + 5, M = 2e5 + 5;

void solve()
{
	int n;
    cin >> n;
    vector<vector<int>> adj(n+1);
    vector<int> fa(n+1,0);
    for (int i = 2; i<=n; i++){
        cin >> fa[i];
        adj[fa[i]].push_back(i);
    }
    vector<int> dep(n+1, 0);
    auto dfs = [&](auto &&dfs, int u, int de)->void
    {
        dep[u] = de;
        for (auto v : adj[u]){
            dfs(dfs, v, de+1);
        }
    };
    dfs(dfs, 1, 0);
    vector<int> sum(n+1, 0), now(n+1, 0);
    vector<int> seq(n+1);
    int ans = 0;
    iota(seq.begin()+1, seq.end(), 1);
    sort(seq.begin()+1, seq.end(), [&](int a, int b){
        return dep[a] < dep[b];
    });
    for (int i = 1; i<=n; i++){
        int x = seq[i];
        if (x == 1 || fa[x] == 1) now[x] = 1;
        else now[x] = (sum[dep[fa[x]]]-now[fa[x]]+mod)%mod;
        sum[dep[x]] = (sum[dep[x]]+now[x]) % mod;
        ans = (ans + now[x])%mod;
    }
    cout << ans << endl;
}

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