A. Red or Not

模拟

#include <bits/stdc++.h>
using namespace std;



int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int a;
    string s;
    cin >> a >> s;
    if(a >= 3200) cout << s << endl;
    else cout << "red" << endl;
    return 0;
}

B. Resistors in Parallel

模拟

#include <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    double res = 0;
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        double x;
        cin >> x;
        res += 1 / x;
    }
    printf("%.6f\n", 1 / res);
    return 0;
}

C. Alchemist

模拟

#include <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n; cin >> n;
    vector<int> a(n); for(int i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(), a.end());
    double ans = a[0];
    for(int i = 0; i < n - 1; i++)
        ans = (ans + a[i + 1]) / 2;
    printf("%.9f", ans);
    return 0;
}

D. Ki

一开始打算直接复制树剖板（肯定能过$O((n + q)logn)$）
然后想想不用修改，直接dfs就行

#include <bits/stdc++.h>
using namespace std;

const int N = 2e5 + 10, M = N << 2;
#define int long long
int h[N], e[M], ne[M], idx;
int n, m;
int  w[N];
void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void dfs(int u, int f)
{
    for(int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(j == f) continue;
        w[j] += w[u];
        dfs(j, u);
    }
}

signed main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n >> m;
    memset(h, -1, sizeof h);
    for(int i = 1; i < n; i++)    
    {
        int x, y;
        cin >> x >> y;
        add(x, y);
        add(y, x);
    }
    for(int i = 0; i < m; i++)
    {
        int x, y;
        cin >> x >> y;
        w[x] += y;
    }
    dfs(1, -1);
    for(int i = 1; i <= n; i++) cout << w[i] << " \n"[i == n];
    return 0;
}