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;
}
E,F后面寄了,今天大失败