giúp với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)
b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)
c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)
d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)
1 correct
2 success => only success
3 was released => released
4 correct
5 focused on => on
6 as a => a
7 correct
8 each others => others
9 satisfy with => satisfy
10 correct
11 correct
1 were - would you play
2 weren't studying - would have
3 had taken - wouldn't have got
4 would you go - could
5 will you give - is
6 recycle - won't be
7 had heard - wouldn't have gone
8 would you buy - had
9 don't hurry - will miss
10 had phoned - would have given
11 were - wouldn't eat
12 will go - rains
13 had known - would have sent
14 won't feel - swims
15 hadn't freezed - would have gone
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
vector<int> edge[maxn];
int f[maxn];
bool visited[maxn];
void dfs(int u) {
visited[u] = true;
f[u] = 1;
for (int v : edge[u]) {
if (!visited[v]) {
dfs(v);
f[u] += f[v];
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
for (int i = 1; i <= m; i++) {
int u, v;
cin >> u >> v;
edge[u].push_back(v);
edge[v].push_back(u);
}
vector<int> components;
for (int i = 1; i <= n; i++) {
if (!visited[i]) {
dfs(i);
components.push_back(f[i]);
}
}
int res = components.front();
int rest = 0;
for (int i = 1; i < components.size(); i++) {
rest = max(rest, components[i]);
}
cout << res + rest << endl;
}