Codeforces Round 479 (Div. 3) 补题 AK
A 模拟
思路
直接模拟就行
代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;
int n, k;
void solve() {
cin >> n >> k;
while (k--) {
if (n % 10 == 0) n /= 10;
else n--;
}
cout << n;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//cin >> T;
T = 1;
while (T--) solve();
return 0;
}
B 哈希 模拟
思路
直接用哈希算每个出现的次数，再统计答案即可代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;
int n;
string s, ans;
umap<string, int> mp;
int cnt;
void solve() {
cin >> n >> s;
s = ' ' + s;
forn(i, 1, n - 1) mp[s.substr(i, 2)]++;
for (auto [i, j] : mp) {
if (j > cnt) {
cnt = j;
ans = i;
}
}
cout << ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//cin >> T;
T = 1;
while (T--) solve();
return 0;
}
C 思维 排序
思路
排序输出第k个就行了，注意特判第k个和k+1是不是相同的然后还有特判k=0的情况
代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;
int n, k;
int a[N];
void solve() {
cin >> n >> k;
forn(i, 1, n) cin >> a[i];
a[0] = 1;
sort(a + 1, a + n + 1);
if (a[k] == a[k + 1]) a[k] = -1;
if (k == 0 && a[1] == 1) a[k] = -1;
cout << a[k];
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//cin >> T;
T = 1;
while (T--) solve();
return 0;
}
D 拓扑排序
思路
代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
const int N = 2e5 + 5;
ll T;
int h[N], idx;
struct node {
int to, ne;
} e[N];
ll n;
ll a[N];
ll in[N];
vector<ll> ans;
void add(int u, int v) {
e[++idx].to = v;
e[idx].ne = h[u];
h[u] = idx;
in[v]++;
}
void topsort() {
queue<ll> q;
forn(i, 1, n) if (in[i] == 0) q.push(i);
while (!q.empty()) {
ll u = q.front();
q.pop();
ans.push_back(u);
for (int i = h[u]; i; i = e[i].ne) {
ll v = e[i].to;
if (--in[v] == 0) q.push(v);
}
}
}
void solve() {
cin >> n;
forn(i, 1, n) cin >> a[i];
forn(i, 1, n) {
forn(j, 1, n) {
if (a[i] * 2 == a[j] || a[j] * 3 == a[i]) add(i, j);
}
}
topsort();
for (auto i : ans) cout << a[i] << ' ';
}
int main() {
//cin >> T;
T = 1;
while (T--) {
solve();
}
return 0;
}
E 思维 搜索
思路
一个圈内所有点的度数均为 2，且大小要大于等于3直接搜索，顺便记录连通块大小和是否入度全为2，等到每个连通块搜索完再判断是否是
代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;
vector<int> g[N];
int ok;
int siz;
bool is[N];
int in[N];
void dfs(int u) {
if (is[u]) return;
is[u] = 1, siz++;
if (in[u] != 2) ok = 0;
for (auto v : g[u]) dfs(v);
}
int n, m;
int ans;
void solve() {
cin >> n >> m;
forn(i, 1, m) {
int u, v;
cin >> u >> v;
in[u]++, in[v]++;
g[u].push_back(v);
g[v].push_back(u);
}
int cnt = 0;
forn(i, 1, n) {
if (is[i]) continue;
cnt++;
ok = 1;
siz = 0;
dfs(i);
if (ok && siz >= 3) ans++;
}
cout << ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//cin >> T;
T = 1;
while (T--) solve();
return 0;
}
F 线性DP
思路
注意本题卡uamp！！！题目有两问，定义一个结构体，表示以i结尾的最大序列长度为v，上一个数的标号是id
这样就能输出答案了，转移方程：x表示a[i]-1的上一个位置，f[i].id = x，f[i].v = f[x].v + 1
注意在结尾更新答案，以及位置，最后再找到最大长度的位置，不断往前找编号，逆序输出即可
代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int n;
int a[N];
map<int, int> mp;
int ans;
struct node {
int v, id;
} f[N];
vector<int> res;
int main() {
cin >> n;
forn(i, 1, n) {
cin >> a[i];
int x = mp[a[i] - 1];
f[i].id = x;
f[i].v = f[x].v + 1;
ans = max(ans, f[i].v);
mp[a[i]] = i;
}
forn_(i, n, i) {
if (f[i].v == ans) {
int x = i;
while (x != 0) res.push_back(x), x = f[x].id;
break;
}
}
cout << ans << endl;
reverse(res.begin(), res.end());
for (auto i : res) cout << i << ' ';
return 0;
}
浙公网安备 33010602011771号