Codeforces Round #764 (Div. 3)
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目录
A - Plus One on the Subset
最大最小值之差
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int N = 2e5 + 10;
int main()
ios::sync_with_stdio(0);
int _; cin >> _; while (_--)
int n; cin >> n;
vector<int> a(n+1);
int ma = 0, mi =1e9;
for (int i = 1; i <= n; i++)
cin >> a[i];
ma = max(ma, a[i]);
mi = min(mi, a[i]);
cout << ma - mi << endl;
B - Make AP
要求a、b、c为等差数列，因为只改动一个数，因此固定两个数即可
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int N = 2e5 + 10;
int main()
ios::sync_with_stdio(0);
int _; cin >> _; while (_--)
ll a, b, c; cin >> a >> b >> c;
ll d = b - a;
if ((b + d) % c == 0 && (b+d > 0))
cout << "YES" << endl;
continue;
d = c - b;
if ((b-d)%a==0 && (b-d > 0))
cout << "YES" << endl;
continue;
d = c - a;
if (d % 2 != 0)
cout << "NO" << endl;
else
if ((c-d/2)%b==0 && (c-d/2 > 0))
cout << "YES" << endl;
else
cout << "NO" << endl;
C - Division by Two and Permutation
数据范围很小，直接上二分图最大匹配
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int N = 2e5 + 10;
vector<int> g[55];
int vis[55];
int match[55];
bool dfs(int x)
for (auto v : g[x])
if (!vis[v])
vis[v] = 1;
if (!match[v] || dfs(match[v]))
match[v] = x;
return true;
return false;
int main()
ios::sync_with_stdio(0);
int _; cin >> _; while (_--)
int n; cin >> n;
for (int i = 1; i <= n; i++) g[i].clear();
vector<int> a(n+1);
memset(match, 0, sizeof match);
memset(vis, 0, sizeof vis);
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = 1; i <= n; i++)
int now = a[i];
while (now)
if (now <= n) g[i].push_back(now);
now /= 2;
int num = 0;
for (int i = 1; i <= n; i++)
memset(vis, 0, sizeof vis);
if (dfs(i))
num++;
// cout << num << endl;
if (num == n) cout << "YES" << endl;
else cout << "NO" << endl;
D - Palindromes Coloring
主要思想就是先放两两相同的，因为求的是最小值，因此最后一层如果两两没铺满就先不放了，全部转化为1去放
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int N = 2e5 + 10;
int main()
ios::sync_with_stdio(0);
int _; cin >> _; while (_--)
int n, k; cin >> n >> k;
string s; cin >> s;
vector<int> cnt(26);
for (int i = 0; i < s.size(); i++)
cnt[s[i]-'a']++;
vector<int> odd, even;
int num = 0;
int idx = 1;
vector<int> len(n+1);
for (int i = 0; i < 26; i++)
// cout << cnt[i] << endl;
while (cnt[i] > 1)
cnt[i] -= 2;
len[idx] += 2;
idx++;
if (idx == k + 1)
idx = 1;
int ans = 0, ma = 0;
for (int i = 1; i <= k; i++)
ma = max(ma, len[i]);
for (int i = 1; i <= k; i++)
if (len[i] == ma)
ans++;
if (ans != k)
for (int i = 1; i <= k; i++)
if (len[i] == ma)
num += 2;
len[i] -= 2;
for (int i = 0; i < 26; i++)
if (cnt[i]) num++;
for (int i = 1; i <= k; i++)
if (num)
len[i]++;
num--;
int mi = 1e9;
for (int i = 1; i <= k; i++)
mi = min(mi, len[i]);
cout << max(mi, 1) << endl;
E - Masha-forgetful
很容易想到，将全部块分为长度2和3的，因为2和3可以组成大于1的所有自然数。接下来就用简单的dp存一下，最后路径返回记录一下答案就可以了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int N = 1e5 + 10;
int main()
ios::sync_with_stdio(0);
int _; cin >> _; while (_--)
int n, m; cin >> n >> m;
map<string, tuple<int, int, int>> mp;
for (int i = 1; i <= n; i++)
string s; cin >> s;
for (int j = 0; j < m; j++)
if (j + 2 <= m) mp.insert(s.substr(j, 2), j + 1, j + 2, i);
if (j + 3 <= m) mp.insert(s.substr(j,3), j+1, j+3, i);
vector<int> dp(m+1);
dp[0] = 1;
string s; cin >> s; s = '#' + s;
for (int i = 1; i <= m; i++)
if (i >= 2)
string tmp = s.substr(i - 1, 2);
if (mp.count(tmp)) dp[i] |= dp[i-2];
if (i >= 3)
string tmp = s.substr(i-2, 3);
if (mp.count(tmp)) dp[i] |= dp[i-3];
if (!dp[m]) cout << -1 << endl;
else
int now = m;
vector<tuple<int,int,int>> ve;
while (now)
if (dp[now-2])
string tmp = s.substr(now-1, 2);
ve.push_back(mp[tmp]);
now -= 2;
else if (dp[now-3])
string tmp = s.substr(now-2, 3);
ve.push_back(mp[tmp]);
now -= 3;
reverse(ve.begin(), ve.end());
cout << ve.size() << endl;
for (auto [l,r,i] : ve)
cout << l << ' ' << r << ' ' << i << endl;
G - MinOr Tree
求或的最小生成树
最开始我们将30位全部置为1，代表全部的边都可取
不难想到通过位去计算，从高位到低位去置0，如果当前边权包含在当前状态里，这条边是可取的，我们只需要判断连通性就好了，复杂度是 O ( 30 ∗ m ∗ l o g n ) O(30*m*logn) O(30∗m∗logn)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const int N = 2e5 + 10;
struct Edge
int u, v, w;
e[N<<1];
int fa[N];
int find(int x) return x == fa[x] ? x : fa[x] = find(fa[x]);
int main()
ios::sync_with_stdio(0);
int _; cin >> _; while (_--)
int n, m; cin >> n >> m;
for (int i = 1; i <= m; i++) cin >> e[i].u >> e[i].v >> e[i].w;
// for (int i = 1; i <= n; i++) fa[i] = i;
int now = (1 << 30) - 1;
for (int i = 29; i >= 0; i--)
int tmp = now - (1 << i);
for (int j = 1; j <= n; j++) fa[j] = j;
for (int j = 1; j <= m; j++)
if ((e[j].w & tmp) == e[j].w)
int u = find(e[j].u);
int v = find(e[j].v);
if (u != v)
fa[u] = v;
int flag = 0;
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