下午训练赛的榜歪得吓人，来补一下题。

CF 1646B Quality vs Quantity

题目链接

双指针两端遍历即可

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;void solve()
{int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; i ++ ) cin >> a[i];sort(a.begin(), a.end());int i = 2, j = n - 2;int sum1 = a[0] + a[1], sum2 = a[n - 1];int cnt1 = 2, cnt2 = 1;while (i <= j){if (sum1 < sum2 && cnt1 > cnt2){cout << "YES\n";return;}else if (sum1 >= sum2){sum2 += a[j];cnt2 ++ ;j -- ;}else if (cnt1 <= cnt2){sum1 += a[i];cnt1 ++ ;i ++ ;}}if (sum1 < sum2 && cnt1 > cnt2){cout << "YES\n";return;}cout << "NO\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1326C Permutation Partitions

题目链接

每个区间分一个前k大的数就行了，剩下就是简单组合数学

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;
const int mod = 998244353;void solve()
{int n, k;cin >> n >> k;vector<PII> a(n);for (int i = 0; i < n; i ++ ){a[i].second = i;cin >> a[i].first;}sort(a.begin(), a.end());int ans = 0;for (int i = n - 1; i >= n - k; i -- ) ans += a[i].first;int cnt = 1;vector<int> tmp;for (int i = n - 1; i >= n - k; i -- ) tmp.push_back(a[i].second);sort(tmp.begin(), tmp.end());for (int i = 1; i < tmp.size(); i ++ ) cnt = cnt * (tmp[i] - tmp[i - 1]) % mod;cout << ans << ' ' << cnt << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

CF 1355D Game With Array

题目链接

前面全取1，后面取S-(n-1)即可

主要思路是让能被替代的数尽可能多，如果是1 2 4 8这种取法，互相之间不能被替代，所以能够组成的数就多了

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;
const int mod = 998244353;void solve()
{int n,s;cin >> n >> s;if (2 * n > s){cout << "NO\n";return;}cout << "YES\n";for (int i = 0; i < n - 1; i ++ ) cout << 1 << ' ';cout << s - (n - 1) << '\n';cout << s - n << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

AT ARC99B Snuke Numbers

题目链接

打表找规律（这么有意思的题到底哪里毒瘤了 ^ _ ^

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;
const int mod = 998244353;void solve()
{int k;cin >> k;int d = 1;int ans = 0;auto sum = [&](int x){string s = to_string(x);int res = 0;for (int i = 0; i < s.size(); i ++ ){res += s[i] - '0';}return res;};for (int i = 0; i < k; i ++ ){if ((ans + d * 10) * sum(ans + d) < (ans + d) * sum(ans + d * 10)) d *= 10;ans += d;cout << ans << '\n';}
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

AT exawizards2019C Snuke the Wizard

题目链接

二分左右边界即可

二分的时候可以修改条件来保证一定能找到你想找到的值，否则可能会出现没有符合条件的点的情况

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;
const int mod = 998244353;void solve()
{int n, q;cin >> n >> q;string s;cin >> s;s = " " + s;vector<char> t(q), d(q);for (int i = 0; i < q; i ++ ) cin >> t[i] >> d[i];auto check = [&](int x){for (int i = 0; i < q; i ++ ){if (t[i] == s[x]){if (d[i] == 'L') x -- ;else x ++ ;if (x <= 0) return 0;else if (x > n) return 1;}}return 2;};int l = 0, r = n + 1;while (l < r){int mid = l + r + 1 >> 1;if (check(mid) == 0) l = mid;else r = mid - 1;}int ansl = r;l = 0, r = n + 1;while (l < r){int mid = l + r >> 1;if (check(mid) == 1) r = mid;else l = mid + 1;}int ansr = r;cout << ansr - ansl - 1 << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

CF 1788F XOR, Tree, and Queries

题目链接

设p[i]是i到根结点的边权值异或和，根据异或的性质可以得到点u和v的边权异或和就是p[u] ^ p[v]

输入边的时候记录点的度数，度数为偶数的可以不用管，因为会相抵消

根据已给的边权信息，可以利用dfs得到一组可行解（设每个连通块遍历的第一个点w的p[w] = 0），同时在遍历的时候发现矛盾直接返回NO

之后就是让所有边权异或和最小了，先记录所有度为奇数的点的异或和为ans，连通块点数为偶数的，异或值会相抵消不用管，只要遇见奇数，只需要让连通块中的点与当前ans取异或就行

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;
const int mod = 998244353;void solve()
{int n, m;cin >> n >> m;vector<PII> edge(n - 1);vector<int> du(n + 1), p(n + 1), size(n + 1), root(n + 1);for (int i = 0; i < n - 1; i ++ ){cin >> edge[i].first >> edge[i].second;du[edge[i].first] ++ , du[edge[i].second] ++ ;}vector<vector<PII>> g(n + 1);for (int i = 0; i < m; i ++ ){int a, b, c;cin >> a >> b >> c;g[a].push_back(make_pair(b, c));g[b].push_back(make_pair(a, c));}function<void(int, int)> dfs = [&](int u, int rt){root[u] = rt;for (int i = 0; i < g[u].size(); i ++ ){int j = g[u][i].first;if (!root[j]){p[j] = (p[u] ^ g[u][i].second);dfs(j, rt);}else if (p[j] != (p[u] ^ g[u][i].second)){cout << "NO\n";exit(0);}}};for (int i = 1; i <= n; i ++ ){if (root[i]) continue;dfs(i, i);}int ans = 0;for (int i= 1; i <= n; i ++ ) {if (du[i] % 2 != 0){ans ^= p[i];size[root[i]] ++ ;}}for (int i = 1; i <= n; i ++ ){if (size[i] % 2 == 0) continue;for (int j = 1; j <= n; j ++ ){if (root[j] == i) p[j] ^= ans;}break;}cout << "YES\n";for (int i = 0; i < n - 1; i ++ ){cout << (p[edge[i].first] ^ p[edge[i].second]) << ' ';}cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}