牛客2025秋季算法编程训练联赛4-提升组

写在前面

已经，无所谓了

C 题没看懂，不写了，F 只有一个过了，也不想写了

A

相当于就是填数字，总共的数字个数是 加号个数 + 1

然后发现要用高精度，完了

#include <bits/stdc++.h>
using namespace std;using i64 = long long;
using ll = long long;struct BigInt{vector<int> d;BigInt(){}BigInt(ll x){*this = x;}BigInt(string &s){ *this = s; }BigInt& operator=(const string &s){d.clear();for (int i = (int)s.size() - 1; i >= 0; --i) d.push_back(s[i] - '0');trim();return *this;}BigInt& operator=(ll x){d.clear();if(x==0){d.push_back(0); return *this;}while(x) {d.push_back(x%10), x/=10;}return *this;}void trim(){while(d.size()>1 && d.back() == 0) d.pop_back();}bool isZero() const {return d.size() == 1 && d[0] == 0;} static int cmp(BigInt A, BigInt B){A.trim(); B.trim();if(A.d.size() != B.d.size()) return A.d.size() < B.d.size() ? -1 : 1;int n = A.d.size(), m = B.d.size();for(int i = min(n,m)-1; i >= 0; --i){if(A.d[i] != B.d[i]) return A.d[i] < B.d[i] ? -1 : 1;}return 0;}
};BigInt operator*(const BigInt A, const BigInt B){if(A.isZero() || B.isZero()) return BigInt(0);BigInt res;res.d.assign(A.d.size() + B.d.size(),0);for(int i = 0; i<A.d.size(); ++i){int j=0, cur = 0, x = A.d[i];for(;j<B.d.size();++j){ll now = res.d[i+j] + x*B.d[j] + cur;res.d[i+j] = now%10, cur = now/10;}while(cur){ll now = res.d[i+j] + cur;res.d[i+j] = now%10;cur = now/10, ++j;}}res.trim();return res;
}BigInt operator+(const BigInt &A, const BigInt &B){BigInt res;int n = A.d.size(), m = B.d.size();int L = max(n, m), now = 0;res.d.resize(L);for (int i = 0; i < L; ++i){int x = now;if(i<n) x += A.d[i];if(i<m) x += B.d[i];res.d[i] = x % 10;now = x / 10;}if (now) res.d.push_back(now);return res;
}void solve(){string s; cin >> s;int num = 1; vector<int> cnt(10);for(auto ch : s){if(ch == '+') num++;else cnt[ch - '0']++;}vector<string> a(num+1);int idx = 1;for(int i = 1; i <= 9; ++i)if(cnt[i]){for(int j = 1; j <= cnt[i]; ++j){a[idx].push_back(char(i + '0'));idx = (idx + 1) % (num+1);if(idx == 0) idx = 1;}}// cerr << num << '\n';// 高精度，得要BigInt ans = 0;for(int i = 1; i <= num; ++i){if(a[i].size())ans = ans + BigInt(a[i]);// cerr << a[i] << '\n';}auto vec = ans.d;for(int i = vec.size() - 1; i >= 0; --i) cout << vec[i];
}signed main(){ios::sync_with_stdio(); cin.tie(0); cout.tie(0);int t = 1;// cin >> t;while(t--){solve();}return 0;
}

B

就是走距离为偶数的路径数量

定义: f[u]：以 u 为根节点的子树中奇数层节点数量，g[u]：以 u 为根节点的子树中偶数层节点数量

然后 dp[u]：表示以 u 为根节点的子树中偶数长度距离的节点对数，然后答案就对所有子树的方案求和完了

发现每个子树的贡献其实是所有点与兄弟节点的组合，

最后，注意是有序对

#include <bits/stdc++.h>
using namespace std;using i64 = long long;constexpr int Maxn = 1e6+10;
vector<int> G[Maxn];// 求一定能胜利的方案数
// 就是路径长度为偶数的路径数量 -> 点的层数奇偶分类，然后对子树 dp 计数完了
void solve(){int n;cin >> n;for(int i = 2; i <= n; ++i){int p; cin >> p;G[p].push_back(i), G[i].push_back(p);}vector<i64> f(n+1), g(n+1), dp(n+1);function<void(int,int)> dfs = [&](int u,int fa) -> void{vector<int> sons;f[u] = 1;int odd = 0, even = 0;for(auto v : G[u]){if(v == fa) continue;dfs(v,u);sons.push_back(v);f[u] += g[v], g[u] += f[v];odd += f[v], even += g[v];}for(auto v : sons){dp[u] += 1ll*f[v]*(odd - f[v]) + 1ll*g[v]*(even - g[v]);}dp[u] += (f[u]-1) * 2;};dfs(1,0);i64 ans = 0;for(int i = 1; i <= n; ++i) ans += dp[i];cout << ans << '\n';
}signed main(){ios::sync_with_stdio(); cin.tie(0); cout.tie(0);int t = 1;// cin >> t;while(t--){solve();}return 0;
}

D

就是询问的颜色区间内左右端点的组合方案数的和

考虑单点修改区间查询的线段树即可解决这个问题

template <class Info>
struct SegmentTree {struct Node {int l, r;Info info;};std::vector<Node> tr;SegmentTree() {};SegmentTree(int n) {init(n);}SegmentTree(std::vector<Info> & info) {init(info);}void init(int n) {tr.assign(n << 2, {});build(0, n - 1);}void init(std::vector<Info> & info) {int n = info.size();tr.assign(n << 2, {});std::function<void(int, int, int)> build = [&](int l, int r, int u) -> void {tr[u] = {l, r, {}};if (l == r) {tr[u].info = info[l];return;}int mid = l + r >> 1;build(l, mid, u << 1); build(mid + 1, r, u << 1 | 1);pushup(u);};build(0, n - 1, 1);}void pushup(int u) {tr[u].info = tr[u << 1].info + tr[u << 1 | 1].info;}void build(int l, int r, int u = 1) {tr[u] = {l, r, {}};if (l == r) {return;}int mid = l + r >> 1;build(l, mid, u << 1); build(mid + 1, r, u << 1 | 1);pushup(u);}void modify(int p, const Info & info, bool set = false) {int u = 1;while (tr[u].l != tr[u].r) {int mid = tr[u].l + tr[u].r >> 1;if (p <= mid) {u = u << 1;} else {u = u << 1 | 1;}}if (set) {tr[u].info = info;} else {tr[u].info = tr[u].info + info;}u >>= 1;while (u) {pushup(u);u >>= 1;}}Info query(int l, int r, int u = 1) {if (l <= tr[u].l && tr[u].r <= r) {return tr[u].info;}int mid = tr[u].l + tr[u].r >> 1;if (r <= mid) {return query(l, r, u << 1);} else if (l > mid) {return query(l, r, u << 1 | 1);}return query(l, r, u << 1) + query(l, r, u << 1 | 1);}
};struct Info{int sum;
};Info operator+(const Info &l, const Info &r){Info res;res.sum = l.sum + r.sum;return res;
}// 其实是颜色区间内不包含 i 位置颜色的三元组对数的和的数量
const int N = 5e5 + 10;
int cntl[N], cntr[N];
void solve(){int n;cin >> n;vector<T> num(n);for(auto &[col, l, r] : num) cin >> col >> l >> r, cntr[col]++;SegmentTree<Info> tr(N + 1);for(auto [col, l, r] : num){cntr[col]--;i64 x = 1ll * cntl[col] * cntr[col];tr.modify(col, {x}, 1);cout << tr.query(l,r).sum << ' ';cntl[col]++;x = 1ll * cntl[col] * cntr[col];tr.modify(col, {x}, 1);}
}

E

显然是 z = 1 类和 z = 0 类进行配对，

先排序 x，然后找 y，每次都找到最大的能够满足的 y 配对就能贪到最大匹配数量

可以用 multiset 完成这个事情

#include <bits/stdc++.h>
using namespace std;using i64 = long long;
using T = tuple<int,int,int>;void solve(){int n;cin >> n;vector<T> point(n);for(auto &[x,y,z] : point) cin >> x >> y >> z;sort(point.begin(), point.end());multiset<int> mst;int ans = 0;for(auto [x,y,z] : point){if(z == 1){auto it = mst.lower_bound(y);if(it != mst.begin()){++ans;--it; mst.erase(it);}}else{mst.insert(y);}}cout << ans << '\n';
}signed main(){ios::sync_with_stdio(); cin.tie(0); cout.tie(0);int t = 1;// cin >> t;while(t--){solve();}return 0;
}