杭电多校（三）2019.7.29--暑假集训
【HDU 6003】
UNSOLVED
【HDU 6004】
SOLVED
【题目大意】有一DAG图，n个节点，m次询问，每次询问两个节点，求令两个节点任意一个和叶节点失去联通的方法数
【思路】支配树，没有听说过于是被当场爆锤（知道了也是被锤的命QwQ）
#include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<string> #include<cstring> #include<climits> #include<cmath> #include<map> #include<set> #include<deque> using namespace std; const int maxn = 200010; vector<int>G[maxn]; int indegree[maxn]; int nex[maxn][30]; int dis[maxn]; int lca_num[maxn]; int N, M; void init(int n) { for (int j = 1; (1 << j) <= N; j++) nex[n][j] = nex[nex[n][j - 1]][j - 1]; } int lca(int a, int b) { if (b == -1) return a; if (dis[a] < dis[b]) swap(a, b); int delta_distance = dis[a] - dis[b]; int base = 0; while (delta_distance) { if (delta_distance & 1) a = nex[a][base]; base++; delta_distance >>= 1; } if (a == b) return a; for (int i = log2(N); i >= 0; i--) { if (nex[a][i] != nex[b][i]) { a = nex[a][i]; b = nex[b][i]; } } return nex[a][0]; } int main() { ios_base::sync_with_stdio(false); int T; cin >> T; while (T--) { memset(lca_num, -1, sizeof(lca_num)); memset(indegree, 0, sizeof(indegree)); memset(dis, 0, sizeof(dis)); memset(nex, 0, sizeof(nex)); cin >> N >> M; for (int i = 0; i <= N; i++) G[i].clear(); for (int i = 1; i <= M; i++) { int u, v; cin >> u >> v; G[v].push_back(u); indegree[u]++; } queue<int>que; for (int i = 1; i <= N; i++) { if (indegree[i] == 0) { lca_num[i] = 0; que.push(i); } } while (!que.empty()) { int n = que.front(); que.pop(); nex[n][0] = lca_num[n]; dis[n] = dis[lca_num[n]] + 1; init(n); for (int to : G[n]) { lca_num[to] = lca(n, lca_num[to]); if (--indegree[to]==0) que.push(to); } } int Q; cin >> Q; for (int i = 1; i <= Q; i++) { int a, b; cin >> a >> b; cout << dis[a] + dis[b] - dis[lca(a, b)] << "\n"; } } }
【HDU 6005】
UNSOLVED
【HDU 6006】
UNSOLVED
【HDU 6007】
UNSOLVED
【HDU 6008】
SOLVED
【思路】循环节
#include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include <iostream> #include <algorithm> using namespace std; using ll = long long int; ll T, x, n, ans; ll a[]={2, 3, 5, 7, 11, 13, 17, 19}; ll pow_mul(ll a, ll b, ll r) { ll ans = 0; while(b) { if(b & 1) ans = (ans + a) % r; a = (a + a) % r; b >>= 1; } return ans % r; } ll pow_mod(ll a, ll b, ll r) { ll ans = 1; while(b) { if(b & 1) ans = pow_mul(ans, a, r) % r; a = pow_mul(a, a, r) % r; b >>= 1; } return ans % r; } bool test(ll a, ll p, ll d) { if(p % 2 == 0) return false; while(!(d & 1)) d >>= 1; ll t = pow_mod(a, d, p); //得到 a^d % p while(d != p - 1) { ll y = pow_mul(t, t, p); // t^2 % p if(y == 1) { if(t != 1 && t != p - 1) return false;//不满足二次探测定理 return true; //满足的话后面的肯定也满足了 因为此时y = 1 } t = y; d <<= 1; } return t == 1; //最后一次测试 a^(p - 1) % p = 1 } bool isprime(ll x) { for(int i = 0; i < 8; i ++) { if(a[i] == x) return true; if(!test(a[i], x, x - 1)) return false; } return true; }ll mul(ll a,ll b,ll mod) { ll ret = 0; while (b) { if (b & 1) ret = (a + ret) % mod; b >>= 1; a = (a + a) % mod; } return ret; } ll quicmod(ll a,ll b,ll mod) { ll ret = 1; while (b) { if (b & 1) { ret = mul(ret,a,mod); } b >>= 1; a = mul(a,a,mod); } return ret; } int main() { int T; scanf("%d",&T); while (T--) { ll p,Q,ret = 0; scanf("%lld",&Q); p = Q - 1; ret = Q - 1; while (isprime(p) == 0) ret = mul(ret,quicmod(p,Q - 2,Q),Q),p--; printf("%lld\n",ret); } return 0; }
【HDU 6009】
SOLVED
【题目大意】一个数列，令1<=i<=n，每个wi都有都要满足到此前缀和小于m，如果大于则需要从前i-1个数中删除，直到符合条件，输出每位最少需要删除的数字个数
【思路】有点类似主席树，建一颗值域线段树，每次遍历动态更新节点，然后保存和，然后二分向下查找
#include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<string> #include<cstring> #include<climits> #include<cmath> #include<map> #include<set> #include<deque> using namespace std; const int maxn = 200010; typedef pair<int, int>P; P arr[maxn]; int rnk[maxn]; struct segment_tree { struct node { int left, right, mid; long long sum, num; }; node tree[4 * maxn]; void update(int k) { tree[k].sum = tree[k << 1].sum + tree[k << 1 | 1].sum; tree[k].num = tree[k << 1].num + tree[k << 1 | 1].num; } void build_tree(int l, int r, int k = 1) { tree[k].left = l; tree[k].right = r; tree[k].mid = (r + l) >> 1; if (l == r) return; const int& m = tree[k].mid; build_tree(l, m, k << 1); build_tree(m + 1, r, k << 1 | 1); } void change_interval(int l, int r, int k = 1) { if (l <= tree[k].left && tree[k].right <= r) { tree[k].sum += arr[tree[k].left].first; tree[k].num++; return; } const int& m = tree[k].mid; if (l <= m) change_interval(l, r, k << 1); if (r > m) change_interval(l, r, k << 1 | 1); update(k); } int query(long long x, int k = 1) { if (tree[k].sum <= x) return 0; int ans = 0; if (tree[k].left == tree[k].right) return 1; if (tree[k << 1].sum <= x) ans += query(x - tree[k << 1].sum, k << 1 | 1); else { ans += tree[k << 1 | 1].num; ans += query(x, k << 1); } return ans; } }my_tree; int main() { ios_base::sync_with_stdio(false); int T; cin >> T; while (T--) { memset(&my_tree, 0, sizeof(my_tree)); int N, K; cin >> N >> K; for (int i = 1; i <= N; i++) { cin >> arr[i].first; arr[i].second = i; } sort(arr + 1, arr + N + 1); for (int i = 1; i <= N; i++) { rnk[arr[i].second] = i; } my_tree.build_tree(1, N); for (int i = 1; i <= N; i++) { cout << my_tree.query(K - arr[rnk[i]].first) << " "; my_tree.change_interval(rnk[i], rnk[i]); } cout << "\n"; } }
【HDU 6010】
UNSOLVED
【HDU 6011】
UNSOLVED
【HDU 6012】
UNSOLVED
【HDU 6013】
UNSOLVED
