10.17
t1
签到题
注意到约数只有 \(O(\sqrt n)\) 级别，暴力找约数即可。
唐人当然有唐做法啦！
分解质因数+dfs搜约数
反正唐就对了。
code
嘻嘻
#include <bits/stdc++.h>
#define int long long
#define pir pair<int, int>
using namespace std;
int n, m, cnt;
int a[210], ans[210];
int cntt[11], num[11];
map<int, int> mp;
pir node[3000000];inline bool cmp(pir a, pir b) { return a.second < b.second; }inline int km(int a, int b)
{int ans = 1;while (b){if (b & 1)ans *= a;a *= a;b >>= 1;}return ans;
}void dfs(int val, int id, int sum)
{if (sum > n)return;if (id == cntt[0] + 1){if (sum > n)return;bool flag = 1;for (int i = 1; i <= cntt[0]; ++i)if (sum == num[i]){flag = 0;break;}if (flag && val % sum == 0)mp[sum]++;return;}for (int i = 0; i <= cntt[id]; ++i){int nxt = km(num[id], i);if (nxt > n)return;dfs(val, id + 1, sum * nxt);}
}inline void chai(int x)
{int val = x;cntt[0] = 0;for (int i = 2; i * i <= x; ++i){if (x % i == 0){mp[i]++;num[++cntt[0]] = i;while (x % i == 0)x /= i, ++cntt[cntt[0]];}}if (x != 1){num[++cntt[0]] = x, cntt[cntt[0]] = 1;if (x <= n)mp[x]++;}int nnum = 1;dfs(val, 1, 1);
}signed main()
{freopen("div.in","r",stdin);freopen("div.out","w",stdout);ios::sync_with_stdio(0);cin.tie(0);cin >> n >> m;for (int i = 1; i <= m; ++i)cin >> a[i], chai(a[i]);sort(a + 1, a + 1 + m);for (int pos = 1; pos <= m; ++pos){if (a[pos] > n)break;if (mp[a[pos]])continue;mp[a[pos]]++;for (int i = pos + 1; i <= m; ++i)if (a[i] % a[pos] == 0)mp[a[pos]]++;}mp[1] = m;for (auto y : mp)node[++cnt] = y;ans[0] = n - cnt;for (int i = 1; i <= cnt; ++i)ans[node[i].second]++;for (int i = 0; i <= m; ++i)cout << ans[i] << "\n";return 0;
}
t2
第一眼：背包版题，第二眼：你值域开玩笑呢？
惊人注意力发现虽然值域很大，但答案很小，于是将 dp 数组的下标与所存内容交换，正常预处理 dp 后将其赋成前缀最小值，每次询问二分即可。
code
哈哈
#include <bits/stdc++.h>
using namespace std;
int n, m;
struct node
{int cst, val, tim;
} a[310];
int ans[310];
int dp[310][90010];inline bool cmp(node a, node b) { return a.tim < b.tim; }signed main()
{freopen("market.in", "r", stdin);freopen("market.out", "w", stdout);ios::sync_with_stdio(0);cin.tie(0);cin >> n >> m;for (int i = 1; i <= n; ++i)cin >> a[i].cst >> a[i].val >> a[i].tim;sort(a + 1, a + 1 + n, cmp);memset(dp, 63, sizeof(dp));dp[0][0] = 0;int l = 1, r = 0;for (int k = 1; k <= 300; ++k){while (r <= n && a[r].tim <= k)++r;int lim = r - 1;for (int i = 0; i <= 90000; ++i)dp[k][i] = dp[k - 1][i];for (int i = l; i <= lim; ++i)for (int j = 90000; j >= a[i].val; --j)dp[k][j] = min(dp[k][j], dp[k][j - a[i].val] + a[i].cst);l = r;}for (int i = 1; i <= 300; ++i)for (int j = 90000; j; --j)if (dp[i][j + 1] < dp[i][j])dp[i][j] = dp[i][j + 1];int t, up;while (m--){cin >> t >> up;int ans = upper_bound(dp[t], dp[t] + 1 + 90000, up) - dp[t];cout << ans - 1 << "\n";}return 0;
}
t3
神秘 dp
改不动了
喝的
详见QED代码
t4
赛时紧急打了个链（还是错的），喜提 8pts ，但是它输出 0 能有 37pts 你逗我呢？？？
听Wy_x 讲的不用线段树写法听懂了，但打的时候就是感觉不对，像 大分 一样，最后还是写了线段树。
我们将边权（边的归属权）下放，拆点，将每个点拆为两个，分别表示边的两种方向。
然后就一堆连边，find ，跳，判无解阿巴阿巴 。
实在不想写了，还是看代码吧。
code
呜呜
#include <bits/stdc++.h>
#define int long long
#define lid (id << 1)
#define rid (id << 1 | 1)
using namespace std;
const int mod = 1e9 + 7;
const int N = 3e5 + 10;
int n, m;
vector<int> e[N];
int Fa[N << 1];
int dep[N], siz[N], fa[N], son[N];
int top[N], tot, id[N], fid[N];
struct tree
{int l, r;bool flag;
} t[N << 3];
#define lid (id << 1)
#define rid (id << 1 | 1)inline int km(int a, int b)
{int ans = 1;while (b){if (b & 1)ans *= a, ans %= mod;a *= a, a %= mod;b >>= 1;}return ans;
}inline int find(int x) { return x == Fa[x] ? x : Fa[x] = find(Fa[x]); }
inline void connect(int x, int y)
{int fx = find(x), fy = find(y);if (fx != fy)Fa[fx] = fy;
}inline void merge(int u, int v, bool opt)
{if (opt)connect(u, v + n), connect(u + n, v);elseconnect(u, v), connect(u + n, v + n);
}void dfs1(int x, int f)
{siz[x] = 1, fa[x] = f, dep[x] = dep[f] + 1;for (auto y : e[x]){if (y == f)continue;dfs1(y, x);siz[x] += siz[y];if (siz[y] > siz[son[x]])son[x] = y;}
}void dfs2(int x, int t)
{top[x] = t;id[x] = ++tot;fid[tot] = x;if (!son[x])return;dfs2(son[x], t);for (auto y : e[x])if (y != fa[x] && y != son[x])dfs2(y, y);
}inline int LCA(int u, int v)
{while (top[u] != top[v]){if (dep[top[u]] < dep[top[v]])swap(u, v);u = fa[top[u]];}return dep[u] < dep[v] ? u : v;
}void build(int id, int l, int r)
{t[id].l = l, t[id].r = r;if (l == r)return;int mid = (l + r) >> 1;build(lid, l, mid);build(rid, mid + 1, r);
}void update(int id, int l, int r)
{if (l > r)return;if (t[id].flag)return;if (t[id].l == t[id].r){merge(fa[fid[t[id].l]], fid[t[id].l], 0);t[id].flag = 1;return;}int mid = (t[id].l + t[id].r) >> 1;if (mid >= l)update(lid, l, r);if (mid < r)update(rid, l, r);t[id].flag = t[lid].flag & t[rid].flag;
}signed main()
{freopen("usmjer.in","r",stdin);freopen("usmjer.out","w",stdout);ios::sync_with_stdio(0);cin.tie(0);cin >> n >> m;for (int i = 1; i <= n; ++i)Fa[i] = i, Fa[i + n] = i + n;for (int i = 1, u, v; i < n; ++i){cin >> u >> v;e[u].push_back(v);e[v].push_back(u);}dfs1(1, 0);dfs2(1, 1);build(1, 1, n);int u, v;while (m--){cin >> u >> v;int lca = LCA(u, v);if (u != lca && v != lca)merge(u, v, 1);while (top[u] != top[lca]){update(1, id[top[u]] + (fa[top[u]] == lca), id[u]);u = fa[top[u]];}update(1, id[lca] + 2, id[u]);while (top[v] != top[lca]){update(1, id[top[v]] + (fa[top[v]] == lca), id[v]);v = fa[top[v]];}update(1, id[lca] + 2, id[v]);}int tot = 0;for (int i = 2; i <= n; ++i){int u = find(i), v = find(i + n);if (u == v){cout << 0;return 0;}tot += (i != u) + (i + n != v);}cout << km(2, n - 1 - (tot >> 1)) << "\n";return 0;
}
不知道为什么感觉很累，大抵是被t4恶心到了。
累了累了，先歇会。
