1112模拟赛
T1：
有一个 \(n\) 个点的无向正权图 \(G\)，这个图是连通的，你知道了这些点两两之间的最短路的长度。
你想要构造一个新的无向正权图 \(G^{\prime}\)，使得新图中两两之间的最短路的长度与原图样，并且边数最少。求最少的边数。
\(1\le n\le 300\)。
最朴素的想法记 \(i,j\) 之间最短路为 \(dis_{i,j}\)，从 \(i\) 到 \(j\) 连一条边，边权 \(dis_{i,j}\)。
然后发现若存在 \(dis_{i,j}=dis_{i,k}+dis_{k,j}\)，则这条边可以不加。
于是跑一遍 Floyd 就完了。考试的时候写了个很难卡的 \(O(n^4)\) 堆乱搞，过了。
#pragma GCC optimize(3)
#include <cstdio>
#include <queue>
#include <cstring>
#define int long long
struct Node {
int u, v, w;
inline bool operator < (const Node a) const {return w > a.w;}
};
std::priority_queue<Node> Q;
int dis[305][305], a[305][305], n, ans;
void update(const int u) {
for (register int i = 1; i <= n; ++ i) {
register int tmp = dis[i][u];
for (register int j = 1; j <= n; ++ j)
if (tmp + dis[u][j] < dis[i][j]) dis[i][j] = dis[u][j] + tmp;
}
}
signed main() {
memset(dis, 0x3f, sizeof dis);
scanf("%lld", &n);
for (int i = 1; i <= n; ++ i) dis[i][i] = 0;
for (int i = 1; i <= n; ++ i)
for (int j = 1; j <= n; ++ j) scanf("%lld", &a[i][j]);
for (int i = 1; i <= n; ++ i)
for (int j = i + 1; j <= n; ++ j) Q.push(Node{i, j, a[i][j]});
while (Q.size()) {
int u = Q.top().u, v = Q.top().v, w = Q.top().w;
Q.pop();
if (dis[u][v] > w) dis[u][v] = dis[v][u] = w, update(u), update(v), ++ ans;
}
printf("%lld", ans);
return 0;
}
T2：
你有一个环，环上有 \(n\) 个正整数，其中 \(a_i\) 表示环上第 \(i\) 个数。你能将环切成 \(k\) 段, 每段包含一个或者多个数字。
对于一个切分方案，优美程度为每段数字和的最大公约数，你想使切分方案的优美程度最大，对于 \(k=1,2...n\) 输出答案。
\(1\le n\le 2000,1\le a_i\le 5\times 10^7\)
首先 gcd 这种东西不好计入状态，考虑枚举最终的答案。
容易证明最终答案一定是 \(\sum a_i\) 的约数，\(\sum a_i\le 10^{11}\)，因此约数个数不超过 \(4032\)。
枚举这个 gcd 后就好做了，计算前缀和 \(\text{mod gcd}\) 的值，看出现次数最多的前缀和，假设出现了 \(i\) 次，那切成 \(1,2...i\) 段的答案都至少是当前枚举的 gcd。
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#define int long long
int a[1000005], d[1000005], ans[1000005], cnt, sum;
std::unordered_map<int, int> mp;
signed main() {
int n;
scanf("%lld", &n);
for (int i = 1; i <= n; ++ i) scanf("%lld", a + i), sum += a[i];
for (int i = 1; i * i <= sum; ++ i)
if (sum % i == 0) {
d[++ cnt] = i;
if (sum / i != i) d[++ cnt] = sum / i;
}
std::sort(d + 1, d + cnt + 1);
for (int i = cnt; i; -- i) {
mp.clear();
for (int j = 1, k = a[1] % d[i]; j <= n; k = (k + a[++ j]) % d[i]) ++ mp[k];
int mx = 0;
for (auto& j : mp) mx = std::max(mx, j.second);
for (int j = 1; j <= mx; ++ j) ans[j] = std::max(ans[j], d[i]);
}
for (int i = 1; i <= n; ++ i) printf("%lld\n", ans[i]);
return 0;
}
T3：
有一棵 \(n\) 个节点的树，每条边都有正的边权。
你要回答 \(q\) 个询问，每次会删掉这个树中的 \(k\) 条边，这棵树被分成 \(k+1\) 个连通块。输出每个连通块中最远点对距离的和。
这里的询问是互相独立的，即每次都是在原树上进行操作。
\(1\le n,q\le 10^5,1\le w\le 10^9,1\le k,\sum k\le 10^5\)。
首先需要知道直径的一个性质，若两棵树直径端点分别为 \(a,b\) 和 \(c,d\)，那么这两棵树合并后直径端点一定是这四个点的其中两个。
所以用 dfs 序把树弄成序列，搞一颗线段树，维护 \(\text{In}\) 值在该区间内所有点构成的树的答案。由于单次合并还有个求两点间距离，线段树复杂度是两只 \(\log\) 的。
然后删除的时候求出删除后每棵树在 dfs 序上对应哪几个区间，把这些区间的答案再合并一下就是答案了。
求每棵树在 dfs 序上对应哪几个区间可以用个单调栈搞，其实不叫单调栈，只是栈下面的元素一定是栈上面的元素的最近父亲。这样就可以求出每颗树的根节点，看看它的哪几个最近子节点被断开了。
时间复杂度 \(O((n+\sum k)(\log n)^2)\)
nkoj 上跑得巨慢，我也不知道为什么。
#include<cstdio>
#include <vector>
#include <stack>
#include <algorithm>
#define int long long
inline int max(const int x, const int y) {return x > y ? x : y;}
std::vector<int> vec[100005];
int In[100005], Out[100005], fa[100005][19], dep[100005], dis[100005];
int ID[100005], from[100005], to[100005], cnt;
struct Edge {
int to, nxt, w;
} e[200005];
int head[100005], tot;
inline void AddEdge(int u, int v, int w) {
e[++ tot].to = v, e[tot].w = w, e[tot].nxt = head[u], head[u] = tot;
}
void dfs(int u) {
dep[u] = dep[fa[u][0]] + 1, ID[In[u] = ++ cnt] = u;
for (int i = 1; i <= 18; ++ i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
for (int i = head[u]; i; i = e[i].nxt)
if (e[i].to != fa[u][0]) dis[e[i].to] = dis[u] + e[i].w, fa[e[i].to][0] = u, dfs(e[i].to);
Out[u] = cnt;
}
int LCA(int u, int v) {
if (dep[u] < dep[v]) u ^= v ^= u ^= v;
int tmp = dep[u] - dep[v];
for (int i = 0; i <= 18; ++ i)
if (tmp & 1 << i) u = fa[u][i];
if (u == v) return u;
for (int i = 18; i >= 0; -- i)
if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
return fa[u][0];
}
inline int distance(int u, int v) {return dis[u] + dis[v] - (dis[LCA(u, v)] << 1);}
struct Diameter {
int x, y, w;
inline Diameter operator + (const Diameter a) const {
int w1 = w, w2 = a.w, w3 = distance(x, a.x), w4 = distance(x, a.y);
int w5 = distance(y, a.x), w6 = distance(y, a.y);
int ans = max(w1, max(w2, max(w3, max(w4, max(w5, w6)))));
if (w1 == ans) return Diameter{x, y, ans};
if (w2 == ans) return Diameter{a.x, a.y, ans};
if (w3 == ans) return Diameter{x, a.x, ans};
if (w4 == ans) return Diameter{x, a.y, ans};
if (w5 == ans) return Diameter{y, a.x, ans};
return Diameter{y, a.y, ans};
}
};
struct Segment_tree {
struct Node {
int l, r;
Diameter d;
} tree[400005];
void build(int O, int L, int R) {
tree[O].l = L, tree[O].r = R;
if (L != R) {
build(O << 1, L, L + R >> 1);
build(O << 1 | 1, (L + R >> 1) + 1, R);
tree[O].d = tree[O << 1].d + tree[O << 1 | 1].d;
} else tree[O].d = Diameter{ID[L], ID[L], 0};
}
Diameter query(int O, int L, int R) {
if (L <= tree[O].l && tree[O].r <= R) return tree[O].d;
int mid = tree[O].l + tree[O].r >> 1;
if (!(L <= mid)) return query(O << 1 | 1, L, R);
if (!(mid < R)) return query(O << 1, L, R);
return query(O << 1, L, R) + query(O << 1 | 1, L, R);
}
} tree;
struct Node {
int x, id;
inline bool operator < (const Node a) const {return In[x] < In[a.x];}
} pnt[100005];
std::stack<int> s;
signed main() {
int n, q;
scanf("%lld", &n);
for (int i = 1; i < n; ++ i) {
int w;
scanf("%lld%lld%lld", from + i, to + i, &w);
AddEdge(from[i], to[i], w);
AddEdge(to[i], from[i], w);
}
dfs(1);
tree.build(1, 1, n);
scanf("%lld", &q);
while (q --) {
int k;
scanf("%lld", &k), pnt[1] = {1, 1};
for (int i = 1, x; i <= k; ++ i) {
scanf("%lld", &x);
int p = dep[from[x]] > dep[to[x]] ? from[x] : to[x];
pnt[i + 1] = Node{p, i + 1};
}
std::sort(pnt + 1, pnt + k + 2);
for (int i = 1; i <= k + 1; ++ i) {
while (s.size() && (Out[s.top()] < In[pnt[i].x] || Out[pnt[i].x] < In[s.top()])) s.pop();
if (s.size()) vec[s.top()].push_back(pnt[i].x);
s.push(pnt[i].x);
}
int sum = 0;
for (int i = 1; i <= k + 1; ++ i) {
int pos = In[pnt[i].x];
bool flag = true;
Diameter ans;
for (int j : vec[pnt[i].x]) {
if (flag) {flag = false; ans = tree.query(1, pos, In[j] - 1); pos = Out[j] + 1;}
else {
if (pos < In[j]) ans = ans + tree.query(1, pos, In[j] - 1);
pos = Out[j] + 1;
}
}
if (pos <= Out[pnt[i].x]) {
if (flag) ans = tree.query(1, pos, Out[pnt[i].x]);
else ans = ans + tree.query(1, pos, Out[pnt[i].x]);
}
sum += ans.w;
}
printf("%lld\n", sum);
while (s.size()) s.pop();
for (int i = 1; i <= k + 1; ++ i) vec[pnt[i].x].clear();
}
return 0;
}
T4：
首先当一个数为 \(2^k\) 时期望 \(2\) 次会变成 \(0\)，这东西大概就是错位相减求出来的，草稿本上推一推就好了。
然后 \(70\) 分就白给了，所以我们分块打表一下，每隔 1e6 打一次表，它就过了!<-邪教
考虑求 \(\sum\limits^n_{i=1} f(i)\)。为了方便讨论，假设 \(i\) 为偶数。首先对于奇数，\(1\) 变换后要不是 \(0\) 要不是 \(2\)，\(3\) 变换后要不是 \(2\) 要不是 \(4\)……以此类推，我们得到
\[\sum\limits^n_{i=1} f(i)=\frac{f(n)}{2}+\frac{n}{2}+2\times \sum\limits^{\frac{n}{2}-1}_{i=1} f(2i)
\]
\(n\) 为奇数也没啥区别。
复杂度 \(O(\log n)\) 或 \(O((\log n)^2)\)。
由于太菜，只写了分块打表。
#include <cstdio>
#include <unordered_map>
#define int long long
const int mod = 998244353, inv = 499122177;
const int sum[] = {0,522685747,277338273,76341087,785887678,507563462,561249401,292985805,806742135,348667879,761228614,186115155,535821676,816237962,927999445,210807679,852451049,496982772,494054984,725742036,275314565,821973791,873464609,855176710,490966226,141516364,720840205,141802434,208538019,901608253,104062712,420195557,951868877,60487854,419137602,537580626,793829194,758616944,306504129,881495657,311730820,806377553,599739544,536091719,262674811,892018762,174468645,947453780,413255326,417987292,390053235,90644589,543044691,483420298,353585759,209584813,780103873,624369078,339088275,473175316,903817131,245350587,536100516,634805983,168460180,819523537,496425335,235608501,280447262,431830064,739667598,49870050,413133261,166666106,152773736,56946961,485272668,216252138,963517307,928118111,404563330,286603268,816770027,437858415,176271050,143978997,177899448,758765495,61339568,127208377,835193308,191219458,832541221,348308796,861907344,233968177,281833526,14967163,280467905,753621794,912126977,81849251,665058852,707045610,213453663,911316639,621297213,857639103,804152409,178888502,589801444,401442992,952991228,760886837,746367702,637088184,241292672,416538095,868374962,910696037,536837263,798425439,223506783,486263444,798910434,277111725,246522759,864194456,630131492,102923511,801263564,57385496,403055944,874214678,855703285,858881532,37066582,223109195,158100545,37829940,180597189,158817127,551266871,889705777,685985748,447703344,622412406,756395968,974576026,624379864,346222388,397791292,880809746,82260439,361909045,75521370,168316254,165433765,835212590,224057937,630228350,671296711,743784970,482045836,878554975,862734113,423019759,760967157,369578415,704959405,471614701,301701936,502759259,400862996,395556671,676480191,700913435,869282163,183978644,292358642,766542400,507814533,518023925,956732641,198197667,82974902,462067705,976552778,737814472,36413576,860071731,436621711,66989926,941336777,150234520,809150974,54429131,459451398,87355188,993875748,9785755,355938939,350813560,35792814,866643025,269529358,701694645,919884390,604515960,162016835,139972430,289500720,950051043,400505142,58571582,697878731,761041356,774196981,610766046,296003335,406990353,394669253,655508417,594453568,356521585,642185537,570830843,383491633,49120597,65234459,285548464,49696353,103701466,341109831,269770319,226573527,719529901,347181146,31874826,770762401,861121880,691347781,3056110,874973314,240819175,826093305,460850376,219413211,388285917,344112518,622694635,717994486,531200664,559151334,26154224,356564144,619229763,881573521,234827040,973111514,829743618,305151415,511351378,504559831,282317162,546756095,267794108,69841334,166404147,750954372,310218894,350248374,616549575,413129311,924041524,322286153,677885860,150785767,173318798,149183,130700724,305327821,532743164,450766666,626816160,411422301,58201815,427585093,305446369,945862549,666563421,310050085,608660653,590257610,214064260,758798271,695691900,61966863,681133629,734245730,1528889,283298684,462719535,144855524,749639549,970718459,638622380,596199602,730222859,753257300,76064761,321961491,652402854,44892055,354459479,447389791,728401548,500093865,376511479,641499877,163314037,154057841,834847210,61908896,740904021,624175159,157542972,720973083,85880518,919500191,259042477,26289545,476342447,310627414,529973791,231190169,840193145,479209454,34368774,817107162,213641756,291396735,23367733,613325457,240234528,131392137,529094309,551782592,754383376,867544461,717205282,93053604,71572463,652745924,120055665,370541064,386519178,312786311,663197976,52033826,719240584,74975423,468021094,826847485,264388506,294840811,767902150,931949179,19754912,485116015,12244858,705272023,782585523,445241477,709611783,756532686,583628728,235199574,864770158,813462630,210790133,439327242,187012066,952174476,731547079,251828183,863234431,251915249,517769234,571077304,171132880,37802084,698595936,996797398,791449290,758395079,850310682,51125119,275286381,549796185,301592017,780777779,65515524,645738974,989742178,131546794,752393946,491258347,373567018,697874596,681988786,647252105,793992715,977250510,234866679,844841344,492396201,244062475,618221003,708585023,698772718,587310517,190050556,200209669,715314350,187133749,555462817,958534504,563925246,97859024,355885794,421749716,782819250,972599792,130168782,398284555,585276781,108440337,31775146,453733933,151368171,891693,79271550,376849754,276394914,317655400,130333771,866601199,273826652,879360042,889564326,223067833,303718855,46244490,996119072,873352015,765115093,604131887,473816011,158553681,695713377,450595541,846313392,779090255,942763407,191996753,657549907,326623897,93234767,964601263,177956444,275458844,539839779,602759804,338828441,683469714,389721110,826364145,250249054,41824615,631446761,590633193,548811870,57957497,529806158,562147340,182967568,61950307,397443959,220643692,236459253,41053957,533024240,206652748,421767795,282581949,689281236,765719080,143337473,874405567,440616102,563192864,960412569,457171179,228312121,146929495,32724496,260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std::unordered_map<int, int> dp;
int dfs(int x) {
if ((x & -x) == x) return 2;
if (dp.count(x)) return dp[x];
return dp[x] = (dfs(x - (x & -x)) + dfs(x + (x & -x)) + 2) * inv % mod;
}
signed main() {
int l, r, ans = 0;
scanf("%lld%lld", &l, &r);
int i = (l - 1) / 1000000 + 1, j = (r - 1) / 1000000 + 1;
if (j - i <= 1) {
for (int i = l; i <= r; ++ i) ans = (ans + dfs(i)) % mod;
} else {
ans = (sum[j - 1] - sum[i]) % mod;
for (int k = l; k <= i * 1000000; ++ k) ans = (ans + dfs(k)) % mod;
for (int k = (j - 1) * 1000000 + 1; k <= r; ++ k) ans = (ans + dfs(k)) % mod;
}
printf("%lld", (ans + mod) % mod);
return 0;
}
