A - Count Takahashi

Problem Statement

You are given $N$ strings.
The $i$-th string $S_i$ $(1\le i\le N)$ is either Takahashi or Aoki.
How many $i$ are there such that $S_i$ is equal to Takahashi?

Constraints

$1\le N\le 100$
$N$ is an integer.
Each $S_i$ is Takahashi or Aoki. $(1\le i\le N)$

Input

The input is given from Standard Input in the following format:

$N$
$S_1$
$S_2$
$⋮$
$S_N$

Output

Print the count of $i$ such that $S_i$ is equal to Takahashi as an integer in a single line.

Sample Input 1

3
Aoki
Takahashi
Takahashi

Sample Output 1

2

$S_2$ and $S_3$ are equal to Takahashi, while $S_1$ is not.
Therefore, print 2.

Sample Input 2

2
Aoki
Aoki

Sample Output 2

0

It is possible that no $S_i$ is equal to Takahashi.

Sample Input 3

20
Aoki
Takahashi
Takahashi
Aoki
Aoki
Aoki
Aoki
Takahashi
Aoki
Aoki
Aoki
Takahashi
Takahashi
Aoki
Takahashi
Aoki
Aoki
Aoki
Aoki
Takahashi

Sample Output 3

7

Solution

具体见文末视频。

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Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int n;cin >> n;int res = 0;for (int i = 1; i <= n; i ++) {string s;cin >> s;if (s == "Takahashi") res ++;}cout << res << endl;return 0;
}

B - Couples

Problem Statement

There are $2N$ people standing in a row, and the person at the $i$-th position from the left is wearing clothes of color $A_i$. Here, the clothes have $N$ colors from $1$ to $N$, and exactly two people are wearing clothes of each color.
Find how many of the integers $i=1,2,\dots ,N$ satisfy the following condition:
There is exactly one person between the two people wearing clothes of color $i$.

Constraints

$2\le N\le 100$
$1\le A_i\le N$
Each integer from $1$ through $N$ appears exactly twice in $A$.
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$
$A_1$ $A_2$ $\dots$ $A_{2N}$

Output

Print the answer.

Sample Input 1

3
1 2 1 3 2 3

Sample Output 1

2

There are two values of $i$ that satisfy the condition: $1$ and $3$.
In fact, the people wearing clothes of color $1$ are at the 1st and 3rd positions from the left, with exactly one person in between.

Sample Input 2

2
1 1 2 2

Sample Output 2

0

There may be no $i$ that satisfies the condition.

Sample Input 3

4
4 3 2 3 2 1 4 1

Sample Output 3

3

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int n;cin >> n;std::vector<int> a(2 * n);for (int i = 0; i < 2 * n; i ++)cin >> a[i];int res = 0;for (int i = 1; i < 2 * n - 1; i ++)if (a[i - 1] == a[i + 1])res ++;cout << res << endl;return 0;
}

C - Tile Distance 2

Problem Statement

The coordinate plane is covered with $2\times 1$ tiles. The tiles are laid out according to the following rules:
For an integer pair $(i,j)$, the square $A_{i,j}=\{(x,y)\mid i\le x\le i+1\wedge j\le y\le j+1\}$ is contained in one tile.
When $i+j$ is even, $A_{i,j}$ and $A_{i+1,j}$ are contained in the same tile.
Tiles include their boundaries, and no two different tiles share a positive area.
Near the origin, the tiles are laid out as follows:

Takahashi starts at the point $(S_x+0.5,S_y+0.5)$ on the coordinate plane.
He can repeat the following move as many times as he likes:
Choose a direction (up, down, left, or right) and a positive integer $n$. Move $n$ units in that direction.
Each time he enters a tile, he pays a toll of $1$.
Find the minimum toll he must pay to reach the point $(T_x+0.5,T_y+0.5)$.

Constraints

$0\le S_x\le 2\times 10^{16}$
$0\le S_y\le 2\times 10^{16}$
$0\le T_x\le 2\times 10^{16}$
$0\le T_y\le 2\times 10^{16}$
All input values are integers.

Input

The input is given from Standard Input in the following format:

$S_x$ $S_y$
$T_x$ $T_y$

Output

Print the minimum toll Takahashi must pay.

Sample Input 1

5 0
2 5

Sample Output 1

5

For example, Takahashi can pay a toll of $5$ by moving as follows:

Move left by $1$. Pay a toll of $0$.
Move up by $1$. Pay a toll of $1$.
Move left by $1$. Pay a toll of $0$.
Move up by $3$. Pay a toll of $3$.
Move left by $1$. Pay a toll of $0$.
Move up by $1$. Pay a toll of $1$.
It is impossible to reduce the toll to $4$ or less, so print 5.

Sample Input 2

3 1
4 1

Sample Output 2

0

There are cases where no toll needs to be paid.

Sample Input 3

2552608206527595 5411232866732612
771856005518028 7206210729152763

Sample Output 3

1794977862420151

Note that the value to be output may exceed the range of a $32$-bit integer.

Solution

具体见文末视频。

---

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int sx, sy, fx, fy;cin >> sx >> sy >> fx >> fy;int res = abs(sy - fy);if ((sx + sy & 1) && fx < sx || ((sx + sy) % 2 == 0) && fx > sx) {res += max(0ll, abs(sx - fx) - abs(sy - fy) >> 1);} else {res += max(0ll, abs(sx - fx) - abs(sy - fy) + 1 >> 1);}cout << res << endl;return 0;
}

D - Avoid K Palindrome

Problem Statement

You are given a string $S$ of length $N$ consisting of characters A, B, and ?.
You are also given a positive integer $K$.
A string $T$ consisting of A and B is considered a good string if it satisfies the following condition:
No contiguous substring of length $K$ in $T$ is a palindrome.
Let $q$ be the number of ? characters in $S$.
There are $2^q$ strings that can be obtained by replacing each ? in $S$ with either A or B. Find how many of these strings are good strings.
The count can be very large, so find it modulo $998244353$.

Constraints

$2\le K\le N\le 1000$
$K\le 10$
$S$ is a string consisting of A, B, and ?.
The length of $S$ is $N$.
$N$ and $K$ are integers.

Input

The input is given from Standard Input in the following format:

$N$ $K$
$S$

Output

Print the answer.

Sample Input 1

7 4
AB?A?BA

Sample Output 1

1

The given string has two ?s.
There are four strings obtained by replacing each ? with A or B:
ABAAABA
ABAABBA
ABBAABA
ABBABBA
Among these, the last three contain the contiguous substring ABBA of length 4, which is a palindrome, and thus are not good strings.
Therefore, you should print 1.

Sample Input 2

40 7
????????????????????????????????????????

Sample Output 2

116295436

Ensure to find the number of good strings modulo $998244353$.

Sample Input 3

15 5
ABABA??????????

Sample Output 3

0

It is possible that there is no way to replace the ?s to obtain a good string.

Sample Input 4

40 8
?A?B??B?B?AA?A?B??B?A???B?BB?B???BA??BAA

Sample Output 4

259240

Solution

具体见文末视频。

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Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int N = 1e3 + 10, M = 11, mod = 998244353;int n, k;
string s;
int f[N][1 << M];bool check(int x) {string a, b;for (int i = 0; i < k; i ++)a += char((x >> i & 1) ^ 48);b = a;reverse(b.begin(), b.end());return a != b;
}signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cin >> n >> k >> s;s = ' ' + s;std::vector<int> avl;avl.push_back(0);for (int i = 1; i <= k; i ++) {std::vector<int> ok;while (avl.size()) {int tmp = avl.back();avl.pop_back();if (s[i] == '?') {ok.push_back(tmp), ok.push_back(tmp | (1ll << i - 1));} else if (s[i] == 'B') {ok.push_back(tmp | (1ll << i - 1));} else {ok.push_back(tmp);}}avl = ok;}for (auto v : avl)if (check(v))f[k][v] = 1;for (int i = k + 1; i <= n; i ++)for (int j = 0; j < 1ll << k; j ++)if (s[i] == '?') {if (check(j) && check(j >> 1)) f[i][j >> 1] = (f[i][j >> 1] + f[i - 1][j]) % mod;if (check(j) && check((j >> 1) | (1ll << k - 1))) f[i][(j >> 1) | (1ll << k - 1)] = (f[i][(j >> 1) | (1ll << k - 1)] + f[i - 1][j]) % mod;} else if (s[i] == 'B') {if (check(j) && check((j >> 1) | (1ll << k - 1))) f[i][(j >> 1) | (1ll << k - 1)] = (f[i][(j >> 1) | (1ll << k - 1)] + f[i - 1][j]) % mod;} else {if (check(j) && check(j >> 1)) f[i][j >> 1] = (f[i][j >> 1] + f[i - 1][j]) % mod;}int res = 0;for (int i = 0; i < 1ll << k; i ++)res = (res + f[n][i]) % mod;cout << res << endl;return 0;
}

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E - Water Tank

Problem Statement

You are given a sequence of positive integers of length $N$: $H=(H_1,H_2,\dots ,H_N)$.
There is a sequence of non-negative integers of length $N+1$: $A=(A_0,A_1,\dots ,A_N)$. Initially, $A_0=A_1=\cdots =A_N=0$.
Perform the following operations repeatedly on $A$:

1. Increase the value of $A _ 0$ by $1$. For $i=1,2,\ldots,N$ in this order, perform the following operation: If $A _ {i-1}\gt A _ i$ and $A _ {i-1}\gt H _ i$, decrease the value of $A _ {i-1}$ by 1 and increase the value of $A _ i$ by $1$.

For each $i=1,2,\ldots,N$, find the number of operations before $A _ i>0$ holds for the first time. ## Constraints

$1\le N\le 2\times 10^5$
$1\le H_i\le 10^9(1\le i\le N)$
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$
$H_1$ $H_2$ $\dots$ $H_N$

Output

Print the answers for $i=1,2,\dots ,N$ in a single line, separated by spaces.

Sample Input 1

5
3 1 4 1 5

Sample Output 1

4 5 13 14 26

The first five operations go as follows.
Here, each row corresponds to one operation, with the leftmost column representing step 1 and the others representing step 2.

From this diagram, $A_1>0$ holds for the first time after the 4th operation, and $A_2>0$ holds for the first time after the 5th operation.
Similarly, the answers for $A_3,A_4,A_5$ are $13,14,26$, respectively.
Therefore, you should print 4 5 13 14 26.

Sample Input 2

6
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 2

1000000001 2000000001 3000000001 4000000001 5000000001 6000000001

Note that the values to be output may not fit within a $32$-bit integer.

Sample Input 3

15
748 169 586 329 972 529 432 519 408 587 138 249 656 114 632

Sample Output 3

749 918 1921 2250 4861 5390 5822 6428 6836 7796 7934 8294 10109 10223 11373

Solution

具体见文末视频。

---

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int N = 2e5 + 10;int n;
int a[N], to[N], f[N];signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cin >> n;stack<int> stk;for (int i = 1; i <= n; i ++) {cin >> a[i];while (stk.size() && a[stk.top()] < a[i]) stk.pop();if (stk.size()) to[i] = stk.top();stk.push(i);}for (int i = 1; i <= n; i ++)f[i] = f[to[i]] + (i - to[i]) * a[i], cout << f[i] + 1 << " ";return 0;
}

F - Tree Degree Optimization

Problem Statement

You are given a sequence of integers $A=(A_1,\dots ,A_N)$. For a tree $T$ with $N$ vertices, define $f(T)$ as follows:
Let $d_i$ be the degree of vertex $i$ in $T$. Then, $f(T)={\sum }_{i=1}^N{d_i}^2{A}_i$.
Find the minimum possible value of $f(T)$.
The constraints guarantee the answer to be less than $2^{63}$.

Constraints

$2\le N\le 2\times 10^5$
$1\le A_i\le 10^9$
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$
$A_1$ $A_2$ $\dots$ $A_N$

Output

Print the answer.

Sample Input 1

4
3 2 5 2

Sample Output 1

24

Consider a tree $T$ with an edge connecting vertices $1$ and $2$, an edge connecting vertices $2$ and $4$, and an edge connecting vertices $4$ and $3$.
Then, $f(T)=1^2\times 3+2^2\times 2+1^2\times 5+2^2\times 2=24$. It can be proven that this is the minimum value of $f(T)$.

Sample Input 2

3
4 3 2

Sample Output 2

15

Sample Input 3

7
10 5 10 2 10 13 15

Sample Output 3

128

Solution

具体见文末视频。

---

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int N = 2e5 + 10;int n;
int a[N], d[N];signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cin >> n;for (int i = 1; i <= n; i ++)cin >> a[i];priority_queue<PII, vector<PII>, greater<PII>> heap;for (int i = 1; i <= n; i ++)d[i] = 1, heap.push({a[i] * (2 * d[i] + 1), i});for (int i = 1; i <= n - 2; i ++) {auto tmp = heap.top(); heap.pop();d[tmp.se] ++, heap.push({a[tmp.se] * (2 * d[tmp.se] + 1), tmp.se});}int res = 0;for (int i = 1; i <= n; i ++) res += d[i] * d[i] * a[i];cout << res << endl;return 0;
}

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G - Sum of Tree Distance

Problem Statement

You are given a tree with $N$ vertices. The $i$-th edge connects vertices $u_i$ and $v_i$ bidirectionally.
Additionally, you are given an integer sequence $A=(A_1,\dots ,A_N)$.
Here, define $f(i,j)$ as follows:
If $A_i=A_j$, then $f(i,j)$ is the minimum number of edges you need to traverse to move from vertex $i$ to vertex $j$. If $A_i\ne A_j$, then $f(i,j)=0$.
Calculate the value of the following expression:

$2\le N\le 2\times 10^5$
$1\le u_i,v_i\le N$
$1\le A_i\le N$
The input graph is a tree.
All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$
$u_1$ $v_1$
$⋮$
$u_{N-1}$ $v_{N-1}$
$A_1$ $A_2$ $\dots$ $A_N$

Output

Print the answer.

Sample Input 1

4
3 4
4 2
1 2
2 1 1 2

Sample Output 1

4

$f(1,4)=2,f(2,3)=2$. For all other KaTeX parse error: Expected 'EOF', got '&' at position 17: …, j\ (1 \leq i &̲lt; j \leq N), we have $f(i,j)=0$, so the answer is $2+2=4$.

Sample Input 2

8
8 6
3 8
1 4
7 8
4 5
3 4
8 2
1 2 2 2 3 1 1 3

Sample Output 2

19

Solution

具体见文末视频。

---

Code

#include <bits/stdc++.h>
#define fi first
#define se second
#define int long longusing namespace std;typedef pair<int, int> PII;
typedef long long LL;const int N = 2e5 + 10, B = 448;int n;
int a[N], dep[N], idx[N], tot[N], cnt[N], acl[N][B + 10], id[N];
int f[N << 1][20], Euler[N << 1], pl[N], m, sub;
std::vector<int> g[N], pos[N], big;void dfs(int u, int fa) {Euler[ ++ m] = u, pl[u] = m;for (auto v : g[u]) {if (v == fa) continue;dep[v] = dep[u] + 1;dfs(v, u), Euler[ ++ m] = u;}
}
void dfs2(int u, int fa) {if (id[a[u]]) acl[u][id[a[u]]] ++;for (auto v : g[u]) {if (v == fa) continue;dfs2(v, u);for (auto i : big) acl[u][i] += acl[v][i];}for (auto i : big) {int sum = 0, res = 0;for (auto v : g[u]) {if (v == fa) continue;res += acl[v][i] * sum, sum += acl[v][i];}sub += res * dep[u] * 2;}if (id[a[u]]) sub += (acl[u][id[a[u]]] - 1) * dep[u] * 2;
}
void build() {for (int j = 0; j <= log2(m); j ++)for (int i = 1; i + (1ll << j) - 1 <= m; i ++)if (!j) f[i][j] = i;else {if (dep[Euler[f[i][j - 1]]] < dep[Euler[f[i + (1ll << j - 1)][j - 1]]]) f[i][j] = f[i][j - 1];else f[i][j] = f[i + (1ll << j - 1)][j - 1];}
}
int query(int l, int r) {int k = log2(r - l + 1);if (dep[Euler[f[l][k]]] < dep[Euler[f[r - (1ll << k) + 1][k]]]) return f[l][k];return f[r - (1ll << k) + 1][k];
}
int lca(int a, int b) {if (pl[a] > pl[b]) swap(a, b);return Euler[query(pl[a], pl[b])];
}signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cin >> n;for (int i = 1; i < n; i ++) {int u, v;cin >> u >> v;g[u].push_back(v), g[v].push_back(u);}for (int i = 1; i <= n; i ++)cin >> a[i], pos[a[i]].push_back(i);dfs(1, -1), build();int res1 = 0, res2 = 0, tmp = 0;for (int i = 1; i <= n; i ++) {if (pos[a[i]].size() <= B) {for (int j = 0; j < idx[a[i]]; j ++)res2 += dep[lca(i, pos[a[i]][j])] * 2;idx[a[i]] ++;} else if (!id[a[i]]) id[a[i]] = ++ tmp, big.push_back(tmp);res1 += tot[a[i]] + cnt[a[i]] * dep[i];tot[a[i]] += dep[i], cnt[a[i]] ++;}dfs2(1, -1);cout << res1 - res2 - sub << endl;return 0;
}

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视频题解

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最后祝大家早日