AtCoder Beginner Contest 373（ABCDEF 题）视频讲解

A - September

Problem Statement

There are $12$ strings $S_1,S_2,\dots ,S_{12}$ consisting of lowercase English letters.
Find how many integers $i$ $(1\le i\le 12)$ satisfy that the length of $S_i$ is $i$.

Constraints

Each $S_i$ is a string of length between $1$ and $100$, inclusive, consisting of lowercase English letters. $(1\le i\le 12)$

Input

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

$S_1$
$S_2$
$⋮$
$S_{12}$

Output

Print the number of integers $i$ $(1\le i\le 12)$ such that the length of $S_i$ is $i$.

Sample Input 1

january
february
march
april
may
june
july
august
september
october
november
december

Sample Output 1

1

There is only one integer $i$ such that the length of $S_i$ is $i$: $9$. Thus, print 1.

Sample Input 2

ve
inrtfa
npccxva
djiq
lmbkktngaovl
mlfiv
fmbvcmuxuwggfq
qgmtwxmb
jii
ts
bfxrvs
eqvy

Sample Output 2

2

There are two integers $i$ such that the length of $S_i$ is $i$: $4$ and $8$. Thus, print 2.

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

	int res = 0;
	for (int i = 1; i <= 12; i ++) {
		string s;
		cin >> s;
		if (s.size() == i) res ++;
	}
	
	cout << res << endl;

	return 0;
}

B - 1D Keyboard

Problem Statement

There is a keyboard with $26$ keys arranged on a number line.
The arrangement of this keyboard is represented by a string $S$, which is a permutation of ABCDEFGHIJKLMNOPQRSTUVWXYZ.
The key corresponding to the character $S_x$ is located at coordinate $x$ $(1\le x\le 26)$. Here, $S_x$ denotes the $x$-th character of $S$.
You will use this keyboard to input ABCDEFGHIJKLMNOPQRSTUVWXYZ in this order, typing each letter exactly once with your right index finger.
To input a character, you need to move your finger to the coordinate of the key corresponding to that character and press the key.
Initially, your finger is at the coordinate of the key corresponding to A. Find the minimal possible total traveled distance of your finger from pressing the key for A to pressing the key for Z. Here, pressing a key does not contribute to the distance.

Constraints

$S$ is a permutation of ABCDEFGHIJKLMNOPQRSTUVWXYZ.

Input

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

$S$

Output

Print the answer.

Sample Input 1

ABCDEFGHIJKLMNOPQRSTUVWXYZ

Sample Output 1

25

From pressing the key for A to pressing the key for Z, you need to move your finger $1$ unit at a time in the positive direction, resulting in a total traveled distance of $25$. It is impossible to press all keys with a total traveled distance less than $25$, so print 25.

Sample Input 2

MGJYIZDKSBHPVENFLQURTCWOAX

Sample Output 2

223

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

	int res = 0;
	unordered_map<char, int> idx;
	for (int i = 1; i <= 26; i ++) {
		char x;
		cin >> x, idx[x] = i;
	}
	for (char i = 'A' + 1; i <= 'Z'; i ++)
		res += abs(idx[i] - idx[i - 1]);
	
	cout << res << endl;

	return 0;
}

C - Max Ai+Bj

Problem Statement

You are given two integer sequences $A$ and $B$, each of length $N$. Choose integers $i,j$ $(1\le i,j\le N)$ to maximize the value of $A_i+B_j$.

Constraints

$1\le N\le 5\times 10^5$
$|A_i|\le 10^9$ $(i=1,2,\dots ,N)$
$|B_j|\le 10^9$ $(j=1,2,\dots ,N)$
All input values are integers.

Input

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

$N$
$A_1$ $A_2$ $\dots$ $A_N$
$B_1$ $B_2$ $\dots$ $B_N$

Output

Print the maximum possible value of $A_i+B_j$.

Sample Input 1

2
-1 5
3 -7

Sample Output 1

8

For $(i,j)=(1,1),(1,2),(2,1),(2,2)$, the values of $A_i+B_j$ are $2,-8,8,-2$ respectively, and $(i,j)=(2,1)$ achieves the maximum value $8$.

Sample Input 2

6
15 12 3 -13 -1 -19
7 17 -13 -10 18 4

Sample Output 2

33

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

	int n;
	cin >> n;
	
	int A = -1e18, B = -1e18;
	for (int i = 1, x; i <= n; i ++) cin >> x, A = max(A, x);
	for (int i = 1, x; i <= n; i ++) cin >> x, B = max(B, x);
	
	cout << A + B << endl;

	return 0;
}

D - Hidden Weights

Problem Statement

You are given a directed graph with $N$ vertices and $M$ edges. The $j$-th directed edge goes from vertex $u_j$ to vertex $v_j$ and has a weight of $w_j$.
Find one way to write an integer between $-10^{18}$ and $10^{18}$, inclusive, to each vertex such that the following condition is satisfied.
Let $x_i$ be the value written on vertex $i$. For all edges $j=1,2,\dots ,M$, it holds that $x_{v_j}-x_{u_j}=w_j$.
It is guaranteed that at least one such assignment exists for the given input.

Constraints

$2\le N\le 2\times 10^5$
1 ≤ M ≤ min ⁡ { 2 × 1 0 5 , N ( N − 1 ) / 2 } 1 \leq M \leq \min\{2 \times 10^5,N(N-1)/2\} 1≤M≤min{2×105,N(N−1)/2}
$1\le u_j,v_j\le N$
$u_j\ne v_j$
If $i\ne j$, then $(u_i,v_i)\ne (u_j,v_j)$ and $(u_i,v_i)\ne (v_j,u_j)$
$|w_j|\le 10^9$
All input values are integers.
There exists at least one assignment satisfying the conditions.

Input

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

$N$ $M$
$u_1$ $v_1$ $w_1$
$u_2$ $v_2$ $w_2$
$⋮$
$u_M$ $v_M$ $w_M$

Output

Let $x_i$ be the integer written on vertex $i$. Print $x_1,x_2,\dots ,x_N$ in this order, separated by spaces, on a single line. If there are multiple solutions, you may print any of them.

Sample Input 1

3 3
1 2 2
3 2 3
1 3 -1

Sample Output 1

3 5 2

By setting $x=(3,5,2)$, we have $x_2-x_1=w_1=2$, $x_2-x_3=w_2=3$, $x_3-x_1=w_3=-1$, satisfying the conditions.
For example, $x=(-1,1,-2)$ is also a valid answer.

Sample Input 2

4 2
2 1 5
3 4 -3

Sample Output 2

5 0 6 3

For example, $x=(0,-5,4,1)$ and $x=(5,0,4,1)$ are also valid answers.

Sample Input 3

5 7
2 1 18169343
3 1 307110901
4 1 130955934
2 3 -288941558
2 5 96267410
5 3 -385208968
4 3 -176154967

Sample Output 3

200401298 182231955 -106709603 69445364 278499365

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

const int N = 2e5 + 10, inf = 0x3f3f3f3f3f3f3f3f;

int n, m;
int h[N], e[N << 1], ne[N << 1], w[N << 1], idx;
int res[N];

void add(int a, int b, int c) {
	e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}
void dfs(int u) {
	for (int i = h[u]; ~i; i = ne[i]) {
		int v = e[i];
		if (res[v] != inf) continue;
		res[v] = res[u] + w[i], dfs(v);
	}
}

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

	memset(h, -1, sizeof h);
	
	cin >> n >> m;
	while (m -- ) {
		int u, v, c;
		cin >> u >> v >> c;
		add(u, v, c), add(v, u, -c);
	}
	
	memset(res, 0x3f, sizeof res);
	for (int i = 1; i <= n; i ++)
		if (res[i] == inf) {
			res[i] = 0;
			dfs(i);
		}
		
	for (int i = 1; i <= n; i ++)
		cout << res[i] << " ";
	cout << endl;

	return 0;
}

E - How to Win the Election

Problem Statement

An election is being held with $N$ candidates numbered $1,2,\dots ,N$. There are $K$ votes, some of which have been counted so far.
Up until now, candidate $i$ has received $A_i$ votes.
After all ballots are counted, candidate $i$ $(1\le i\le N)$ will be elected if and only if the number of candidates who have received more votes than them is less than $M$. There may be multiple candidates elected.
For each candidate, find the minimum number of additional votes they need from the remaining ballots to guarantee their victory regardless of how the other candidates receive votes.
Formally, solve the following problem for each $i=1,2,\dots ,N$.
Determine if there is a non-negative integer $X$ not exceeding $K-\sum _{i=1}^N{A}_i$ satisfying the following condition. If it exists, find the minimum possible such integer.
If candidate $i$ receives $X$ additional votes, then candidate $i$ will always be elected.

Constraints

$1\le M\le N\le 2\times 10^5$
$1\le K\le 10^{12}$
$0\le A_i\le 10^{12}$
$\sum _{i=1}^N{A}_i\le K$
All input values are integers.

Input

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

$N$ $M$ $K$
$A_1$ $A_2$ $\dots$ $A_N$

Output

Let $C_i$ be the minimum number of additional votes candidate $i$ needs from the remaining ballots to guarantee their victory regardless of how other candidates receive votes. Print $C_1,C_2,\dots ,C_N$ separated by spaces.
If candidate $i$ has already secured their victory, then let $C_i=0$. If candidate $i$ cannot secure their victory under any circumstances, then let $C_i=-1$.

Sample Input 1

5 2 16
3 1 4 1 5

Sample Output 1

2 -1 1 -1 0

$14$ votes have been counted so far, and $2$ votes are left.

The $C$ to output is $(2,-1,1,-1,0)$. For example:
Candidate $1$ can secure their victory by obtaining $2$ more votes, while not by obtaining $1$ more vote. Thus, $C_1=2$.
Candidate $2$ can never (even if they obtain $2$ more votes) secure their victory, so $C_2=-1$.

Sample Input 2

12 1 570
81 62 17 5 5 86 15 7 79 26 6 28

Sample Output 2

79 89 111 117 117 74 112 116 80 107 117 106

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

const int N = 2e5 + 10;

int n, m, k;
int a[N], b[N], c[N];

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

	cin >> n >> m >> k;
	int sum = 0;
	for (int i = 1; i <= n; i ++)
		cin >> a[i], b[i] = a[i], sum += a[i];
	if (n == 1) {
		cout << 0 << endl;
		return 0;
	}
	if (n == m) {
		for (int i = 1; i <= n; i ++)
			cout << 0 << " ";
		cout << endl;
		return 0;
	}
	sort(b + 1, b + 1 + n);
	for (int i = 1; i <= n; i ++)
		c[i] = c[i - 1] + b[i];
	
	for (int i = 1; i <= n; i ++) {
		auto check = [&](int x) -> bool {
			int now = a[i] + x, rest = k - x - sum, pos = upper_bound(b + 1, b + 1 + n, now) - b - 1;
			if (a[i] < b[n - m + 1]) {
				if (pos < n - m + 1) return 0;
				return (pos - n + m) * (now + 1) - c[pos] + c[n - m] > rest;
			} else {
				if (pos < n - m) return 0;
				// if (a[i] > b[pos]) return (pos - n + m + 1) * (now + 1) - c[pos] + c[max(0ll, n - m - 1)] > rest;
				return (pos - n + m) * (now + 1) - c[pos] + c[max(0ll, n - m - 1)] + a[i] > rest;
			}
		};
		int lo = 0, ro = k - sum, res = -1;
		while (lo <= ro) {
			int mid = lo + ro >> 1;
			if (check(mid)) ro = mid - 1, res = mid;
			else lo = mid + 1;
		}
		cout << res << " ";
	}
	cout << endl;

	return 0;
}

F - Knapsack with Diminishing Values

Problem Statement

There are $N$ types of items. The $i$-th type of item has a weight of $w_i$ and a value of $v_i$. Each type has $10^{10}$ items available.
Takahashi is going to choose some items and put them into a bag with capacity $W$. He wants to maximize the value of the selected items while avoiding choosing too many items of the same type. Hence, he defines the happiness of choosing $k_i$ items of type $i$ as $k_i{v}_i-k_i^2$. He wants to choose items to maximize the total happiness over all types while keeping the total weight at most $W$. Calculate the maximum total happiness he can achieve.

Constraints

$1\le N\le 3000$
$1\le W\le 3000$
$1\le w_i\le W$
$1\le v_i\le 10^9$
All input values are integers.

Input

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

$N$ $W$
$w_1$ $v_1$
$w_2$ $v_2$
$⋮$
$w_N$ $v_N$

Output

Print the answer.

Sample Input 1

2 10
3 4
3 2

Sample Output 1

5

By choosing $2$ items of type $1$ and $1$ item of type $2$, the total happiness can be $5$, which is optimal.
Here, the happiness for type $1$ is $2\times 4-2^2=4$, and the happiness for type $2$ is $1\times 2-1^2=1$.
The total weight is $9$, which is within the capacity $10$.

Sample Input 2

3 6
1 4
2 3
2 7

Sample Output 2

14

Sample Input 3

1 10
1 7

Sample Output 3

12

Solution

具体见文末视频。

Code

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

const int N = 3e3 + 10;

int n, m;
priority_queue<int> heap[N];
int dp[N];

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

	cin >> n >> m;
	for (int i = 1; i <= n; i ++) {
		int w, v;
		cin >> w >> v, heap[w].push(v - 1);
	}
	
	for (int i = 1; i <= m; i ++)
		for (int j = 1; j <= m / i && heap[i].size(); j ++) {
			int v = heap[i].top();
			heap[i].pop();
			for (int k = m; k >= i; k --) dp[k] = max(dp[k], dp[k - i] + v);
			heap[i].push(v - 2);
		}
	
	cout << *max_element(dp + 1, dp + 1 + m) << endl;

	return 0;
}

视频题解

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