AtCoder Beginner Contest 070 (A B C D)

A - Palindromic Number

Problem Statement

You are given a three-digit positive integer N.
Determine whether N is a palindromic number.
Here, a palindromic number is an integer that reads the same backward as forward in decimal notation.

Constraints

• 100≤N≤999
• N is an integer.

---

Input

Input is given from Standard Input in the following format:

N

Output

If N is a palindromic number, print Yes; otherwise, print No.

---

Sample Input 1

Copy

575

Sample Output 1

Copy

Yes

N=575 is also 575 when read backward, so it is a palindromic number. You should print Yes.

 

求n是否回文，水题。

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    if(n/100 == n%10) printf("Yes\n");
    else printf("No\n");
    return 0;
}

 

B - Two Switches

Problem Statement

Alice and Bob are controlling a robot. They each have one switch that controls the robot.
Alice started holding down her button A second after the start-up of the robot, and released her button B second after the start-up.
Bob started holding down his button C second after the start-up, and released his button D second after the start-up.
For how many seconds both Alice and Bob were holding down their buttons?

Constraints

• 0≤A<B≤100
• 0≤C<D≤100
• All input values are integers.

---

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the length of the duration (in seconds) in which both Alice and Bob were holding down their buttons.

---

Sample Input 1

Copy

0 75 25 100

Sample Output 1

Copy

50

Alice started holding down her button 0 second after the start-up of the robot, and released her button 75 second after the start-up.
Bob started holding down his button 25 second after the start-up, and released his button 100 second after the start-up.
Therefore, the time when both of them were holding down their buttons, is the 50 seconds from 25 seconds after the start-up to 75 seconds after the start-up.

求ab与cd的相同区间，没有就是0.

#include <bits/stdc++.h>
using namespace std;

int main() {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    printf("%d\n",max(0,min(b,d)-max(a,c)));
    return 0;
}

 

C - Multiple Clocks

Problem Statement

We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.
Initially, the hand of every clock stands still, pointing directly upward.
Now, Dolphin starts all the clocks simultaneously.
In how many seconds will the hand of every clock point directly upward again?

Constraints

• 1≤N≤100
• 1≤T_i≤10¹⁸
• All input values are integers.
• The correct answer is at most 10¹⁸ seconds.

---

Input

Input is given from Standard Input in the following format:

N
T1
:  
TN

Output

Print the number of seconds after which the hand of every clock point directly upward again.

---

Sample Input 1

Copy

2
2
3

Sample Output 1

Copy

6

We have two clocks. The time when the hand of each clock points upward is as follows:

• Clock 1: 2, 4, 6, … seconds after the beginning
• Clock 2: 3, 6, 9, … seconds after the beginning

Therefore, it takes 6 seconds until the hands of both clocks point directly upward.

求n个数的最小公倍数。

#include <bits/stdc++.h>
#define ll long long
using namespace std;

ll gcd(ll x, ll y) {
    return y ? gcd(y, x%y): x;
}
ll low(ll x, ll y) {
    return y/gcd(x, y) * x;
}
int main() {
    int n;
    cin >> n;
    ll x = 1, y;
    while(n--) {
        cin >> y;
        x = low(x, y);
    }
    cout << x << endl;
    return 0;
}

D - Transit Tree Path

Problem Statement

You are given a tree with N vertices.
Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N−1 edges, where N is the number of its vertices.
The i-th edge (1≤i≤N−1) connects Vertices a_i and b_i, and has a length of c_i.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

• find the length of the shortest path from Vertex x_j and Vertex y_j via Vertex K.

Constraints

• 3≤N≤10⁵
• 1≤a_i,b_i≤N(1≤i≤N−1)
• 1≤c_i≤10⁹(1≤i≤N−1)
• The given graph is a tree.
• 1≤Q≤10⁵
• 1≤K≤N
• 1≤x_j,y_j≤N(1≤j≤Q)
• x_j≠y_j(1≤j≤Q)
• x_j≠K,y_j≠K(1≤j≤Q)

---

Input

Input is given from Standard Input in the following format:

N  
a1 b1 c1  
:  
aN−1 bN−1 cN−1
Q K
x1 y1
:  
xQ yQ

Output

Print the responses to the queries in Q lines.
In the j-th line j(1≤j≤Q), print the response to the j-th query.

---

Sample Input 1

Copy

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5

Sample Output 1

Copy

3
2
4

The shortest paths for the three queries are as follows:

• Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3
• Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2
• Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4

求x到k加y到k的最小距离，最短路问题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
const ll INF = 1LL<<60;
struct Nod {
    ll to, cost;
};
vector<Nod> G[N];
ll dist[N];
typedef pair<ll, ll> P;
void dij(ll x) {
    priority_queue<P, vector<P>, greater<P> > que;
    for(int i = 0; i < N; i ++) dist[i] = INF;
    dist[x] = 0;
    que.push(P(0, x));
    while(!que.empty()) {
        P p = que.top();
        que.pop();
        ll v = p.second;
        if(dist[v] < p.first) continue;
        for(ll i = 0; i < G[v].size(); i ++) {
            Nod e = G[v][i];
            if(dist[e.to] > dist[v] + e.cost) {
                dist[e.to] = dist[v] + e.cost;
                que.push(P(dist[e.to], e.to));
            }
        }
    }
}
int main() {
    int n, q, cnt = 0;
    scanf("%d", &n);
    for(int i = 0; i < n-1; i ++) {
        ll v, u, w;
        scanf("%lld %lld %lld", &v, &u, &w);
        G[v].push_back((Nod){u,w});
        G[u].push_back((Nod){v,w});
    }
     ll k;
     scanf("%d %lld", &q, &k);
     dij(k);
     while(q--) {
         ll x, y;
         scanf("%lld %lld", &x, &y);
         printf("%lld\n",dist[x] + dist[y]);
     }
    return 0;
}