题目链接:http://abc070.contest.atcoder.jp/assignments

A - Palindromic Number


Time limit : 2sec / Memory limit : 256MB

Score : 100 points

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.


Sample Input 2

Copy
123

Sample Output 2

Copy
No

N=123 becomes 321 when read backward, so it is not a palindromic number. You should print No.


Sample Input 3

Copy
812

Sample Output 3

Copy
No

题解:水水
 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;
#define ll long long
const int N=;
const int mod=1e9+;
const int maxn=1e7;
int main()
{
int n;
while(cin>>n){
int a=n/;
int b=(b-a*)/;
int c=n%;
if(a==c) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return ;
}

B - Two Switches


Time limit : 2sec / Memory limit : 256MB

Score : 200 points

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.


Sample Input 2

Copy
0 33 66 99

Sample Output 2

Copy
0

Alice and Bob were not holding their buttons at the same time, so the answer is zero seconds.


Sample Input 3

Copy
10 90 20 80

Sample Output 3

Copy
60

题解:分情况讨论即可
 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;
#define ll long long
const int N=;
const int mod=1e9+;
const int maxn=1e7;
int main()
{
int a,b,c,d;
while(cin>>a>>b>>c>>d){
if(c>b||a>d) cout<<<<endl;
else if(a<=c&&d<=b){
cout<<d-c<<endl;
}
else if(c<=a&&b<=d){
cout<<b-a<<endl;
}
else if(a<=c&&c<=b&&b<=d){
cout<<b-c<<endl;
}
else if(c<=a&&a<=d&&d<=b){
cout<<d-a<<endl;
}
}
return ;
}

C - Multiple Clocks


Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

We have N clocks. The hand of the i-th clock (1≤iN) rotates through 360° in exactly Ti 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≤Ti≤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 1246 seconds after the beginning
  • Clock 2369 seconds after the beginning

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


Sample Input 2

Copy
5
2
5
10
1000000000000000000
1000000000000000000

Sample Output 2

Copy
1000000000000000000

题解:就是求最小公倍数 (如果只有一个数直接输出)
 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;
#define ll long long
const int N=;
const int mod=1e9+;
const int maxn=1e7;
ll a[];
ll gcd(ll x,ll y)
{
ll c;
ll m=x,n=y;
while(y!=){
c=x%y;
x=y;
y=c;
}
return m/x*n;
}
int main()
{
int n;
cin>>n;
ll m,s,l;
cin>>m;
if(n==) cout<<m<<endl;
else if(n>){
cin>>l;
s=gcd(l,m);
for(int i=;i<n-;i++){
cin>>m;
s=gcd(s,m);
}
cout<<s<<endl;
}
return ;
}

D - Transit Tree Path


Time limit : 2sec / Memory limit : 256MB

Score : 400 points

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≤iN−1) connects Vertices ai and bi, and has a length of ci.

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

  • find the length of the shortest path from Vertex xj and Vertex yj via Vertex K.

Constraints

  • 3≤N≤105
  • 1≤ai,biN(1≤iN−1)
  • 1≤ci≤109(1≤iN−1)
  • The given graph is a tree.
  • 1≤Q≤105
  • 1≤KN
  • 1≤xj,yjN(1≤jQ)
  • xjyj(1≤jQ)
  • xjK,yjK(1≤jQ)

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≤jQ), 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

Sample Input 2

Copy
7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

Copy
5
14
22

The path for each query must pass Vertex K=2.


Sample Input 3

Copy
10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

Copy
17000000000

题解:dfs
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int maxn = 1e5+;
const int N = ;
typedef long long ll;
int t,head[maxn];
ll dist[maxn],w;
bool vis[maxn];
struct Edge
{
int from,to,nxt;
ll cost;
}e[*maxn];
void addedge(int u,int v,int w)
{
e[t].from=u;
e[t].to=v;
e[t].cost=w;
e[t].nxt=head[u];
head[u]=t++;
}
void dfs(int u,int fa)
{
for(int i=head[u];i!=-;i=e[i].nxt){
int to=e[i].to;
if(to==fa) continue;
dist[to]=dist[u]+e[i].cost;
dfs(to,u);
}
}
int main()
{
int n;
while(cin>>n){
t=;
memset(head,-,sizeof(head));
memset(dist,,sizeof(dist));
int u,v;
for(int i=;i<n;i++){
cin>>u>>v>>w;
addedge(u,v,w);
addedge(v,u,w);
}
int q,k;
cin>>q>>k;
dfs(k,-);
int x,y;
for(int i=;i<q;i++){
cin>>x>>y;
cout<<dist[x]+dist[y]<<endl;
}
}
return ;
}

AtCoder Beginner Contest 070 ABCD题的更多相关文章

  1. AtCoder Beginner Contest 068 ABCD题

    A - ABCxxx Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement This contes ...

  2. AtCoder Beginner Contest 053 ABCD题

    A - ABC/ARC Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Smeke has ...

  3. AtCoder Beginner Contest 069 ABCD题

    题目链接:http://abc069.contest.atcoder.jp/assignments A - K-City Time limit : 2sec / Memory limit : 256M ...

  4. AtCoder Beginner Contest 057 ABCD题

    A - Remaining Time Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Dol ...

  5. AtCoder Beginner Contest 051 ABCD题

    A - Haiku Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement As a New Yea ...

  6. AtCoder Beginner Contest 052 ABCD题

    A - Two Rectangles Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement The ...

  7. AtCoder Beginner Contest 054 ABCD题

    A - One Card Poker Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Ali ...

  8. AtCoder Beginner Contest 058 ABCD题

    A - ι⊥l Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Three poles st ...

  9. AtCoder Beginner Contest 050 ABC题

    A - Addition and Subtraction Easy Time limit : 2sec / Memory limit : 256MB Score : 100 points Proble ...

随机推荐

  1. dedecms清空所有文章怎么操作?sql语句如何写?

    小C新建了一个站,确切的说是复制,出于seo考虑,决定清空所有文章,那么dedecms清空所有文章怎么操作?sql语句如何写呢?特别提醒:修改之前一定要先做好备份,以防万一!下面的语句在迫不得已的情况 ...

  2. MathWorks.MATLAB.NET.Arrays.MWArray”的类型初始值设定项引发异常 解决方法

    原因 用的是matlab7运行时,后面又安装了matlab11,后面又重新安装了matlab7运行时,c盘下就有二个运行时的版本了,程序引用了后面的那个,编译后就出上面的问题 解决方法 1重新引用上面 ...

  3. wx鼠标拖动事件

    #coding:UTF- import wx app = wx.App() def dragEVT(event): if event.ButtonDown(): panel1.SetPosition( ...

  4. [QGLViewer]鼠标取点后回调

    纠结的一天:QGLViewer控件重载鼠标事件(AxMapControl类),如何在点击鼠标之后执行一个回调,通知主界面Viewer类执行一个操作. 先是考虑直接使用C风格的回调函数,在AxMapCo ...

  5. 怎样使用EOS.JS的API

    https://medium.com/coinmonks/how-to-use-eosjs-api-1-770b037b22ad https://blog.csdn.net/mongo_node/ar ...

  6. centos7部署fabric-ca错误解决

    1.fabric-ca 编译错误:ltdl.h: no such file 在fabric-ca目录中使用make编译时,会出现如下错误: 解决方案: 如果在ubunt操作系统中,只需安装:apt i ...

  7. [LeetCode] 586. Customer Placing the Largest Number of Orders_Easy tag;SQL

    Query the customer_number from the orders table for the customer who has placed the largest number o ...

  8. MSSqlServer 发布/订阅配置(主从同步)

    背景: 1.单个独立数据库的吞吐量是有瓶颈的,那么如何解决这个瓶颈? 2.服务器直接数据如何复制.并具备一致性.可扩展性? 资源: 官方资源:https://technet.microsoft.com ...

  9. Reveal使用

    本人手机是7.1的,reveal2.0以后不支持8.0以下,没办法,这里只能使用reveal1.6. 这里提醒一下,reveal2.0以后libReavel.lib改名了,. https://reve ...

  10. cocos2dx 3.x 网络循环接收数据(RakNet::Packet* packet)单步网络接收

    void FriendFightLayer::update(float dt) { dealWithPacket(dt); if (m_isNeedSwitchToLobby) { PublicMet ...