https://beta.atcoder.jp/contests/abc103

A - Task Scheduling Problem

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100100 points

Problem Statement

You have three tasks, all of which need to be completed.

First, you can complete any one task at cost 00.

Then, just after completing the ii-th task, you can complete the jj-th task at cost |Aj−Ai||Aj−Ai|.

Here, |x||x| denotes the absolute value of xx.

Find the minimum total cost required to complete all the task.

Constraints

  • All values in input are integers.
  • 1≤A1,A2,A3≤1001≤A1,A2,A3≤100

Input

Input is given from Standard Input in the following format:

A1A1 A2A2 A3A3

Output

Print the minimum total cost required to complete all the task.


Sample Input 1 Copy

Copy
1 6 3

Sample Output 1 Copy

Copy
5

When the tasks are completed in the following order, the total cost will be 55, which is the minimum:

  • Complete the first task at cost 00.
  • Complete the third task at cost 22.
  • Complete the second task at cost 33.

Sample Input 2 Copy

Copy
11 5 5

Sample Output 2 Copy

Copy
6

Sample Input 3 Copy

Copy
100 100 100

Sample Output 3 Copy

Copy
0

最大的减去最小的。
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main { public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int [] a = new int[3];
for(int i = 0;i < 3;i ++) {
a[i] = in.nextInt();
}
Arrays.sort(a);
System.out.println(a[2] - a[0]);
}
}

B - String Rotation


Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 200200 points

Problem Statement

You are given string SS and TT consisting of lowercase English letters.

Determine if SS equals TT after rotation.

That is, determine if SS equals TT after the following operation is performed some number of times:

Operation: Let S=S1S2...S|S|S=S1S2...S|S|. Change SS to S|S|S1S2...S|S|−1S|S|S1S2...S|S|−1.

Here, |X||X| denotes the length of the string XX.

Constraints

  • 2≤|S|≤1002≤|S|≤100
  • |S|=|T||S|=|T|
  • SS and TT consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

SS
TT

Output

If SS equals TT after rotation, print Yes; if it does not, print No.


Sample Input 1 Copy

Copy
kyoto
tokyo

Sample Output 1 Copy

Copy
Yes
  • In the first operation, kyoto becomes okyot.
  • In the second operation, okyot becomes tokyo.

Sample Input 2 Copy

Copy
abc
arc

Sample Output 2 Copy

Copy
No

abc does not equal arc after any number of operations.


Sample Input 3 Copy

Copy
aaaaaaaaaaaaaaab
aaaaaaaaaaaaaaab

Sample Output 3 Copy

Copy
Yes
排着遍历拼接。
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main {
static boolean check(String a,String b) {
if(a.length() != b.length())return false;
for(int i = 0;i < a.length();i ++) {
if(a.charAt(i) == b.charAt(0) && (a.substring(i, a.length()) + a.substring(0, i)).equals(b)) {
return true;
}
}
return false;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String a = in.nextLine();
String b = in.nextLine();
System.out.println(check(a,b) ? "Yes" : "No");
}
}

C - Modulo Summation

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300300 points

Problem Statement

You are given NN positive integers a1,a2,...,aNa1,a2,...,aN.

For a non-negative integer mm, let f(m)=(m mod a1)+(m mod a2)+...+(m mod aN)f(m)=(m mod a1)+(m mod a2)+...+(m mod aN).

Here, X mod YX mod Y denotes the remainder of the division of XX by YY.

Find the maximum value of ff.

Constraints

  • All values in input are integers.
  • 2≤N≤30002≤N≤3000
  • 2≤ai≤1052≤ai≤105

Input

Input is given from Standard Input in the following format:

NN
a1a1 a2a2 ...... aNaN

Output

Print the maximum value of ff.


Sample Input 1 Copy

Copy
3
3 4 6

Sample Output 1 Copy

Copy
10

f(11)=(11 mod 3)+(11 mod 4)+(11 mod 6)=10f(11)=(11 mod 3)+(11 mod 4)+(11 mod 6)=10 is the maximum value of ff.


Sample Input 2 Copy

Copy
5
7 46 11 20 11

Sample Output 2 Copy

Copy
90

Sample Input 3 Copy

Copy
7
994 518 941 851 647 2 581

Sample Output 3 Copy

Copy
4527
一个数对多个数去摸和要最大。
对于m,m%a1最大值是a1 - 1,也就是说令m = t1 * a1 - 1(t1 >= 1)时m%a1最大,同样如果对于a1~n都满足m = ti * ai - 1,总和就会最大,实际上就是满足m = t1 * a1 - 1 = t2 * a2 - 1 = ...
= tn * an - 1,即t1 * a1 = t2 * a2 = ... = tn * an = m + 1,所以m + 1是a1~n的公倍数,是存在的。
答案就显而易见是所有属的和-n。
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main {
static boolean check(String a,String b) {
if(a.length() != b.length())return false;
for(int i = 0;i < a.length();i ++) {
if(a.charAt(i) == b.charAt(0) && (a.substring(i, a.length()) + a.substring(0, i)).equals(b)) {
return true;
}
}
return false;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int d;
int ans = -n;
for(int i = 0;i < n;i ++) {
d = in.nextInt();
ans += d;
}
System.out.println(ans);
}
}
D - Islands War

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400400 points

Problem Statement

There are NN islands lining up from west to east, connected by N−1N−1 bridges.

The ii-th bridge connects the ii-th island from the west and the (i+1)(i+1)-th island from the west.

One day, disputes took place between some islands, and there were MM requests from the inhabitants of the islands:

Request ii: A dispute took place between the aiai-th island from the west and the bibi-th island from the west. Please make traveling between these islands with bridges impossible.

You decided to remove some bridges to meet all these MM requests.

Find the minimum number of bridges that must be removed.

Constraints

  • All values in input are integers.
  • 2≤N≤1052≤N≤105
  • 1≤M≤1051≤M≤105
  • 1≤ai<bi≤N1≤ai<bi≤N
  • All pairs (ai,bi)(ai,bi) are distinct.

Input

Input is given from Standard Input in the following format:

NN MM
a1a1 b1b1
a2a2 b2b2
::
aMaM bMbM

Output

Print the minimum number of bridges that must be removed.


Sample Input 1 Copy

Copy
5 2
1 4
2 5

Sample Output 1 Copy

Copy
1

The requests can be met by removing the bridge connecting the second and third islands from the west.


Sample Input 2 Copy

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

Sample Output 2 Copy

Copy
2

Sample Input 3 Copy

Copy
5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5

Sample Output 3 Copy

Copy
4
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import static java.lang.Math.*;
class bridge{
public int a;
public int b;
public bridge() {
a = b = 0;
}
}
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
bridge [] s = new bridge[k];
for(int i = 0;i < k;i ++) {
s[i] = new bridge();
s[i].a = in.nextInt();
s[i].b = in.nextInt();
}
Arrays.sort(s,new Comparator<bridge>() {
public int compare(bridge x,bridge y) {
if(x.a == x.b)return x.b - y.b;
return x.a - y.a;
}
});
int ans = 0,r = 0;
for(int i = 0;i < k;i ++) {
if(s[i].a >= r) {
ans ++;
r = s[i].b;
}
else if(r > s[i].b)r = s[i].b;
}
System.out.println(ans);
}
}

AtCoder Beginner Contest 103的更多相关文章

  1. AtCoder Beginner Contest 100 2018/06/16

    A - Happy Birthday! Time limit : 2sec / Memory limit : 1000MB Score: 100 points Problem Statement E8 ...

  2. AtCoder Beginner Contest 052

    没看到Beginner,然后就做啊做,发现A,B太简单了...然后想想做完算了..没想到C卡了一下,然后还是做出来了.D的话瞎想了一下,然后感觉也没问题.假装all kill.2333 AtCoder ...

  3. AtCoder Beginner Contest 053 ABCD题

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

  4. AtCoder Beginner Contest 136

    AtCoder Beginner Contest 136 题目链接 A - +-x 直接取\(max\)即可. Code #include <bits/stdc++.h> using na ...

  5. AtCoder Beginner Contest 137 F

    AtCoder Beginner Contest 137 F 数论鬼题(虽然不算特别数论) 希望你在浏览这篇题解前已经知道了费马小定理 利用用费马小定理构造函数\(g(x)=(x-i)^{P-1}\) ...

  6. AtCoder Beginner Contest 076

    A - Rating Goal Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Takaha ...

  7. AtCoder Beginner Contest 079 D - Wall【Warshall Floyd algorithm】

    AtCoder Beginner Contest 079 D - Wall Warshall Floyd 最短路....先枚举 k #include<iostream> #include& ...

  8. AtCoder Beginner Contest 064 D - Insertion

    AtCoder Beginner Contest 064 D - Insertion Problem Statement You are given a string S of length N co ...

  9. AtCoder Beginner Contest 075 D - Axis-Parallel Rectangle【暴力】

    AtCoder Beginner Contest 075 D - Axis-Parallel Rectangle 我要崩溃,当时还以为是需要什么离散化的,原来是暴力,特么五层循环....我自己写怎么都 ...

随机推荐

  1. CSS解决无空格太长的字母,数字不会自己主动换行的问题

    事实上非常easy,代码例如以下所看到的,注意 Style: <div class="detail_title" style="word-break: break- ...

  2. springboot+async异步接口实现和调用

    什么是异步调用? 异步调用是相对于同步调用而言的,同步调用是指程序按预定顺序一步步执行,每一步必须等到上一步执行完后才能执行,异步调用则无需等待上一步程序执行完即可执行. 如何实现异步调用? 多线程, ...

  3. Java设计模式(九)责任链模式 命令模式

    (十七)责任链模式 责任链模式的目的是通过给予多个对象处理请求的机会,已解除请求发送者与接受者之间的耦合关系.面对对象的开发力求对象之前保持松散耦合,确保对象各自的责任最小化.这种设计能够使得系统更加 ...

  4. Redis单台的安装部署及集群部署

    Redis是一种高级key-value数据库.它跟memcached类似,不过数据可以持久化,而且支持的数据类型很丰富.有字符串,链表,集 合和有序集合.支持在服务器端计算集合的并,交和补集(diff ...

  5. 【文献阅读】Stack What-Where Auto-encoders -ICLR-2016

    一.Abstract 提出一种新的autoencoder -- SWWAE(stacked what-where auto-encoders),更准确的说是一种 convolutional autoe ...

  6. 有状态的EJB对象和无状态的EJB对象

    一,定义有状态Bean和无状态Bean 有状态Bean: @Stateful @Remote public class StatefulEjbBean implements StatefulEjb{ ...

  7. programming review (c++): (1)vector, linked list, stack, queue, map, string, bit manipulation

    编程题常用知识点的review. most important: 想好(1)详尽步骤(2)边界特例,再开始写代码. I.vector #include <iostream> //0.头文件 ...

  8. 【BZOJ3721】PA2014 Final Bazarek 贪心

    [BZOJ3721]PA2014 Final Bazarek Description 有n件商品,选出其中的k个,要求它们的总价为奇数,求最大可能的总价. Input 第一行一个整数n(1<=n ...

  9. 九度OJ 1044:Pre-Post(先序后序) (n叉树、递归)

    时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:701 解决:398 题目描述: We are all familiar with pre-order, in-order and post- ...

  10. 限制线程数 Limit the number of threads started by colly 随机延迟

    random delay | Colly http://go-colly.org/docs/examples/random_delay/