A - Rating Goal


Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

Takahashi is a user of a site that hosts programming contests.
When a user competes in a contest, the rating of the user (not necessarily an integer) changes according to the performance of the user, as follows:

  • Let the current rating of the user be a.
  • Suppose that the performance of the user in the contest is b.
  • Then, the new rating of the user will be the avarage of a and b.

For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000.

Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest.
Find the performance required to achieve it.

Constraints

  • 0≤R,G≤4500
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

R
G

Output

Print the performance required to achieve the objective.


Sample Input 1

Copy
2002
2017

Sample Output 1

Copy
2032

Takahashi's current rating is 2002.
If his performance in the contest is 2032, his rating will be the average of 2002 and 2032, which is equal to the desired rating, 2017.


Sample Input 2

Copy
4500
0

Sample Output 2

Copy
-4500

Although the current and desired ratings are between 0 and 4500, the performance of a user can be below 0.

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<map>
using namespace std; const int N = + ; int c, n;
struct node{
int total_time = , total_solve = , num;
map<int, int> state;
}Node[N]; int main(){
double a, b;
cin >> a >> b;
cout << b * - a <<endl;
}

B - Addition and Multiplication


Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value:

  • Operation A: The displayed value is doubled.
  • Operation B: The displayed value increases by K.

Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.

Constraints

  • 1≤N,K≤10
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the minimum possible value displayed in the board after N operations.


Sample Input 1

Copy
4
3

Sample Output 1

Copy
10

The value will be minimized when the operations are performed in the following order: A, A, B, B.
In this case, the value will change as follows: 1 → 2 → 4 → 7 → 10.


Sample Input 2

Copy
10
10

Sample Output 2

Copy
76

The value will be minimized when the operations are performed in the following order: A, A, A, A, B, B, B, B, B, B.
In this case, the value will change as follows: 1 → 2 → 4 → 8 → 16 → 26 → 36 → 46 → 56 → 66 → 76.

By the way, this contest is AtCoder Beginner Contest 076.

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<map>
using namespace std; int main(){
int n, k;
int a = ;
cin >> n >> k;
for(int i = ; i <= n; i++){
if(a + k < * a) a += k;
else a *= ;
}
cout << a << endl;
}
#include<bits/stdc++.h>

using namespace std;

int n, k;

int DFS(int n, int val){
if(n == ) return val;
int a = DFS(n - , val * );
int b = DFS(n - , val + k);
return a > b ? b : a;
}
int main(){
cin >> n >> k;
cout << DFS(n, ) << endl;
}

C - Dubious Document 2


Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

E869120 found a chest which is likely to contain treasure.
However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.
He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with ?.

One more thing he found is a sheet of paper with the following facts written on it:

  • Condition 1: The string S contains a string T as a contiguous substring.
  • Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.

Print the string S.
If such a string does not exist, print UNRESTORABLE.

Constraints

  • 1≤|S'|,|T|≤50
  • S' consists of lowercase English letters and ?.
  • T consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T'

Output

Print the string S.
If such a string does not exist, print UNRESTORABLE instead.


Sample Input 1

Copy
?tc????
coder

Sample Output 1

Copy
atcoder

There are 26 strings that satisfy Condition 1: atcoderbtcoderctcoder,..., ztcoder. Among them, the lexicographically smallest is atcoder, so we can say S=atcoder.


Sample Input 2

Copy
??p??d??
abc

Sample Output 2

Copy
UNRESTORABLE

There is no string that satisfies Condition 1, so the string S does not exist.

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<map>
#include<set>
using namespace std;
string s, t;
set<string> S; void Work(){
int lens = s.size();
int lent = t.size();
for(int i = ; i < lens; i++){
if(s[i] == '?' || s[i] == t[]){
string tmp = s;
for(int j = ; j < i; j++) if(tmp[j] == '?') tmp[j] ='a';
bool can = false;
for(int j = ; j < lent; j++){
if(tmp[i+j]!= '?' && tmp[i+j] != t[j]) break;
tmp[i+j] = t[j];
if(j == lent - ) can = true;
}
if(can){
for(int j = i + lent; j < lens; j++) if(tmp[j] == '?') tmp[j] = 'a';
S.insert(tmp);
}
}
}
if(S.size() == ) cout << "UNRESTORABLE" << endl;
else cout << *S.begin() << endl;
}
int main(){
cin >> s >> t;
Work();
}

AtCoder Beginner Contest 076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 079 D - Wall【Warshall Floyd algorithm】

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

  7. AtCoder Beginner Contest 064 D - Insertion

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

  8. 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 ...

  9. AtCoder Beginner Contest 075 C bridge【图论求桥】

    AtCoder Beginner Contest 075 C bridge 桥就是指图中这样的边,删除它以后整个图不连通.本题就是求桥个数的裸题. dfn[u]指在dfs中搜索到u节点的次序值,low ...

随机推荐

  1. AtCoder AGC036C GP 2 (组合计数)

    题目链接 https://atcoder.jp/contests/agc036/tasks/agc036_c 题解 终于有时间补agc036的题了. 这题其实不难的来着--我太菜了考场上没想出来 首先 ...

  2. MergeActors技巧

    打开界面 MergeActors的界面在Window>DeveloperTools>MergeActors 启用合并材质的方法 使用Use specific LOD Level 一种特殊情 ...

  3. Spring Boot教程(三十二)多数据源配置与使用

    之前在介绍使用JdbcTemplate和Spring-data-jpa时,都使用了单数据源.在单数据源的情况下,Spring Boot的配置非常简单,只需要在application.propertie ...

  4. 分布式-信息方式-ActiveMQ静态网络连接信息回流功能

    “丢失”的消息 有这样的场景, broker1和 broker2通过 netwoskconnector连接,一些 consumers连接到 broker1,消费 broker2上的消息.消息先被 br ...

  5. java正则表达式详细总结

    Java 提供了功能强大的正则表达式API,在java.util.regex 包下.本教程介绍如何使用正则表达式API. 正则表达式 一个正则表达式是一个用于文本搜索的文本模式.换句话说,在文本中搜索 ...

  6. Redis的高可用详解:Redis哨兵、复制、集群的设计原理,以及区别

    谈到Redis服务器的高可用,如何保证备份的机器是原始服务器的完整备份呢?这时候就需要哨兵和复制. 哨兵(Sentinel):可以管理多个Redis服务器,它提供了监控,提醒以及自动的故障转移的功能. ...

  7. D3(没写完

    说在博客前 这篇博客有许多使用到 STL 的地方,由于本人实在是记不全,所以我也参考了北大的一些教材,就别说我黈力了 QwQ 数据结构 今天讲的是数据结构啦(也是我这个蒟蒻最喜欢的 一些天天见面的好盆 ...

  8. 【汇总】Windows linux 敏感目录 路径汇总

    日期:2019-08-02 10:53:52 更新:2019-08-19 15:48:01 作者:Bay0net 介绍:中间件.套件等等敏感信息,做个记录. 0x01. 基本信息 遇到文件包含.任意文 ...

  9. tmux 学习

    这几天学习了一下 tmux的使用 tmux 可以同时打开多个窗口 关于使用技巧 复制文章一下  哈哈 感谢网友 ================================华丽的分割线====== ...

  10. 使用 Blender* 重新拓扑 VR 和游戏素材

    本文介绍如何将网格重新拓扑成一个整洁的低密度模型,然后 UV 解包该网格,以便将纹理贴添加至新模型.本文还将探讨如何使用免费工具,比如 Blender* 及其 Bsurface 插件,重新拓扑雕塑的 ...