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. JavaWeb_Servlet生命周期

    菜鸟教程 传送门 Servlet生命周期 package com.Gary.servlet; import java.io.IOException; import javax.servlet.Serv ...

  2. Unity3D_(游戏)卡牌02_主菜单界面

      启动屏界面.主菜单界面.选关界面.游戏界面 卡牌01_启动屏界面 传送门 卡牌02_主菜单界面 传送门 卡牌03_选关界面 传送门 卡牌04_游戏界面    传送门 主菜单界面 (选择左边图标或选 ...

  3. Unity3D_(游戏)跳一跳超简单制作过程

    跳一跳 工程文件界面 游戏界面 脚本 using DG.Tweening; using System.Collections; using System.Collections.Generic; us ...

  4. wannafly 挑战赛9 D 造一造 (卡特兰数)

    链接:https://www.nowcoder.com/acm/contest/71/D 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K 64b ...

  5. jQuery 全选和反选demo

    前段时间做了一个全选和反选的功能,最近不忙了,做了一个简化版的demo. 全部代码如下: <!DOCTYPE html> <html> <head> <tit ...

  6. C++入门经典-例6.20-修改string字符串的单个字符

    1:使用+可以将两个string 字符串连接起来.同时,string还支持标准输入输出函数.代码如下: // 6.20.cpp : 定义控制台应用程序的入口点. // #include "s ...

  7. Android Studio设置国内镜像代理

    点击主面板右下角的Configure –> settings –> Appearance & Behavior –> System Settings –> HTTP P ...

  8. android studio中方法和类被调用多次,但是AS显示灰色,解决办法

    Android Studio里面的一些类及方法,明明有被其他的类或者方法调用,但是去看的时候显示灰色,鼠标放上面的时候显示:Class ‘XXX’ is never used或者Method ‘XXX ...

  9. 阶段3 2.Spring_08.面向切面编程 AOP_3 spring基于XML的AOP-编写必要的代码

    新建项目 先改打包方式 导包,就先导入这俩包的坐标 aspectjweaver为了解析切入点表达式 新建业务层接口 定义三个方法 看返回和参数的区别.为了把这三类方法表现出来,并不局限于方法干什么事 ...

  10. 四十五:数据库之SQLAlchemy之subquery实现复杂查询

    子查询让多个查询变成一个查询,只需要查找一次数据库,性能相对来讲更高效,不用写多个SQL语句就可以实现一些复杂的查询,在SQLAlchemy中要实现一个子查询,应该使用以下步骤:1.将子查询按照传统方 ...