Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2)E - Buy Low Sell High

25-03-12 10

在这篇文章中，我们将带领您了解CodeforcesRound#437(Div.2,basedonMemSQLStart[c]UP3.0-Round2)E-BuyLowSellHigh的全貌，同时，我们

在这篇文章中，我们将带领您了解Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2)E - Buy Low Sell High的全貌，同时，我们还将为您介绍有关(AB)Codeforces Round #528 (Div. 2, based on Technocup 2019 Elimination Round、Codeforces Round #380 (Div. 2, Rated, Based on Technocup 2017 - Elimination Round 2) 题解、Codeforces Round #412 (rated, Div. 1, based on VK Cup 2017 Round 3) D. Perishable Roads、Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)的知识，以帮助您更好地理解这个主题。

本文目录一览：

Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2)E - Buy Low Sell High
(AB)Codeforces Round #528 (Div. 2, based on Technocup 2019 Elimination Round
Codeforces Round #380 (Div. 2, Rated, Based on Technocup 2017 - Elimination Round 2) 题解
Codeforces Round #412 (rated, Div. 1, based on VK Cup 2017 Round 3) D. Perishable Roads
Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)

Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2)E - Buy Low Sell High

总结

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(AB)Codeforces Round #528 (Div. 2, based on Technocup 2019 Elimination Round

A. Right-Left Cipher

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

polycarp loves ciphers. He has invented his own cipher called Right-Left.

Right-Left cipher is used for strings. To encrypt the string $s = s_{1} s_{2} \dots s_{n}$

he writes down $s_{1}$
he appends the current word with $s_{2}$
he prepends the current word with $s_{3}$
he appends the current word with $s_{4}$
he prepends the current word with $s_{5}$
and so on for each position until the end of $s$

For example,if $s$

Given string $t$

Input

The only line of the input contains $t$

Output

Print such string $s$

Examples

input

ncteho

output

techno

input

erfdcoeocs

output

codeforces

input

output

z
我好菜啊...脑袋都锈住了!

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdlib>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 int main(){
 9     string str{"0"};
10     string out{"0"};
11     //memset(s,‘\0‘,sizeof(s));
12     //memset(out,sizeof(out));
13     while(cin>>str){
14         int len=str.size();
15         out=str;
16         if(len==1 || len==2){
17             cout<<str<<endl;
18             continue;
19         }
20         int tmp=0;
21         int len_right=0;
22         int len_left=0;
23         if(len%2==1){
24             tmp=(len-1)/2;
25         }else{
26             tmp=len/2-1;
27         }
28         out[0]=str[tmp];
29         out[1]=str[tmp+1];
30         for(int i=tmp+2,j=3;i<len;i++,j++,j++){
31             out[j]=str[i];
32         }
33         for(int i=tmp-1,j=2;i>=0;i--,j++){
34             out[j]=str[i];
35         }
36         cout<<out<<endl;
37         //memset(str,sizeof(str));
38         //memset(out,sizeof(out));
39 
40 
41     }
42 
43 
44 
45 
46     return 0;
47 }

View Code

B. Div Times Mod

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya likes to solve equations. Today he wants to solve $(x d i v k) \cdot (x mod k) = n$

Input

The first line contains two integers $n$

Output

Print a single integer $x$

Examples

input

6 3

output

input

1 2

output

input

4 6

output

Note

The result of integer division $a d i v b$

In the first sample, $11 d i v 3 = 3$

思路:让找一个最小的x,使得(x/k)*(x%k)==n,如果对x暴力枚举肯定会超时啊,所以可以从x%k这里下手,x%k的值一定>=0 且<k,又因为n不可能为0,所以x%k是大于0的.所以在1~(k-1)之间枚举k.

再设x%k=i,上式可以变成(x-i)/k * i =n,所以x=n/i * k +i.

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

int main(){
    LL n,k;
    while(cin>>n>>k){
        LL x(LONG_MAX);
        for(LL i=1;i<k;i++){
            if(n%i!=0) continue;
            LL tmp=n/i*k+i;
            x=(x<tmp?x:tmp);
        }
        cout<<x<<endl;
    }    
    return 0;
}

View Code

Codeforces Round #380 (Div. 2, Rated, Based on Technocup 2017 - Elimination Round 2) 题解

总结

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Codeforces Round #412 (rated, Div. 1, based on VK Cup 2017 Round 3) D. Perishable Roads

总结

以上是小编为你收集整理的Codeforces Round #412 (rated, Div. 1, based on VK Cup 2017 Round 3) D. Perishable Roads全部内容。

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Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)

总结

以上是小编为你收集整理的Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)全部内容。

如果觉得小编网站内容还不错，欢迎将小编网站推荐给好友。

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