在这篇文章中，我们将为您详细介绍CodeforcesRound#437(Div.2,basedonMemSQLStart[c]UP3.0-Round2)B-Savetheproblem!的内容。此外，

在这篇文章中，我们将为您详细介绍Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2) B - Save the problem!的内容。此外，我们还会涉及一些关于(AB)Codeforces Round #528 (Div. 2, based on Technocup 2019 Elimination Round、codeforces cf educatonal round 57(div2) D. Easy Problem、Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution —— 异或、Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem —— 贪心的知识，以帮助您更全面地了解这个主题。

• Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2) B - Save the problem!
• (AB)Codeforces Round #528 (Div. 2, based on Technocup 2019 Elimination Round
• codeforces cf educatonal round 57(div2) D. Easy Problem
• Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution —— 异或
• Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem —— 贪心

Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2) B - Save the problem!

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

数据技术都能四世同堂，凭什么开发 30 岁就要被干掉？

Examples

input

ncteho

output

techno

input

erfdcoeocs

output

codeforces

input

z

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,''\0'',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++,j++){
34             out[j]=str[i];
35         }
36         cout<<out<<endl;
37         //memset(str,''\0'',sizeof(str));
38         //memset(out,''\0'',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 div k)⋅(xmodk)=n(x div k)⋅(xmodk)=n, where divdiv and modmod stand for integer division and modulo operations (refer to the Notes below for exact definition). In this equation, kk and nn are positive integer parameters, and xx is a positive integer unknown. If there are several solutions, Vasya wants to find the smallest possible xx. Can you help him?

Input

The first line contains two integers nn and kk (1≤n≤1061≤n≤106, 2≤k≤10002≤k≤1000).

Output

Print a single integer xx — the smallest positive integer solution to (x div k)⋅(xmodk)=n(x div k)⋅(xmodk)=n. It is guaranteed that this equation has at least one positive integer solution.

Examples

input

6 3

output

11

input

1 2

output

3

input

4 6

output

10

Note

The result of integer division a div ba div b is equal to the largest integer cc such that b⋅c≤ab⋅c≤a. aa modulo bb (shortened amodbamodb) is the only integer cc such that 0≤c<b0≤c<b, and a−ca−c is divisible by bb.

In the first sample, 11 div 3=311 div 3=3 and 11mod3=211mod3=2. Since 3⋅2=63⋅2=6, then x=11x=11 is a solution to (x div 3)⋅(xmod3)=6(x div 3)⋅(xmod3)=6. One can see that 1919 is the only other positive integer solution, hence 1111 is the smallest one.

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 cf educatonal round 57(div2) D. Easy Problem

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
string s;
ll a[100010];
ll dp[5];
int main()
{
    int n;
    scanf("%d",&n);
    cin>>s;
    s=‘0‘+s;
    for(int i=1;i<=n;i++)
       scanf("%lld",&a[i]); 
    for(int i=1;i<=n;i++)
    {
        if(s[i]==‘h‘)
        {
            dp[1]=dp[1]+a[i];
        }
        else if(s[i]==‘a‘)
        {
            dp[2]=min( dp[1],dp[2]+a[i] );
        }
        else if(s[i]==‘r‘)
        {
            dp[3]=min( dp[2],dp[3]+a[i] );
        }
        else if(s[i]==‘d‘)
        {
            dp[4]=min( dp[3],dp[4]+a[i] );
        }
    }
    printf("%lld",dp[4]);
}

Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution —— 异或

Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem —— 贪心

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