Given string tt — the result of encryption of some string ss. Your task is to decrypt it, i.e. find the string ss.

The only line of the input contains tt — the result of encryption of some string ss. It contains only lowercase Latin letters. The length of tt is between 11 and 5050, inclusive.

Print such string ss that after encryption it equals tt.

ncteho

techno

erfdcoeocs

codeforces

z

z
我好菜啊...腦袋都鏽住了!

 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 }

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?

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

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.

6 3

11

1 2

3

4 6

10

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:讓找一個最小的x,使得(x/k)*(x%k)==n,如果對x暴力列舉肯定會超時啊,所以可以從x%k這裡下手,x%k的值一定>=0 且<k,又因為n不可能為0,所以x%k是大於0的.所以在1~(k-1)之間列舉k.

再設x%k=i,上式可以變成(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;
}