H. Fake News (medium)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Thanks to your help, Heidi is confident that no one can fool her. She has now decided to post some fake news on the HC² Facebook page. However, she wants to be able to communicate to the HC² committee that the post is fake, using some secret phrase hidden in the post as a subsequence. To make this method foolproof, she wants the phrase to appear n

times in the post. She is asking you to design a post (string) s and a hidden phrase p such that p appears in s as a subsequence exactly n times.

Input

The first and only line of input contains a single integer n (1 ≤ n ≤ 1 000 000).

Output

The output should contain two nonempty strings s and p separated by a single space. Each string should be composed of letters (a

-z and A-Z: both lowercase and uppercase are allowed) and have length at most 200. The number of occurrences of p in s as a subsequence should be exactly n. If there are many possible solutions, output any of them. It is guaranteed that at least one solution exists.

ExamplesInputCopy

2

OutputCopy

hHheidi Hei

InputCopy

4

OutputCopy

bbbba ba

InputCopy

6

OutputCopy

aaabb ab

Note

An occurrence of p as a subsequence in s should be thought of as a set of positions in s such that the letters at these positions, in order, form p. The number of occurences is thus the number of such sets. For example, ab appears 6 times as a subsequence in aaabb, for the following sets of positions: {1, 4}, {1, 5}, {2, 4}, {2, 5}, {3, 4}, {3, 5} (that is, we should choose one of the a's and one of the b's).

題意：給出一個數n 要你構造字串a,b 使得b作為子序列出現在a中恰好n次。

解題思路： 一開始想因數分解，發現並不可行，遇到素數就GG。 看了別人的程式碼才知道怎麼寫。 假設對於 p，s串，如果p已經在s中出現了k次 思考是否可以以此為基礎，倍增的構造出其他出現次數的字串對。

For k = 1, we can just have s = p = a. Then: • To go k → 2k + 1, suppose that for k we have a pair (s,p) with s = pu. Then generate p0 = px, where x is a new letter, and s0 = pxuxx. • To go k → 2k + 2, generate the same p0 and take s0 = pxxuxx.

然後就很好寫了 遞迴找一下就完事了

#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6+10;
int getlen(int b) {
    int ans =0;
    while(b) {
        ans++;
        b>>=1;
    }
    return ans ;
}
string p,s;
void slove(int now,int deep){
    string cnt;
    cnt+=((char)('a'+deep));
    if(now==1){
        p=s=cnt;
        return;
    }else if(now==2){
        p+=((char)('a'+deep+1))+cnt;
        s+=((char)('a'+deep+1))+cnt+cnt;
        return;
    }
    if(now&1){
        slove(now>>1,deep+1);
        int lenp=p.size();
        int lens=s.size();
        s=s.substr(0,lenp)+cnt+s.substr(lenp,lens-lenp)+cnt+cnt;
        p+=cnt;
    }else{
        slove((now>>1)-1,deep+1);
        int lenp=p.size();
        int lens=s.size();
        s=s.substr(0,lenp)+cnt+cnt+s.substr(lenp,lens-lenp)+cnt+cnt;
        p+=cnt;
    }
}
int ans=0;
void dfs(int lenp,int lens){
    if(lenp==p.size()){
        ans++;
        return;
    }
    if(lens==s.size()){
        return;
    }
    if(s[lens]==p[lenp]){
        dfs(lenp+1,lens+1);
    }
    dfs(lenp,lens+1);
}
int main() {
    int n;
    while(~scanf("%d",&n)){
        s.clear();
        p.clear();
        slove(n,0);
        cout<<s<<" "<<p<<endl;
//        ans=0;
//        dfs(0,0);
//        cout<<ans<<endl;
    }
    return 0;
}