Codeforces 395 B2. iwiwi

구조
B2. iwiwi
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
The Codeforces users sometimes have very interesting handles! For example, you can meet a grandmaster from Japan named iwiwior an international grandmaster from Taiwan by name 0O0o00OO0Oo0o0Oo.
To be honest, Polycarpus prefers handle iwiwi to 0O0o00OO0Oo0o0Oo.
Polycarpus noticed that in many fonts the width of letter w is exactly twice wider that the width of letter i. Now he wants to write all sorts of strings containing letters i and w on a board exactly as wide as n letters i. He doesn't want the problem to be too easy, so Polycarpus added another limit: each consequence string must be different from the previous one:
either by replacing two consecutive letters i by one letter w,
or be replacing letter w by two consecutive letters i.
For example, string iii can follow after string iw, as the second string results after replacing w by ii in the first one. String wi can follow after string iii, as it is obtained by replacing ii by w.
Help Polycarpus, print all strings of width of exactly n letters i in the described manner. For each pair of consecutive strings there must be exactly one replacement by exactly one of the given rules. The written sequence doesn't have to be cycled, that is, the first string isn't considered consequent for the last string.
Input
The single line of the input contains a positive integer n. The limits for n are different for the subproblems:
for subproblem B1, the following inequation fulfills 1 ≤ n ≤ 5,
for subproblem B2, the following inequation fulfills 1 ≤ n ≤ 20.
Output
Print all strings of letters i and w, such that i + 2w = n, where i — the number of letters i, and w — the number of letters w. Print the strings in the order that is described in the problem.
If there are many answers, print any of them.
Sample test(s)
input

3

output

iw
iii
wi

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

vector<string> vs[30];

int main()
{
    vs[1].push_back("i");
    vs[2].push_back("ii");
    vs[2].push_back("w");

    int n;
    scanf("%d",&n);

    if(n==1) puts("i");
    else if(n==2) puts("ii
w");
    else
    {
        int s;
        for(int i=3;i<=n;i++)
        {
            s=vs[i-1].size();
            for(int j=s-1;j>=0;j--)
            {
                vs[i].push_back("i"+vs[i-1][j]);
            }
            s=vs[i-2].size();
            for(int j=s-1;j>=0;j--)
            {
                vs[i].push_back("w"+vs[i-2][j]);
            }
        }
        s=vs[n].size();
        for(int i=0;i<s;i++)
        {
            cout<<vs[n][i]<<endl;
        }
    }

    return 0;
}