codeforces 395 B1. iwiwi(待续)-白红宇

codeforces 395 B1. iwiwi(待续)

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发布时间：2019-05-24

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1、

2、题目：

The Codeforces users sometimes have very interesting handles! For example, you can meet a grandmaster from Japan namediwiwi or an international grandmaster from Taiwan by name0O0o00OO0Oo0o0Oo.

To be honest, Polycarpus prefers handle iwiwi to0O0o00OO0Oo0o0Oo.

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 asn 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 letterw,

or be replacing letter w by two consecutive lettersi.

For example, string iii can follow after stringiw, as the second string results after replacingw by ii in the first one. Stringwi can follow after string iii, as it is obtained by replacing ii byw.

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, wherei — 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

Output

iwiiiwi

3、wrong answer

#include
   
    #include
    
     #include
     
      using namespace std;int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int i=1;        int w=(n-i)/2;        int flag=(n-i)%2;        printf("i");        for(int j=1; j<=w; j++)            printf("w");        if(flag)            printf("i");        printf("\n");        // printf("***********\n");        for(int k=1; k<=w; k++)        {            printf("i");            for(int j=1; j<=k; j++)                printf("ii");            for(int j=k+1; j<=w; j++)                printf("w");            if(flag)                printf("i");            printf("\n");        }        for(int k=1; k<=n/2; k++)        {            for(int j=1; j<=k; j++)                printf("w");            for(int p=1; p<=n-k*2; p++)                printf("i");            printf("\n");        }        for(int k=1; k<=(n-1)/2; k++)        {            if(n/2-k>0)            {                for(int j=1; j<=k; j++)                    printf("ii");                for(int j=1; j<=n/2-k; j++)                    printf("w");                if(n%2==1)                    printf("i");                printf("\n");            }        }    }    return 0;}

4、附AC代码：（参见网上代码）

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;vector
         
           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\nw"); 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

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