String of concatenation, rotating the matrix, turning binary matrix

string concatenation

A string is called a pString if it can be represented as p concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1String, a 2String and a 4String, but it is not a 3String, a 5String, or a 6String and so on. Obviously any string is a 1String.

You are given a string s, consisting of lowercase English letters and a positive integer p. Your task is to find if it is possible to reorder the letters in the string s in such a way that the resulting string is a pString.

Input Format

The first input line contains integer p.
The second line contains s, all characters in s are lowercase English letters.

Constraints

1 ≤ p ≤ 1000
1 ≤ |s| ≤ 1000

Output Format

Print "YES" if it is possible to rearrange the letters in string s in such a way that the result is a pString. Print the result on a single output line. If it is not possible print "NO". (without quotes).

Sample Input 0

2
aazz

Sample Output 0

YES

Explanation 0

aazz can be rearranged to azaz which is a 2String

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  
    int n;
    scanf("%d",&n);
    char s[1010];
    scanf("%s",s);
    int h[26]={0},i;
    for(i=0;s[i]!='\0';i++)
        h[s[i]-'a']++;
    for(i=0;i<26;i++)
    {
        if(h[i]!=0 )
        {
            if(h[i]%n != 0)
            {
                printf("NO\n");
            return 0;
            }
        }
           
        }
    printf("YES");
    return 0;
}

=============

rotating the matrix

Thomas, a computer programmer, is led to fight an underground war against powerful computers who now rule the world with a system called 'The Matrix'.
To win against the system, he must rotate the Matrix clockwise by an angle of 90 degrees.
Unfortunately the system has wiped Thomas' memory and it is now in your hands to save the world.
The matrix can be considered as a 2Dimensional Array of size NxN.

SEE THE SAMPLE INPUT OUTPUT FOR CLARITY

Input Format

The first line contains the size of the matrix N.
The next N lines contain N integers each, together denoting the square matrix of size NxN.

Constraints

1 <= N <= 10

Output Format

Output the final matrix after rotation.

Sample Input 0

2
1 2
3 4

Sample Output 0

3 1
4 2

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int n;
    scanf("%d",&n);
    int a[n][n],b[n][n],i,j;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            scanf("%d",&a[i][j]);
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            b[i][j]=a[n-j-1][i];
    for(i=0;i<n;i++){
        for(j=0;j<n;j++)
            printf("%d ",b[i][j]);
        printf("\n");
    }
    return 0;
}

Turning binary matrix

Consider a binary matrix A of size N X N.
Now, consider the following matrices :
A90 - obtained by rotating A clockwise by 90 degrees.
A180 - obtained by rotating A clockwise by 180 degrees.
A270 - obtained by rotating A clockwise by 270 degrees.

Note : Binary matrix implies that every element will be either 0 or 1.

Your task is to construct another binary matrix B of size N X N such that :
B(i,j) = 1 iff either A(i,j) = 1 OR A90(i,j) = 1 OR A180(i,j) = 1 OR A270(i,j) = 1
B(i,j) = 0 otherwise

INPUT

First line contains the size of the matrix N (1 ≤ N ≤ 100)
Next N lines contain N integers each (Only 0 or 1) denoting the matrix A

OUTPUT

Print N X N integers, denoting the matrix B.






Sample Input 0




4
0 0 0 0
0 0 0 0
0 0 1 0
1 0 0 0








Sample Output 0




1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int n;
    scanf("%d",&n);
    int a[n][n],b[n][n],i,j;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            scanf("%d",&a[i][j]);
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            b[i][j]=a[i][j]|a[n-j-1][i]|a[n-i-1][n-j-1]|a[j][n-i-1];//for 360,90,180,270 rotations
    for(i=0;i<n;i++){
        for(j=0;j<n;j++)
            printf("%d ",b[i][j]);
        printf("\n");
    }
    return 0;
}