The Text Splitting CodeForces 612A

Description
You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string "Hello" for p = 2, q = 3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input
The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output
If it's impossible to split the string s to the strings of length p and q print the only number "-1".

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Sample Input
Input
5 2 3
Hello
Output
2
He
llo
Input
10 9 5
Codeforces
Output
2
Codef
orces
Input
6 4 5
Privet
Output
-1
Input
8 1 1
abacabac
Output
8
a
b
a
c
a
b
a

c

#include<stdio.h>
#include<string.h>
int main()
{
	int k,p,q,i,j,b,c,d,ans;
	char a[110];
	while(scanf("%d%d%d",&k,&p,&q)!=EOF)
	{
		ans=-1;
		getchar();
		scanf("%s",&a);
		getchar();
		for(i=0;i<=k/p;i++)
        {   
            for(j=0;j<=k/q;j++)
                if(i*p+j*q==k)<span style="white-space:pre">			</span>//要判斷是否 i個 p 加上 j個 q 也等於 K
                {
                    ans=i+j;
                    c=i;
                    d=j;
                }
   	 }
		if(ans==-1)
		{
			printf("-1\n");
		}
		else
		{
				printf("%d\n",ans);
				for(i=0;i<c;i++)
				{
					for(j=(0+p*i);j<(p+p*i);j++)
					printf("%c",a[j]);
					printf("\n");
				}
					for(i=0;i<d;i++)
				{
					for(j=(c*p+q*i);j<(c*p+q+q*i);j++)
					printf("%c",a[j]);
					printf("\n");
				}
		}
	}
	return 0;
}