codeforces 612A The Text Splitting

The Text Splitting
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

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
代码：
#include<stdio.h>
#include<string.h>
int main()
{
int n, p, q;
scanf("%d%d%d",&n,&p,&q);
char str[110];
scanf("%s",str);
bool flag = false;
for(int i = 0; i <= n; i++)
{
for(int j = 0; j <= n; j++)
{
if(i * p + j * q == n)
{
printf("%d\n",i+j);
int k = 0;
for(int k = 0; k < i; k++)
{
for(int l = k*p; l < (k+1)*p; l++)
printf("%c", str[l]);
printf("\n");
}
for(int k = 0; k < j; k++)
{
for(int l = i*p + k*q; l < (k+1)*q + i*p; l++)
printf("%c", str[l]);
printf("\n");
}
flag = true;
break;
}
}
if(flag)
break;
}
if(!flag)
printf("-1\n");
return 0;
}