CodeForces 612A The Text Splitting

E - 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

题意：给出一个长度为N的字符串，把他截取为长度为p，q的字符串

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n,p,q,i,j,k,l,m;
char s[110];
while(scanf("%d%d%d",&n,&p,&q)!=EOF)
{
scanf("%s",s);
int flag=0;
for(i=0;i<=100;i++)
{
for(j=0;j<100;j++)
{
if(p*i+q*j==n)
{
k=i;
l=j;
flag=1;
break;
}
}
}
if(flag==0)
{
printf("-1\n");
continue;
}

printf("%d\n",k+l);
for(i=0;i<k*p;i=i+p)
{
for(j=i;j<i+p;j++)
printf("%c",s[j]);
printf("\n");
}
for(i=k*p;i<n;i=i+q)
{
for(j=i;j<i+q;j++)
printf("%c",s[j]);
printf("\n");
}
}
return 0;
}