CodeForces 612A The Text Splitting

A. The Text Splitting

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

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

这题主要就是切分……

分对了就可以了。具体看代码：

#include<stdio.h>
#include<string.h>
char ch[111];
int main()
{
int n,p,q;
int i,j,a,b;
while(~scanf("%d%d%d",&n,&p,&q))
{
int flag=0;
scanf("%s",ch);
int l=strlen(ch);
a=b=-1;
for(i=0;i<=n;i++)
{
if(flag)
break;
for(j=0;j<=n;j++)
{
if((i!=0||j!=0)&&i*p+j*q==n)
{
flag=1;
a=i;b=j;
break;
}
}
}
if(a==b&&a==-1)
printf("-1\n");
else
{
if(a==-1) a=0;
if(b==-1) b=0;
printf("%d\n",a+b);
i=0;
for(;i<a*p;i++)
{
if(i%p==0&&i!=0)
printf("\n");
printf("%c",ch[i]);
}
for(;i<l;i++)
{
if((i-a*p)%q==0&&i!=0)
printf("\n");
printf("%c",ch[i]);
}
printf("\n");
}
}
return 0;
}