【CodeForces】612A - The Text Splitting（枚举）

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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

p * x + q * y = n   枚举解 x，y 就行了。

代码如下：

#include <cstdio>
int main()
{
int l,p,q;
char s[111];
int ant1,ant2; //两个循环次数
while (~scanf ("%d %d %d",&l,&p,&q))
{
scanf ("%s",s+1);
if (l % p == 0)
{
printf ("%d\n",l / p);
for (int i = 1 ; i <= l ; i++)
{
printf ("%c",s[i]);
if (i % p == 0)
printf ("\n");
}
}
else if (l % q == 0)
{
printf ("%d\n",l / q);
for (int i = 1 ; i <= l ; i++)
{
printf ("%c",s[i]);
if (i % q == 0)
printf ("\n");
}
}
else
{
bool ans = false;
for (int i = 1 ; i <= 100 ; i++)
{
if (i * p > l)
{
ans = false;
break;
}
for (int j = 1 ; j <= 100 ; j++)
{
if (j * q > l)
break;
if (i * p + j * q == l)
{
ant1 = i;
ant2 = j;
ans = true;
}
}
if (ans)
break;
}
if (ans)
{
printf ("%d\n",ant1+ant2);
int pos = 1;
for (int i = 1 ; i <= ant1 ; i++)
{
for (int j = 1 ; j <= p ; j++)
printf ("%c",s[pos++]);
printf ("\n");
}
for (int i = 1 ; i <= ant2 ; i++)
{
for (int j = 1 ; j <= q ; j++)
printf ("%c",s[pos++]);
printf ("\n");
}
}
else
printf ("-1\n");
}
}
return 0;
}