【Codeforces】-612A-The Text Splitting（数学）

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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

n可以整除p或q是比较好写的情况。

还有就是x*p+y*q=l,虽然情况可能很多但是写出一种就好，求出一个x，y就出来了。

#include<cstdio>
#include<cstring>
int main()
{
int n,p,q;
char a[110];
while(~scanf("%d %d %d",&n,&p,&q))
{
scanf("%s",a);
int k=1,num=0;
for(k=1;k<=n/p;k++)
{
if((n-p*k)%q==0)
break;
else
num++;
}
if(num!=n/p)
{
printf("%d\n",k+(n-k*p)/q);
int i=0;
for(int z=0;z<k;z++)
{
for(int j=1;j<=p;j++)
{
printf("%c",a[i]);
i++;
}
printf("\n");
}
int ant=(n-k*p)/q;
for(int z=0;z<ant;z++)
{
for(int j=1;j<=q;j++)
{
printf("%c",a[i]);
i++;
}
printf("\n");
}
}
else if(n%p==0)
{
printf("%d\n",n/p);
int i=0;
for(int k=0;k<n/p;k++)
{
for(int j=1;j<=p;j++)
{
printf("%c",a[i]);
i++;
}
printf("\n");
}
}
else if(n%q==0)
{
printf("%d\n",n/q);
int i=0;
for(int k=0;k<n/q;k++)
{
for(int j=1;j<=q;j++)
{
printf("%c",a[i]);
i++;
}
printf("\n");
}
}
else
printf("-1\n");
}
}