codeforces 612A The Text Splitting(扩展欧几里得)

A. The Text Splitting

 

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 test(s)

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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
void exgcd(int a,int b,int& d,int& x,int& y)
{
    if(!b)d=a,x=1,y=0;
    else exgcd(b,a%b,d,y,x),y-=x*(a/b);
}
int main()
{
    char s[105];
    int n,p,q,d,x,y;
    bool ok;
    scanf("%d%d%d",&n,&p,&q);
    scanf("%s",s);
    int t=0;
    if(n%p==0)
    {
        n/=p;
        printf("%d
",n);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<p;j++)
                printf("%c",s[t++]);
            puts("");
        }
    }
    else if(n%q==0)
    {
        n/=q;
        printf("%d
",n);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<q;j++)
                printf("%c",s[t++]);
            puts("");
        }
    }
    else
    {
        ok=1;
        exgcd(p,q,d,x,y);
        if(n%d!=0)
        {
            puts("-1");
            return 0;
        }
        else
        {
            x*=n/d,y*=n/d;
            int a=p/d,b=q/d;
            int tx=x,ty=y;
            while(x<0||y<0)
            {
                x+=b;
                y-=a;
                if(tx*x<0&&ty*y<0)
                {
                    while(x<0||y<0)
                    {
                        x-=b;
                        y+=a;
                        if(tx*x<0&&ty*y<0)
                        {
                            ok=0;
                            break;
                        }
                    }
                    break;
                }
            }
        }
        if(ok)
        {
            printf("%d
",x+y);
            for(int i=0;i<x;i++)
            {
                for(int j=0;j<p;j++)
                    printf("%c",s[t++]);
                puts("");
            }
            for(int i=0;i<y;i++)
            {
                for(int j=0;j<q;j++)
                    printf("%c",s[t++]);
                puts("");
            }
        }
        else
            puts("-1");
    }
    return 0;
}