E - The Text Splitting

E - The Text Splitting

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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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

---

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
    int n,p,q,i,j,k,cnt;
    char str[333]; 
    while(cin>>n>>p>>q)
    {
        cin>>str;

        cnt=0;
        for(i=0;i<=n;++i)
        {
            for(j=0;j<=n;++j)
            {
                if(n==i*p+j*q)
                {
                    cnt=1;
                    break;
                }

            }

            if(cnt==1)
            break;
        }

        if(cnt==0)
        {
              cout<<"-1"<<endl;
        }
        else
        {
            cout<<i+j<<endl;

            for(k=0;k<i*p;++k)
            {
                cout<<str[k];
                if((k+1)%p==0)
                cout<<endl;
             } 
             for(i=k;k<n;++k)
             {
                cout<<str[k];
                if((k+1-i)%q==0)
                cout<<endl;
             }
        }
    }
    return 0;
}