2015_12_26 A. The Text Splitting

一貼題：

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.

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

代碼：/*

/*
ID: Ben biss
PROG: #####
LANG: C++
*/

#include <iostream>
using namespace std;
#include<cstring>
#include <fstream>
#include<cmath>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<stack>
#include<vector>
#include<map>
#include<queue>
#define FOR(i,n) for(i=0;i<n;i++)
//#define cin fin
//#define cout fout

char s[1000];

int main()
{
    //ofstream fout ("#####.out");
    //ifstream fin ("#####.in");
    int sum,b,c,i,j,z;
    cin>>sum>>b>>c;
    cin>>s;
    if(sum==(b+c))
    {
        cout<<2<<endl;
        for(i=0;i<b;i++)cout<<s[i];
        cout<<endl;
        for(;i<sum;i++)cout<<s[i];
    }
    else{
        if(sum%b==0)
        {
            cout<<sum/b<<endl;
            for(i=0;i<sum/b;i++)
            {
                for(j=0;j<b;j++)
                {
                    cout<<s[i*b+j];
                }
                cout<<endl;
            }
        }else if(sum%c==0){
            cout<<sum/c<<endl;
            for(i=0;i<sum/c;i++)
            {
                for(j=0;j<c;j++)
                {
                    cout<<s[i*c+j];
                }
                cout<<endl;
            }
        }else
        {
            int flag=0,pb,pc;
            FOR(i,100)
            {
                FOR(j,100)
                if(i*b+j*c==sum)
                {
                    pb=i,pc=j,flag=1;
                    cout<<pb+pc<<endl;
                    break;
                }
                if(flag==1)break;
            }

            if(flag==1)
            {
                for(i=0;i<pb;i++)
                {
                    for(j=0;j<b;j++)
                    {
                        cout<<s[i*b+j];
                    }
                    cout<<endl;
                }
                for(z=0;z<pc;z++)
                {
                    for(j=0;j<c;j++)
                    {
                        cout<<s[z*c+j+pb*b];
                    }
                    cout<<endl;
                }
            }
            else cout<<-1<<endl;
        }

    }

	return 0;
}