【Codeforces 612A】The Text Splitting

The Text Splitting

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

可以分成任意数量的 p,q,题目简单不多解释，看代码：

#include<stdio.h>int main(){    char ch[1010];    int n,p,q,i,x,y,j,t;    while(scanf("%d%d%d",&n,&p,&q)!=EOF)    {        scanf("%s",ch);        if(n%p==0)        {            printf("%d\n",n/p);            for(i=1;i<=n;i++)            {            printf("%c",ch[i-1]);            if(i%p==0)            printf("\n");               }        }        else if(n%q==0)        {            printf("%d\n",n/q);            for(i=1;i<=n;i++)            {            printf("%c",ch[i-1]);            if(i%q==0)            printf("\n");               }        }        else        {            t=0;        for(i=0;i<=n;i++)        for(j=0;j<=n;j++)        {            if(i*p+j*q==n)            {                x=i;                y=j;                t=1;                break;            }        }        if(!t)        printf("-1\n");        else         {            printf("%d\n",x+y);            for(i=1;i<=x*p;i++)            {            printf("%c",ch[i-1]);            if(i%p==0)            printf("\n");            }            for(i=x*p+1;i<=n;i++)            {                printf("%c",ch[i-1]);                if((i-x*p)%q==0)                printf("\n");            }        }        }    }    return 0;}