#UPCF Round1 A - The Text Splitting

CodeForces 612A
*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*

思路：枚举

#include <iostream>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <cstdio>#include <vector>using namespace std;int main(){    int i,j,m,n;int a,b,c;    scanf("%d%d%d",&a,&b,&c);getchar();    char d[a+10];gets(d);    for(i=0;b*i<=a;i++){        for(j=0;j*c<=a;j++){            if(i*b+j*c==a)break;        }if(j*c<=a)break;    }int q=0;    if(b*i>a)printf("-1\n");    else {            printf("%d\n",i+j);        for(m=0;m<i;m++){            for(n=0;n<b;n++)printf("%c",d[q++]);printf("\n");        }        for(m=0;m<j;m++){            for(n=0;n<c;n++)printf("%c",d[q++]);printf("\n");        }    }    //system("pause");    return 0;}