Educational Codeforces Round 4 A. The Text Splitting(判断下标）***

最新推荐文章于 2023-03-13 16:03:25 发布

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最新推荐文章于 2023-03-13 16:03:25 发布

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本文链接：https://blog.youkuaiyun.com/qq_36424540/article/details/79660972

本文介绍了一个解决特定字符串按长度p和q拆分问题的算法。该算法通过动态规划确定是否可以将输入字符串成功拆分为指定长度的子串，并输出所有可能的拆分方案。

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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.

Example

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>
#define bug(x) printf("%d****\n",x)
typedef long long ll;
using namespace std;

const int maxn=110;
int val[maxn];
char str[maxn];
int ans[maxn];
/*
WA了一次，TL了一次，因为没有判断下表越界 
但是我的编译器竟然完美的运行我的程序，感觉
很难受 
*/

int main(){
	int n,p,q;
	scanf("%d %d %d",&n,&p,&q);
	scanf(" %s",str);
	val[0]=val[p]=val[q]=1;
	for(int i=min(p,q);i<=n;i++){
		if(i-p>=0&&val[i-p]) val[i]=1;
		if(i-q>=0&&val[i-q])val[i]=1;
	}
	
	if(!val[n]){
		printf("-1\n");
		return 0;
	}

	int cnt=0;
	for(int i=n;i>0;){
		if(i-p>=0&&val[i-p]){
			ans[cnt++]=p;
			i=i-p;	
		}
		else if(i-q>=0&&val[i-q]){
			ans[cnt++]=q;
			i=i-q;
		}
	}
	printf("%d\n",cnt);
	int now=0;
	for(int i=0;i<cnt;i++){
		for(int j=0;j<ans[i];j++){
			printf("%c",str[now++]);
		}
		printf("\n");
	}
	return 0;
}