【codeforces】A. The Text Splitting（思维题-水题）

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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.

Examples

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

题意：给出字符串长度，给出p和q两种切割方式，任选其中一种，把字符串分割输出结果。

题解：先进行判断，p和q是否能整个的分割n，利用p和q的函数关系判断（见代码），再计算有几个p几个q，再进行输出即可。

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>
#include<stack>
#define N 1000000
typedef long long LL;
using namespace std;
int main(){
double n,p,q;
while(cin>>n>>p>>q){
char str[110];
cin>>str;
double temp;
int flag=0;
for(double i=0;p*i<=n;i++){//判断p和q是否能整块分割n
temp=(n-p*i)/q;//p和q 的函数关系
if(temp-int(temp)==0){
flag=i+1;//记录有几个p
break;
}
}
if(flag){
int v=flag-1;//几个p
int c=(n-v*p)/q;//几个q
int point=0;
flag=0;
cout<<c+v<<endl;
for(int i=0;i<v;i++){
for(int j=i*p;j<p+i*p;j++){
cout<<str[j];
point=j;
flag=1;
}
cout<<endl;
}
for(int i=0;i<c;i++){
for(int j=i*q+point+flag;j<q+i*q+point+flag;j++){
cout<<str[j];
}
cout<<endl;
}
}
else
cout<<"-1"<<endl;
}
return 0;
}