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  • Educational Codeforces Round 4 A. The Text Splitting 水题

    A. The Text Splitting

    题目连接:

    http://www.codeforces.com/contest/612/problem/A

    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

    5 2 3

    Hello

    Sample Output

    2

    He

    llo

    Hint

    题意

    给你一个字符串,让你分割成长度为p,或者长度为q的串

    问你如何分割

    题解:

    数据范围很小,所以直接暴力枚举就好了

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    string s;
    int main()
    {
        int n,p,q;
        cin>>n>>p>>q;
        cin>>s;
        for(int i=0;i<150;i++)
        {
            for(int j=0;j<150;j++)
            {
                if(i*p+j*q==s.size())
                {
                    cout<<i+j<<endl;
                    for(int k=0;k<i;k++)
                    {
                        for(int t=0;t<p;t++)
                            cout<<s[k*p+t];
                        cout<<endl;
                    }
                    for(int k=0;k<j;k++)
                    {
                        for(int t=0;t<q;t++)
                            cout<<s[i*p+k*q+t];
                        cout<<endl;
                    }
                    return 0;
                }
            }
        }
        return puts("-1");
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5081772.html
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