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  • The Text Splitting (将字符串分成若干份,每份长度为p或q)

    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
     1 #include<cstdio>
     2 char s[1000000];
     3 int main()
     4 {
     5     int n,p,q,k,l1,l2,l,g;
     6     while(scanf("%d %d %d",&n,&p,&q)!=EOF)
     7     {
     8         scanf("%s",&s);
     9         g=1;
    10         for(int i = 0; i <= n ; i++)
    11             {
    12                  for( int j = 0; j <= n ; j++)
    13                 {
    14                     if(i*p+j*q == n)
    15                     {
    16                         printf("%d
    ",i+j);
    17                         k=-1;
    18                         while(i--)
    19                         {
    20                             l=p;
    21                             while(l--)
    22                             {
    23                                 printf("%c",s[++k]);
    24                             }
    25                             printf("
    ");
    26                         }        
    27                         while(j--)
    28                         {
    29                             l=q;
    30                             while(l--)
    31                             {
    32                                 printf("%c",s[++k]);
    33                             }
    34                             printf("
    ");
    35                         }
    36                         g=0;
    37                         break;
    38                     }
    39                 }
    40                 if(g == 0) break;
    41             }
    42             if(g != 0) printf("-1
    ");
    43             
    44     }
    45     
    46 }
    ——将来的你会感谢现在努力的自己。
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  • 原文地址:https://www.cnblogs.com/yexiaozi/p/5692699.html
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