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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).
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.
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
2
He
llo
10 9 5
Codeforces
2
Codef
orces
6 4 5
Privet
-1
8 1 1
abacabac
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 if(n % p == 0 || n % q == 0) 11 { 12 l1=(n%p == 0)?p:q; 13 l2=n/l1; 14 k=-1; 15 printf("%d\n",l2); 16 while(l2--) 17 { 18 l=l1; 19 while(l--) 20 { 21 printf("%c",s[++k]); 22 } 23 printf("\n"); 24 } 25 } 26 else 27 { 28 for(int i = 1; i < n ; i++) 29 { 30 for( int j = 1; j < n ; j++) 31 { 32 if(i*p+j*q == n) 33 { 34 printf("%d\n",i+j); 35 k=-1; 36 while(i--) 37 { 38 l=p; 39 while(l--) 40 { 41 printf("%c",s[++k]); 42 } 43 printf("\n"); 44 } 45 46 while(j--) 47 { 48 l=q; 49 while(l--) 50 { 51 printf("%c",s[++k]); 52 } 53 printf("\n"); 54 } 55 g=0; 56 break; 57 } 58 } 59 if(g == 0) break; 60 } 61 if(g != 0) printf("-1\n"); 62 } 63 64 65 } 66 67 }
The Text Splitting (将字符串分成若干份，每份长度为p或q）
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原文地址：http://www.cnblogs.com/yexiaozi/p/5692699.html
