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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.
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 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 void exgcd(int a,int b,int& d,int& x,int& y) 6 { 7 if(!b)d=a,x=1,y=0; 8 else exgcd(b,a%b,d,y,x),y-=x*(a/b); 9 } 10 int main() 11 { 12 char s[105]; 13 int n,p,q,d,x,y; 14 bool ok; 15 scanf("%d%d%d",&n,&p,&q); 16 scanf("%s",s); 17 int t=0; 18 if(n%p==0) 19 { 20 n/=p; 21 printf("%d\n",n); 22 for(int i=0;i<n;i++) 23 { 24 for(int j=0;j<p;j++) 25 printf("%c",s[t++]); 26 puts(""); 27 } 28 } 29 else if(n%q==0) 30 { 31 n/=q; 32 printf("%d\n",n); 33 for(int i=0;i<n;i++) 34 { 35 for(int j=0;j<q;j++) 36 printf("%c",s[t++]); 37 puts(""); 38 } 39 } 40 else 41 { 42 ok=1; 43 exgcd(p,q,d,x,y); 44 if(n%d!=0) 45 { 46 puts("-1"); 47 return 0; 48 } 49 else 50 { 51 x*=n/d,y*=n/d; 52 int a=p/d,b=q/d; 53 int tx=x,ty=y; 54 while(x<0||y<0) 55 { 56 x+=b; 57 y-=a; 58 if(tx*x<0&&ty*y<0) 59 { 60 while(x<0||y<0) 61 { 62 x-=b; 63 y+=a; 64 if(tx*x<0&&ty*y<0) 65 { 66 ok=0; 67 break; 68 } 69 } 70 break; 71 } 72 } 73 } 74 if(ok) 75 { 76 printf("%d\n",x+y); 77 for(int i=0;i<x;i++) 78 { 79 for(int j=0;j<p;j++) 80 printf("%c",s[t++]); 81 puts(""); 82 } 83 for(int i=0;i<y;i++) 84 { 85 for(int j=0;j<q;j++) 86 printf("%c",s[t++]); 87 puts(""); 88 } 89 } 90 else 91 puts("-1"); 92 } 93 return 0; 94 }
codeforces 612A The Text Splitting(扩展欧几里得)
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原文地址：http://www.cnblogs.com/homura/p/5086641.html
