Anfisa the Monkey

mac2026-04-16  2

Anfisa the Monkey

Time limit 2000 ms

Memory limit 262144 kB

Anfisa the monkey learns to type. She is yet unfamiliar with the “space” key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into k lines not shorter than a and not longer than b, for the text to resemble human speech more. Help Anfisa.

Input

The first line contains three integers k, a and b (1 ≤ k ≤ 200, 1 ≤ a ≤ b ≤ 200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.

Output

Print k lines, each of which contains no less than a and no more than b symbols — Anfisa’s text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print “No solution” (without quotes).

Examples

Input 3 2 5 abrakadabra Output ab rakad abra Input 4 1 2 abrakadabra Output No solution

#include<stdio.h> #include<string.h> int main() { int k,a,b,d[200],i,j,x; char p[201]; while(~scanf("%d %d %d",&k,&a,&b)) { getchar(); gets(p); x=strlen(p); if(k>200||x==0) break; if(x>b*k||x<a*k) printf("No solution\n"); else { for(i=0;i<k;i++) { d[i]=a; } x-=(a*k); while(x) { for(i=0;i<k;i++) { d[i]++; x--; if(x==0) { break; } } } j=0; for(i=0;i<k;i++) { while(d[i]--) { printf("%c",p[j]); j++; } printf("\n"); } } } return 0; }
My Ideas

Make a judgment on whether it exists. If so, then each group has at least ‘a’ characters, so give each group a initial value ‘a’. Next, add the rest to the array through a loop.

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