Uva 129 Krypton Factor(困难的串)

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运用回溯法，是否包含两个相邻的的字串只需要判断当前串的后缀

代码如下：

#include <algorithm>  #include <iostream>  #include <cstdlib>  #include <cstring>  #include <string>#include <cstdio>  #include <climits>#include <cmath> using namespace std;int n,l,cnt,times=0;int A[99];int dfs(int cur){                                        //回溯法if(cnt++==n){int times=0;for(int i=0;i<cur;i++){printf("%c",'A'+A[i]-1); times++;if(times%4==0){if(times<cur&×!=64)   printf(" ");}if(times==64&&cur>64){printf("\n");}}printf("\n%d\n",cur);return 0;} for(int i=0;i<l;i++){                 int OK=1;A[cur]=i+1;    for(int j=1;j*2<=cur+1;j++){int equal=1;for(int z=0;z<j;z++)if(A[cur-z]!=A[cur-z-j]){equal=0;break;}if(equal){OK=0;break;}}if(OK)   if(!dfs(cur+1)) return 0;}return 1;}int main(){#ifdef DIDfreopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);#endifwhile(scanf("%d%d",&n,&l)!=EOF&&(n||l)){cnt=0;memset(A,0,sizeof(A));    int kk=dfs(0);}  return 0;}

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