uva 129

暴力求解

大致题意 如果一个字符串含有相邻的重复字串称为容易的串，反之为非容易

求字典序第n困难的串……

大致思路，暴力如果是容易的串停过，然后困难的串继续求解tot++

总之先记着吧……

最后输出格式……

uva 129

#include <iostream>

#include <cstdio>

#include <cmath>

#include <cstring>

#include <algorithm>

#include <cstdlib>

#include <stack>

#include <cctype>

#include <string>

#include <malloc.h>

#include <queue>

#include <map>



using namespace std;

const int INF = 0xffffff;

const double esp = 10e-8;

const double Pi = 4 * atan(1.0);

const int Maxn = 200 + 10;

const int mod = 10000007;

const int dr[] = {1,0,-1,0,-1,1,-1,1};

const int dc[] = {0,1,0,-1,1,-1,-1,1};

//int dir2[8][2] = {{-1,0},{0,-1},{-1,1},{1,-1},{-1,-1},{1,0},{0,1},{1,1}};



int n,L;

int a[1000];

int tot;



bool dfs(int cur){

  if(tot++ == n){

    for(int i = 0;i < cur;i++){

        if(i && i % 4 == 0){

            if(i % (16*4) == 0){

                cout << endl;

                //cout << "$$$$";

            }

            else

                cout << ' ';

        }

        cout << char(a[i] + 'A');

    }

    cout << endl;

    cout << cur << endl;

    return 0;

  }

  for(int i = 0;i < L;i++){

    a[cur] = i;

    bool ok = 1;

    for(int j = 1;j * 2 < cur+2;j++){

        bool flag = 1;

        for(int k = 0;k < j;k++){

            if(a[cur-k] != a[cur-k-j]){

                flag = 0;

                break;

            }

        }

        if(flag){

            ok = 0;

            break;

        }

    }

    if(ok){

        if(!dfs(cur+1)){

            return 0;

        }

    }

  }

  return 1;

}



int main()

{

#ifndef ONLINE_JUDGE

    freopen("inpt.txt","r",stdin);

#endif

    while(cin >> n >> L){

        if(!n && !L)

            break;

        tot = 0;

        dfs(0);

    }

    return 0;

}

View Code

回家一直在玩，好难过……！！！！

以后要加油……！！！