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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……
#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; }
回家一直在玩,好难过……!!!!
以后要加油……!!!
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原文地址:http://www.cnblogs.com/hanbinggan/p/4284296.html