ACM/ICPC 之 暴力打表(求解欧拉回路)-编码（POJ1780）

///找到一个数字序列包含所有n位数(连续)一次且仅一次
///暴力打表
///Time:141Ms    Memory:2260K
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

#define MAX 1000010

bool v[MAX];
char num[6][MAX];

int main()
{
    //freopen("in.txt", "r", stdin);
    for(int t = 1; t <= 6; t++)
    {
        //初始化
        memset(v, false, sizeof(v));
        int len = t-1, mod = 1;
        for(int i = 1; i <= t; i++)
            mod *= 10;  //mod:取模数
        len += mod; //输出总数
        for(int i = 0; i < t; i++)
            num[t-1][i] = '0';

        int cur = 0;
        v[cur] = true;
        for(int i = t; i < len; i++)
        {
            int j = 0;  //向后一位试探0-9
            for(; j < 10; j++)
            {
                int last = (cur * 10 + j) % mod;
                if(!v[last]){   //标识
                    cur = last;
                    v[cur] = true;
                    break;
                }
            }
            if(j == 10){    //试探未果
                num[t-1][--i]++;    //回推一位+1
                v[cur] = false;     //取消已有标识
                while(num[t-1][i] == '0' + 10 || v[cur+1])  //超出十进制或已标识
                {
                    if(num[t-1][i] != '0' + 10 && v[cur + 1]){  //仅仅已标识
                        num[t-1][i]++;
                        cur++;
                        continue;
                    }
                    i--;    //否则继续回推+1
                    cur = ((num[t-1][i-t+1] - '0') * mod + cur)/10;
                    v[cur] = false;
                    num[t-1][i]++;
                }
                cur++;
                continue;
            }
            else num[t-1][i] = '0' + j;
        }
    }

    int n;
    while(scanf("%d", &n), n)
        printf("%s\n", num[n-1]);
    return 0;
}