poj 3208 Apocalypse Someday(数位DP，4级)

Apocalypse Someday

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 1385		Accepted: 632

Description

The number 666 is considered to be the occult “number of the beast” and is a well used number in all major apocalypse themed blockbuster movies. However the number 666 can’t always be used in the script so numbers such as 1666 are used instead. Let us call the numbers containing at least three contiguous sixes beastly numbers. The first few beastly numbers are 666, 1666, 2666, 3666, 4666, 5666…

Given a 1-based index n, your program should return the nth beastly number.

Input

The first line contains the number of test cases T (T ≤ 1,000).

Each of the following T lines contains an integer n (1 ≤ n ≤ 50,000,000) as a test case.

Output

For each test case, your program should output the nth beastly number.

Sample Input

3
2
3
187

Sample Output

1666
2666
66666

Source

POJ Monthly--2007.03.04, Ikki, adapted from TCHS SRM 2 ApocalypseSomeday

思路：数位ＤＰ　dp[位数][状态] 合法的含666的个数

             0 无六 1最后一位是6 2最后两位是6 3已经合法

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define LL long long
using namespace std;
int dp[15][4];///0,1,2,3have how many 6
int turn[4][10];
LL ans;
void init()
{
  clr(dp,0);clr(turn,0);
  turn[0][6]=1;
  turn[1][6]=2;
  turn[2][6]=3;
  FOR(i,0,9)turn[3][i]=3;
  dp[0][3]=1;///终态
  FOR(i,1,14)
  FOR(j,0,3)FOR(k,0,9)
  dp[i][j]+=dp[i-1][ turn[j][k] ];
}
LL get(int n)
{ int len,staue=0;
  FOR(i,1,14)
  {
    if(dp[i][0]>n)
    { //cout<<i<<endl;
      len=i;break;
    }
  }
  int add;LL ans=0;
  for(int i=len-1;i>=0;--i)
  {
    for(int j=0;j<10;++j)
    {
      if(dp[i][ turn[staue][j] ]<n)
        n-=dp[i][ turn[staue][j] ];
      else {add=j;break;}

    }
    ans=ans*10+add;
    staue=turn[staue][add];
  }
  return ans;
}
int main()
{
  int cas,n;
  init();
  scanf("%d",&cas);
  while(cas--)
  {
    scanf("%d",&n);
    printf("%I64d\n",get(n));
  }
}