[POJ 3208] Apocalypse Someday

[题目链接]

          http://poj.org/problem?id=3208

[算法]

        数位DP

[代码]

        

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h> 
using namespace std;

int T;
long long n;
long long f[25][5];

inline void dp()
{
        int i,j;
        f[0][0] = 1;
        for (i = 0; i <= 20; i++)
        {
                for (j = 0; j < 3; j++)
                {
                        f[i + 1][j + 1] += f[i][j];
                        f[i + 1][0] += f[i][j] * 9; 
                }
                f[i + 1][3] += f[i][3] * 10;
        }
        
}
inline void calc(long long n)
{
        int i,j,k,len,now;
        long long sum;
        for (len = 3; f[len][3] < n; len++);
        now = 0;  
        sum = 0;
        for (i = len; i >= 1; i--)
        {
                for (j = 0; j <= 9; j++)
                {
                        if (j == 6 || now == 3)
                        {
                                sum = f[i - 1][3];
                                for (k = max(3 - now - (j == 6),0); k < 3; k++) sum += f[i - 1][k];
                                if (sum < n) n -= sum;
                                else 
                                {
                                        printf("%d",j);
                                        if (now < 3 && j == 6) now++;
                                        break;
                                }
                        } else
                        {
                                if (f[i - 1][3] < n)
                                        n -= f[i - 1][3];
                                else 
                                {
                                        now = 0;
                                        printf("%d",j);
                                        break;
                                }
                        }
                }    
      }                               
        printf("\n");                                                                                                                                                                                                                                                                                                                                                                                       
}

int main() 
{
        
        dp();
        scanf("%d",&T);
        while (T--)
        {
                scanf("%lld",&n);
                calc(n);    
        }
        
        return 0;
    
}