[POJ 2282] The Counting Problem

[题目链接]

http://poj.org/problem?id=2282

[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 i;
long long a,b;
long long f[][][];
 
inline void dp(long long m)
{
        long long i,j,k,x;
        memset(f,,sizeof(f));
        f[][][] = ;
        for (i = ; i <= ; i++)
        {
                for (j = ; j <= ; j++)
                {
                        for (k = ; k <= i; k++)
                        {
                                if (j != m)
                                {
                                        for (x = ; x <= ; x++)
                                                f[i][j][k] += f[i - ][x][k];
                                } else if (k >= )
                                {
                                        for (x = ; x <= ; x++)
                                                f[i][j][k] += f[i - ][x][k - ];
                                }
                        }
                }
        }
}
inline long long calc(long long x,long long t)
{
        long long i,j,k,len = ;
        long long res = ;
        long long cnt = ;
        long long a[];
        memset(a,,sizeof(a));
        while (x != )
        {
                a[++len] = x % ;
                x /= ;
        }
        reverse(a + ,a + len + );
        for (i = ; i <= len; i++)
        {
                for (j = ; j <= ; j++)
                {
                        for (k = ; k <= len - i + ; k++)
                                res += f[len - i + ][j][k] * k;
                }
        }
        for (i = ; i <= len; i++)
        {
                for (j = ; j < a[i]; j++)
                {
                        if (i ==  && !j) continue;
                        for (k = cnt; k <= len; k++)
                        {
                                res += f[len - i + ][j][k - cnt] * k;
                        }
                }
                if (a[i] == t) cnt++;
        }
        return res;
}
 
int main()
{
 
        while (scanf("%lld%lld",&a,&b) && (a || b))
        {
                if (a > b) swap(a,b);
                for (i = ; i < ; i++)
                {
                        dp(i);
                        printf("%lld ",calc(b + ,i) - calc(a,i));
                }
                dp();
                printf("%lld\n",calc(b + ,) - calc(a,));
        }
 
        return ;
 
}