[POJ 2282] The Counting Problem

Posted 2020-12-25 evenbao

[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[15][10][15];

inline void dp(long long m)
{
        long long i,j,k,x;
        memset(f,0,sizeof(f));
        f[0][0][0] = 1;
        for (i = 1; i <= 12; i++)
        {
                for (j = 0; j <= 9; j++)
                {
                        for (k = 0; k <= i; k++)
                        {
                                if (j != m) 
                                {
                                        for (x = 0; x <= 9; x++)
                                                f[i][j][k] += f[i - 1][x][k];
                                } else if (k >= 1)
                                {
                                        for (x = 0; x <= 9; x++)
                                                f[i][j][k] += f[i - 1][x][k - 1];
                                }
                        }
                }
        }
}
inline long long calc(long long x,long long t)
{
        long long i,j,k,len = 0;
        long long res = 0;
        long long cnt = 0;
        long long a[12];
        memset(a,0,sizeof(a));
        while (x != 0)
        {
                a[++len] = x % 10;
                x /= 10;
        }
        reverse(a + 1,a + len + 1);
        for (i = 2; i <= len; i++)
        {
                for (j = 1; j <= 9; j++)
                {
                        for (k = 1; k <= len - i + 1; k++)
                                res += f[len - i + 1][j][k] * k;
                }
        }
        for (i = 1; i <= len; i++)
        {
                for (j = 0; j < a[i]; j++)
                {
                        if (i == 1 && !j) continue;
                        for (k = cnt; k <= len; k++)
                        {
                                res += f[len - i + 1][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 = 0; i < 9; i++) 
                {
                        dp(i);
                        printf("%lld ",calc(b + 1,i) - calc(a,i));
                }
                dp(9);
                printf("%lld
",calc(b + 1,9) - calc(a,9));
        }
        
        return 0;
    
}