Codeforces Round #157 (Div. 2) D. Little Elephant and Elections(数位DP+枚举)

数位DP部分，不是很难。DP[i][j]前i位j个幸运数的个数。枚举写的有点搓。。。

#include <cstdio>
#include <cstring>
using namespace std;
#define LL __int64
#define MOD 1000000007
int dp[21][21];
int p[21],num;
int o[21];
int dfs(int pos,int pre,int bound)
{
    int ans,i,end;
    ans = 0;
    if(pos == -1)
    return pre == 0;
    if(!bound&&dp[pos][pre] != -1)
    return dp[pos][pre];
    end = bound ? p[pos]:9;
    for(i = 0;i <= end;i ++)
    {
        if(i == 4||i == 7)
        ans += dfs(pos-1,pre-1,bound&&i == end);
        else
        ans += dfs(pos-1,pre,bound&&i == end);
    }
    if(!bound)
    dp[pos][pre] = ans;
    return ans;
}
void CL(int m)
{
    num = 0;
    while(m)
    {
        p[num++] = m%10;
        m /= 10;
    }
}
int main()
{
    int i,m;
    int a1,a2,a3,a4,a5,a6,sum;
    LL ta1,ta2,ta3,ta4,ta5,ta6;
    LL ans,t1,t2;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&m);
    CL(m);
    for(i = 0;i <= 9;i ++)
    {
        o[i] = dfs(num-1,i,1);
    }
    o[0] --;
    ans = 0;
    for(a1 = 0;a1 <= 9;a1 ++)
    {
        ta1 = o[a1];
        o[a1] -- ;
        for(a2 = 0;a2 <= 9&&o[a1] >= 0;a2 ++)
        {
            ta2 = o[a2];
            o[a2] -- ;
            for(a3 = 0;a3 <= 9&&o[a2] >= 0;a3 ++)
            {
                ta3 = o[a3];
                o[a3] -- ;
                for(a4 = 0;a4 <= 9&&o[a3] >= 0;a4 ++)
                {
                    ta4 = o[a4];
                    o[a4] -- ;
                    for(a5 = 0;a5 <= 9&&o[a4] >= 0;a5 ++)
                    {
                        ta5 = o[a5];
                        o[a5] -- ;
                        for(a6 = 0;a6 <= 9&&o[a5] >= 0;a6 ++)
                        {
                            ta6 = o[a6];
                            o[a6] -- ;
                            sum = a1 + a2 + a3 + a4 + a5 + a6;
                            if(sum <= 9&&o[a6] >= 0)
                            {
                                t1 = ta1;
                                t1 = t1*ta2%MOD;
                                t1 = t1*ta3%MOD;
                                t1 = t1*ta4%MOD;
                                t1 = t1*ta5%MOD;
                                t1 = t1*ta6%MOD;
                                t2 = 0;
                                for(i = sum+1;i <= 9;i ++)
                                t2 += o[i];
                                ans = (ans + t1*t2)%MOD;
                            }
                            o[a6] ++ ;
                        }
                        o[a5] ++ ;
                    }
                    o[a4] ++ ;
                }
                o[a3] ++ ;
            }
            o[a2] ++;
        }
        o[a1] ++ ;
    }
    printf("%I64d
",ans);
    return 0;
}