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[][];
 int p[],num;
 int o[];
 int dfs(int pos,int pre,int bound)
 {
     int ans,i,end;
     ans = ;
     if(pos == -)
     return pre == ;
     if(!bound&&dp[pos][pre] != -)
     return dp[pos][pre];
     end = bound ? p[pos]:;
     for(i = ;i <= end;i ++)
     {
         if(i == ||i == )
         ans += dfs(pos-,pre-,bound&&i == end);
         else
         ans += dfs(pos-,pre,bound&&i == end);
     }
     if(!bound)
     dp[pos][pre] = ans;
     return ans;
 }
 void CL(int m)
 {
     num = ;
     while(m)
     {
         p[num++] = m%;
         m /= ;
     }
 }
 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,-,sizeof(dp));
     scanf("%d",&m);
     CL(m);
     for(i = ;i <= ;i ++)
     {
         o[i] = dfs(num-,i,);
     }
     o[] --;
     ans = ;
     for(a1 = ;a1 <= ;a1 ++)
     {
         ta1 = o[a1];
         o[a1] -- ;
         for(a2 = ;a2 <= &&o[a1] >= ;a2 ++)
         {
             ta2 = o[a2];
             o[a2] -- ;
             for(a3 = ;a3 <= &&o[a2] >= ;a3 ++)
             {
                 ta3 = o[a3];
                 o[a3] -- ;
                 for(a4 = ;a4 <= &&o[a3] >= ;a4 ++)
                 {
                     ta4 = o[a4];
                     o[a4] -- ;
                     for(a5 = ;a5 <= &&o[a4] >= ;a5 ++)
                     {
                         ta5 = o[a5];
                         o[a5] -- ;
                         for(a6 = ;a6 <= &&o[a5] >= ;a6 ++)
                         {
                             ta6 = o[a6];
                             o[a6] -- ;
                             sum = a1 + a2 + a3 + a4 + a5 + a6;
                             if(sum <= &&o[a6] >= )
                             {
                                 t1 = ta1;
                                 t1 = t1*ta2%MOD;
                                 t1 = t1*ta3%MOD;
                                 t1 = t1*ta4%MOD;
                                 t1 = t1*ta5%MOD;
                                 t1 = t1*ta6%MOD;
                                 t2 = ;
                                 for(i = sum+;i <= ;i ++)
                                 t2 += o[i];
                                 ans = (ans + t1*t2)%MOD;
                             }
                             o[a6] ++ ;
                         }
                         o[a5] ++ ;
                     }
                     o[a4] ++ ;
                 }
                 o[a3] ++ ;
             }
             o[a2] ++;
         }
         o[a1] ++ ;
     }
     printf("%I64d\n",ans);
     return ;
 }