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 ;
}