Codeforces Round #157 (Div. 1)B. Little Elephant and Elections 数位dp好

B. Little Elephant and Elections

There have recently been elections in the zoo. Overall there were 7 main political parties: one of them is the Little Elephant Political Party, 6 other parties have less catchy names.

Political parties find their number in the ballot highly important. Overall there are m possible numbers: 1, 2, …, m. Each of these 7 parties is going to be assigned in some way to exactly one number, at that, two distinct parties cannot receive the same number.

The Little Elephant Political Party members believe in the lucky digits 4 and 7. They want to evaluate their chances in the elections. For that, they need to find out, how many correct assignments are there, such that the number of lucky digits in the Little Elephant Political Party ballot number is strictly larger than the total number of lucky digits in the ballot numbers of 6 other parties.

Help the Little Elephant Political Party, calculate this number. As the answer can be rather large, print the remainder from dividing it by 1000000007(109 + 7) .
Input

A single line contains a single positive integer m (7 ≤ m ≤ 109) — the number of possible numbers in the ballot.
Output

In a single line print a single integer — the answer to the problem modulo 1000000007(109 + 7) .
Sample test(s)
Input

7

Output

0

Input

8

Output

1440

思路：首先 数位dp计算出包含n个7或4的数个有多少个，然后dfs一边找出来就行了。dp[cur][S][K]表示处理到当前为止，有S个7或者4，一共有K个7或者4的有多少个

#include
using namespace std;
typedef long long LL;
const int maxn=20;
const int MOD=1e9+7;
int M;
int dig[maxn];
LL dp[maxn][maxn][maxn];
LL sum[maxn];
int len;
LL dfs(int cur,int e,int S,int K)
{
    if(cur<0)return S==K;
    if(!e&&dp[cur][S][K]!=-1)
        return dp[cur][S][K];
    int end=(e?dig[cur]:9);
    LL ans=0;
    for(int i=0;i<=end;i++)
    {
        ans+=dfs(cur-1,e&&i==end,S+(i==4||i==7),K);
        ans%=MOD;
    }
    if(!e)dp[cur][S][K]=ans;
    return ans;
}
void solve(int x)
{
    len=0;
    memset(dp,-1,sizeof(dp));
    while(x)
    {
        dig[len++]=x%10;
        x/=10;
    }
    for(int i=0;i<=len;i++)
        sum[i]=dfs(len-1,1,0,i);
    sum[0]--;//把0也算进去了
}
LL solve(int num,int pos)
{
    if(pos<=0)return 1;
    LL ans=0;
    for(int i=0;i<=num;i++)
    {
        if(sum[i]<=0)continue;
        sum[i]--;
        ans=(ans+(1+sum[i])*solve(num-i,pos-1))%MOD;
        sum[i]++;
    }
    return ans;
}
int main()
{
    scanf("%d",&M);
    solve(M);
    LL ans=0;
    for(int i=1;i<=len;i++)
        ans=(ans+sum[i]*solve(i-1,6))%MOD;
    cout<return 0;
}