Codeforces 258B. Little Elephant and Elections【数位DP，DFS】

B. Little Elephant and Elections

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 7parties 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 by1000000007 (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).

题目大意：

规定4,7为lucky数。给一个数m，求1~m中的七个数字的排列中，满足下列条件的排列有多少个：

第七个数中含有的lucky数的数量大于前六个含有的lucky数的总和。（比如说1234567，该数字有一个4，一个7，则他的lucky值为2）。

因为m最大位数为10，在这个范围内的所有书的lucky值最多为10，如此，我们可以预处理出这样一个数组lucky[i]，代表1~m中lucky值为i的数有多少个。暴力的寻找所有可能情况，计算结果。

首先，我们该怎么预处理lucky数组呢？

方法是用dfs，递归的算出该数组，过程中利用数位dp数组记忆化搜索。具体实现见代码和注释。

怎么暴力？

首先确定第七个数的lucky值，然后分别枚举其他六个数的情况。因为对于该题来说，我们只需要知道这个数含有多少个lucky值，而不需要知道具体值，所以具有相同lucky值的数应该是等价的。确定了第七个数，我们就可以利用dfs深搜出每种符合条件的结果并且将结果加到ans中。这样算起来，虽然数很大，但是复杂度其实很小。

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define mod 1000000007ll
long long bt,bit[12],lucky[12],dp[12][12],ans;
void DFS(long long now,long long Max,long long number,long long cur)//计算结果，now，当前前六个数lucky值的总和；Max：第七个数的lucky值，number确定了多少个数，cur当前方案的方案数
{
    if(now>=Max) return ;//如果超过，那么不用再算
    if(number==7)//确定了7个，加到结果中
    {
        ans=(ans+cur)%mod;
        return ;
    }

    for(long long i=0;i<bt;i++)
    {
        if(lucky[i]){
            lucky[i]--;
            DFS(now+i,Max,number+1,cur*(lucky[i]+1)%mod);
            lucky[i]++;
        }
    }
}
long long dfs(long long pos,long long res,bool limit)//计算lucky数组，pos：代表当前的数位；res：当前还剩多少个lucky值需要得到；limit：此次取数有没有限制
{
    long long sum=0;
    if(pos==0) return res==0;
    if((!limit)&&dp[pos][res]!=-1) return dp[pos][res];//如果当前没限制，且dp值已经计算过，那么不用再计算

    long long tail=limit?bit[pos]:9;//计算当前遍历的上届
    for(long long i=0;i<=tail;i++)
    {
        sum+=dfs(pos-1,res-(i==4||i==7),limit&&(i==bit[pos]));
    }
    if(!limit) dp[pos][res]=sum;//没有限制那么存下来该值
    return sum;
}
int main()
{
    long long m;
    memset(dp,-1,sizeof(dp));
    scanf("%I64d",&m);
    while(m)
    {
        bit[++bt]=m%10;
        m/=10;
    }
    for(long long i=0;i<=bt;i++)
    {
        lucky[i]=dfs(bt,i,1);
    }
    lucky[0]--;//减去数字0（不在讨论范围中)
    for(long long i=1;i<=bt;i++)
    {
        DFS(0,i,1,lucky[i]);//首先确定第七个值。
    }
    printf("%I64d\n",ans);
    return 0;
}