Codeforces Round #157 (Div. 2)D（数位DP+组合数）

D. 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).

Sample test(s)

input

7

output

0

input

8

output

1440

题意：RT

思路：首先要预处理出含4或7的数字有多少个，用dp[i]表示含4或7的个数为i的数字有多少个，dfs递推求解即可

            有了这个dp数组就可以开始求了，如果 Little Elephant Political Party选了含i个4或7的数字，则要乘以dp[i]

         然后是给剩下的6个party选择数字，要求它们的总的4和7的个数比i少，这就是一个背包问题了，还是用动态规划求解，感觉题目还不错~

#include 
   
   
    
    
#include 
    
    
     
     
#include 
     
     
      
      
#include 
      
      
       
       
#include 
       
       
         #include 
        
          #include 
         
           using namespace std; typedef long long ll; const int MOD = (int)1e9+7; int dig[12]; int dp[12]; int dp2; int top; int add(int x,int y) { x+=y; if(x>=MOD)x-=MOD; return x; } int sub(int x,int y) { x-=y; if(x<0)x+=MOD; return x; } void dfs(int cent,int val,int vis,int i,int j) { if(i+j>top)return; if(cent==top){ dp[i+j] = add(dp[i+j],val); return; } if(vis==0){ dfs(cent+1,(ll)val*8,0,i,j); dfs(cent+1,val,0,i,j+1); dfs(cent+1,val,0,i+1,j); } else { if(dig[cent]!=7 && dig[cent]!=4)dfs(cent+1,val,1,i,j); if(dig[cent]>=7){ if(dig[cent]==7)dfs(cent+1,val,1,i,j+1); else dfs(cent+1,val,0,i,j+1); } if(dig[cent]>=4){ if(dig[cent]==4)dfs(cent+1,val,1,i+1,j); else dfs(cent+1,val,0,i+1,j); } if(dig[cent]>7)dfs(cent+1,(ll)val*(dig[cent]-2),0,i,j); else if(dig[cent]>4)dfs(cent+1,(ll)val*(dig[cent]-1),0,i,j); else if(dig[cent]>0)dfs(cent+1,(ll)val*dig[cent],0,i,j); } } void dfs2(int cent,int total,int val,int mx) { if(cent==6){ dp2= add(dp2,val); return; } for(int i=0;i+total<=mx;i++){ if(dp[i]==0)continue; dp[i]--; dfs2(cent+1,total+i,(ll)val*(dp[i]+1)%MOD,mx); dp[i]++; } } int main() { int m; scanf("%d",&m); if(m==7){ printf("0\n"); } else { top=0; while(m){ dig[top++]=m%10; m/=10; } for(int i=0;i