Lucky Sum （dfs打表）

Posted 2021-01-10 1013star

Lucky Sum 

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l)?+?next(l?+?1)?+?...?+?next(r?-?1)?+?next(r). Help him solve this problem.

Input

The single line contains two integers l and r (1?≤?l?≤?r?≤?10⁹) — the left and right interval limits.

Output

In the single line print the only number — the sum next(l)?+?next(l?+?1)?+?...?+?next(r?-?1)?+?next(r).

Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2 7

Output

33

Input

7 7

Output

7

Note

In the first sample: next(2)?+?next(3)?+?next(4)?+?next(5)?+?next(6)?+?next(7)?=?4?+?4?+?4?+?7?+?7?+?7?=?33

In the second sample: next(7)?=?7

值得学习之处：

　　如何将luck number打表是本题的精华，果然还是要多做题啊

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long ll;
 7 const int maxn=1e5+10;
 8 ll mp[maxn];
 9 int id=1;
10 void dfs(ll x,int cur)
11 {
12     if(cur>10)
13         return;
14     mp[id++]=x;
15     dfs(x*10+4,cur+1);
16     dfs(x*10+7,cur+1);
17 }
18 ll solve(ll x)
19 {
20     ll sum=0;
21     if(x==0)
22         return 0;
23     for(int i=1;i<=id;i++)
24     {
25         if(x>=mp[i])
26         {
27             sum+=mp[i]*(mp[i]-mp[i-1]);
28         }
29         else
30         {
31             sum+=mp[i]*(x-mp[i-1]);
32             break;
33         }
34     }
35     return sum;
36 }
37 int main()
38 {
39     ll l,r;
40     dfs(4,0);
41     dfs(7,0);
42     sort(mp+1,mp+id+1);
43    /* for(int i=0;i<20;i++)
44         printf("%d ",mp[i]);*/
45     scanf("%lld%lld",&l,&r);
46     printf("%lld
",solve(r)-solve(l-1));
47     return 0;
48 }