Codeforces 121A Lucky Sum

Lucky Sum

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 121A
64-bit integer IO format: %I64d      Java class name: (Any)

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 %I64dspecificator.

Sample Input

Input

2 7

Output

33

Input

7 7

Output

7

Hint

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

Source

Codeforces Beta Round #91 (Div. 1 Only)

解题：只包含4或者7的数字很少的。搜一遍就是了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 20000;
18 LL d[maxn];
19 int tot = 1;
20 void dfs(LL num,int cur) {
21     d[tot++] = num;
22     if(cur > 11) return;
23     dfs(num*10+4,cur+1);
24     dfs(num*10+7,cur+1);
25 }
26 LL sum(LL x) {
27     if(x == 0) return 0;
28     LL temp = 0;
29     for(int i = 1; i < tot; i++) {
30         if(x >= d[i])
31             temp += d[i]*(d[i]-d[i-1]);
32         else {
33             temp += d[i]*(x-d[i-1]);
34             break;
35         }
36     }
37     return temp;
38 }
39 int main() {
40     dfs(4,0);
41     dfs(7,0);
42     sort(d+1,d+tot);
43     LL lt,rt;
44     while(~scanf("%I64d %I64d",&lt,&rt))
45         printf("%I64d\n",sum(rt)-sum(lt-1));
46     return 0;
47 }