Description

Lucky Sum

time limit per test: 2 seconds

memory limit per test: 256 megabytes

input: standard input

output: standard output

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 expressionnext(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.

Sample test(s)

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

题解：题意很好理解，重点难点是把lucky numbers存到一个数组里(dfs)，再根据题意求和。

          其中sum(x) = next(1) + next(2) + next(3) + ... + next(x), 所以next(l) + next(l+1) + ... + next(r) = sum(r) - sum(l-1)。

代码：

1 #include 
     
       2 #include 
      
        3 #include 
       
         4  5 using namespace std; 6 typedef long long LL; 7  8 const int maxLength = 10000; 9 LL luck[maxLength];10 int index = 0;11 12 void dfs(LL x, int cursor) {13     if(cursor > 10) {14         return;15     }16     luck[index++] = x;17     dfs(x*10+4, cursor+1);18     dfs(x*10+7, cursor+1);19 }20 //sum(x) = next(1)+next(2)+...next(x)21 LL sum(LL x) {22     LL sum = 0;23     if(x == 0) {24         return 0;25     }26     for(int i=0; i
        
         = luck[i]) {28             sum += luck[i]*(luck[i] - luck[i-1]);29         }else {30             sum += luck[i]*(x-luck[i-1]);31             break;32         }33     }34     return sum;35 }36 int main()37 {38     LL l, r;39     dfs(4, 0);40     dfs(7, 0);41     sort(luck, luck+index);42     cin >> l >> r;43     cout << sum(r)-sum(l-1) << endl;44     return 0;45 }
        
       
      
     

 

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