(Jizhong) 2175. Lucky number (sum)

(File IO): input: sum.in output: sum.out
time limit: 1000 ms space constraints: 131072 KB specific restrictions
Goto ProblemSet

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Title Description
$4$ and $7$ is a lucky number flavor taste. Lucky number is a positive integer of only those lucky numbers. ¥ 47,4774 is as lucky number, and $45，17，417$ is not a lucky number.
definition $next(x)$ is greater than or equal to $x$ minimum number of lucky.
Taste taste very interested in the value of the expression:
$next(L)+next(L+1)+...+next(R-1)+next(R)。$
Now tell you the value of L and R, and I hope you can help flavor taste calculated value of the expression.

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Enter the
input file $sum.in$ only line containing two positive integers and L $R(1≤L≤R≤10^9 )$， $L$ and $R$ separated by a space between the values.

Output
Output file $sum.out$ only one line an integer representing the value of the expression.

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Sample input
[1] Input Sample
27

Sample input [2]
77

Sample output
[output 1] Sample
33

[2] Output Sample
7

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Data range limits
for $20$ % of the data, $1≤L≤R≤1000$
For $40$ % of the data, $1≤L≤R≤10^6$
Another $20$ % of the data, $L=R$
for $100$ % of the data, $1≤L≤R≤10^9$

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Tips
[description] Sample 1
$next(2)+next(3)+next(4)+next(5)+next(6)+next(7)=4+4+4+7+7+7=33$
[description] Sample 2
$next(7)=7$

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Problem-solving ideas
first dfs a lucky number, (because dfs is sequentially stored lucky number, so no need to sort it again). Then simulation.

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Code

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
long long l,r,ans,t,s,t1,x,ii;
long long d[5000];
void dfs(long long x,long long y)
{
    if(x==1)
    {
        d[++t1]=y;
        return;
    }
    dfs(x-1,y*10+4);
    dfs(x-1,y*10+7);
}
int main()
{
	freopen("sum.in","r",stdin);
      freopen("sum.out","w",stdout);
    scanf("%lld%lld",&l,&r);
    for(int i=1; i<=12; i++)
        dfs(i,0);
    t=1,ii=l;
    while(l>d[t])
        t++;
    while(ii<=r)
    {
        x=min(d[t]-ii,r-ii)+1; 
		s+=d[t]*x;
        ii+=x;
        t++;
    }
    printf("%lld",s);
    return 0;
}