https://ac.nowcoder.com/acm/contest/5086/E
题意：一串数字只有7和4，称之为幸运数字。给一个区间（包括端点）和一个函数next（x),即大于x的第一个幸运数字。求区间内的next(x)的和。
思路：打表，然后运用前缀和进行优化。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
ll num[1000100],a[1000100],num_cnt=1,sum[1000100];
void bfs(){
ll cur;
queue<ll>sui;
sui.push(4);
sui.push(7);
while(1){
cur=sui.front();
sui.pop();
if(cur>4444444444) return ;
num[num_cnt++]=cur;
sui.push(cur*10+4);
sui.push(cur*10+7);
}
}
int main(){
ll r,l;
cin>>r>>l;
bfs();
for(int i=1;i<=num_cnt;i++){
sum[i]+=sum[i-1]+(num[i]-num[i-1])*num[i];
}
ll r1=lower_bound(num+1,num+1+num_cnt,r)-num;
ll l1=lower_bound(num+1,num+1+num_cnt,l)-num;
ll sum1=(num[r1]-r)*num[r1]+sum[l1]-sum[r1]-(num[l1]-l)*num[l1]+num[r1];
cout<<sum1<<endl;
return 0;
}