牛客练习赛13B 幸运数字2

题目链接：https://ac.nowcoder.com/acm/contest/70/B

题目大意：

　　略

分析：

　　先DFS求出所有幸运数，然后暴力即可

代码如下：

 #pragma GCC optimize("Ofast")
 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define pii pair<int,int>
 #define piii pair<pair<int,int>,int>
 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef set< int > SI;
 typedef vector< int > VI;
 const double EPS = 1e-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;

 LL l, r, ans;
 LL lucky[], n; 

 // 求所有幸运数
 inline void dfs(LL x, int cnt) {
     if(cnt > ) return;
     lucky[n++] = x;
     dfs(x* + , cnt + );
     dfs(x* + , cnt + );
 }

 int main(){
     INIT();
     cin >> l >> r;
     lucky[n++] = ;
     dfs(, );
     sort(lucky, lucky + n);

     int j = lower_bound(lucky, lucky + n, l) - lucky;

     while(l <= r) {
         ans += (min(lucky[j], r) - l + ) * lucky[j];
         l = lucky[j++] + ;
     }

     cout << ans << endl;
     return ;
 }

 /*
 1 1000000000
 1394672350065645019

 1 1000
 1397683

 447 447477474
 168389348342066109

 1 436278568
 163403864955643707

 1 4328955
 16190029435407

 4328956 4444444
 513284393116

 4328956 436278568
 163387674926208300

 2354 543262
 231508617956

 999999999 1000000000
 8888888888
 */