codeforces Lucky Sum

http://codeforces.com/contest/122/problem/C    

C. 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：用队列（bfs）处理出含有4,7 组合的数有多少个？？（这个地方可以思考一下）

让后对区间分段处理即可

code：

#include<cstdio>#include<cstring>#include<queue>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;#define maxn 1023LL dx[] = {4,7};int cnt ;LL a[maxn];struct node{    LL x;};void bfs(){    a[0] = 4;a[1] = 7;    cnt = 2;    queue<node> Q;    node p,q;    p.x = 4;Q.push(p);    p.x = 7;Q.push(p);    while(!Q.empty()){        q = Q.front();Q.pop();        for(int i = 0;i < 2;i++)        {            p.x = q.x * 10 + dx[i];            a[cnt ++] = p.x;            Q.push(p);             if(p.x > 1e9) return;        }    }}int main(){    bfs();    LL r,l,n,m;    while(cin >> l >> r){        LL sum = 0;        while(l <= r){            int k = lower_bound(a,a+maxn,l) - a;            if(l == a[k] ) {l++;sum += a[k];}            else {                n = r - l;                m = a[k] - l;                if(n >= m) sum += a[k] * (m+1);                else sum += a[k] * (n + 1);                l = a[k] + 1;            }        }        cout << sum << endl;    }}