Codeforces 122 C. Lucky Sum（分块）

Description

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 expression next(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 ≤ 10^9) — 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.

Examples input

2 7

Examples output

33

题意

我们定义 $last(i)$ 代表大于等于 $i$ 最小的幸运数字，求 $\sum_{i=l}^rlast(i)$ 。

思路

显然分块可行，于是我们只需要写一个查找某个数字之后最小的幸运数字即可。

AC 代码

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);

using namespace std;
typedef long long LL;
const int maxn = 1e5+10;

LL last(LL x)
{
    int len = int(log10(x)) + 1;    //数字位数
    LL ans = 0,cnt = 0;
    for(int i=0; i<len; i++)        //同等位数最大最小幸运数
        ans = ans*10+4,cnt = cnt*10+7;
    if(x>cnt)                       //位数+1
        return ans*10+4;
    while(ans<x)
    {
        LL res = cnt;
        for(int i=0; i<1<<len; i++)
        {
            LL tmp = 0;
            for(int j=0; j<len; j++)
            {
                if(i&(1<<j))
                    tmp = tmp*10+7;
                else
                    tmp = tmp*10+4;
            }
            if(tmp>=x)
                res = min(res,tmp);
        }
        ans = res;
    }
    return ans;
}

int main()
{
    LL l,r;
    cin>>l>>r;
    LL now = l,ans = 0;
    while(true)
    {
        LL la = last(now);
        if(la>r)
        {
            ans+= la * (r-now+1);
            break;
        }
        else
            ans += la * (la-now+1);
        now = la+1;
    }
    cout<<ans<<endl;
    return 0;
}