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 ∑ri=llast(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;
}