HUST 1600 Lucky Numbers

1600 - Lucky Numbers

时间限制：2秒 内存限制：64兆

401 次提交 109 次通过

题目描述

Isun loves digit 4 and 8 very much. He thinks a number is lucky only if the number satisfy the following conditions:

1. The number only consists of digit 4 and 8.

2. The number multiples 48.

One day, the math teacher gives Isun a problem:

Given L and R(1 <= L <= R <= 10^15), how many lucky numbers are there between L and R. (i.e. how many x satisfy L <= x <= R, x is a lucky number).

输入

Multiple test cases. For each test case, there is only one line consist two numbers L and R.

输出

For each test case, print the number of lucky numbers in one line.

Do use the %lld specifier or cin/ cout stream to read or write 64-bit integers in С++.

样例输入

1 48

1 484848

样例输出

1

7

提示

用深搜把每个符合条件的数字找出来，打表，然后就好了

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
long long int l,r;
long long int ans[200005];
int cot;
int count(int x)
{
int num=0;
while(x>0)
{
num++;
x/=10;
}
return num;
}
void dfs(long long int x,int cnt)
{
if(cnt>16)
return;
if(x%48==0)
ans[cot++]=x;
dfs(x*10+8,cnt+1);
dfs(x*10+4,cnt+1);
}
int cmp(long long int x,long long int y)
{
return x<y;
}
int main()
{
cot=0;
memset(ans,0,sizeof(ans));
dfs(4,1);
dfs(8,1);

while(scanf("%lld%lld",&l,&r)!=EOF)
{

sort(ans,ans+cot,cmp);
int pos=0;
for(int i=0;i<cot;i++)
{
if(l<=ans[i])
{
pos=i;
break;
}

}
int res=0;

for(int i=pos;i<cot;i++)
{
if(ans[i]<=r)
res++;
else
break;

}
printf("%d\n",res);

}

return 0;
}