HUST-1600 - Lucky Numbers 深搜

1600 - Lucky Numbers

时间限制：2秒 内存限制：64兆

416 次提交 115 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

来源

Problem Setter : Yang Xiao

题意：输入两个数字，表示一个范围，求这个范围内能被48整除，并且只包含4或者8的数字的个数。

解析：构造由4和8组成的数字，判断是否能被48整除并且在输入要求的范围之内，符合条件则计数。当数字超出上界m时，需返回。

根据以上的思路，然后写成代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int cnt;
void f(long long int n,long long int m,long long int sum)
{
if(sum%48==0&&sum<=m&&sum>=n) {cnt++;}
if(sum>m) return;
for(int i=1;i<=2;i++)
{
sum=sum*10+4*i;
f(n,m,sum);
sum=(sum-4*i)/10;
}
}
int main()
{
long long int n,m,s1,s2;
while(~scanf("%lld%lld",&n,&m))
{
cnt=0;
f(n,m,0);
printf("%d\n",cnt);
}
return 0;
}