611B. New Year and Old Property【模拟位运算】

最新推荐文章于 2024-01-29 11:49:29 发布

The_Dawn_Star

最新推荐文章于 2024-01-29 11:49:29 发布

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本文链接：https://blog.youkuaiyun.com/wlx65003/article/details/50443420

本文探讨了一种算法来计算特定区间内具有唯一二进制零的年份数量，通过位运算和数组模拟解决该问题。

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The year 2015 is almost over.

Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510 = 111110111112. Note that he doesn't care about the number of zeros in the decimal representation.

Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?

Assume that all positive integers are always written without leading zeros.

Input

The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 1018) — the first year and the last year in Limak's interval respectively.

Output

Print one integer – the number of years Limak will count in his chosen interval.

Sample test(s)

input
5 10

output
2

input
2015 2015

output
1

input
100 105

output
0

input
72057594000000000 72057595000000000

output
26

Note

In the first sample Limak's interval contains numbers 510 = 1012, 610 = 1102, 710 = 1112, 810 = 10002, 910 = 10012 and1010 = 10102. Two of them (1012 and 1102) have the described property.

模拟一下位运算即可，当时本来还试了下真正的位运算，但总是这里不对那里不对，最后直接用数组模拟了。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

#define ULL unsigned long long
int bin[64+10];
ULL cal(int n)
{
	ULL ans=0;
	for(int i=1 ; i<=n ; ++i)
	{
		ans+=(ULL)bin[i]<<(i-1);
	}
	return ans;
}
int main(void)
{
	ULL a,b,cnt=0;
	cin>>a>>b;
	
	for(int i=0 ; i<=64 ; ++i)
		bin[i]=1;
	for(int i=2 ; i<=64 ; ++i)
	{
		for(int j=1 ; j<i ; ++j)
		{
			bin[j]=0;
			ULL ans=cal(i);
			if(ans>=a && ans<=b)
				cnt++;
			bin[j]=1;
		}
	}
	cout<<cnt<<"\n";
	return 0;
}