CF Good Bye 2015 B- New Year and Old Property(CF611B)

http://codeforces.com/contest/611/problem/B

B. New Year and Old Property

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 and 1010 = 10102.
Two of them (1012 and 1102)
have the described property.

题意：给你两个数a，b，让你算出来从a到b一共b-a+1个数中一共有几个数的二进制中只有一个0（比如5的二进制101,6的二进制110都是）。

思路：一开始想的比较复杂想要找到最小的大于等于a的二进制只有一个零的数，然后不停的将0后移，如果0到了最后一个则进位，直到大于b为止，后来看了别人的代码，算了算时间复杂度，发现直接把所有的二进制只有一位零的数遍历一遍也不会超时间...

我的代码：

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<queue>
#include<math.h>

using namespace std;

unsigned long long INF=0x8000000000000000;
//-9223372036854775808

void pr(long long n)
{
long long i,k,x[100];
for(i=0;n!=0;i++)// 判断条件 n!=0{
x[i]=n%2;
n=n/2;
}
for(k=i-1;k!=(-1);k--)// 判断条件 k!=-1printf("%d",x[k]);// 输出x[k]，不是x[i]
}

int main()
{
long long a,b,temp;
int ans=0;
cin>>a>>b;
int n=0,l,s;
temp=a;

while(1)
{
if(temp&INF)
{
s=n;
break;
}
temp=temp<<1;
n++;
}
temp=temp<<1;
n++;
while(1 && n<=64)
{
if(!(temp&INF))
{
l=n;
break;
}
temp=temp<<1;
n++;
}
if(n<63)
{
temp=temp<<1;
n++;
while(n<64)
{
if(!(temp&INF))
{
a+=(1LL<<(63-n));
}
temp=temp<<1;
n++;
}
}

else if(n==64)
{
a+=(1LL<<(64-s));
a-=(1LL<<(63-s));
l=s;
s--;
}

while(1)
{
if(a<=b)
{
ans++;
}
else if(a>b) break;

if(l<63)
{
a+=(1LL<<(63-l));
l++;
a-=(1LL<<(63-l));
}
else if(l<=63)
{
a++;
a+=(1LL<<(64-s));
a-=(1LL<<(63-s));
l=s;
s--;
}
}
cout<<ans<<endl;
return 0;
}

别人的代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <stack>
#include <map>

using namespace std;

long long a,b,rs,x;
int main(){
cin >> a >> b;
for(int i = 1; i < 63; i++)
for(int j = 0; j < i-1; j++)
{
x = ((1LL << i) - 1 - (1LL << j));
if(a <= x && x <=b ) rs++;
}
cout<<rs;
return 0;
}