Good Bye 2015-New Year and Old Property（二进制暴力枚举）

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.

思路：

    这题其实就是暴力题，因为最多才18位，肯定是不会超时的。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cmath>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
typedef unsigned __int64 ll;
#define T 2000005
#define mod 1000000007

ll m[100],val[T];
int cnt = 0;

void play_table()
{
ll c = 0;
while(c<63)
{
m[c] = ((1LL)<<c);
/*printf("%lld\n",m[c]);*/
c++;
}
/*printf("+---------------+\n");*/
for(int i=2;i<63;++i){
ll tmp = m[i]-1;
for(ll j=0;j<(i-1);++j){
val[cnt++] = tmp^((1LL)<<j);
/* printf("%lld\n",tmp^((1LL)<<j));*/
}
}
/*printf("%d\n",cnt);*/
}

int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif

play_table();
ll u,v,i;
while(~scanf("%I64d%I64d",&u,&v))
{
ll k = 0;
for(i = 0;i<T;++i){
if(val[i]>=u&&val[i]<=v)k++;
}
printf("%I64d\n",k);
}

return 0;
}