# 动态规划 | 数位 | Round Numbers

## Description

The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone’ (also known as ‘Rock, Paper, Scissors’, ‘Ro, Sham, Bo’,and a host of other names) in order to make arbitrary decisions such as who gets to be milked first. They can’t even flip a coin because it’s so hard to tossusing hooves.

Theyhave thus resorted to “round number” matching. The first cow picks aninteger less than two billion. The second cow does the same. If the numbers areboth “round numbers”, the first cow wins, otherwise the second cowwins.

A positive integer N is said to be a “round number” if the binary representation of N has as many or more zero es than it has ones. For example,the integer 9, when written in binary form, is 1001. 1001 has two zero es and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zero es and three ones, it is not a round number.

Obviously,it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many “round numbers” are in a given range.

Helpher by writing a program that tells how many round numbers appear in theinclusive range given by the input (1 < Start < Finish <2,000,000,000).

## Input

Line 1: Two space-separated integers, respectively Start and Finish.

## Output

Line 1: A single integer that is the count of round numbers in the inclusive range Start..Finish

## Sample Input

2 12


## Sample Output

6


No Hint

## Codes

#include <cstdio>
#include <iostream>
#include <cstring>
#define maxn 41
using namespace std;
typedef long long LL;
LL dp[maxn][maxn][maxn];
int spr[maxn];
LL dfs(int x,int c1,int c0,bool lead,bool lim){
if(x==0) return c0>=c1;
if(!lim && dp[x][c1][c0]!=-1) return dp[x][c1][c0];
LL ans=0; int imax=lim?spr[x]:1;
for(int i=0;i<=imax;i++){
if(i==1)
else if(i==0)
}
if(dp[x][c1][c0]==-1) dp[x][c1][c0]=ans;
return ans;
}
inline LL res(int x)
{
memset(spr,0,sizeof(spr));
int cnt=0;
while(x>0){
spr[++cnt]=x&1;
x>>=1;
}
return dfs(cnt,0,0,1,1);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("t2.in","r",stdin);
freopen("t2.out","w",stdout);
#endif
memset(dp,-1,sizeof(dp));
LL a,b;
cin>>a>>b;
cout<<res(b)-res(a-1);
return 0;
}