题意：求一个区间内满足化为二进制后0多于1的数的数量

#pragma comment(linker, "/STACK:10240000,10240000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<set>
#include<vector>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
const double R=0.5772156649015328606065120900;
const int N=1e5+5;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
typedef long long ll;
int dp[35][66];
int a[66];
int dfs(int pos,int sta,bool lead,bool limit)
{
    if(pos==-1)
        return sta>=32;//转化成32-64内，分小于32和大于32不再分小于大于0
    if(!limit && !lead && dp[pos][sta]!=-1) return dp[pos][sta];
    int up=limit?a[pos]:1;
    int ans=0;
    for(int i=0;i<=up;i++)
    {//lead处理前导零问题;//注意体会lead&&i==0
        if(lead && i==0) ans+=dfs(pos-1,sta,lead,limit && i==a[pos]);//有前导零就不统计在内
        else ans+=dfs(pos-1,sta+(i==0?1:-1),lead && i==0,limit && i==a[pos]);
    }
    if(!limit && !lead ) dp[pos][sta]=ans;
    return ans;
}
int solve(int x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x&1;
        x>>=1;//处理进制
    }
    return dfs(pos-1,32,true,true);//刚开始并没有前导0
}
int main()
{
    memset(dp,-1,sizeof dp);
    int a,b;
    while(~scanf("%d%d",&a,&b))
    {
        printf("%d\n",solve(b)-solve(a-1));
    }
    return 0;
}