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POJ3252:Round Numbers 数位DP+前导0


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 toss using hooves.

They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.

A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes 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.

Help her by writing a program that tells how many round numbers appear in the inclusive 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


Source

​​USACO 2006 November Silver​​

Time Limit: 2000MS

 

Memory Limit: 65536K

Total Submissions: 16045

 

Accepted: 6600

算法分析:


题意:

给出区间[a,b],求其中二进制表示时0的数量>=1的数量的数个数(不能算前导零)

分析:

dp[pos][num],到当前数位pos,0的数量减去1的数量不少于num的方案数,一个简单的问题,中间某个pos位上num可能为负数(这不一定是非法的,因为我还没枚举完嘛,只要最终的num>=0才能判合法,中途某个pos就不一定了),这里比较好处理,Hash嘛,最小就-32吧(好像),直接加上32,把32当0用。这题主要是要想讲一下lead的用法,这次要包含lead表示是否有前导0,因为 00001 和 10000 是不一样的,而他们在状态表示的时候却一样,这样子就状态重叠了,显然我要统计0的数量,前导零是有影响的。至于!lead&&!limit才能dp,都是类似的,自己慢慢体会吧。

代码实现:

#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;
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++)
{
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;
}
return dfs(pos-1,32,true,true);
}
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;
}

 

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