POJ3252:Round Numbers(数位DP+记忆化DFS)

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

这道题还是花了不少的时间啊，本人愚笨，看了大神的代码没看懂，经过一个半小时的推敲理解，将搜索过程列出之后总算理解了他的思想

题意：算出区间内二进制中0的个数大于等于1的个数的数字有多少个

思路：数位DP确实是一个很巧妙的东西，在搜索过程中，我们要先确定首位1的放置，然后再根据这个首位去对后面所有位置的放置情况进行一次搜索即可

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int dp[50][50][50],bit[50];
//dp[k][i][j]，k为长度，i，为0的个数，j为1的个数

int dfs(int pos,int num0,int num1,int work,int first)
{
    if(!pos)
    {
        if(first)
            return 1;
        if(num0>=num1)
            return 1;
        return 0;
    }
    if(!first && !work && dp[pos][num0][num1]!=-1)//已经被访问过了
        return dp[pos][num0][num1];
    int end = work?bit[pos]:1;//判断此位能否为1
    int ans = 0;
    for(int i = 0; i<=end; i++)
    {
        //doing && end==i是看此位是不是算完
        if(first)//first是看第一位是否放下
        {
            if(i==0)
                ans+=dfs(pos-1,0,0,work&&end==i,1);//第一位还未放下，继续往后面看
            else
                ans+=dfs(pos-1,num0,num1+1,work&&end==i,0);//在这个选定的位置放下1，作为次二进制的初始
        }
        else
        {
            if(i==0)
                ans+=dfs(pos-1,num0+1,num1,work&&end==i,0);//在二进制后面放0
            else
                ans+=dfs(pos-1,num0,num1+1,work&&end==i,0);//放1
        }
    }
    if(!work && !first)
        dp[pos][num0][num1] = ans;//记录，防止重复访问，节省时间
    return ans;
}

int solve(int n)
{
    int len = 1;
    while(n)
    {
        bit[len++] = n&1;
        n>>=1;
    }
    int ans = dfs(len-1,0,0,1,1);
    return ans;
}

int main()
{
    memset(dp,-1,sizeof(dp));
    int l,r;
    while(~scanf("%d%d",&l,&r))
    {
        printf("%d
",solve(r)-solve(l-1));
    }

    return 0;
}