poj3252 數位dp（所有比n小的二進位制位0的個數不少於1的個數）記憶化搜尋

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).

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

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

2 12

6

/**
poj 3252   數位dp（所有比n小的二進位制位0的個數不少於1的個數）記憶化搜尋
題目大意：求出區間內二進位制表示是0的個數不小於1的個數的數的個數
解題思路：用記憶化搜尋dfs(len,num0,num1,flag,first)len表示二進位制的位置，num0表示0的個數，flag標記是不是訪問到上限，
           first表示第一個1的位置，對於每個這樣的位置開始以後看做一個次二進位制數
           http://blog.csdn.net/libin56842/article/details/10037607
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int dp[50][50][50],n,m,bit[50];

int dfs(int len,int num0,int num1,int flag,int first)
{
    if(len<0)return num0>=num1;
    if(flag==0&&dp[len][num0][num1]!=-1)return dp[len][num0][num1];
    int ans=0;
    int end=flag?bit[len]:1;
    for(int i=0; i<=end; i++)
    {
        int t=first&&(i==0);
        ans+=dfs(len-1,t?0:num0+(i==0),t?0:num1+(i==1),flag&&i==end,t);
    }
    if(flag==0)dp[len][num0][num1]=ans;
    return ans;
}

int solve(int n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%2;
        n>>=1;
    }
    return dfs(len-1,0,0,1,1);
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(dp,-1,sizeof(dp));
        printf("%d\n",solve(m)-solve(n-1));
    }
    return 0;
}