poj3252 Round Numbers

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 rangeStart.. Finish

Sample Input

2 12

Sample Output

6

找到给定区间内，一个数转换成二进制数0的个数大于等于1的个数的数的个数。

嗯，感觉数位dp都是差不多一个套路，这一次就是换成二进制来搜素，注意前导0的情况。

搜索的时候dp[i][j1][j2]表示长度为i，0的个数为j1，1的个数为j2的数的个数。

//#include <bits/stdc++.h>#include <cstdio>#include <cstring>using namespace std;typedef long long LL;const int MAXN=1e5+10;int l,r;int dp[50][50][50];int digit[20];int dfs(int len,int num0,int num1,bool up,bool zero){    if(!len)return num0 >= num1;    if(!up && dp[len][num0][num1] != -1)return dp[len][num0][num1];    int res = 0;    int n = up?digit[len] : 1;    for(int i = 0 ; i <= n ; ++i)    {        if(zero)        {            if(i)res += dfs(len-1,0,1,up && i==n,0);            else res += dfs(len-1,0,0,up && i==n,1);        }        else        {            if(i)res += dfs(len-1,num0,num1 + 1,up && i==n,0);            else res += dfs(len-1,num0 + 1,num1,up && i==n,0);        }    }    if(!up)dp[len][num0][num1] = res;    return res;}int cal(int x){    int len = 0;    while(x)    {        digit[++len] = x & 1;        x >>= 1;    }    return dfs(len,0,0,1,1);}int main(){    memset(dp,-1,sizeof dp);    scanf("%d%d",&l,&r);    printf("%d\n",cal(r) - cal(l-1));    return 0;}